I zakon termodinamike dU =dq+dw+ dw =dU+ pdVMaksvelove relacije V V T U T P P T ∂ ∂ − ∂ ∂ = ∑ = ∆ = − =− n i S meš S S R n i x i 1 ... Iz jednačine adijabate se

V

VT

UC

=partpart

edwdwdqdU ++=

P

PT

HC

=partpart

pdVdUdH +=

dPP

HdTCdH

T

p

partpart

+=dVPdTCdVV

UdTCdU uV

T

V +=

+=partpart

CPm ndash CVm = R

CP ndash CV = nR

Izotermski reverzibilni zapreminski rad gasa u IGS

Vpw ∆minus= Izotermski revetzibilni zapreminski rad isparavanja p-napon pare

Izotermski procesi

I zakon termodinamike

2

1

1

2 lnlnP

PRT

V

VnRTw minus=minus=

minus

=

minus=minus= 11)(

2

11

1

2112

VC

nR

VVVadV

VTC

T

TTCTTCw

Adijabatski zapreminski rad

1

2

1

2

2

1

1

2

1

2

2

1

P

P

T

T

V

V

T

T

V

V

T

T R

C

R

CCR PVV

=

=

=

γ

γ

===

1

2

2

1

V

V

P

PconstPVPV VP CC

Adijabatski procesi

Jednačina adijabate

RTnUVPUH gas∆+∆=∆+∆=∆

2

12

2 T

TT

q

w minus=

minus=η Efikasnost toplotne mašine

0gege dST

dqdS II zakon termodinamike-

Reverzibilni i ireverzibilni

1

2

1

212 lnln

V

VR

T

TCSSS V +=minus=∆

oksistotok

i

iirev

sis SSST

HSS

T

qS ∆+∆=∆

∆minus=∆==∆ sumν Ireverzibilni

1

2

1

212 lnln

P

PR

T

TCSSS p minus=minus=∆

tr

trtr

T

HS

∆=∆ Fazni prelazi

V

VV

V T

STCodnosno

T

C

T

S

partpart

==

partpart

( ) int int==minus=2

1

2

1

ln12

T

T

T

TVVV

TdCT

dTCSSS∆


1

2

1

ln12

T

T

T

TPPP

TdCT

dTCSSS∆

VT T

P

V

S

partpart

=

partpart

PT T

V

P

S

partpart

minus=

partpart

int

partpart

minus=minus=∆ dPT

VSSS

P

12int

partpart

=minus=∆ dVT

PSSS

V

12

Maksveloverelacije

TV V

U

T

PTP

partpart

minus

partpart

=

sum=

minus=minus=∆n

i

iimeš xnRSSS1

12 ln Entropija mešanja

TP P

H

T

VTV

partpart

+

partpart

=

Termodinamičke jednačinestanja

WkS ln= k-Bolcmanova konstanta

k=RNA=138middot10-23 JK

Helmholcova i Gipsova slobodnaenergija

TV V

S

T

P

=

partpart

partpart

PV

A

T

minus=

partpart

ST

A

V

minus=

partpart

dA=minus PdV minus SdT dG = VdP ndash SdT

TP P

S

T

V

minus=

partpart

partpart

ST

G

P

minus=

partpart

VP

G

T

=

partpart

pT

GTHG

+=

part∆part

∆∆2

)(

T

H

T

TG

p

∆part∆part

minus=

0

0 lnP

PRTGG mmm +=

ZADATAK 1

Izračunati rad potreban da ptica mase 120g uzleti do visine

od 50m od

a) Površine Zemlje

b) Površine Meseca (g=16ms-2)

Rešenje

a)

b) Jmsmkgmghw

Jmsmkgmghw

695061120

865850819120

2

2

minus=sdotsdotminus=minus=


ZADATAK 2

Hemijska reakcija se dešava u cilindru preseka 100cm2 Kao

rezultat klip se pomerio za 20cm nasuprot pritiska od 1 atm

Koliki je rad

Rešenje

JmPaVPw 65202102000101325 36 minus=sdotsdotminus=∆minus= minus

Zadatak 3

U ciklusu 1 mol gasa u idealnom gasnom stanju vrši rad od 4186J

Koliko je q

Rešenje

w=-4186J ∆U=0 q=4186J

Zadatak 4

Rad (u J) koji izvrši gas u IGS pri širenju nasuprot pritisku od 1 atm i od 10 do 30 L je

a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje

JmPaw 5202610)1030(101325 33 minus=sdotminusminus= minus

ZADATAK 5

Jedan mol gasa prelazi iz stanja sa pritiskom P0 i zapreminom V0 u

stanje sa duplo većom zapreminom i pritiskom Na p-V dijagramu taj

proces je predstavljen pravom linijom Koliki rad izvrši gas u tom

procesu

a)PoV0 b) PoV02 c) 3 PoV02

d) 2 PoV0 e) 3 PoV0 f) ne znam

P

VVo 2Vo

Po

2Po

1

2

Rešenje

ooooo VPVPP

w2

3

2

)2(=

+=

ZADATAK 6

U termodinamičkom procesu promena unutrašnje energije

sistema je ∆U = -300 J sistem prima toplotu od 100 J i širi se

nasuprot pritiska od 1 bar Kolika je promena zapremine (L)

a) 2 b) 3 c) 4

d) 5 e) 1 f) ne znam

Rešenje

LdmmV

JVm

JVPw

JJJwwqU

410104100000

400

400101

400100300

3333

3

5

=sdotsdot=minus

minus=∆

minus=∆sdotminus=∆minus=

minus=minusminus=+=∆

minus

Zadatak 7bull Električni grejač snage 15W greje masu od 12g

vode tokom jednog minuta Ako je početna temperatura vode 35oC kolika će biti krajnja temperatura vode (K)Specifični toplotni kapacitet vode je 1 calstepg

bull Rešenje

Snaga grejača je P=∆Ut Energija koju

grejač oslobantildea je∆U=Psdott=15Js sdot60s=900J

KKT

CCK

CgJ

gJ

C

U

UCCgJCgJ

g

J

m

UU

k

o

pko

o

vs

s

svs

o

vss

07326921715308

925292173592179217

1864

75

18647512

900

=+=

=+=∆+====

∆=∆

∆∆

====∆

=∆

θθθθ

θ

Zadatak 8bull 1 mol vode isparava Kolika je promena entalpije ako je

pritisak 1 bar Promena unutrašnje energije pri isparavanju je 407kJmol

molkJmolmPamolJH

molmPa

KKmolJ

P

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

3

5

=sdotsdot+=∆

=sdot

==

rangrangminus=∆∆+∆=∆

ZADATAK 9

Izračunati rad širenja pri elektrolizi 50 g vode pri konstantnom

pritisku i temperaturi od 25oC

Rešenje

2H2O(t)rarr2H2(g)+O2(g)

w=-Psp∆V=-Psp(Vk-Vp)asymp-PspVk= nRTP

nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći

bull Koliko energije treba dovesti masi od 135kg vode da bi se zagrejala od 20oC do

bull temperature ključanja Pretpostaviti da je Cvs=4186JgK

Zadatak 11

Jedan mol gasa u IGS u početku na pritisku od 1 atm i 300K

se reverzibilno zagreva do 400 K pri konstantnoj zapremini

Izračunati krajnji pritisak i promenu unutrašnje energije

toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12

bull Izračunati napon pare vode na 35oC ako se vodena para širi reverzibilno i izotermski od zapremine 5 cm3 do zapremine 100 cm3 ako se pri tome vrši rad od 534 mJ

