Giai Bai Tap Nhiet

7/30/2019 Giai Bai Tap Nhiet

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Khoa Vt L, trng H Khoa Hc, H Thi Nguyn

Phn B. Nhit hc

Chng: M u

0-1. C 40g kh O2 chim th tch 3l p sut 10at.a.Tnh nhit ca khb.Cho khi kh gin n ng p ti th tch 4l. Hi nhit ca khi

kh sau khi gin n.

Giia.Phng trnh Mendeleev Crapayron

pV=m/ RTNhit khi kh T1=p1V1/R=292,5K.

b.Qu trnh ng p: V/T=constNhit khi kh T2=T1V2/V1=390K

0-2. C 10g kh H2 p sut 8,2at ng trong mt bnh th tch 20l.a.Tnh nhit ca khi khb.H nng ng tch khi kh ny n p sut ca n bng 9at. Tnh

nhit ca khi kh sau khi h nng

Giia.Nhit khi kh T1=p1V1/R=388K.b.Qu trnh ng tch: p/T=constNhit khi kh T2=T1p2/p1=425K (ly 1at=9,81Pa)

0-3. C 10g kh ng trong mt bnh, p sut 107Pa. Ngi ta ly bnh ramt lng kh cho ti khi p sut ca kh cn li trong bnh bng2,5.106Pa. Coi nhit kh khng i. Tm lng kh ly ra

GiiPhng trnh Mendeleev Crapayron cho khi kh trc v sau khi ly kh

p1V=m1/ RT, p2V=m2/ RT,

21

21

2

2

1

1

mm

pp

m

p

m

p

==

Khi lng kh ly:

kg5,7mp

p1mmm 1

1

221 =

==

0-4. C 12g kh chim th tch 4l 7oC. Sau khi h nng ng p, khilng ring ca n bng 6.10-4g/cm3. Tm nhit ca khi kh sau khih nng.

Gii


2/54


Trc khi h nngpV=m/ RT1 (1)

Sau khi h nng pV=m/ RT2p=RT2/ (2)

Ly (1)/(2) ( ) K1400273tV

mT

V

mT 112 =+==

0-5. C 10 g kh Oxy nhit 10oC, p sut 3at. Sau khi h nng ng p,khi kh chim th tch 10l. Tm:

a. Th tch khi kh trc khi gin n.b. Nhit khi kh sau khi gin n.c. Khi lng ring khi kh trc khi gin n.d. Khi lng ring khi kh sau khi gin n.

Giia. Th tch kh trc khi gin n: 4,2RT/pV 11 = b. Nhit kh sau khi gin n: T2=T1V2/V1 1170Kc. Khi lng ring ca kh trc khi gin n: 3

1

11 m/kg14,4

V

m==

d. Khi lng ring ca kh sau khi gin n: 32

11 m/kg1

V

m==

0-6. Mt bnh cha mt kh nn 27oC v p sut 40at. Tm p sut cakh khi c mt khi lng kh thot ra khi bnh v nhit hxung ti 12oC.

GiiPhng trnh Mendeleev Crapayron

( )at19p

T2

TTp

TTR2/m

Vp

RTmVp

12

2

1

=

=

=

0-7. Mt kh cu c th tch 300m3. Ngi ta bm vo kh cu kh hyr 200C di p sut 750mmHg. Nu mi giy bm c 25g th sau baolu th bm xong?

GiiKhi lng kh cn bm

RTPVm =

Thi gian cn bm

mRT

pV

m

mt

=

=


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Thay s p=750mmHg=1,05Pa, T=273+20=293K, 3m300V = ,R=8,31J/molK, =2g, m=25g. Nhn c t 990s

0-8. Cho tc dng H2SO4 vi vi thu c 1320cm3 kh CO2 nhit 22oC v 1000mmHg. Hi lng vi tham gia phn ng.

GiiPhn ng

OHCOCaSOSOHCaCO 224423 +++ S mol 2CO sinh ra bng s mol ca 3CaCO tham gian phn ng. Khilng ca 3CaCO tham gian phn ng:

100RT

pV100.nM.nm

233 COCaCOCaCO===

Thay s p=1000mmHg= Pa10.33,1 5 , 33m10.32,1V = g18,7m

0-9. C hai bnh cu c ni vi nhau bng mt ng c kho, cha cngmt cht kh. p sut bnh th nht bng 2.105Pa, bnh th hai l106Pa. M kha nh nhng hai bnh thng nhau sao cho nhit kh khng i. Khi cn bng, p sut hai bnh l 4.105Pa. Tm thtch ca bnh cu th hai , bit th tch ca bnh th nht l 15l.

GiiTng s mol kh trc v sau khi m kha khng i (v nhit cng khngi) nn:

( )

RT

VVp

RT

Vp

RT

Vp 212211 +=+

Vy, th tch ca bnh cu th hai.3

1

2

12 dm5V

ppppV =

=

0-10. C hai bnh cha hai th kh khc nhau thng vi nhau bng mt ngthy tinh c kha. Th tch ca bnh th nht l 2 lt, ca bnh th hail 3 lt. Lc u ta ng kha, p sut hai bnh ln lt l 1 at v3at. Sau m kha nh nhng hai bnh thng nhau sao cho nhit vn khng thay i. Tnh p sut ca cht kh trong hai bnh khkhi thng nhau.

GiiTng t bi tp 0-9, ta c:( )

RT

VVp

RT

Vp

RT

Vp 212211 +=+

at6,1VV

VpVpp

21

2211 =+

+=

24cm20cm


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0-11. Mt ng thy tinh tit din u, mt u kn mt u h. Lc ungi ta nhng u h vo mt chu nc sao cho nc trong v ngoing bng nhau, chiu cao cn li ca ng bng 20cm. Sau ngi tart ng ln mt on 4cm (hnh 0-1). Hi mc nc trong ng dngln bao nhiu, bit rng nhit xung quanh khng i v p sut khquyn l 760mmHg.

GiiGi cao ct nc trong ng l x

p sut trong ng sau khi nng ln( )( )OcmHxpp 2o =

nh lut Bil - Marit cho khi kh b giam( ) ( )( )x4lxpx4lplp oo +=+=

Thay s: OcmH1033mmHg760p 2o == , cm20l =

cm95,3x04132x1057-x2 ==+ ; ( cm1053x = >l+4 loi)

0-12. Trong ng phong v biu c mt t khng kh, do iu kin bnh thng n li ch p sut l 750mmHg.Tm khi lng ring ca khng kh trong ng

Giip sut kh bn trong phong v biu

Pa1360mmHg10pp'p o === Khi lng ring ca kh

( ) 3

o

o m/g17273.31,8

1360.29

RT

pp=

=

0-13. C 8g kh xy hn hp vi 22g kh ccbonnc (CO2). Xc nh khilng ca 1 kilmol hn hp .

GiiKhi lng ca 1 mol hn hp

( ) ( ) kmol/kg4044/2232/8

228kmol/kg

mm

mmmol/g

n

m

2

2

1

1

21 =+

+=

+

+==

0-14. Mt hn hp kh c 2,8kg Nit v 3,2kg xy nhit 17oC v p sut4.105N/m2. Tm th tch ca hn hp .GiiTh tch hn hp

x

nh B.1

l


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( )3

5

2

2

1

1

m2,110.4

17273.31,8.32

3200

28

2800

p

RTmm

p

nRTV

+

+

=

+

==

0-15. Kh n l mt hn hp gm mt phn khi lng hy v tm phnkhi lng xy. Hy xc nh khi lng ring ca kh n iukin thng.

GiiTheo bi 13, khi lng mol ca cht n

mol/g1232/82/1

81

m/m1

m/m1

mm

mm 8m/m

2

12

1

12

2

2

1

1

2112

=+

+=

+

+=

+

+=

=

Khi lng ring ca hn hp3

5

o

o m/g534273.31,8

10.01,1.12

RT

p==

Chng 8: Nguyn l th nht ca nhit ng lc hc8-1. 160g kh oxy c nung nng t nhit 50 oC n 60oC. Tm nhit

lng m kh nhn c v bin thin ni nng ca khi kh tronghai qa trnh

a. ng tch; b. ng pGii:

a. Qu trnh ng tch( ) cal250J10405060.31.8.

2

5.

32

160TR

2

5mTC

mUQ V ====

b. Qu trnh ng p bin thin ni nng

cal250TCm

U V ==

Nhit lng kh nhn vo

( ) TR2

7mTRC

mVpTC

mAUQ VV =+=+=+=

Thay s

Q ( ) cal350J14545060.31.8.27

.32

160

=

8-2. Tm nhit dung ring (gam) ng tch ca mt cht kh a nguyn t,bit rng khi lng ring ca kh iu kin chun l =7,95.10-4

kg/cm3.


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GiiVi kh a nguyn t, nhit dung ring mol ng tch ( )molK/JR3CV =

iu kin tiu chun

o

oooo

p

RTRT

mVp

==

Nhit dung ring gam ng tch

kgK/J1400

T

p3

RT

CpCc oVoVV ===

8-3. Tm nhit dung ring (gam) ng p ca mt cht kh, bit rng khilng ca mt kilmol kh l =30kg/kmol. H s Potxng (ch son nhit) =1,4.

Gii:Nhit dung ring mol ng p:

RCCVp

+=

ViV

p

C

C= =>

1

=

RCp

Nhit dung ring gam ng p:

( ) ( )kgKJ

RCc

p

p/5,969

14,1.10.30

31,8.4,1

13

=

=

==

8-4. Mt bnh kn cha 14g kh Nit p sut 1at v nhit 270C. Sau khih nng, p sut trong bnh ln ti 5at. Hi:

a. Nhit ca kh sau khi h nng?b. Th tch ca bnh?c. tng ni nng ca kh?

