Ejercicios Segunda Ley-Pablo Chango

8/18/2019 Ejercicios Segunda Ley-Pablo Chango

1/15

UNIVERSIDAD CENTRAL DEL ECUADORFACULTAD DE INGENIERÍA EN GEOLOGÍA, MINAS, PETRÓLEOS Y AMBIENTAL

ESCUELA DE PETRÓLEOSFISICOQUIMCA Y TERMODINAMICA

PROFERSOR: Ing. Bolivar Enriquez FECHA: 2013-07-12

NOMBRE: Pablo Antonio Chango DEBER: Nº 02CURSO: Cuarto Petróleos CALIFICACION:TEMA: Eer!i!ios "egun#a $e%

Lib!: Fi"i#!$%i&i#' D( L%i" R!&!

)* U+ &! -( .'" i-(' &!+!'/&i#! $%( ("/1 ' 03C 4 $%( !#%5' ) -& "( (65'+-( 7'"/' )00 -&*

C'#%' ( #'&bi! -( (+/!58' DATOS& ' 30 (C)1 ' 10 #*

3

)2' 100 #*3

2* U+' (9i.('-!' $%( 9%+#i!+' (+/( 03C 4 23C /i(+( 0*0 -( (9i#'#i'* C'#%': '; ( /'b'

5'' #!+.(' )00 .'&!" -( '.%' 4 b; ( #'! $%( 5i(-( ' 23C';DATOS&1 ' 0(C&2 ' 2+(C*agua ' 100g

b;DATOS&1 ' 2+(C

* P'' ( #i#! -( C'+!/ "( /'=' ( -i'.'&' P > V, "(?'' !" "i.+!" -( V, T, P, @U, @H 4 @S 5''

#'-' %+! -( !" #%'/! 5'"!"*

∆ S= R ln (V 2V 1 )

∆ S=8.3144J

molK ln ( 1001 )

∆ S=38.289226J

mol K Q=mc∆T

Q=100 g∗4.186J

g ° C ∗25° C

Q=10465J

L=∈∗Q=0.50∗10465 J Q=

L∗T 2T 2−T 1

Q=5232.5J ∗298.15

298.15−273.15


2/15

A > B: El gas se en!uentra en el esta#o A a P1, )1 a la te*eratura &2, se ean#e isot/r*i!a*ente hasta

B P2, )2, &2 #e )1 a )2 !on la absor!ión #el !alor 2 que es igual al trabao *i*o $2.

Al au*entar el volu*en se ro#u!e un au*ento en el #esor#en #e las *ol/!ulas #el gas, en tal situa!ión

P2˃P1

∆U =∫T 1

T 2

CvdT

∆ H =∫T 1

T 2

CpdT

"ien#o evi#ente que 45 % 46 son iguales a !ero.B C: El gas en el esta#o B P2, )2, &2 se asila a#iabti!a*ente !on el in #e re#u!ir a !ero el luo #e

!alor # ' 0, #e B a C se ro#u!e una eansión a#iabti!a #e )2 a )3 !on la #is*inu!ión #e la

te*eratura &1, !o*o !onse!uen!ia el gas ier#e una !anti#a# #e energ8a igual a $ 2, en tal !aso se

ro#u!e una #is*inu!ión #e 45 % 46, "eg9n la segun#a le% #e la ter*o#in*i!a " ' # ara siste*a

reversible, en tal situa!ión #" ' 0.C D: El gas en el esta#o C P3, )3, &1 so*eti#o a la !onversión, se !o*ri*e isot/r*i!a*ente a &1

!on la re#u!!ión #e volu*en #e )3 a ): !on una /r#i#a #e !alor ;1 '


3/15

>' 0.22 100

22*

* I-(+/i9i$%( %+ 5!#("! 5'' ' #%' U 0, H 0 4 S 0 4 (65i$%( "%" ("5%("/'"*

En un ro!eso isoter*al reversible resenta 6' 0, 5' 0 uesto queD #6'Cv.


4/15

* A ) &! -( '.%' #!+.('-' ' 0ºC $%( ("/1 (+ %+ 9'"#! -( D(' "( '?'-( )0 &!(" -( '.%' '

2ºC* E 7i(! "( 9%+-( 4 (+98' ' '.%'* L' (+/'58' -( 9%"i+ -( '.%' (" ∆H f$%=5941.3J

mol

* C'#%' K'; ∆ H & Kb; T final 4 K#; ∆ S 5'' ( 5!#("!*

a

¿C ¿ p=dH dT

Q=n∫T 1

T 2

C pdT

C p=f (T )