Rešenje

w=-p(V2-V1) w=-0534J p=0534Nm95sdot10-6m3=5621Pa

ZADATAK 13

Izračunati ∆U za reakciju sagorevanja 10 mola propana na 25oC ako

je ∆H=-2058 kJ

Rešenje

C3H8(t)+5O2(g)rarr3CO2(g)+4H2O(t)

∆ng=3-5=-2

∆U=∆H-∆ngRT=-2058000J+2sdot83145sdot298=-2053kJ

ZADATAK 14

Rad reverzibilnog izotermskog širenja jednog mola gasa u

idealnom gasnom stanju od zapremine V1 do zapremine V2 je

dat izrazom

a) ndashp(V2 -V1) b) ndashp(V1 -V2) c)

d) CV(T2-T1) e) f) ne znam

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15

Dva mola idealnog gasa podleže izotermalnoj reverzibilnoj ekspanziji

od početne zapremine V1 do krajnje zapremine 10V1 i vrši rad od

41860J Ako je početni pritisak 100 bar kolika je početna zapremina

(u L) i temperatura (u K)

Rešenje

w=nRTlnV2V1=nRTmiddot23 P1V1=nRT=w23=4186023=18200JV1=18200100middot105=182middot10-3m3=182 L

nRT=18200J T= 182002middot8314=1093K

Zadatak 16

Uzorak argona mase 656g zauzima zapreminu od 185dm3

Na 305K a) Izračunati rad koji bi se izvršio kada se gas širi izotermski nasuprot konstantnog spoljašnjeg pritiska od 77kPa dok mu se zapremina ne poveća na 25 dm3b) Izračunati rad koji bi se izvršio kada bi se ista ekspanzija vršila reverzibilno

Rešenjea)

b)

JmPaVPw sp 251910521077 333 minus=sdotsdotsdotminus=∆minus= minus

J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=

=sdotsdotsdotminus=minus=

Zadatak 17

Kada se 3 mola kiseonika zagreje pri konstantnom pritisku od 325atm njihova temperatura poraste od 260 K do 285 KAko je CPm=204JKmol izračunati ∆H q i ∆U

Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203

=sdotsdotminus=∆minus∆=∆

=minussdot=∆sdot==∆

ZADATAK 18

Rad reverzibilnog adijabatskog širenja jednog mola gasa u idealnom

gasnom stanju od zapremine V1 pri temperaturi T1 do zapremine V2

pri temperaturi T2 je dat izrazom


d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19

Jedan mol idealnog gasa na 300 K se širi adijabatski i reverzibilno

od 20 bar do 1 bar Koja je temperatura u krajnjem stanju gasa

pretpostavljajući da je CV=(32) R

a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20

Izračunati krajnji pritisak argona (bar) posle reverzibilnog i adijabatskog

širenja pri kome se zapremina poveća dva puta Početni pritisak je

iznosio 100 kPa a γAr=53

Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ

Iz jednačine adijabate se dobija

ZADATAK 21

Izračunati rad i promenu unutrašnje energije pri adijabatskom širenju

02 mol Ar od 05 do 10L Početna temperatura je izosila 25oC a

molarni toplotni kapacitet Ar na konstantnoj zapremini iznosi 1248

JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=

minus=minussdotsdot=

=sdot

=

=

minusminus

ZADATAK 22 Domaći

Dva mola idealnog gasa za koji je Cvm=5R2 je reverzibilno

zagrevan do 356K na konstantnoj zapremini Početni pritisak i

temperatura su bili P1=111kPa i T1=277K Izračunati krajnji

pritisak ∆U q i w

Rešenje

Jedan mol idealnog monoatomskog gasa u početku na 10 atm pritisku i temperaturi od 0oC širi se izotermski nasuprot pritiska od 1 atm Uslovi su takvi da je konačna zapremina 10 puta veća od početne krajnji pritisak je jednak spoljašnjem pritisku

(a) Izračunati početnu i krajnju zapreminu(b) Izračunati q w ∆U za proces

Zadatak 23

Kako je V1=224 L to je V2=224 LUkupni proces je izotermski pa je ∆U=0Izvršeni rad je -P∆V=-1013105Nm2 x(00224-000224)m3=w=-20422J q=20422J

ZADATAK 24

Na vrlo niskoj temperaturi toplotni kapacitet čvrstih supstancija se

može uzeti da je proporcionalan sa T3 pa se može pisati da je C=aT3

Kolika je promena entalpije takve supstancije pri zagrevanju od 0 do

temperature T (koja je bliska 0)

Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25

bull Gas se pokorava jednačini stanja

a) Odrediti reverzibilni rad koji se vrši pri zagrevanju gasa od T1 do T2 pri konstantnom pritisku

b) Odrediti reverzibilni izotermski rad pri širenju od V1 do V2

pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα

minusminusminusminus=minusminus=

+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26

Uzorak od 5mol ugljendioksida pri zapremini od 15dm3 na 280K podleže adijabatskoj ekspanziji nasuprot konstantnog pritiskaod 785kPa dok mu se zapremina poveća za faktor 4 Izračunatiq w ∆T ∆U i ∆H (CP=3711JKmol)

Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333

minus=minussdotsdot+minus=∆

∆+∆=∆+∆=∆+∆=∆

minus=+=∆minus=∆

minussdotminus

==∆∆sdot=minus=

sdotminussdotsdotsdotminus=∆minus== minus

Zadatak 27Jedan mol idealnog monoatomskog gasa na početnim P1=2 atm i T1=27315K je preveden na pritisak P2=4atm reverzibilnim putem definisanim sa PV=const Izračunati V1 V2 i T2 ∆U ∆H q i w

RešenjeIz jedn id g stanja je V1=112 L Pošto je PV=const to je V2=224 L Kombinovanjem PV=const sa PV=RT dobija se TV2=const pa je T2=4T1=1092K∆U=CV∆T=(3R2)819=1022kJ ∆H=CP∆T=17032kJ Da bi se dobilo w treba odrediti w=intPdV Iz početnih uslovaPV=const=2112=0178atmL pa je w=-0178intVdV=0089(V2