Gii:a. Qu trnh ng tch, nhit khi kh sau khi h nng l T2

K1500Tp

pT

T

p

T

p1

1

22

2

2

1

1 ===

b. Th tch bnhl72,12

p

mRTV

1

1 ==

c. tng ni nng ca khi kh :( ) kJ46,12T1

p

pR

2

5mTTC

mU 1

1

212V =

==

( 2N l kh lng nguyn t i=5, 2/R5CV = )

8-5. Nn ng tch 3l khng kh p sut 1at. Tm nhit ta ra bit rngth tch cui cng bng 1/10 th tch ban u.


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GiiNguyn l th nht ca nhit ng lc hc, nhit lng m khi kh nhnc

UAQ += ' Qu trnh ng nhit nn U= =TnCV 0Nhit lng m khi kh nhn c

JQ

V

VVp

V

dVpVpdVAQ

V

V

V

V

67610

1ln10.3.10.81,9

ln'

34

1

211

2

1

2

1

=

====

Du - ch ra rng qu trnh thc s ta nhit.

8-6. Mt bnh kn th tch 2l, ng 12g kh nit nhit 10oC. Sau khi hnng, p sut trung bnh ln ti 104mmHg. Tm nhit lng m khikh nhn c, bit bnh gin n km.

GiiBnh gin n km, th tch ca bnh khng i, qu trnh l ng tch.

A=0Nguyn l I nhit ng lc hc

( )

=

=

==+=

1212

12

RTm

Vp2

iRT

mRT

m

2

iQ

TT2

iRmUUAQ

( 2N l kh lng nguyn t i=5, 2/R5CV = )Thay s p2=104mmHg=1,33.106Pa, V=2.10-3m3, T1=283K.

kJ1,4Q =

8-7. H nng 16 gam kh xy trong mt bnh kh gin n km nhit 370C, t p sut 105 N/m2 ln ti 3.105N/m2. Tm:

a. Nhit ca khi kh sau khi h nng;b. Nhit lng cung cp cho khi kh.

Gii:a. Bnh kn, gin n km, qu trnh ng tch, nhit khi kh sau khi hnng l T2

( ) K9303727310

10.3T

p

pT

T

p

T

p5

5

1

1

22

2

2

1

1 =+===

b. Nhit lng cung cp cho kh bng nhit lng m kh nhnc trong qu trnh ng tch trn


8/54


( )

( ) kJ4,6110

10.337273.31,8.

2

5

32

16Q

1p

pRT

2

imTTRC

mUQ

5

5

1

2

112V

+=

===

8-8. Sau khi nhn c nhit lng Q=150cal, nhit ca m=40,3g khOxi tng t t1= 16

oC ti t2=40oC. Hi qu trnh h nng c tin

hnh trong iu kin no?

GiiNhit lng m kh nhn c

( )12xx

ttm

QCTC

mQ

==

( )molK/J77,20

1640.3,40

18,4.150.32Cx =

=

Nhit dung ring mol ng tch ca Oxi:

xV CmolK/J77,20

2

31,8.5

2

iRC ====

Nh vy Vx CC = , qu trnh l ng tch.

8-9. 6,5g hyr nhit 27oC, nhn nhit lng gin n gp i, trongiu kin p sut khng i. Tnh

a. Cng m kh sinh ra.b. bin thin ni nng ca khi kh.c. Nhit lng cung cp cho khi kh.

Gii:

a.

Cng sinh ra( ) ( )

( ) J10.1,827273.31,8.2

5,6A

RTm

VV2pVVpA

3

11112

+=

===

b. bin thin ni nng ca khi kh:( ) ( )

( ) JU

RTmi

pVpVi

RTm

RTmi

TTCm

UV

3

1111212

10.2,2027273.31,8.2

5,6.2

5

22

22

+=

==

==

c. Nhit lng cung cp cho khi kh chnh xc bng nhit lngm kh nhn c. Theo nguyn l IJ10.3,2810.2,2010.1,8UAQ

333 =+=+= (i vi nguyn t hyr (lng nguyn t) s bc t do nguyn ti=5)


9/54


8-10. 10g kh oxy 10oC, p sut 3.105Pa. Sau khi h nng ng p, th tchkh tng n 10l. Tm:

a. Nhit lng m khi kh nhn cb. Ni nng ca khi kh trc v sau khi h nng

Giia. Theo nguyn l I, nhit lng m khi kh nhn c trong qa

trnh ng p

( )

( ) J10.9,710273.31,8.32

1010.10.10.3

2

25Q

RTmpV2

2iRTmRTm

2

2iTTCmUAQ

335

121212p

+

+=

+=

+==+=

b. Ni nng ca khi kh trc khi h nng

( ) J10.8,110273.31,8.2

5.

32

10U

RT2

imTC

mU

3

1

11V1

+=

==

Ni nng ca khi kh sau khi h nng

J10.5,710.10.10.3.2

5U

pV2

iRT

2

imTC

mU

335

2

222V2

==

===

(i vi nguyn t oxy (lng nguyn t) s bc t do nguyn t i=5)

8-11. Mt thy li chuyn ng trong nc nh khng kh nn trong bnhcha ca thy li pht ra pha sau. Tnh cng do kh sinh ra. Bit rngth tch ca bnh cha l 5lt, p sut ca khng kh nn t p sut

100atm gim ti 1atm.

GiiKh pht ra pha sau l mi trng nc rt ln v c nhit coi nh

khng i. Do qu trnh gin n kh ca thy li trong nc coi l qutrnh ng nhit (gn ng l thun nghch).

Cng do kh sinh ra: J10.26,2100ln10.5.10.81,9.1p

plnVpA 334

2

111 ==

.

8-12. 2 kmol kh ccbonic c h nng ng p cho n khi nhit tngthm 50oC. Tm

a. bin thin ni nng ca khi khb. Cng do kh gin n sinh rac. Nhit lng truyn cho kh

Gii


10/54


a. bin thin ni nng ca khi khkJ250050.

2

31,8.610.2T

2

iRmU 3 ==

(kh CO2 l kh a nguyn t (chnh xc l 3) nn s bc t do ca phnt l 6)

b. Cng do kh gin n sinh ra( ) ( )

1212 TTRm

VVpA ==

kJ83050.31,8.10.2A 3 = c. Nhit lng truyn cho kh bng nhit lng m kh nhn c

kJ33308302500AUQ =+=+=

8-13. 7 gam kh ccbonic c h nng cho ti khi nhit tng thm 10oCtrong iu kin gin n t do. Tm cng do kh sinh ra v bin thinni nng ca n.

GiiGin n t do c ngha l ng p (gin n trong kh quyn, p sut bng p

sut kh quyn)Cng do kh sinh ra khi gin n

( ) ( )121212 TTRm

RTm

RTm

VVpA ===

J2,1310.31,8.44

7A =

bin thin ni nng ca khi kh

J7,3910.2

31,8.6.

44

7T

2

iRmU ==

(kh CO2 l kh a nguyn t (chnh xc l 3) nn s bc t do ca phn t l

6)8-14. 10g kh oxy p sut 3at v nhit 10oC c h nng ng p vgin n ti th tch 10l. Tm:

a. Nhit lng cung cp cho khi kh.b. bin thin ni nng ca khi kh.c. Cng do kh sinh ra khi gin n.

Gii.a. Nhit lng cung cp cho kh bng nhit lng m kh nhn vo

+=

+==+= 121212p RT

mpV

2

2iRT

mRT

m

2

2i)TT(C

mUAQ

( ) J10.8,710273.31,8.321010.10.10.81,9.3

225Q 334 =

++=

b. bin thin ni nngJ10.5,5Q

2i

i)TT(C

C

Cm)TT(C

mU 312p

p

V12V =

+===

c. Cng do kh sinh ra khi gin n


11/54


J10.3,2UQA3==

8-15. Mt cht kh ng trong mt xilanh t thng ng c pittng khilng khng ng k di ng c. Hi cn phi thc hin mt cngbng bao nhiu nng pittng ln cao thm mt khong cmh 101 = nuchiu cao ban u ca ct khng kh l cmho 15= , p sut kh quyn l

at1po = , din tch mt pittng2

10cmS= . Nhit ca kh coi l khng

i trong sut qu trnh.Gii:Cng do kh sinh ra

0

1ooo

o

1ooo

h

hhlnVp

V

VlnVpA

+==

Hay, khi bin i kh nhn vo mt cng :

+=

1o

oooo

hh

hlnVpA

Cng ca p sut kh quyn : 1ok ShpA =

Cng cn thc hin bao gm cng truyn cho kh v cng thng kh quyn

+==

o

1o1ok

h

h1lnhhSpAA'A

J3,215

101ln.10.1510.1010.10.10.8,9.1'A

2244

+=

8-16. 2m3 kh gin n ng nhit t p sut p=5at n p sut 4at. Tnhcng do kh sinh ra v nhit lng cung cp cho kh trong qu trnhgin n.