Q p=nC p (T 2−T 1 )

nC p(T 2−T −1)=¿Q p=0

Comoel %i%temae% adia'(tico :∑ ¿

b; ( #'! ' #(! .'-!" (":

' 1*ol +:1.3J

mol

El hielo gana !alor #es#e 273.1+F@

'1*ol 7+.:GJ

K mol (T m=273.15 )

$as 10 *oles #e agua ier#en !alor 23.1+

' 10*ol 7+.:GJ

K mol (T m=298.15 )

Q1+Q2=Q3

nC p=(T m−273.15 )+nC p8298.15−T m¿=5941.3

T m=288.6 !K

#; C'&bi! -( (+/!58'

S1=∆H

273.15=

5941.3

273.15=21.75

J

) mol

S2 %er( el cam'iodeentropía dedeel p$ntode f$%i*n+a%tala temperat$radelame,cla


5/15

* ) &! -( .'" i-(' &!+!'/&i#! $%( ("/1 ' 200 º "( (65'+-( '-i'b1/i#'&(+/( -( dm3

'

0 dm3

* C'#%' ( #'&bi! -( (+/!58' 4 ' /(&5('/%' -( .'"*

a

T ∗V ( - −1)=con%tante Por Poisson

T 1

∗V 1

(- −1)=T 2

∗V 2

( - −1)

T 2

T 1

=(V 1V 2 )

( - −1 )

T 2=(V 1V 2 )( - −1)

∗T 1

T 2=( 550 )(

5

3−1)∗300

T 2=61.633!K

b

S= R∗C V ∗lnT 2

T 1

S=8.314∗5

3 ∗ln

61.633

300

S=−19.15 J mol.K

* D(iJ' '" (#%'#i!+(" ( /S/T )V =Cp

T ( /S/T ) =1

T ∗ ( / H / )T −

~V &

Cv∗lnT 2

T 1+ R∗ln

V 2

V 1=0

dV =dQ− . dV 1ra le% #e la ter*o#in*i!a

dV =T.dS− . dV 2#a le% #e la ter*o#in*i!a (dQ=T.dS)

dS=dV

T +

. dV

T


6/15

( /S/T )V =Cv

T &o*an#o la #eriva#a ar!ial

P! !/! '-!:5 ' &,P

dH =

(/ H

/T

)

∗dT +

(/ H

/

)T

∗d

dH =dV + .dV +V.dodH =dQ− . dV + . dV +V .d

dH

T =dS+

V

T ∗d

"i el siste*a es reversible

dS=

( / H /T ) ∗dT T +

( / H / )T ∗dT −

V

T ∗d Igualan#o

dS=Cp∗dT

T +

[( /H / )T −V ]∗dT

Heran#o

( /S/ )∗T = 1T ∗[( / H / )∗T −V ]

Si ( .'" (" i-('

dS=Cv∗dT

T +

T ∗dV

dS=Cv∗dT

T +

R

V ∗dV

Integran#o

S= R∗C V ∗lnT 2

T 1+ R∗ln

V 2

V 1 si S=0

R∗C V ∗lnT 2

T 1+ R∗lnV 2

V 1=0

)0* L' #'5'#i-'- #'!89i#' -( '&!+i! "( -(9i+( &(-i'+/( ' (#%'#i+:

C5 2,7+ = 2+,1010-3 &-1,++&-2J*ol-1@-1

Cal!ular a 5 b "

a "e re!ono!e queD


7/15

5' ∫T 1

T 2

CpdT

5'

¿¿

∫T 1

T 2

¿ 2,7+ = 2+,1010-3 &-1,++&-2#&

ue reresenta el 5 #el a*onia!o.

5'

¿¿

∫300

1000

¿ 2,7+ = 2+,1010-3 &-1,++&-2#&

5' G7+,:+J*ol-1@-1

b "' ∫T 1

T 2

CpdT K&

"'

¿¿

∫T 1

T 2

¿ 2,7+ = 2+,1010-3 &-1,++&-2#&K&

"'

¿¿

∫300

1000

¿ 2,7+&-1

= 2+,1010-2

&-1,++&-1

#&

5' 3.1 J*ol-1@-1

))* ) &! -( '.%' "( #'i(+/' ' ) '/&"9(' -( 5("i+ ' 0ºC 7'"/' ( ("/'-! -( J'5! ' )00ºC* E#'! ("5(#89i#! -( '.%' (" ,)0 &!)) 4 ( #'! -( J'5!i='#i+ (" 220 . ) *C'#%' (#'&bi! -( (+/!58'*

H2OKL;K2,); H2OKL;K,) º;

En tal !asoD

"&' nCIn&2KL1= 5!K&

"&' 1*ol7G,G J*ol-1@-1In373,1+K273,1+=22M0JKg1GK1*ol 373,1+ F@

"&' 23,02=10,2

"&'132,0: JF@*ol-1

)2* L' (#%'#i+ -( ("/'-! -( %+ .'" (" PKVb;RT "i(+-! CJ'bT#T2 -(iJ' %+' (#%'#i+ 5''#'#%' ( #'&bi! -( (+/!58' 5'' ( #'"! (+ $%( ) &! -( .'" "( "!&(/( ' %+ #'&bi! -( T) ' T2

4 -( V) ' V2.