2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28

Toplotni kapacitet gasa u IGS varira sa temperaturom prema izrazu

za 1 mol Izračunati q w ∆U i ∆H za jedan mola gasa kada

temperatura raste od 0oC do 100oC

a) Na konstantnom pritisku

b) Na konstantnoj zapremini

)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273

==minus=∆minus∆=∆

minus=sdotsdotminus=∆minus=∆minus=

=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29

Jedan mol idealnog gasa sa CV=2093Jmol step u početku pri

standardnim uslovima prolazi kroz sledeći reverzibilni ciklus

A izohorsko zagrevanje do dvostruke vrednosti početne temperature

od stanja 1 do 2 B adijabatsko širenje od 2 do 3 do početne temperature i

C izotermalnu kompresiju od 3 do 1 Izračunati q w ∆U i ∆H za

korake A B i C

Rešenje

Korak A V=const T=27315K P=1bar w=0 ∆U=qv =CV ∆T=

2093middot(2T1-T1)=2093middot27315=5717Jii) Korak B q=0 ∆U=w=CV ∆T=2093(T1-2T1)=-5717Jmol∆H=-7988Jmoliii) Korak C ∆U=0 T2T3=2 V3V2=(T2T3)CVR V2=V1

w=RT1lnV3V1=RT1ln(T2T3) CVR=CVT1sdot2303log2=

2093middot27315middot2303middotlog2=396346Jmol q=-396346 Jmol

30 Jedan mola gasa u idealnom gasnom stanju prolazi kroz termodinamički ciklus

koji se sastoji od reverzibilnih promena (koraka) A B i C i stanja 1 2 i 3 i koji

je prikazan na slici Popuni Tablice 1 i 2 za dati ciklus

21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]

Stanje P Pa V m3 T K

1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546

Korak Ime procesa q J w J ∆∆∆∆U J

A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0

31 Jedan mol monoatomskog gasa u idealnom gasnom stanju prolazi kroz ciklus

koji se sastoji iz tri procesa što je prikazano na slici Ispuniti tablice 1 i 2

P atm

V L

1

2

3

A B (q=0)

C1

2

224

Stanje P Pa V m3sdotsdotsdotsdot10-3 T K

1 101325 2242 2026503 101325

Proces Tip procesa q J w J ∆∆∆∆U J

A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći

Jedan mola gasa u idealnom gasnom stanju prolazi

kroz termodinamički ciklus koji se sastoji od

a) izotermske kompresije od 2 atm i 10L do 20 atm i

1L

b) izobarske ekspanzije kojom se gas vraća do

zapremine od 10 L a temperatura menja od T1 do T2

c) izohorskog hlantildeenja do početnog stanja

Nacrtati grafik i popuniti tablicu vrednostima q w i

∆∆∆∆U za procese i ciklus

Stepen korisnog dejstva mašine koja hladnjaku preda jednu trećinu količine toplote uzete od grejača je

a) 025 b) 035 c) 067

d) 05 e) ne znam

Izračunati entropiju topljenja (S) u Jmol K za KCl čija je tačka topljenja 7700C Promena entalpije topljenja 268 kJmol

a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34

Za sledeću reakciju na 250CCuO(č)+H2(g)rarrCu(č)+H2O(g)

vrednosti standardnih entropija su S0

CuOč=4263 JKmol S0H2g=13068 JKmol S0

Cuč=3315 JKmol i S0

H2Og=18883 JKmol Odrediti da li će se reakcija odigravati spontano sa aspekta sistema

RešenjeStandardna promena entropije u reakciji je

Za gornju reakciju promena standardne entropije je

Pošto je promena entropije za sistem pozitivna to je reakcija spontana sa aspekta sistema

)tan()( 000 tireakSproduktiSS minus=∆

( )KmolJ

KmolJSSSSS gHcCuOgOHcCu

6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=

=minusminus+=minusminus+=∆

Zadatak 35

36Grafit i dijamant su dve alotropske modifikacije ugljenika Izračunati ∆Ssis ∆Sok i ∆Stot za hemijsku reakciju u kojoj grafit i gasoviti vodonik grade metan

C(graf)+2H2(g)rarrCH4(g) ∆Ho298m=-7481kJmol

574 130684 18626 Somf (JKmol)

Rešenje∆Ssis=18626-2x130684-574=-80848JKmol∆Sok=74810298=251JKmol∆Stot=-80848+251=17019JKmol

Koja od sledećih reakcija je praćena najpozitivnijom promenom entropijea) 2 CO(g) + O2(g)rarr 2 CO2(g)

b) N2(g) + O2(g) rarr 2 NO(g)

c) 2 CH4(g) + O2(g) rarr 2 CH3OH(t)

d) 2 H2O2(t) + N2H4(t) rarr N2(g) + 4 H2O(g)

e) C(č grafit) + H2O(g) rarr CO(g) + H2(g)

Zadatak 37

Zadatak 38Za reakciju CHCl3(t)+Cl2(g)rarrCCl4(t)+HCl(g)na 250C standardne entropije suS0

298(CHCl3(t))=20302 JKmol

S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol

S0298(HCl(g))=18691 JKmol

a toplotni kapaciteti suC0

p(CHCl3(t))=11548 JKmol

C0p(Cl2(g))=3436 JKmol

C0p(CCl4(t))=13263 JKmol

C0p(HCl(g))=2884 JKmol

Odrediti standardnu promenu entropije reakcije na 50oC

Rešenje

sumsum minus=∆j

jTj

i

iTiT tireakSnproduktiSnS )tan()( 0

0

0

KmolJKmolJS 8324)0722322039118653214(0

298 minus=minusminus+=∆

sumsum minus=∆i

iPi

i

iPiP tireakCnproduktiCnC )tan()( 0

0

0

( ) KmolJKmolJCP 63113634481158428631320 =minusminus+=∆

KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0

323 minus=+minus=+minus=∆ int

Zadatak 39

Izračunati promenu entropije kada se idealan gas čiji je CVm=5R2 komprimuje do jedne trećine svoje početne zapremine i istovremeno zagreje do tri puta veće temperature od početne

Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40

Koliki je porast entropije kada zagrevamo 1 mol hloroforma CHCl3 od 240 do 330K ako je Cp=(9147+75sdot10-2T) JmolK

Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41

U sistemu se odigrava proces u kome se entropija menja za551JK Za vreme procesa 15 kJ toplote dodato je sistemu na 350K Da li je proces termodinamićki reverzibilan

Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928

Proces nije termodinamički reverzibilan-ireverzibilan je

Zadatak 42Uzorak bakra (M=63546gmol) mase 275kg i toplotnog kapaciteta 2444JKmol se hladi na konstantnom pritisku od 330 K do 275 K Izračunati a)energiju koja se mora razmeniti u vidu toplote i b) promenu entropije sistemaRešenje

a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus

sdot=∆==∆ minusminus

minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint

Zadatak 43Uzorak azota mase 35 g na 230 K i 211 atm širi se izotermalno do pritiska od 43atm Izračunati promenu entropije gasa

Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆

Zadatak 44Uzorak idealnog gasa u početku na 270K 120 atm i 110Lkomprimuje se izotermalno Do koje zapremine treba da se komprimuje da bi se entropija smanjila za 30JK

Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus

Zadatak 45Jedan mol čvrstog bakra se širi izotermski od 200 bar do 1 bar Izračunati promenu entropije sistema za proces u kome je

a gustina ρ=896sdot103kgm-315106611 minusminussdot=

partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus

minusminusminusminus sdot=minus

sdot

sdotsdotsdotsdotminus

=minusminus=minus=

partpart

minus=∆ intint αα

Zadatak 46Jedan mol CO2(g) na 273 K se hladi do CO2(t) na 1944 K Hlantildeenje se vrši reverzibilno i ireverzibilno stavljanjem uzorka u tečni vodonik na 1396 K Izračunati promenu entropije za proces ako je standardna entalpija isparavanja 231752 kJmol-1 na 1944 K i ako je Cp=3222+(2218sdot10-3)T+(-2347sdot10-6)T2

RešenjeMože se razmatrati proces u dva koraka Prvi je hlantildeenje gasa na konstantnom pritisku do tačke ključanja i drugi prevontildeenje u tečno stanje

KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+

sdotminus+sdot+=∆minus

+=∆

minusminus

minusminusintint

Stoga je ukupna promena entropije za reverzibilan proces

0=∆+∆=∆ oksisiz SSS

Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆

Ako se promena vrši na irevrzibilan način toplota preneta u okolini je

kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak

Toplotna mašina radi izmentildeu 1200K i 500K

a) Kolika je maksimalna efikasnost mašine

b) Izračunati maksimalan rad izvršen za svaki 1kJ toplote uzete iz izvora

c) Koliko toplote se oslobantildea u utok u reverzibilnom procesu za svakih 1kJ uzete toplote iz izvora

47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη

kJkJqw 5830158302 =sdot==η

kJkJkJwqqqqw 4170583012112 =minus=minus=minus=

Zadatak 48

Dat je proces za koji je a) ∆U=0 b) ∆H=0 c) ∆A=0 d) ∆G=0 i e) ∆S=0 Objasniti u kojim procesima je ovo ispunjeno

Rešenjea) U izohorsko-adijabatskimb) U izobarsko-adijabatskimc) U izotermsko-izohorskimd) U izotermsko-izobarskime) U adijabatskim

Zadatak 49

Jedan mol gasa u idealnom gasnom stanju u početku pri normalnim uslovima širi se izotermski i ireverzibilno do 448 L pod takvim uslovom da je rad w=4186J Izračunati ∆S i ∆G

RešenjeP1=1atm T1=27315K V1=224LP2=05atm T2=27315K V2=448L

JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50

Jedan mol gasa u IGS u početku pri zapremini od 5L pritisku P1 i temperaturi od 298K prolazi kroz sledeće reverzibilne promene A)izotermsku kompresiju do polovine početne zapremine B) izohorsko hlantildeenje dok se pritisak ne vrati do početne vrednosti P1 pri temperaturi T3

a) Izračunati P1P2 i T3 kao i q w ∆U ∆H ∆S i ∆G za korake A i B odvojeno

b) Da li su vrednosti za ∆U q i w za korak C (od stanja 1 do 3) jednake vrednostima ovih veličina za sumu koraka A i B

Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆

minus==∆minus=minus=∆

minus=minus=∆=∆=

==∆minus==∆

minus=minus===∆=∆

sdot=====

sdot===

Zadatak 51Promena Gipsove energije za neki izobarski proces može

da se prikaže izrazom

Izračuneti promenu entropije i entalpije za taj proces=minusminus=∆ )(342786 KTJG

RešenjePoznato je da je

( )

JTTSTGH

JTTT

GS

p

786342342786

342342786

minus=+minusminus=∆+∆=∆

=minusminuspartpart

minus=

part∆part

minus=∆

Zadatak 52Kada se dva mola gasa na 330 K i 35 atm izotermski

komprimuje entropija opadne za 250JK Izračunati krajnji

pritisak gasa i promenu Gipsove energije za kompresiju

Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2

=minussdotminus=∆minus=∆

=sdotsdot

sdotsdot==∆

Zadatak 53Kada pritisak 35 g uzorka tečnosti raste izotermalno od

1atm do 3000 atm Gibsova energija raste za 12 kJ Izračunati

gustinu tečnosti (u gcm3)

33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54

Jedan mola idealnog monoatomskog gasa se prevodi iz početnog stanja (224 L 273 K1atm S=20 calK) u krajnje stanje ( 124L 2 atm 303K) Izračunati ∆U ∆H ∆S i ∆G za ovu termodinamičku promenu

Rešenje

bull Označićemo stanja gasa

bull Početno Krajnje

V=224L=224middot10-3m3 V=124L=124middot10-3m T=273 K T=303 K

P= 1 atm=1013middot105Pa P=2atm=2026middot105Pa

S= 20 calKmol=8372JKmol

Pošto je gas u idealnom gasnom stanju to je

∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J

bull Promena entropije i Gipsove funkcije će se odrediti razmatrajući promenu kroz dva stupnja