GiiTheo nguyn l I

0U

UAQ

=

+=

2

111

VpVp

1

211

V

V p

plnVp

V

VlnVppdVAQ

22112

1

=

====

J10.2,24

5ln.10.81,9.5.2AQ

54 ===

8-17. Mt khi kh N2

p sut p1=1at c th tch V

1=10l c gin n ti th

tch gp i. Tm p sut cui cng v cng do kh sinh ra nu gin n l:

a. ng p.b. ng nhitc. on nhit


12/54


Giia. Qu trnh ng pp sut cui p2=p1=1at.Cng do kh sinh ra

( ) J98010.1010.10.2.10.81,9.1VpA 3341 ==

b. Qu trnh ng nhitp sut cui p2:

at5,0pV

VpVpVp 1

2

122211 ===

Cng do kh sinh ra

=== 2ln.10.10.10.81,9.1V

VlnVpA

34

1

211 680J

c. Qu trnh on nhit+ p sut

==

2

1122211V

VppVpVp

i vi N2,4,1

2

25

i

2i=

+=

+=

Nnat38,02/1p 4,1

2 == + Cng do kh sinh ra

( )

( )

( ) J5902110.10.10.81,92

5A

V

V1Vp

2

iV

V

VVp

2

iVpVp

2

iA

RTm

RTm

2

iTTC

mUAUAQ

4,034

1

2

1112

2

1112211

2121V

=

=

==

===+=

8-18. Nn 10g kh oxy t iu kin tiu chun ti th tch 4l. Tm:a. p sut v nhit ca khi kh sau mi qu trnh nn ng

nhit v on nhitb. Cng cn thit nn kh trong mi trng hp. T , suy ra

nn nn theo cch no th li hn.

GiiTh tch kh ban u

l74,22.32

10V1

==

a. Qu trnh nn ng nhit:- p sut cui qu trnh l p2:


13/54


2

1122211V

VppVpVp ==

Pa10.7,14

7.10p

55

2 =

Hoc c th tnh nh phng tnh trng thi:

Pa10.7,110.4.32

273.31,8.10

p

V

mRTpRT

mVpVp

5

32

2

1212211

=

===

- Nhit kh khng i K273TT 12 == - Cng nn kh bng v ngc du vi cng kh sinh ra

JA

V

VRT

m

V

VVpAA

3977

4ln.273.31,8.

32

10

lnln

1

1

2

1

1

2

112

=

=

==

b.- p sut p2:

Pa10.2,24710

VVppVpVp 5

4,1

5

2

1122211 =

=

==

- Nhit T2K341

4

7273

V

VTTVTVT

14,1

2

112

1

22

1

11

=

==

1

- Cng nn kh bng v ngc du vi cng kh sinh ra1

14,1351

2

111

2 4394

71

14,1

10.7.101

1AJ

V

VVpA >

=

=

Vy nn ng nhit th tt hn

8-19. Ngi ta mun nn 10 lt khng kh n th tch 2 lt. Hi nn nnng nhit hay nn on nhit?

GiiCng nn kh theo qu trnh ng nhit (bng v ngc du vi vi cngm kh sinh ra):

=

=

2

111

1

2111

V

VlnVp

V

VlnVpA (1)

Tng t, i vi qu trnh on nhit:UA0UAQ ==+=

Cng nn kh trong trng hp ny, tng t nh lm vi bi 8.17 ta c:

===

1

2

1112

V

V1Vp

2

iUAA

(2)

T (1) v (2)


14/54


( )

( )

( )

( )14,1

2/10ln

12/10.2

5

V/Vln

1V/V.2

i

A

A14,1

21

1

21

1

2 >

=

=

Vy nn theo qu trnh ng nhit tn t cng hn, do li hn.

8-20. Gin on nhit mt khi khng kh sao cho th tch ca n tng gpi. Hy tnh nhit khi khng kh cui qu trnh, bit rng lc n c nhit 0oC.

GiiPhng trnh cho qu trnh on nhit

( ) ( )

K2072

1273

V

VTT

VTVTVVpVVpVpVp

14,11

2

112

1

22

1

11

1

222

1

1112211

=

=

===

8-21. 7,2 lt kh oxy c nn on nhit n th tch 1 lt, lc p sut cakh nn l 16at. Hi p sut ban u?

GiiPhng trnh (xem ph lc) cho qu trnh on nhit

at12,7

1.16

V

VppVpVp

4,1

2

1122211

=

==

8-22. 1kg khng kh nhit 30oC v p sut 1,5at c gin on nhitn p sut 1at. Hi:

a. Th tch khng kh tng ln bao nhiu ln?b. Nhit khng kh sau khi gin?c. Cng do khng kh sinh ra khi gin n?

Giia. T phng trnh

33,11

5,1

p

p

V

VVpVp

4,1/1/1

2

1

1

22211

=

==

Th tch tng khong 1,33 lnb. Phng trnh cho qu trnh an nhit

===

1

2

112

1

22

1

112211p

pTTpTpTVpVp

( ) K2701

5,130273T

4,14,11

2

+=

c. Cng do kh sinh ra( )

21 TT2

iRmUA ==


15/54


i vi khng kh =29g/mol, bc t do ca phn t i=5

( ) J10.4,2270303.2

31,8.5

29

10A

43

=

8-23. Chng minh rng i vi mt kh l tng xc nh c phng trnh:Ui

2pV =

U l ni nng ca khi kh y, i l bc t do.

GiiNi nng kh l tng

nRT2

iU =

Phng trnh Mendeleev CrapayronnRTpV =

Do

Ui

2pV =

8-24. Mt kilmol kh N2 (=28kg/kmol) iu kin tiu chun gin onnhit sao cho th tch ca n tng ln 5 ln. Tm:

a. Cng do kh thc hin.b. bin thin ni nng ca khi kh.

Giia. Nhit kh sau khi nn l T2:

1

22

1

112211 VTVTVpVp

==

K4,143

5

1.273

V

VTT

14,11

2

112

=

=

Cng do kh thc hin

( )

( ) J10.7,24,1432732

31,8.5.10A

TT2

iRmUA

63

21

=

==

b. bin thin ni nng ca khi kh bng v ngc du vi cngdo kh sinh ra

J10.7,2AU6==

8-25. Khng kh trong xilanh ca mt ng c t trong c nn an nhitt p sut 1at n p sut 35at. Tnh nhit ca n cui qu trnhnn bit rng nhit ban u ca n l 40 Co

GiiPhng trnh cho qu trnh on nhit


16/54


===

1

2

112

1

22

1

112211p

pTTpTpTVpVp

( ) C592K86535

140273T

o4,1

4,11

2 =

+=

8-26. Mt khi kh gin n on nhit, th tch ca n tng gp i, nhngnhit tuyt i ca n gim i 1,32 ln. Tm s bc t do ca phnt kh .

GiiT phng trnh

( )

( )21

121

22

1

112211V/Vln

T/Tln1VTVTVpVp ===

( )

( )

( )

( )5

32,1/1ln

2/1ln.2

T/Tln

V/Vln2i

12

21i/21 === =

S bc t do kh l 5.

8-27. Mt cht kh lng nguyn t c th tch lV 5,01 = , p sut atmp 5,01 = bnn on nhit ti th tch V2 v p sut p2. Sau ngi ta ginguyn th tch 2V v lm lnh n ti nhit ban u. Khi p sutca kh l atmp

o 1= a.V th ca qu trnh .b. Tm th tch 2V v p sut 2p

Giia. th ca qu trnh:

b. Qu trnh 3 1 ng nhit nn :21

o

13113o Vl25,0V

p

pVVpVp ====

Qu trnh 1 2 on nhit nn:

at32,15,0.2p

pp

pp

V

VpVpVp

4,1

2

1

1

o1

2

122211

=

=

==

p

V

2

13

p2

p1

po

V V


17/54


(Kh lng nguyn t i=5 nn

4,15

25

i

2i=

+=

+= )

8-28. Khi nn on nhit 1kmol kh lng nguynt, ngi ta tn cng 146kJ. Hi nhit ca kh tng ln bao nhiu?

GiiKh nhn mt cng A=146kJ (sinh cng A=-146J), tng ni nng kh bng cng nhn vo ca kh

T2

iRmAU ==

K710.31,8.5

10.146.2

/iRm

A2T

3

3

==

Nhit kh tng 7 Co .

8-29. Mt lng kh oxy chim th tch V1=3l nhit 27 Co v p sutp1=8,2.105Pa. trng thi th hai, kh c cc thng s V2=4,5l vp2=6.105Pa (hnh 8.1). Tm nhit lng m kh sinh ra khi gin n, v bin thin ni nng ca khi kh. Gii bi ton trong trng hpbin i khi kh t trng thi 1 ti trng thi 2 theo hai con ng:

a.ACBb.ADB

Giia. Qu trnh ACB

- AC ng tch:( )

=== ACACAC RTmRTmiTTiRmUQ

22

( ) ( ) JVppiQAC 165010.310.2,810.62

5

2

355

112 ===

- Qu trnh CB ng p: RCC Vp += ( )( )

+=+= CBCBVCB RT

mRT

miTTRC

mQ

2

2

( ) ( ) JVVpi

QCB 31501035,410.62

25

2

2 35122 =

+=

+=

- C qu trnhJ150031501650QQQ CBACACB =+=+=

Qu trnh ACB kh nhn lng nhit JQACB 1500= bin thin ni nng:

p

VV1 V2o

P1

P2

Hnh 8-1

C

A D

B


18/54


( ) ( )

( ) J60010.3.10.2,810.5,4.10.62

5U

VpVp2

iTTC

mU

3535

AB

1122ABVAB

==

==

Cng kh thc hin trong qu trnh bin i:( )

( ) J90010.310.5,4.10.6AVVpAA

335

ACB

122CBACB

==

==

b. Qu trnh ADBNhit

- Qu trnh AD ng p( )( )

+=+= ADADVAD RT

mRT

m

2

2iTTRC

mQ

( ) ( ) J43051035,410.2,82

25VVp

2

2iQ

35

121AD =+

=+

=

- DB ng tch:( )

=== CADBDB RTm

RTm

2

iTT

2

iRmUQ

( ) ( ) J247510.5,410.2,810.625Vpp2iQ355

212DB === - C qu trnh

J183024754305QQQ CBACADB ==+= bin thin ni nng:

( ) ( )

( ) J60010.3.10.2,810.5,4.10.62

5U

VpVp2

iTTC

mU

3535

AB

1122ABVAB

==

==

Cng kh thc hin trong qu trnh:( )

( ) J123010.310.5,4.10.2,8AVVpAA

335

ADB

121ADADB

==

==

8-30. Mt kmol kh (khi lng mol ) thc hin mt chu trnh ABCD nhhnh di, trong AB, CD l hai qu trnh ng nhit, ng vi nhit 1T v 2T , BC v DA l hai qua trnh ng tch ng vi hai th tch 2V v 1V.