8/15

"e !ono!e queD

#'#6=P#)

"i el ro!eso es reversible #'" % #$'P#)

&ene*osD

dU =TdS0dV dS=dU

T +

T dV

Enton!esD

dS=CvdT

T +

T dV

Integran#o

∆ S=∫T 1

T 2

(a+'T +cT 2)dT

T + R∫

V 1

V 2dV

V −'

T 2

(¿¿2−T 12)2

+ R lnV 2−'V 1−'

∆ S=a lnT 2

T 1+' (T 2−T 1 )+c ¿

)* )00 . -( ("/'?! KC52,2 &!)); ' )003C "( '?'-( ' )00. -( '.%' KC5, &! )); ' 2

3C (+ %+ 9'"#! '-i'b1/i#!* C'#%' K'; ' /(&5('/%' 9i+' -( "i"/(&' 4 Kb; !" #'&bi!" -((+/!58' 5'' ( ("/'?! 4 ( '.%'*

K'; En un siste*a a#iabti!o #'0

∑mi"Cpi"Ti

Calor !e#i#o or el estaNo 'Calor gana#o or el agua

100g " (25,52/118,69)(100−Tm)=100 g "(75,35 /18)(Tm−25)

&' 2G,MM(C

Kb; El !a*bio #e entro8a

#" ' nC#&K&

C es !onstante

∆ SSn= 100g

118,7gmol−1

"25,52J

mol°K ln

301,81

373,15


9/15

∆ SSn=−4,55J

mol° K

∆ S H 21= 100 g

18 gmol−1

"75,3J

mol° K ln

301,8

298,15

∆ S H 21=5,1J

mol° K

)*) &! -( .'" i-(' &!+!'/&i#! i+i#i'&(+/( ' ) '/&!"9(' -( 5("i+ 4 00 "( "!&(/( ' %+' /'+"9!&'#i+ #8#i#' (J("ib( $%( i+#%4( !" "i.%i(+/(" 5'"!": '; #!&5("i+ i"!/&i#' ' 2'/&!"9('" b; i+#(&(+/! -( /(&5('/%' ' 00 ' 5("i+ #!+"/'+/( 4 #; /(&i+'#i+ -( #i#! '("/'-! i+i#i' &(-i'+/( ( #'&i+! "(?''-! 5! 5'bT* T'=' %+ -i'.'&' 4 #'#%'

∆U 2 ∆H 2 ∆ S .

P

2 ------------------

1 ------------------

1 2 &(@

P'' ∆U P'' S

6 '0 "'Cln&2K&1-LlnP1KP2

6 ' Cv &2-&1 "'+.7MJK*ol(@

6'3

2(8.314 ) (400−300)

"'Cln&2K&1-0

6'12:7.1JK*ol "'5

2(6.314) ln( 400300 )

6'-12:7.1JK*ol "' +.7 JK*ol(O

"' Cln&2K&1-LlnP1KP2


10/15

"'5

2 Rln( 300400 )− Rln( 21 )

P'' H "'-11.73JK*ol(O

5'0

5'C &

5' 2GG+JK*ol

5' -2GG+JK*ol

P'' ( #i#! #!&5(/!: 5 ' 0 6 ' 0 " ' 0.

)* C'#%' ( #'&bi! -( (+/!58' -( ) &! -( b(+#(+! ' 23C #%'+-! ' 5(J(+#i+ "(

i+#(&(+/' -( ) b' ' )000 b'* E #!(9i#i(+/( i"!b1i#! -( (65'+"i+ /&i#' ("∝=1.237 "10−3 ) −1 4 ' -(+"i-'- (" -( 0* .#&

"e re!ono!e queD

/ %

/ p ¿

' / v

/T ¿

. ∝=

1

vo(/V

/)

/%

/ p ¿=∝Vo 3=

m

vV =

m

3

#s ' 1.236 "10

−3

°

) 70 g

mol

0.873g cm−3 G0-0.