bull 1 00224m3 273 K 1013middot105Pa rarr 002486 m3

303 K 1013middot105Pa ∆S1=CplnT2T1=(5R2)ln111=2167JK (Skrajnje=2169JK)

∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J

bull 2 002486 m3 303 K 1013middot105Pararr 00124 m3 303 K 2026middot105Pa

∆S2=RlnP1P2=8314x23log05=-5763JK

∆G2=RTlnP2P1=174614J

Ukupne promene su ∆S=2167-5763=-3596JK

∆G=-2544651+174614=-798511J

minus

=

minus=minus= 11)(

2

11

1

2112

VC

nR

VVVadV

VTC

T

TTCTTCw

Adijabatski zapreminski rad

1

2

1

2

2

1

1

2

1

2

2

1

P

P

T

T

V

V

T

T

V

V

T

T R

C

R

CCR PVV

=

=

=

γ

γ

===

1

2

2

1

V

V

P

PconstPVPV VP CC

Adijabatski procesi

Jednačina adijabate

RTnUVPUH gas∆+∆=∆+∆=∆

2

12

2 T

TT

q

w minus=


0gege dST



1

2

1

212 lnln

V

VR

T

TCSSS V +=minus=∆

oksistotok

i

iirev

sis SSST

HSS

T

qS ∆+∆=∆


1

2

1

212 lnln

P

PR

T


tr

trtr

T

HS


V

VV

V T

STCodnosno

T

C

T

S

partpart

==

partpart


1

2

1

ln12

T

T

T

TVVV

TdCT

dTCSSS∆


1

2

1

ln12

T

T

T

TPPP

TdCT

dTCSSS∆

VT T

P

V

S

partpart

=

partpart

PT T

V

P

S

partpart

minus=

partpart

int

partpart

minus=minus=∆ dPT

VSSS

P

12int

partpart

=minus=∆ dVT

PSSS

V

12

Maksveloverelacije

TV V

U

T

PTP

partpart

minus

partpart

=

sum=

minus=minus=∆n

i

iimeš xnRSSS1


TP P

H

T

VTV

partpart

+

partpart

=





TV V

S

T

P

=

partpart

partpart

PV

A

T

minus=

partpart

ST

A

V

minus=

partpart


TP P

S

T

V

minus=

partpart

partpart

ST

G

P

minus=

partpart

VP

G

T

=

partpart

pT

GTHG

+=

part∆part

∆∆2

)(

T

H

T

TG

p

∆part∆part

minus=

0

0 lnP

PRTGG mmm +=

ZADATAK 1


od 50m od

a) Površine Zemlje


Rešenje

a)

b) Jmsmkgmghw

Jmsmkgmghw

695061120

865850819120

2

2



ZADATAK 2



Koliki je rad

Rešenje


Zadatak 3


Koliko je q

Rešenje

w=-4186J ∆U=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


ZADATAK 5




procesu



P

VVo 2Vo

Po

2Po

1

2

Rešenje

ooooo VPVPP

w2

3

2

)2(=

+=

ZADATAK 6




a) 2 b) 3 c) 4


Rešenje

LdmmV

JVm

JVPw

JJJwwqU

410104100000

400

400101

400100300

3333

3

5

=sdotsdot=minus

minus=∆



minus



bull Rešenje



KKT

CCK

CgJ

gJ

C

U

UCCgJCgJ

g

J

m

UU

k

o

pko

o

vs

s

svs

o

vss

07326921715308

925292173592179217

1864

75

18647512

900

=+=

=+=∆+====

∆=∆

∆∆

====∆

=∆

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJ

P

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

3

5

=sdotsdot+=∆

=sdot

==


ZADATAK 9



Rešenje



nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

2

12

2 T

TT

q

w minus=


0gege dST



1

2

1

212 lnln

V

VR

T

TCSSS V +=minus=∆

oksistotok

i

iirev

sis SSST

HSS

T

qS ∆+∆=∆


1

2

1

212 lnln

P

PR

T


tr

trtr

T

HS


V

VV

V T

STCodnosno

T

C

T

S

partpart

==

partpart


1

2

1

ln12

T

T

T

TVVV

TdCT

dTCSSS∆


1

2

1

ln12

T

T

T

TPPP

TdCT

dTCSSS∆

VT T

P

V

S

partpart

=

partpart

PT T

V

P

S

partpart

minus=

partpart

int

partpart

minus=minus=∆ dPT

VSSS

P

12int

partpart

=minus=∆ dVT

PSSS

V

12

Maksveloverelacije

TV V

U

T

PTP

partpart

minus

partpart

=

sum=

minus=minus=∆n

i

iimeš xnRSSS1


TP P

H

T

VTV

partpart

+

partpart

=





TV V

S

T

P

=

partpart

partpart

PV

A

T

minus=

partpart

ST

A

V

minus=

partpart


TP P

S

T

V

minus=

partpart

partpart

ST

G

P

minus=

partpart

VP

G

T

=

partpart

pT

GTHG

+=

part∆part

∆∆2

)(

T

H

T

TG

p

∆part∆part

minus=

0

0 lnP

PRTGG mmm +=

ZADATAK 1


od 50m od

a) Površine Zemlje


Rešenje

a)

b) Jmsmkgmghw

Jmsmkgmghw

695061120

865850819120

2

2



ZADATAK 2



Koliki je rad

Rešenje


Zadatak 3


Koliko je q

Rešenje

w=-4186J ∆U=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


ZADATAK 5




procesu



P

VVo 2Vo

Po

2Po

1

2

Rešenje

ooooo VPVPP

w2

3

2

)2(=

+=

ZADATAK 6




a) 2 b) 3 c) 4


Rešenje

LdmmV

JVm

JVPw

JJJwwqU

410104100000

400

400101

400100300

3333

3

5

=sdotsdot=minus

minus=∆



minus



bull Rešenje



KKT

CCK

CgJ

gJ

C

U

UCCgJCgJ

g

J

m

UU

k

o

pko

o

vs

s

svs

o

vss

07326921715308

925292173592179217

1864

75

18647512

900

=+=

=+=∆+====

∆=∆

∆∆

====∆

=∆

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJ

P

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

3

5

=sdotsdot+=∆

=sdot

==


ZADATAK 9



Rešenje



nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

V

VV

V T

STCodnosno

T

C

T

S

partpart

==

partpart


1

2

1

ln12

T

T

T

TVVV

TdCT

dTCSSS∆


1

2

1

ln12

T

T

T

TPPP

TdCT

dTCSSS∆

VT T

P

V

S

partpart

=

partpart

PT T

V

P

S

partpart

minus=

partpart

int

partpart

minus=minus=∆ dPT

VSSS

P

12int

partpart

=minus=∆ dVT

PSSS

V

12

Maksveloverelacije

TV V

U

T

PTP

partpart

minus

partpart

=

sum=

minus=minus=∆n

i

iimeš xnRSSS1


TP P

H

T

VTV

partpart

+

partpart

=





TV V

S

T

P

=

partpart

partpart

PV

A

T

minus=

partpart

ST

A

V

minus=

partpart


TP P

S

T

V

minus=

partpart

partpart

ST

G

P

minus=

partpart

VP

G

T

=

partpart

pT

GTHG

+=

part∆part

∆∆2

)(

T

H

T

TG

p

∆part∆part

minus=

0

0 lnP

PRTGG mmm +=

ZADATAK 1


od 50m od

a) Površine Zemlje


Rešenje

a)