19/54


a. Chng minh rngC

D

B

A

p

p

p

p=

b. Tnh cng v nhit trong c chu trnh.Gii:

a.p dng lin tip cc phng trnh ca cc qu trnh ng nhit:C

D

D

C

A

B

B

A

p

p

V

V

V

V

p

p=== (pcm)

b. Cng ca chu trnh bng cng trn cc qu trnh AB v CD, ccqu trnh cn li cng bng khng.

( )1

212

2

1

1

2 lnlnlnV

VTTR

m

V

VVp

V

VVpAAA

RTm

pV

DDAACDAB =+=+=

=

Nhit kh nhn trong c chu trnh1:

( )1

212 ln

VVTTRmAQ ==

8-31. Mt khi kh thc hin mt chu trnh nh hnh v di, trong 1-2 v3-4 l hai qu trnh ng nhit ng vi cc nhit 1T v 2T , 2-3 v 3-4l cc qu trnh on nhit. Cho lVlV 5,2 21 == , lV 83 = , atmp 71 = . Tm:

1 Trong mt chu trnh kn U=0, do AUAQ =+=

V

ApA

pD

pB

pCC

V1 V2

BD

p


20/54


a. 24432 ,,,, TVppp b. Cng kh thc hin trong tng qu trnh v trong ton chu trnh.c. Nhit m khi kh nhn c hay ta ra trong tng qu trnh

ng nhit.

Gii:

a. atmpV

Vp 7,21

2

12 == , coi khng kh l kh lng nguyn t: i=5, ta c:

atmpV

Vp 45,12

3

23 =

=

atmT

TppK

V

VTT 6,3;331

1

1

214

1

3

212 =

==

=

lVp

pV 2,334

34 ==

b. Cng thc hin trn tng qu trnh:J

V

VVpA 1300ln

1

21112 ==

JT

TVpA 6201

1 1

22223 =

=

JV

VVpA 1070ln

3

42234 ==

JT

TVpA 6201

1 2

12241 =

=

Cng kh thc hin trong c chu trnh:JAAAAA 23041342312 =+++=

c. Nhit m kh nhn trong tng qu trnh on nhit:

p

VV1

1p1

p3

p4

3

2

4

V4 V2 V3

p2


21/54


JAQ 13001212 == , kh nhn nhit.JAQ 10703434 == , kh nh nhit.

8-32. Trong mt bnh c 20g N2 v 32g oxy. Tm bin thin ni nng cahn hp kh khi lm lnh n xung 28oC.

Gii gim ni nng

T2

iRmT

2

iRmU

N

N

O

O +=

( ) J10000282732

31,8.5

28

20

32

32U +

+=

8-33. Gin cng tc theo l thuyt ca mt my nn c v trn hnh 8 4. (gin thc nghim c cc gc trn hn). on AB ng vi qutrnh nn ng nhit khng kh, BC qu trnh y khng kh vo bnhcha (p sut khng i); CD gim t ngt p sut trong xilanh camy nn khi ng van thot v m van np; DA cho khng kh vo p sut 1at. Hy chng minh rng cng ca my nn sau mt chutrnh bng cng i vi qu trnh ng nhit v c biu din bngdin tch ABGF.

GiiCng A ca my nn bng cng kh nhn c v bng din tch ABCD

( ) ( )

( ) ( ) ( ) ( )AFGHdtVVpVpVpVppBCDHBdt

BCDHBdtABHAdt)ABCDA(dtA

2112111212 ====

+==

( ) ( ) ( )ABGFdtAFGHdtABHAdtA =+= (pcm)

8-34. V cc th ca nhng qu trnh ng tch, ng p, ng nhit voon nhit ca gin

a. T,pb. T,V c. T,Ud.V,U

Gii

p

V

P2C B

AD

V2 V1Hnh 8-4

H

G F

P1


22/54


a. Gin T,p

Qu trnh ng tch : p/T=const, c th biu din l ngthng qua gc t (ng 1)

Qu trnh ng p: p sut khng i, c th biu din lng thng song song vi OT (ng 2)

Qu trnh ng nhit: nhit khng i, c th biu din lng thng song vi Op.

Qu trnh on nhit. S ph thuc p vo T cho bi phng trnh1

T.constp =

(v do ( )( )

( ) 11111

T.constpp

TnR

p

pVcconstcpV

=====

)

Phng trnh ny c th (4) l mt ng cong i qua gc ta b. Gin T,V

Qu trnh ng tch: th tch khng i, c th biu din l ngthng song song vi OT (ng 1)

Qu trnh ng p : V/T=const, c th biu din l ng thng quagc to (ng 2) Qu trnh ng nhit: T=const, c th biu din l ng thng song

vi OV. Qu trnh on nhit. S ph thuc V vo T cho bi phng trnh

.T.constV 11

=

T

p

1

3

T

V3

4


23/54


(v do ( ) ( ) ( ) 11

11T.constVTVnRVpVcconstcpV

===== , >1)

Phng trnh ny c th (4) l mt ng cong dng hypecbol timcn vi hai trc ta (ng 4).

c. Trong mt qu trnh bt k : RTmU

= , cc qu trnh ng tch, ng

p, on nhit ng biu din l ng thng qua gc ta (ng1,2,4), qu trnh ng nhit c cho bi ng nm ngang (ng 3).

d. Gin U,V (U khc T mt hng s VCm

, do ta ch cn ko di

thm mt t s VCm

i vi trc T th T,V s nhn c th

U,V )

U

V

23

1

4

T

U1,2,4

3


24/54


Chng 9: Nguyn l th hai ca nhit ng lc hc

9-1. Mt my hi nc c cng sut 14,7kW, tiu th 8,1kg than trongmt gi. Nng sut ta nhit ca than l 7800kcal/kg. Nhit cangun nng 200oC, nhit ca ngun lnh l 58 Co . Tm hiu sutthc t ca my. So snh hiu sut vi hiu sut l tng camy nhit lm vic theo chu trnh Ccn vi nhng ngun nhit k

trn.

GiiHiu sut thc t ca my

%20%10018,4.7800.1,8

3600.7,14%100

Q

Qh

toanphan

coich ==

Hiu sut l tng theo chu trnh Ccn

%30%100273200

58200%100

T

TTh

n

lnlt

+

=

=

Hay

lth3

2h =

9-2. Cc ngoi lc trong my lm lnh l tng thc hin mt cng bngbao nhiu ly i mt nhit lng 105J t bung lm lnh, nunhit ca bung l 263K, cn nhit ca nc lm lnh l285K.

GiiH s lm lnh ca ng c

AQ2=

Nu my chy theo chu trnh Ccn ngc th:

21

2

TT

T

=

Suy ra

( )J8365101263

285Q1

T

TA

5

2

2

1

=

=


25/54


9-3. Mt ng c nhit l tng chy theo chu trnh Ccn, nh chongun lnh 80% nhit lng m n thu c ca ngun nng. Nhitlng thu c trong mt chu trnh l 1,5kcal. Tm:a. Hiu sut ng c.b. Cng m ng c sinh ra trong mt chu trnh

Gii

a. Hiu sut ca ng c( ) %20%1008,01%100

Q

Q1%100

Q

A

1

2

1

==

==

c. Cng m ng c sinh ra trong mt chu trnhA=Q1=0,2.1,5=0,3kcal=12,54kJ

9-4. Mt ng c nhit lm vic theo chu trnh Ccn, sau mi chu trnhsinh mt cng A=7,35.104J. Nhit ca ngun nng l 100 Co ,nhit ca ngun lnh l 0oC. Tm:

a.

Hiu sut ng c.b. Nhit lng nhn c ca ngun nng sau mt chu trnh.c. Nhit lng nh cho ngun lnh sau mt chu trnh.