* ∆ S=−10.36J

mol°)

)* L' (#%'#i+ -( ("/'-! -( %+ .'" (" V =nRT −n

2

a/V

2

-(iJ' ' (#%'#i+ $%( -(9i+(%+ #'&bi! i"!/&i#! -( (+/!58' #!+ ' 5("i+*


11/15

"e re!ono!e queD

(/S // )T =(/V //T )

"ea la e!ua!iónD

V =nRT −n2a/V 2

dV =Vd=nRdT −[−2n2aVdV /V 4 ]

(/V //T ) =nR−2n2a /V 3

(/V //T ) =nR / −(2n2a/V 3 )

)* D(iJ' '" "i.%i(+/(" (#%'#i!+("

( / S/V ) =Cv

TV4 +

4

5 6U +6S ¿ =T − [ /T / ]S

¿

/S=CvdT

T +

dV

T

( / S/V ) =Cv

T ( /T /V ) +

T

"eg9n la teor8a #e los gases i#eales

4 = 1

V o( / v/T ) Q 5=

1

V o( /V / )T

4 = 1T Q 5= 1


12/15

&o*an#o la #eriva#a !on rese!to a " % *antenien#o la te*eratura !onstante se obtieneD

( /U /S ) =T − (/V

/ S ) Por otro la#o

dH =Td%+Vdp

( /H / S ) =T &(/ H

/ )S=V

Enton!es

[ // ( /H /S ) ]S=[/

/S ( /H / )S]

(/T / )S=(

/V /S )

Lee*lazan#o la e!ua!ión b

(/V //S ) =T − (/T // )S

)* D'-' ' ('##i+:1

2 7 2 ( g )+3

2 H 2 (g )= 7 H 3 ( g ) , 5'' ' $%( '" #'5'#i-'-(" #'!89i#'"

"!+:

C5KN2; 27,28+3,26 "10−3

T +0,50 " 105T −2

C5KH2; 28,58+3,76 "10−3

T −0,50 "105T −2

C5KNH; 29,75+25,10 "10−3

T −1,55 " 105T −2

D(iJ' ' (#%'#i+ -( S 9 KT; #'#%' S ' 003, 003, 003 4 003 4 /'=' ' #%J' ' #%J' -( S > T*

S=∫T 1

T 2

CpdT

T

S=(a+'T +cT −2 ) dT T


13/15

S=( aT +'+ cT 3 )dT

S=a ln (T 2−T 1 )+' (T 2−T 1 )−c (T 2−2−T 1−2 )/2

$os valores #e a, b % ! sonD

a ' -2G,11

b ' 0,01G1

! ' -1,++ 10+

Lealizan#o las oera!ionesD

S300=−18,0807J

molK

S500=−136,88 J mol K

S600=−146,37J

mol K

S700=−152,79J

mol K

)* C'#%' ( #'&bi! -( (+/!58' -( ) &! -( #i5/+ $%( "( "!&(/( ' ' /'+"9!&'#i+: KT ) 00 , P) 0 b';

8 KT2 200 , P2 )0 b';*

"e !ono!e queD

S=Cp ln ( T 2T 1 )− R ln( 2 1 )

S=2,5 R ln

(

200

300

)− R ln(

98,7

49,35

)

S=2,5∗8,314∗(−0,4 )−8,314∗0,69

S=−14,25J

mol K


14/15

20* S( #'i(+/' 0 .'&!" -( '%&i+i! -( 003C ' 003C* E 5%+/! -( 9%"i+ (" 03C 4 ' (+/'58'

-( 9%"i+ (" *.)* A-(&1", '" #'5'#i-'-(" #'!89i#'" -( 8$%i-! 4 "!i-! "!+: * 4 )* *

)*&!)* C'#%' ( #'&bi! -( E+/!58'*

D'/!"

*' +0gr n'm

9

&1'M00 (C G73(@ n'50 gr

26.982 gr /mol

&2'700 (C 73(@ n' 1.G+3 *ol

&3'MM0 (C 33(@

5R' 3:J

gr 26.982 gr

1mol ' 10M30.0G JK*ol

Cs' 31.G JK(@.*ol

Cl' 3:.3 JK(@.*ol"'?

P!#("! )

AKS; 003C ' AKS; 03C

"' n Cs lnT 3

T 1

"' 1.G+3 31.G JK(@.*ol ln933° K

873° K

"1' 3.17 JK(@.*ol

P!#("! 2

AKS; 03C ' AK; 03C

"' H:

T 3

"'10630.908 J /mol

933° K

"2' 11.3: JK(@.*ol

P!#("!

AKS; 003C ' AKS; 03C

"' n Cs lnT 3

T 1

"' 1.G+ 3:.3 JK(@.*ol ln973° K

933° K

"3' 2.MM: JK(@.*ol


15/15

"&' "1 = "2 = "3

ST )* 3*&!

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Ejercicios Segunda Ley-Pablo Chango