b) Jmsmkgmghw

Jmsmkgmghw

695061120

865850819120

2

2



ZADATAK 2



Koliki je rad

Rešenje


Zadatak 3


Koliko je q

Rešenje

w=-4186J ∆U=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


ZADATAK 5




procesu



P

VVo 2Vo

Po

2Po

1

2

Rešenje

ooooo VPVPP

w2

3

2

)2(=

+=

ZADATAK 6




a) 2 b) 3 c) 4


Rešenje

LdmmV

JVm

JVPw

JJJwwqU

410104100000

400

400101

400100300

3333

3

5

=sdotsdot=minus

minus=∆



minus



bull Rešenje



KKT

CCK

CgJ

gJ

C

U

UCCgJCgJ

g

J

m

UU

k

o

pko

o

vs

s

svs

o

vss

07326921715308

925292173592179217

1864

75

18647512

900

=+=

=+=∆+====

∆=∆

∆∆

====∆

=∆

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJ

P

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

3

5

=sdotsdot+=∆

=sdot

==


ZADATAK 9



Rešenje



nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

TV V

U

T

PTP

partpart

minus

partpart

=

sum=

minus=minus=∆n

i

iimeš xnRSSS1


TP P

H

T

VTV

partpart

+

partpart

=





TV V

S

T

P

=

partpart

partpart

PV

A

T

minus=

partpart

ST

A

V

minus=

partpart


TP P

S

T

V

minus=

partpart

partpart

ST

G

P

minus=

partpart

VP

G

T

=

partpart

pT

GTHG

+=

part∆part

∆∆2

)(

T

H

T

TG

p

∆part∆part

minus=

0

0 lnP

PRTGG mmm +=

ZADATAK 1


od 50m od

a) Površine Zemlje


Rešenje

a)

b) Jmsmkgmghw

Jmsmkgmghw

695061120

865850819120

2

2



ZADATAK 2



Koliki je rad

Rešenje


Zadatak 3


Koliko je q

Rešenje

w=-4186J ∆U=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


ZADATAK 5




procesu



P

VVo 2Vo

Po

2Po

1

2

Rešenje

ooooo VPVPP

w2

3

2

)2(=

+=

ZADATAK 6




a) 2 b) 3 c) 4


Rešenje

LdmmV

JVm

JVPw

JJJwwqU

410104100000

400

400101

400100300

3333

3

5

=sdotsdot=minus

minus=∆



minus



bull Rešenje



KKT

CCK

CgJ

gJ

C

U

UCCgJCgJ

g

J

m

UU

k

o

pko

o

vs

s

svs

o

vss

07326921715308

925292173592179217

1864

75

18647512

900

=+=

=+=∆+====

∆=∆

∆∆

====∆

=∆

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJ

P

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

3

5

=sdotsdot+=∆

=sdot

==


ZADATAK 9



Rešenje



nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J


TV V

S

T

P

=

partpart

partpart

PV

A

T

minus=

partpart

ST

A

V

minus=

partpart


TP P

S

T

V

minus=

partpart

partpart

ST

G

P

minus=

partpart

VP

G

T

=

partpart

pT

GTHG

+=

part∆part

∆∆2

)(

T

H

T

TG

p

∆part∆part

minus=

0

0 lnP

PRTGG mmm +=

ZADATAK 1


od 50m od

a) Površine Zemlje


Rešenje

a)

b) Jmsmkgmghw

Jmsmkgmghw

695061120

865850819120

2

2



ZADATAK 2



Koliki je rad

Rešenje


Zadatak 3


Koliko je q

Rešenje

w=-4186J ∆U=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


ZADATAK 5




procesu



P

VVo 2Vo

Po

2Po

1

2

Rešenje

ooooo VPVPP

w2

3

2

)2(=

+=

ZADATAK 6




a) 2 b) 3 c) 4


Rešenje

LdmmV

JVm

JVPw

JJJwwqU

410104100000

400

400101

400100300

3333

3

5

=sdotsdot=minus

minus=∆



minus



bull Rešenje



KKT

CCK

CgJ

gJ

C

U

UCCgJCgJ

g

J

m

UU

k

o

pko

o

vs

s

svs

o

vss

07326921715308

925292173592179217

1864

75

18647512

900

=+=

=+=∆+====

∆=∆

∆∆

====∆

=∆

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJ

P

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

3

5

=sdotsdot+=∆

=sdot

==


ZADATAK 9



Rešenje



nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

ZADATAK 1


od 50m od

a) Površine Zemlje


Rešenje

a)