Giia. Hiu sut ca ng c

ngc

nhit

1Q

2Q

A

Ngun lnh

Ngun nng

Ngun nng

Mylnh

1Q

2Q

A

Ngun lnh

T1

T2


26/54


%8,26%100273100

2731%100

T

T1

1

2

+=

=

b. Nhit lng nhn c ca ngun nng sau mt chu trnh( )J10.42,27268,0/10.35,7/AQ 441 ==

c. Nhit lng nh cho ngun lnh sau mt chu trnh.( )J10.07,2010.35,710.42,27AQQ 44412 ===

9-5.Nhit ca hi nc t l hi vo my hi nc l t 1=227 C

o

, nhit ca bnh ngng l t2=27 Co . Hi khi tn mt nhit lng Q=1kcal

th thu c mt cng cc i theo l thuyt bng bao nhiu?GiiCng cc i theo l thuyt thu c khi ng c lm vic theo chu trnhCcn thun nghch vi hiu sut l tng

1

2

T

T1=

Mt khc

( ) ( )kJ672,1kcal4,01273227

273271Q

T

T1A

Q

A

1

2 ==

+

+=

==

9-6. Mt chu trnh Ccn thc hin gia hai my iu nhit nhit t1=400oC, t2=20oC. Thi gian thc hin chu trnh l =1s. Tmcng sut (sinh cng) lm vic ca ng c theo chu trnh y, bittc nhn l 2kg khng kh, p sut cui qu trnh gin ng nhitbng p sut u qu trnh nn on nhit. Cho khng kh c =29kg/kmol.

Gii

Nhit lng nhn c ca ng c trong mt chu trnh l nhit nhn ctrong qu trnh 1-2 (hnh v)

2

111

p

plnRT

mQ

=

Qu trnh 4-1 on nhit nn4

1

41

1

14411 TpTpVpVp ==

p

VV1

1p1

3

2

4

V4 V2 V3

p2

T1

T2


27/54


Theo gi thit p2=p4, T4=T21

2

1

2

1

T

T

p

p

=

Do

2

111

T

TlnRT

1

mQ

=

Cng sinh ra trong mt chu trnh

( )2

1211

1

21TTT

1T

TlnTTR

1

mQ

T

TTQA

1

21

=

==

=

( ) ( )kJ63427320

273400ln.20400.31,8

14,1

4,1.

29

2000A

+

+

=

Cng sut ca ng c

( )kW634A

P ==

9-7. Mt my lm lnh lm vic theo chu trnh Cacn nghch, tiu thcng sut 36800W. Nhit ca ngun lnh l -10

o

C, nhit ngun nng l 17oC. Tnh:a. H s lm lnh ca my.b. Nhit lng ly c ca ngun lnh trong 1s.c. Nhit lng nh cho ngun nng trong 1 giy.

Giia. H s lm lnh ca my

( )74,9

1017

27310

TT

T

A

Q

21

22 =

+=

==

b.

Nhit lng ly c ca ngun lnh trong 1s( ) cal86000J10.6,31.36800.74,9PtA'Q 52 === c. Nhit lng nh cho ngun nng trong 1 giy

( ) ( ) cal10.5,9J10.41.36800174,9Pt1'QAQ 4521 +=+=+=

9-8. Khi thc hin chu trnh Ccn, kh sinh cng 8600J v nh nhit2,5kcal cho ngun lnh. Tnh hiu sut ca chu trnh.

GiiHiu sut ca chu trnh

%45%10018,4.10.5,28600

8600%100

QA

A%100

Q

A3

21+=+==

9-9. Khi thc hin chu trnh Ccn, kh nhn c nhit lng 10kcal tngun nng v thc hin cng 15kJ . Nhit ca ngun nng l100 Co . Tnh nhit ca ngun lnh


28/54


GiiHiu sut ca chu trnh Cc n

( ) K23910027318,4.10.10

10.151T

Q

A1T

Q

A

T

T1

3

3

12

1

2 =+

=

==

9-10. Mt my nhit l tng, chy theo chu trnh Ccn, c ngun nng nhit 117oC v ngun lnh nhit 27oC. My nhn ca ngunnng l 63000cal/s. Tnh:a. Hiu sut ca my.b. Nhit lng nh cho ngun lnh trong mt giy.c. Cng sut ca my.

Giia. Hiu sut ca my l hiu sut ca chu trnh Ccn

%23273117

273271

T

T1

1

2 +

+==

b. Nhit lng nh cho ngun lnh trong mt giy l Q2s/cal4800063000273117

27327

QT

T

QT

T

1Q

Q

1 11

2

21

2

1

2

+

+

=== c. Cng sut my l P bng cng my sinh ra trong mt giy

kW6318,41

4800063000QQAP 21

=

==

9-11. Mt my lm lnh l tng, chy theo chu trnh Ccn ngc lynhit t ngun lnh 0 Co nh cho bnh nc si 100 Co . Tnh lngnc cn lm ng ngun lnh c th bin 1kg nc thnh hi bnh si. Cho bit nhit nng chy ring ca nc l=3,35.105J/kg, v nhit ha hi ring ca nc l L=2,26.106J/kg.

GiiC th hnh dung my lnh ny nh s nu phn tm tt l thuyt.Nhit nhn t ngun lnh 2Q , nh ra ngun nng l 1Q :

1

1

22

21

2

21

2 QT

TQ

TT

T

QQ

Q=

=

(1)

Nhit lng cn lm bay hi nc:LmQ1 = (2)

Khi lng nc cn lm nng chy l m:

'mQ2 = (3)T (1) (2) v (3) ta c:

kg93,41.373

273.

10.35,3

10.26,2m

T

TL'm

5

6

1

2 ==


29/54


9-12. Mt kmol kh l tng thc hin mt chu trnh gm 2 qu trnhng tch v hai qu trnh ng p. Khi th tch ca kh thay it V1=25m

3 n V2=50m3 v p sut t p1=1at n p2=2at. Hi cng

thc hin bi chu trnh ny nh hn bao nhiu ln cng thc hinbi chu trnh Ccn c cc ng ng nhit ng vi nhit lnnht v nh nht ca chu trnh ni trn, nu khi gin ng nhitth tch tng ln gp i?

GiiCng thc hin trong c chu trnh: ( )( )1212 VVppA = Trong chu trnh Ccn, nhit ngun nng ng vi im (V2,p2), ngun lnh

vi (V1,p1). Trong mt chu trnh tc nhn nhn nhit1

222

'V

'VlnVp vi hiu sut:

22

1122

2

12

Vp

VpVp

T

TT =

=

Cng kh sinh ra trong mt chu trnh:

( )1

211221

'V

'VlnVpVpQ'A ==

( )

( )( )1,2

VVpp

'V

'VlnVpVp

A

'A

1221

1

21122

=

=

9-13. Mt my hi nc lm vic theo chu trnh nh hnh v 9-1

a. Thot tin hi nc t ni hi vo xilanh, p sut hi nc tngt po ti p1, th tch khng i v bng Vo (nhnh AB).

b. Hi nc tip tc i vo, pittng chuyn ng t tri sang phi(nhnh BC) vi p sut hi khng i l p1 v th tch tng ln

V1.c. Xilanh ng van li, pittng chuyn ng tip tc sang phi khi xy ra qu trnh gin on nhit (Nhnh CD);

d. Khi n v tr cui cng bn phi, th hi nc trong xilanh i vongun lnh, khi p sut hi gim xung po, cn th tch khngi bng V2, (nhnh DE).

p

P1

Po

B C

DA

Vo V1 V2Hnh 9-1

E


30/54


e. Pittng chuyn ng ngc li, y hi nc cn li trong xilanhra ngoi, khi p sut khng i bng po, th tch gim t V2 ti

Vo (nhnh EA).Hy tnh cng m my nhit sinh ra mi chu trnh, nu V o=0,5l;

V1=1,5l;V2=3l;po=1at;p1=12at v h s on nhit l =1,33.

GiiCng kh thc hin trn tng qu trnh ring bit:

+ Qu trnh A-B v D-E ng tch, kh khng sinh cng0AA DEAB ==

+ Qu trnh BC, kh gin n ng p, sinh cng( ) ( ) ( )J117610.5,05,110.8,9.12VVpA 34o11BC ===

+ Qu trnh CD kh gin n an nhit, theo cng thc (P.6) phn ph lc,cng sinh ra trong qu trnh gin n on nhit:

( )J10933

5,11

133,1

10.5,1.10.8,9.12

V

V1

1

VpA

133,1341

2

111CD

=

=

+ Qu trnh EA kh bin i ng p, cng thc hin

( ) ( ) ( )J24510.35,010.8,9VVpA34

2ooEA ==

Cng m my nhit sinh ra trong mi chu trnh lm vic chnh bng cngthc hin bi kh

( )J20242450109311760AAAAAA EADECDBCAB =+++=++++=

9-14. Hnh v 9-2 trnh by gin l thuyt ca ng c t trong bnk.a. Trong qu trnh u tin, hn hp chy c np vo xilanh, khi

po=const v th tch tng t V2 ti V1. (nhnh AB);b. Trong qu trnh th hai (nhnh BC), hn hop chy c nn

on nhit t th tch V1 ti V2.Khi nhit tng t To n T1v p sut t po n p1;

p

V

P1

Po B

E

D

V1V2

Hnh 9-2

A

C


31/54


c. Tip theo l qu trnh t chy nhanh hn hp chy bng tie lain; khi p sut tng t p1 ti p2, th tch khng i v bng

V2 (nhnh CD), nhit tng ti T2;d. Tip theo l qu trnh gin on nhit t th tch V2 ti V1

(nhnh DE), nhit gim xung T3;e. cui cng ca pittng (im E), van m, kh thot ra ngoi, p

sat gim nhanh ti po, th tch khng i v bng V1. (nhnhEB).

f. Cui cng l qu trnh nn ng p p sut po (nhnh BA).Hy tnh hiu sut ca chu trnh nu h s nn =V1/V2=5 v h s onnhit l =1,33.

GiiNhit tc nhn nhn trong c chu trnh chnh bng nhit tc nhn nhntrong qu trnh CD:

( )CDVCD1 TTnCQQ ==

Trn EB tc nhn ta nhit (nhit nhn vo s c du m):( )EBV2 TTnCQ =

Hiu sut ca ng c:( )

( ) 12

4o

2

1

2122

141o

CD

EB

CD

EB

1

2

pp

pp

V

V1

VpVp

VpVp1

TTnR

TTnR1

TT

TT1

Q

Q1

+=

+=

+=

+=+= (1)

Mt khc:( ) ( )

1o421214221o21 VppVppVpVp;VpVp ===

=

=

1

2

12

4o

V

V

pp

pp(2)

Thay (2) vo (1):%2,41511 33,111 ===

9-15. Tm hiu sut ca ng c t trong, cho bit h s on nhit l1,33 v h s nn bng:a.V1/V2=4; b. V1/V2=6; c. V1/V2=8;

GiiTheo bi 9-14 ta tnh c ln lt cho cc qu trnh c th

a. %7,36411 33,111 === b. %6,44611 33,111 === c. %6,49811 33,111 ===

9-16. Chu trnh ca ng c iezen bn k c trnh by trn hnh 9-3


32/54


a. Nhnh AB ng vi qu trnh np khng kh , p sut po=1at;b. Nhnh BC khng kh c nn on nhit ti p sat p1.c. cui k nn, nhin liu c phun vo xilanh, nhin liu chy

trong khng kh nng, khi pittng chuyn ng sang phi, utin l ng p (nhnh CD), sau l on nhit (nhnh DE);

d. cui qu trnh on nhit, van thot m, p sut gim xung po(nhnh EB);

e. Nhnh BA ng vi qu trnh y kh ra khi xilanh.Tnh hiu sut ca ng c diezen.

GiiTrong mt chu trnh, tc nhn ch nhn nhit trn qu trnh CD:

( )CDpCD1 TTnCQQ ==

Nh nhit:( )EBVEB2 TTnCQQ ==

Hiu sut :

1V/V

p/pp/p

V

V11

VpVp

VpVp11

TT

TT11

Q

Q1

13

121o

1

2

1131

222o

CD

EB

1

2

+=

+=

+=+=

1

p

p

p

p

1 1

2

1

o

+=

(1)

Trong 12 V/V= Mt khc

=

==

2

1

1

o

112oV

V

p

pVpVp ;

=

=

==

2

1

1

3

2

3

1

23122

V

V

V

V

V

V

p

pVpVp (2)

Vi1

3

V

V=

Thay (2) vo (3) v bin i ta nhn c:

V1

p1

po

V3 V2

p2

C D

E

BA

p

VHnh 9-3


33/54


( )1

11

1

=

9-17. Mt my hi nc chy theo chu trnh stilin gm hai qu trnh ngnhit v hai qu trnh ng tch nh hnh 9-4. Tnh hiu sut cachu trnh . So snh hiu sut vi hiu sut chu trnh Ccn ccng nhit ca ngun nng v ngun lnh.

GiiNhit tc nhn nhn c trong mt chu trnh lm vic bao gm qa trnh2 3 v 3 4

( )21V23 TTnCQ =

==

1

2

13434V

VlnnRTAQ

Cng tc nhn sinh ra (bng cng ng co sinh ra) trong mt chu trnh lmvic

( )

=+

++

=+++=

1

2

21

1

2

1

2

1

241342312V

VlnTTnR0

V

VlnnRT0

V

VlnnRTAAAAA

Hiu sut ca ng c

( )

( )

Carnot

1

21

12

21V

1

21

3423 T

TT

V/Vln

TT

R

CT

TT

QQ

A =

T2)Cng thc hin trong chu trnh bng hiu ca nhit nhn vo thc s vnhit ta ra thc s trong mt chu trnh (chnh bng tng nhit lng nhnvo trong c chu trnh)

21 QQA += Trong chu trnh Carnot

2121

21

2

2

1

1

1

21

1

21

TT

A

TT

QQ

T

Q

T

Q

T

TT

Q

QQ

=

==

=

+=

Trong qu trnh ng nhit (gia hai qu trnh on nhit), bin thinentropy l

2

2

1

1

T

Q

T

Q

T

QS ===

Nn( ) ( ) ( )kJ418kcal100STTA 21 ===

9-26. B 100g nc 0oC vo 400g nc 30oC trong mt bnh c vcch nhit l tng. Tnh bin thin entrpy ca h trong qu

trnh trao i nhit. T suy ra rng nhit ch truyn t vt nngsang vt lnh. Cho bit nhit nng chy ring ca nc 0oC l=80kcal/kg; nhit dung ring ca nc l 1kcal/kg.

Gii


37/54


Nhit cn bng T ca h sau khi trao i nhit xc nh t phng trnhcn bng nhit

( ) ( )( )

( )21

11122

22111mmc

mtmtmctttcmttcmm

+

+==+

( )

( )( ) ( )K281C8

4001001

100.800.10030.4001t

o ==+

+=

i vi nc , tng entropy bao gm tng do nng chy v tng dotng nhit

+=+

=+=

1

1

1

1T

T1

121

1T

Tlncm

T

m

T

dTcm

T

Q

T

Q

T

QS

1

Vi T =281(K) l nhit cn bng ca h.i vi nc b lnh i, entropy s gim, bin thin khi ny l

===

1

2

T

T22

T

Tlncm

T

dTcm

T

QS

2

bin thin entropy ca h l

( )( )K/kcal002,0

27330

281ln.4,0.1

2730

281ln.1,0.1

2730

1,0.80S

T

Tlncm

T

Tlncm

T

mSSS

2

2

1

1

1

121

++

++

+=

+

+=+=

Ta thy S>0, iu chng t nhit ch c th truyn t vt nng sang vtlnh* chng minh nhit ch c th truyn t vt nng sang vt lnh ta c thlm nh sau: Xt h hai vt c lp, nng lng ca h bo ton (nu qutrnh ta xt ch lin quan n s truyn nhit th nhit c bo ton)

2121 QQconstQQQ ==+= Trong 1Q , 2Q l bin thin nhit lng ca vt 1 v 2.

bin thin entropy

2

2

1

1

21T

Q

T

QdSdSdS

+=+=

Theo (*) ta c( )

( ) 0QTT0TT

QTT

T

1

T

1QdS 112

21

112

21

1 >>

=

=

Nu T2>T1 th Q1>0 tc l vt 1 nhn nhit hay nhit truyn t vt 2sang vt 1Nu T2


38/54


a. ng ACBb. ng ADB

Cho bit: V1=3l; p1=8,31.105N/m2;V2=4,5l; t1=27 Co , p2=6.105N/m2

Gii bin thin entropy khng ph thuc vo qu trnh bin i nh th no,m ch ph thuc vo trng thi u v trng thi cui, nn:

+

=+=+= C

Bp

A

CV

B

Cp

C

AVCBAC

TTlnnC

TTlnnC

TdTnC

TdTnCSSS

A - C v C B ng p nn

1

2

A

C

p

p

T

T= ;

1

2

C

B

V

V

T

T=

Do

++

=

++

=

1

2

1

2

1

11

1

2

1

2

V

Vln

2

2i

p

pln2

i

T

Vp

V

VlnR

2

2in

p

plnR

2

inS

( )

( )K/J4,5

3

5,4ln

2

26

10.31,8

10.6ln

2

6

27327

10.3.10.31,8S

5

535

++

+

=

9-28. C hai bnh kh, bnh th nht c th tch V1=2l cha kh Nit psut p1=1at, bnh th hai c th tch V2=3l cha kh CO p sutp2=5at. Cho hai bnh thng vi nhau v t chng trong mt v cchnhit l tng. Tnh bin thin entrpy ca h khi hai kh trnln vo nhau, bit nhit ban u trong hai bnh bng nhau vbng 27oC.

Gii

Khi gin n vo nhau cc cht kh khng sinh cng, nhit li b cch nn qutrnh t c trng thi cui cng c nhit khng i (li , y khngphi l qu trnh on nhit thun nghch). Entropy thay i mt lng (tnhbng con ng ng nhit)

+=+= 2121 TQ

T

QSSS

Qu trnh ng nhit

P

VV1 V2O

P1

P2

Hnh 9-5

C

A D

B


39/54


1

2

V

VlnpVAQ ==

( )K/J1,33

32ln

27273

10.3.10.8,9.5

2

32ln

27273

10.2.10.8,9S

V

VVln

T

Vp

V

VVln

T

VpS

3434

2

2122

1

2111

+

++

+

+=

++

+=

9-29. 200g st 100oC c b vo mt nhit lng k cha 300g nc 12 Co . Entrpy ca h ny thay i nh th no khi cn bng nhit?GiiSau khi trao i nhit h s cn bng nhit t oC. Phng trnh cn bngnhit

( ) ( )2211

222111

222111mcmc

tmctmctttmcttmc

+

+==

( )C183,0.41802,0.460

12.3,0.4180100.2,0.460t

o+

+=

bin thin entropy ca h bao gm s gim entropy ca ming st v stng entropy ca khi nc

+

=+=

2

22

1

112211T

Tlnmc

T

Tlnmc

T

dTmc

T

dTmcS

( )K/J3,327312

27318ln.3,0.4180

273100

27318ln.2,0.460S

+

++

+

+=

Vy sau khi cn bng nhit entropy ca h tng ln mt lng 3,3(J/K)