b) Jmsmkgmghw

Jmsmkgmghw

695061120

865850819120

2

2



ZADATAK 2



Koliki je rad

Rešenje


Zadatak 3


Koliko je q

Rešenje

w=-4186J ∆U=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


ZADATAK 5




procesu



P

VVo 2Vo

Po

2Po

1

2

Rešenje

ooooo VPVPP

w2

3

2

)2(=

+=

ZADATAK 6




a) 2 b) 3 c) 4


Rešenje

LdmmV

JVm

JVPw

JJJwwqU

410104100000

400

400101

400100300

3333

3

5

=sdotsdot=minus

minus=∆



minus



bull Rešenje



KKT

CCK

CgJ

gJ

C

U

UCCgJCgJ

g

J

m

UU

k

o

pko

o

vs

s

svs

o

vss

07326921715308

925292173592179217

1864

75

18647512

900

=+=

=+=∆+====

∆=∆

∆∆

====∆

=∆

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJ

P

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

3

5

=sdotsdot+=∆

=sdot

==


ZADATAK 9



Rešenje



nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

ZADATAK 2



Koliki je rad

Rešenje


Zadatak 3


Koliko je q

Rešenje

w=-4186J ∆U=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


ZADATAK 5




procesu



P

VVo 2Vo

Po

2Po

1

2

Rešenje

ooooo VPVPP

w2

3

2

)2(=

+=

ZADATAK 6




a) 2 b) 3 c) 4


Rešenje

LdmmV

JVm

JVPw

JJJwwqU

410104100000

400

400101

400100300

3333

3

5

=sdotsdot=minus

minus=∆



minus



bull Rešenje



KKT

CCK

CgJ

gJ

C

U

UCCgJCgJ

g

J

m

UU

k

o

pko

o

vs

s

svs

o

vss

07326921715308

925292173592179217

1864

75

18647512

900

=+=

=+=∆+====

∆=∆

∆∆

====∆

=∆

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJ

P

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

3

5

=sdotsdot+=∆

=sdot

==


ZADATAK 9



Rešenje



nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 3


Koliko je q

Rešenje

w=-4186J ∆U=0 q=4186J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


ZADATAK 5




procesu



P

VVo 2Vo

Po

2Po

1

2

Rešenje

ooooo VPVPP

w2

3

2

)2(=

+=

ZADATAK 6




a) 2 b) 3 c) 4


Rešenje

LdmmV

JVm

JVPw

JJJwwqU

410104100000

400

400101

400100300

3333

3

5

=sdotsdot=minus

minus=∆



minus



bull Rešenje



KKT

CCK

CgJ

gJ

C

U

UCCgJCgJ

g

J

m

UU

k

o

pko

o

vs

s

svs

o

vss

07326921715308

925292173592179217

1864

75

18647512

900

=+=

=+=∆+====

∆=∆

∆∆

====∆

=∆

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJ

P

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

3

5

=sdotsdot+=∆

=sdot

==


ZADATAK 9



Rešenje



nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 4


a) 22000 b) -101325 c)-20265

d) -20 e) 20288 f) 4056

Rešenje


ZADATAK 5




procesu



P

VVo 2Vo

Po

2Po

1

2

Rešenje

ooooo VPVPP

w2

3

2

)2(=

+=

ZADATAK 6




a) 2 b) 3 c) 4


Rešenje

LdmmV

JVm

JVPw

JJJwwqU

410104100000

400

400101

400100300

3333

3

5

=sdotsdot=minus

minus=∆



minus



bull Rešenje



KKT

CCK

CgJ

gJ

C

U

UCCgJCgJ

g

J

m

UU

k

o

pko

o

vs

s

svs

o

vss

07326921715308

925292173592179217

1864

75

18647512

900

=+=

=+=∆+====

∆=∆

∆∆

====∆

=∆

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJ

P

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

3

5

=sdotsdot+=∆

=sdot

==


ZADATAK 9



Rešenje



nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

ZADATAK 5




procesu



P

VVo 2Vo

Po

2Po

1

2

Rešenje

ooooo VPVPP

w2

3

2

)2(=

+=

ZADATAK 6




a) 2 b) 3 c) 4


Rešenje

LdmmV

JVm

JVPw

JJJwwqU

410104100000

400

400101

400100300

3333

3

5

=sdotsdot=minus

minus=∆



minus



bull Rešenje



KKT

CCK

CgJ

gJ

C

U

UCCgJCgJ

g

J

m

UU

k

o

pko

o

vs

s

svs

o

vss

07326921715308

925292173592179217

1864

75

18647512

900

=+=

=+=∆+====

∆=∆

∆∆

====∆

=∆

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJ

P

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

3

5

=sdotsdot+=∆

=sdot

==


ZADATAK 9



Rešenje



nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

ZADATAK 6




a) 2 b) 3 c) 4


Rešenje

LdmmV

JVm

JVPw

JJJwwqU

410104100000

400

400101

400100300

3333

3

5

=sdotsdot=minus

minus=∆



minus



bull Rešenje



KKT

CCK

CgJ

gJ

C

U

UCCgJCgJ

g

J

m

UU

k

o

pko

o

vs

s

svs

o

vss

07326921715308

925292173592179217

1864

75

18647512

900

=+=

=+=∆+====

∆=∆

∆∆

====∆

=∆

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJ

P

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

3

5

=sdotsdot+=∆

=sdot

==


ZADATAK 9



Rešenje



nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J



bull Rešenje



KKT

CCK

CgJ

gJ

C

U

UCCgJCgJ

g

J

m

UU

k

o

pko

o

vs

s

svs

o

vss

07326921715308

925292173592179217

1864

75

18647512

900

=+=

=+=∆+====

∆=∆

∆∆

====∆

=∆

θθθθ

θ



molkJmolmPamolJH

molmPa

KKmolJ

P

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

3

5

=sdotsdot+=∆

=sdot

==


ZADATAK 9



Rešenje



nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J



molkJmolmPamolJH

molmPa

KKmolJ

P

RTV

VVVVVVPUH

isp

p

tptp

843031010140700

031010

3733148

35

3

5

=sdotsdot+=∆

=sdot

==


ZADATAK 9



Rešenje



nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

ZADATAK 9



Rešenje



nRTP

sp

sp minus=minus

kJJ

KKmolJmolg

gmolmolw

1010329

152983145818

5051

minus=minus=

=sdotsdotsdotminus=

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 10Domaći



Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 11




toplotu i rad

Rešenje

JqU

JKKmolJmoldTCnqw

PaK

KPa

T

TPP

T

T

v

11247

1124710031485110

135100300

400101325

2

1

1

212

==∆

=sdotsdotsdot===

=sdot=sdot=

int

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 12


Rešenje


ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

ZADATAK 13


je ∆H=-2058 kJ

Rešenje


∆ng=3-5=-2


ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

ZADATAK 14



dat izrazom



1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 15





Rešenje


nRT=18200J T= 182002middot8314=1093K

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 16



Rešenjea)

b)


J

dm

dmKKmolJ

molg

g

V

VnRTw

7152

518

21ln3053148

40

566ln

3

3

1

2

minus=


Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 17


Rešenje

JKKmolJmolJTnRHU

JKKmolJTnCqH pmP

4590625314831530

1530)260285(4203



ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

ZADATAK 18





d) CV(T1-T2) e) f)

1

2lnV

VRTminus

minus

1

2

1

1

VC

R

V V

VTC

minus

1

2

12

R

C

V

V

V

VTC

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

ZADATAK 19




a) 1055 b) 905 c) 2056

d)996 e) 1110 f) ne znam

Rešenje

KTRT

R

RRCCP

PR

T

TC VPP

59020

1ln

300ln

2

5

2

5lnln

22

1

2

1

2

==

=+==

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

ZADATAK 20




Rešenje

barkPakPaPPV

VP 31031100

2

1 3

5

21

2

12 ==

=

=

γ


ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

ZADATAK 21




JK-1mol-1

Rešenje

JUq

JKmolJKmolw

KKL

LT

V

VT

V

VT

VmV C

R

C

nR

182750

18275)152988187(481220

91871529801

50

11

6660

2

11

2

12

minus=∆=


=sdot

=

=

minusminus

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

ZADATAK 22 Domaći




pritisak ∆U q i w

Rešenje



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J



Zadatak 23


ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

ZADATAK 24





Rešenje

4

4

0

3 aTdTaTH

T

==∆ int

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 25




pTRTpV )(α+=

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Rešenjea)

b)

[ ])()()()(

)()(

121212

111222

TTpTTRVVpw

pTRTpVpTRTpV

αα

αα


+=+=

)(

)(ln

)(

)(ln

)( 2

1

1

2

2

1

2

1TV

TVRT

TV

TVRTdV

TV

RTpdVw

V

V

V

Vαα

αα

α minusminus

=minusminus

minus=minus

minus=minus= intint

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 26


Rešenje

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

JKmolKJmolJH

TnRUnPVUPVUH

JwqUKT

KmolJmol

J

C

wTTCwJw

mPaVPwq

m

V

V

4551)524(3148553532

)(

53532524

)31481137(5

535353532

10)15154(105780 333


∆+∆=∆+∆=∆+∆=∆


minussdotminus

==∆∆sdot=minus=




2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J



2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J


2-V12)

=-335Latm ili w=33944Jq=∆U-w=136144J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

ZADATAK 28






)(400101720)( KTKJC p +=

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Rešenje

a) P=const

V=const

kJJTnRHU

JKKmolJmolTnRVPw

JqH

dTdTCHq

P

PP

114831411448312314946

483110031481

2314946)1527315373(200050)273373(1720

)4002001720(

22

373

273

373

273



=minus+minus==∆

+==∆= int int

qkJHUw ==∆=∆= 1140

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 29






korake A B i C

Rešenje








21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J




21

3

A

BC

273 546

224

448

V[[[[dm3]]]]