40/54


Chng 10: Cht kh

10-1. C 10g kh He chim th tch 100cm3 p sut 108N/m2. Tm nhit ca kh trong hai trng hpa. Coi kh He l l tngb. Coi kh He l kh thc

Giia. Kh He l tng, nhit c xc nh t phng trnh

Meneleev Crapayron

( ) ( )( )K481

31,8.4/10

10.100.10

R/m

pVT

68

==

b. Kh He kh thc, nhit c xc nh t phng trnh Van deWalls

+==

+ b

mV

V

am

m

p

R

1TRT

mb

mV

V

amp

222

2

( )K20510.3,2

41010.100

10.10010.1,4

410

1010.4

31,81T 5626

48

+=

i vi Hekmol/m2,3.10b;kmol/Jm4,121.10a 3-523-4 ==

10-2. Trong mt bnh th tch 10lt cha 0,25kg kh nit nhit 27oC.a. Tm t s gia ni p v p sut do kh tc dng ln thnh bnhb. Tm t s gia cng tch v th tch ca bnh

Gii

Cc hng s Van de Walls ca kh Nit 2mol/m10.92,3b;mol/Jm0,141a

3523 == Phng trnh Van de Walls

RTm

bm

VV

amp

22

2

=

+ (1)

a. T s gia ni p v p sut do kh tc dng ln thnh bnhNi p

22

2

V

am'p

=

Chia hai v ca (1) cho p ta c:

=

==

+ b

mVTRV

ma

p

'p

bm

Vam

TRV

'p

p

am

RTVb

mV1'p

p2

22

(2)

%9,410.92,3.28

25001,0.

300.01,0.31,8.28

141,0.250

p

'p 52

=

2 N.I.Kosin, M.G. Sirkevich, S tay vt l c s, NXB cng nhn k thut H Ni 1980, trang 106.


41/54


b. T s gia cng tch v th tch ca bnhCng tch

bm

'V

=

T s

%5,301,0.28

10.92,3.250

V

mb

V

'V5

===

10-3. Tm p sut ca kh cacbonic 3oC nu bit khi lng ring ca n nhit l 550kg/m3.

GiiPhng trnh Van de Walls

( )

2

2

2

2

22

2

a

b/

RTp

RTb1a

pRTm

bm

VV

amp

=

=

+=

+

Thay s( )

( )( )Pa10.4,1141,0.

028,0

550

10.92,3550/028,0

3273.31,8p

8

2

5

+=

10-4. Th tch ca 4g kh oxy tng t 1 n 5 dm3. Xem kh oxy l thc.Tm cng ca ni lc trong qu trnh gin n .

GiiNi p

2

2

V

am'p

=

Cng ca ni lc

( )J7,1005,0

1

001,0

1

32

4138,0'A

V

1

V

1ma

V

adVmdV'p'A

2

21

2V

V 2

2

2

1

=

=

==

10-5. Tnh ni p ca kh ccbonic lc khi lng ring ca n l550kg/m

3

. Cho bit i vi kh cacbonic c: Tk=304K vpk=7,4.106N/m2

GiiNi p ca kh Cacbonic


42/54


k

k

p

RT

V

amp

64

27'

2

2

2

22

2

==

Nhng do

=ak

k

p

RT

64

272

Nn( )Pa10.8,6

10.4,7.64

304.31,8.27

044,0

550'p

6

6

22

=

10-6. Tnh khi lng nc cn cho vo mt ci bnh th tch 30cm3 khiun nng ti trng thi ti hn n chim ton b th tch ca bnh.

GiiGi khi lng nc cn cho vo bnh l m. Khi un nng ti trng thi tihn, th tch ca bnh l th tch ti hn, nn

b3

Vmb3

mV

mVV okk

====

Hng s Van de Walls ca nc b=30,5.10-6

m3

/mol, ta tnh c m=5,9g10-7. Xc nh khi lng ring ca hi nc im ti hn theo gi tr

cng tch b=0,03m3/kmol.

GiiCng tch

( )33

k

kokk m/kg20010.03,0.3

018,0

b3V

mb3

mV

mV ======

10-8.i vi kh cacbonic : a=3,64.10

5

Jm

3

/kmol

2

, b=0,043m

3

/kmol. Hi:a. 1g ccbonic lng c th tch ln nht l bao nhiu?b.p sut hi bo ha ln nht l bao nhiu?c. CO2 lng c nhit cao nht l bao nhiu?d. Cn phi nn kh CO2 vi p sut bng bao nhiu thnh CO2

lng nhit 31oC v 50oC.

Giia. Th tch ln nht ca ccbonc lng ng vi trnh thi ti hn

(suy ra t cc ng ng nhit Van de Walls)

kg/m10.93,210.44

10.043,0.3bm3V

mV

33

3

3

okk

== (b=0,043m3/kmol =0,043.10-3m3/mol)

b.p sut hi bo ho cc i ng vi im ba (suy ra t cc ngng nhit Van de Walls)


43/54


( )( )Pa10.4,7

0,043.1027

0,364

b27

ap

6

23-2k ==

(a=3,64.105Jm3/kmol2=0,364Jm3/mol)c. Nhit cao nht m nit cn th lng ng vi nhit im ba

( )( )!C31K30410.043,0.31,8.27

364,0.8

Rb27

a8T

o

3k===

d. Ccbonc lng 31oC cn nn ti p sut bng p sut ti hn( )Pa10.4,7p 6k =

cng l nhit ln nht m ccbonc th lng mi psut. Vi nhit 51oC l khng th thc hin ho lng vi bt c p sut no

10-9. nghin cu trng thi ti hn nh vt l hc Nga A. Vnariuytdng mt ci hp trong c ng mt ng cha ete c hn kn.H nng hp quan st trng thi ti hn.a. 20oC, te nc phi chim mt th tch bng bao nhiu phn

trm th tch ca ng khi n nhit ti hn, ng cha yte trng thi ti hn? Bit rng khi lng 1 kmol te l74kg/kmol, khi lng ring ca te 20oC bng 714kg/m3. i

vi te Tk=193oC, pk=35,9.105N/m2;b. Nu th tch ca ng ln hay nh hn th tch te ti hn th s

xy ra hin tng g khi nhit nng ln?

Giia. Gi th tch v khi lng te vo ng l V v m, th tch ca

n trng thi ti hn l Vk (do cng l th tch ca ng). Tac

k

k

okkp8

RTb3

mb3

mV

mV

===

( )%25

193273.31,8.714.3

10.9,34.074,0.8

RT3

p8

V

VmV

5

k

k

k

+

===

b. Khi th tch ca ng nh hn th tch Vk th cha un ete ln titrng thi ti hn ete chim y ng.

c. Khi th tch ca ng ln hn th tch Vk th cha un ete ln titrng thi ti hn ete bay hi ht.


44/54


Chng 11: Cht lng

11-1. Xc nh cng cn thit bin mt git nc 1g thnh sng m(ngha l tch git nc thnh nhng git nh) ng knh0,2m. Din tch b mt ca git nc lc u coi nh khng ngk so vi tng din tch b mt ca git sng m.

GiiKhi lng ca mt git sng m

3r

3

4m = (r l bn knh git sng)

S git sng c to thnh t git nc

3r4

M3

m

MN

==

Din tch b mt ca mt git 2r4s = , vy din tch b mt tng cng

d

M6

r

M3sNS

=== (d=2r l ng knh git sng)

Cng cn thit bin git thnh sng m ti thiu bng nng lng mtngoi

d

M6SA

==

( )J19,210.2.1000

001,0.073,0.6A

6==

11-2. Hai git thy ngn vi bn knh mi git l 1mm nhp li thnhmt git ln. Hi nhit ca git thy ngn tng ln bao nhiu?

Cho bit thy ngn c sut cng mt ngoi =0,5N/m, khi lngring =13,6.103kg/m3, nhit dung ring c=138J/kg.

GiiGi bn knh ca git nh l r, ca git ln c to thnh l R. Ta c:

333 2rRr3

42R

3

4==

Nhp lm mt, din tch mt ngoi ca git ln s nh hn tng din tch mtngoi ca hai git nh, nng lng b mt s gim. gim nng lng bmt ny s bng nhit lng ca git ln nhn c.

( ) tmcR4r4.222

= ;3

r3

4

2m = Do

( )C10.65,1001,0.6,13.138

5,0.3

2

41

rc

3

2

41t

o433

=

=


45/54


11-3. Tnh cng cn thc hin thi mt bong bng x phng t nbn knh r=7cm. Sut cng mt ngoi ca nc x phng l =4.10-2N/m. p sut kh quyn po=1,01.10

5N/m2.

GiiTa coi nhit ca kh trong qu trnh thi l khng i (qu trnh ngnhit). Cng cn thi bong bng bng cng to ra mt ngoi (bng nng lngmt ngoi) A1, v cng nn ng nhit A2 ca mt lng kh ng bng lng

kh trong bong bng cui qu trnh t kh quyn vo.221 r8r4.2A ==

o

2p

plnpVA =

po l p sut kh quyn, p l p sut kh trong bong bng tnh p, ta ch mng x phng gm hai mt phn cch, mi mt phncch (mt khum) s gy ra ra p sut ph p vo tm vi gi tr tnh theocng thc Laplace

r/2pp = . p sut ph tng cng do mng x phng gy ra cho kh bn trong

bao gm p sut do mt cong pha ngoi v mt cong pha trong (vi bn knh

xp x bn knh mt cong ngoi) v p sut kh quyn cng li:

opr

4p +=

Do

+

+=

+

+=

rp

41ln

rp

41pr

3

4

p

r/4plnr

3

4

r

4pA

oo

o

3

o

o3

o2

V do 1rp

4x

o


46/54


GiiLc ti thiu ko AB bng lc cng mt ngoi tc dng ln AB, cng ca lcny (cng chnh bng nng lng mt ngoi c tng ln do tng dintch b mt mng)

( )J10.4,504,0.15,0.045,0xlFsA 4====

11-5. C mt khung hnh ch nht chiu di l=10cm. on dy AB linhng chia khung thnh hai khung nh hnh vung (h 11-4). Hion AB s dch chuyn v pha no v dch chuyn mt on bng

bao nhiu nu hai khung hnh vung c ph bng hai mngcht lng khc nhau c sut cng mt ngoi tng ng l: 1-=0,06N/m v 2=0,04N/m.