T[[[[K]


1 224sdot10-3 273

2 224sdot10-3 546

3 448sdot10-3 546


A

B

C

Ciklus

Tablica 2

Tablica 1

1013sdot105

2026sdot105

1013sdot105

izohorski

izotermski

izobarski

340458 0 340458

-314650 0314650

-56743 226912 -34045887678 -87738 0



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J



P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 2026503 101325


A

B

C

ciklus

Tablica 1

Tablica 2

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

31 Rešenje

P atm

V L

1

2

3

A B (q=0)

C1

2

224


1 101325 2242 202650 2243 101325 3395


A

B

C

ciklus

Tablica 1

Tablica 2

izohorski

adijabatski

izobarski

34046 340460

0 -16488 -16488

-29263 11703 -17558

4783 -4785 0

273546

41379

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

32 Domaći




1L







a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J


a) 025 b) 035 c) 067

d) 05 e) ne znam


a) 348 b) 0035 c)257 d) 0026e)4879 f) ne znam

Zadatak 33

Zadatak 34










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J










( )KmolJ


6748

6813063428318815330

)(

0

)(

0

)(

0

)(

0

22

=


Zadatak 35



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J



574 130684 18626 Somf (JKmol)







Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J






Zadatak 37



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J



S0298(Cl2(g))=22307 JKmol

S0298(CCl4(t))=21453 JKmol








Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Rešenje

sumsum minus=∆j

jTj

i


0

0

KmolJKmolJS 8324)0722322039118653214(0


sumsum minus=∆i

iPi

i


0

0


KmolJdTT

S 8923298

323ln63118324

63118324

323

298

0


Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 39


Rešenje

molKnJnRnRRn

nRRnV

VnR

T

TnCS

VVTTRC

mV

mv

7133ln2

33ln3ln

2

5

3

1ln3ln

2

5lnln

33

2

5

1

2

1

2

1212

==minussdot

=+sdot=+=∆

===

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 40


Rešenje

[ ] [ ] molKJTT

T

dTCSSS p

883575613291057ln4791

T

dTT)1075(9147

330

240

2330

240

330

240

2-

330

240

240330

=+=sdot+=

=sdot+==minus=∆

minus

intint

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 41


Rešenje

revrev

revrev

sis

qqJqJq

JTSqT

qS

ne==

=sdot∆==∆

519281500

51928



a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J


a)

b)

kJJ

KmolJKgmol

gTnCqH mpp

25810825

)55)(4424(54663

10752

4

11

1

3

minus=sdotminus=

=minus


minus

KJKJKmolJmol

T

dTnC

T

dTC

T

qS

T

T

mp

T

T

p

T

T

rev

193109281)330275ln()4424(343 2

2

1

2

1

2

1

minus=sdotminus==

====∆ intintint


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J


Rešenje

KJKmolJmolg

g

p

pnRS 516)

34

121ln()3148(

01428

35ln

2

1 =

==∆


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J


Rešenje

molKKmolLatm

Latm

RT

PVn 5960

2700820

1121=

sdotsdot

==

LL

molJKmolJKLnRSVV

V

V

nR

S

V

V

nR

S

6546011

)3148)(5960(03exp()11()exp(

)exp(ln

111

12

1

2

1

2

=sdot

=minus=∆=

=∆

=∆

minusminusminus



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J



partpart

= KT

V

V P

α

Rešenje

KJbarkgm

kgmolmolK

PPVdPVdPT

VsistemS

P

PP

10342)2001(10968

1054663110661

)()(

3

33

1315

12

2

1

minusminus


sdot


=minusminus=minus=

partpart




KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J



KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

KJS

KJ

dTTTT

H

T

dTCS

ok

T

T klj

isp

Psis

47131

46131211192512

4194

223175))273()4194((

2

)104723()2734194)(101822(

273

4194ln2232

4194

223175)104723()101822(

2232

226

3

4194

273

63

02

1

=∆

minus=minusminus=

=minus

+minussdotminus

+minussdot+

=

=minus

+


+=∆

minusminus

minusminusintint



Reverzibilan proces

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

KJS

KJK

J

T

qS

tot

ok

71731461311863

18639613

26010

=minus=∆

===∆


kJ

Jmol

HdTCq ispP

0126

)223175())273()4194((3

104723))273()4194((

2

101822)2734194(2232

)(

336

223

4194

273

0

=

=

minusminus

sdotminusminus

sdot+minusminus=

=∆minusminus=

minusminus

int

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

47 Zadatak





47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

47 Rešenje

a)

b)

c)

58301200

5001200

2

12 =minus

=minus

=T

TTη



Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 48



Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 49



JP

PRTG

KJV

VRS

11574ln

765ln

1

2

1

2

minus==∆

==∆

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 50




Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Rešenje

P1

AB

C

VL

P

P 2

25 5

CumanjewiqistoUb

STSTHG

KJT

TRSJTTRH

JKKRTCUwB

JP

PRTGKJ

V

VRS

JV

VRTwqHUA

PaPLVVKTT

PaPLVKTa

v

)

)(

765ln03097)(2

5

181858)298149(2

30)

321717ln765ln

321717ln0)

10929522298

109645298)

2233

1

323

1

2

1

2

1

2

5

21221

5

111

∆

minusminus∆=∆



==∆minus==∆


sdot=====

sdot===





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J





( )

JTTSTGH

JTTT

GS

p

786342342786

342342786


=minusminuspartpart

minus=

part∆part

minus=∆




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J




Rešenje

PaKmolJmol

KJPaP

nR

SPP

nR

S

P

P

P

PnRS

65

2

12

2

1

2

1

10595131482

25exp1054553

expexpln

sdot=

sdotminus

minussdot=

∆minus=

∆==∆

JKJKSTG

JP

PnRTG

825025330

882461054553

105941ln33031482ln

5

6

1

2


=sdotsdot

sdotsdot==∆




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J




33

53

8860886

12000

10013129991035

cmgmkg

J

Pakg

G

Pm

V

mPVG

==

sdotsdotsdotsdot=

∆∆

==∆=∆minus

ρ

ρρ

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Zadatak 54


Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Rešenje







∆U=Cv∆T=(3R2)x30=37413J ∆H=Cp∆T=(5R2)x30=62355J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J




∆G1=∆H 1-(T2S2-T1S1)=62355-(303x85887-273x8372)=

-2544651 J


∆S2=RlnP1P2=8314x23log05=-5763JK



∆G=-2544651+174614=-798511J

Documents

I zakon termodinamike dU =dq+dw+ dw =dU+ pdVMaksvelove relacije V V T U T P P T ∂ ∂ − ∂ ∂ = ∑ = ∆ = − =− n i S meš S S R n i x i 1 ... Iz jednačine adijabate se