GiiXt v mt nng lng, nng lng mt ngoi ca h ban u l ( )S21 + (Sl din tch na hnh ch nht). H cn bng v tr sao cho nng lng mtngoi ca h hai mng x phng l nh nht, c ngha l dy AB s chuynng v pha lm mng x phng c sut cng mt ngoi ln hn. Kt qu ldy AB s chuyn ng n tn cng bn pha mng c sc cng mt ngailn hn

11-6. xc nh lc cng mt ngoi ca ru ngi ta lm nh sau: choru trong mt ci bnh chy nh git ra ngoi theo mt ng nhthng ng c ng knh d=2mm. Thi gian git ny ri theo gitkia l =2 giy. Ngi ta thy rng sau thi gian t=780 giy th cm=10 gam ru chy ra. Tnh sut cng mt ngoi ca ru. Coich tht ca git ru khi n bt u ri c ng knh bng ngknh ca ng nh git.

Gii:

Khi lng ca mt git ru

t

mm

= (1)

Git (bt u) nh xung khi:

A

B

l

A

B

nh 11-3 nh 11-4


47/54


( )

td

mg

t

mgmgd

=

= =

1

Thay s m/N04,0780.10.2.14,3

8,9.10.10.23

3

=

11-7. Mt si dy bc ng knh d=1mm, c treothng ng. Khi lmnng chy c 12 git bc th si dy bc ngn i mt onh=20,5cm. Xc nh sut cng ca mt ngoi bc th lng? Chobit khi lng ring ca bc th lng l =9300kg/m3 v xem rngch tht ca git bc khi n bt u ri c ng knh bng ngknh ca si dy bc.

Gii

Khi lng ca bc ho lng h4

dM

2=

Khi lng ca mt git bc lng

k4

hd

k

Mm

2== (1)

Git (bt u) nh xung khi:

( )

k4

gdhg

k4

hdmgd

21

== =

Thay s

( )m/N39,012.4

205,0.10.8,9.9300

k4

gdh3

==

11-8. C mt ng mao dn ng knh ngoi d=3mm, mt u c btkn. u bt kn ng mt t thy ngn (h 11-5). Khi lng cang mao dn v thy ngn l 0,2g. u bt kn ca ng mao dntrong nc. Xem nc lm t hon ton ng mao dn v nc csut cng mt ngoi =0,073N/m, c khi lng ring =103kg/m3.


48/54


GiiH (ng mao dn+ Thu ngn) chu tc dng ca lc cng mt ngoi v trnglc hng xung, lc Acsimet hng ln, khi cn bng th

( )

( )

( )cm8,3

003,0.8,9.1000.14,3

8,9.0002,0003,0.073,0.14,34h

gdmgd4hh

4dmgd

2

2

2

+

=

+==+

11-9. chng minh lc cng mt ngoi, ngi ta nc vo mt cidy bng li st m cc si li c ph mt lp parafin. Cc lca li st c dng hnh trn ng knh d=0,2mm. Hi chiu caoln nht ca mc nc vo dy m nc chy ra theo cc l ?

GiiNc cn ng c trn dy l do lc cng mt ngoi xut hin ti mtthong ni tip xc ca vng dy v nc. Nc s nh xung khi sc cngmt ngoi khng cn sc gi n na. Khi

( )cm150002,0.10.1000

073,0.4

gd

4hd4/dgh

2 =

11-10.Trong mt ng mao dn h t thng ng, ng knh trong 1mmc mt git nc. Hi khi lng ca git nc phi nh th no mt khum bn di ca git nc l: mt lm, mt phng, mt li?

GiiC th hiu nh tnh l git nc s i dn ti pha y ng mao dn. khum (lm) ca mt bn di s km hn mt pha trn (mt pha trn lunlun l mt cu vi ng knh bng ng knh ca ng mao dn). Khi khilng ca git nc cng ln cong mt khum bn di cng gim, n khi

d

h

Hnh 11 5


49/54


khi lng git nc t ti gi tr mo no mt khum ny s l mt phng,ln hn gi tr ny mt khum tr thnh mt li. Ta c

kg10.34,28,9

001,0.073,0.14,3

g

dmdgm

5

oo

===

Vy nn: Khi m< kg10.34,2 5 mt khum bn di l mt lm Khi m= kg10.34,2 5 mt khum bn di l mt phng Khi m> kg10.34,2 5 mt khum bn di l mt li11-11.Hai ng mao dn c ng knh trong ln lt l 0,5mm v 1mm

nhng trong mt bnh ng cht lng. Tnh hiu cc mc cht lngtrong hai ng mao dn nu:a. Cht lng l nc.b. Cht lng l thy ngn

Giia. Khi nhng hai ng trong nc, nc s dng ln. dng ln ca

nc trong tng ng l

gd

4h

=

Hiu cc mc cht lng trong hai ng mao dn l

( )cm3001,0

1

0005,0

1

8,9.1000

073,0.4

d

1

d

1

g

4h

12

=

=

b. Khi nhng hai ng trong thu ngn, thu ngn trong ng s hxung. h xung ca thu ngn trong tng ng l h. Cn bnggia p sut thu tnh v p sut ph cho im bn trong chtlng ngay ti mt khum ta c

gd

4

h2/d

2

gh

== Hiu cc mc cht lng trong hai ng mao dn l

( )cm5,1001,0

1

0005,0

1

8,9.13600

5,0.4

d

1

d

1

g

4h

12

=

=

11-12.Mt ng c nhng thng ng trong mt bnh dng cht lng. Hichiu cao ca ct nc trong ng thay i nh th no nu ng maodn v bnh c nng ln nhanh dn u vi gia tc a=g? H xungnhanh dn u vi gia tc a=g/2?

Gii:Khi bnh v ng c nng ln vi gia tc a, p sut ca im bn trong

ng ngang mt thong cht lng ngoi ng bao gm p sut kh quyn, psut thu tnh ca ct nc, p sut ph gy bi mt khum v p sut gy ra


50/54


do lc qun tnh3. p sut ny c gi tr bng p sut ti nhng im ngangbng vi n ngoi ng

( )

+=++=

g

phhghppp

p

poo

Khi ng mao dn v bnh khng chuyn ng ( 0= ) th

g

ph

p

o

=

Do :+

=g

g

h

h

o

Khi nng ln g=

2

1

h

h

o

=

Khi h xung: 2/g=

2

h

h

o

=

11-13.C hai ng mao dn lng vo nhau, ng trc, nhng thng ngvo mt bnh nc. ng knh trong ca ng mao dn nh, bng brng ca khe to nn gia hai ng mao dn. B qua b dy ca ngmao dn trong. Hi mc cht lng trong ng no cao hn, cao hnbao nhiu ln?

Gii dng ca nc trong ng mao dn trong

gd

4h1

= (d l ng knh trong ca ng mao dn trong)

ng knh trong ca ng mao dn ngoi theo gi thit ta tnh c l 3d. dng ca nc trong ng mao dn ngoi l h2. Khi cn bng tng lc cng mtngoi hai ng tip xc gia mt thong ca nc vi cc ng mao dntrong v ngoi (thng ng hng ln) bng trng lc ca ct nc dng ln(thng ng hng xung).

( )( )gd

2hgh4/d4/d3d3d 22

22

==+

Do

2h

h

2

1 =

3 Trong h quy chiu gn vi bnh+ng mao dn, p sut do lc qun tnh gy ra c hiu thng thng bng p lcca n hS ln din tch phn tip xc ca n S vi tit din y hay pqt= h


51/54


Vy nc dng ln trong ng mao dn trong cao hn 2 ln so vi dngca khe mao dn gia hai ng.

11-14.C hai tm thy tinh phng t song song cch nhau mt khongd=0,2mm, nhng thng ng vo trong mt cht lng. Xc nhkhi lng ring ca cht lng nu bit rng chiu cao ca khicht lng gia hai tm thy tinh dng ln mt on h=3,2cm. Sutcng mt ngoi ca cht lng l 0,027N/m. Xem cht lng lm t

hon ton thy tinh.

Gii dng ca mc cht lng trong ng l h, ta c

( )3m/kg8610002,0.032,0.8,9

027,0.2

ghd

2ghdbb2 ===

11-15.Hiu mc thy ngn trong hai nhnh ca ng mao dn hnh ch Uc ng knh trong d1=1mm v d2=2mm l h=1cm. Xc nh sutcng mt ngoi ca thy ngn. Cho bit khi lng ring ca thy

ngn l 13,6.103

kg/m3

.

GiiCn bng p sut cho im ngay di mt phn cch ca nhnh ln ta c

( )12

21

21 dd4

dhdghg

d

4

d

4

=+=

( )( )m/N67,0

001,0002,04

002,0.001,0.01,0.8,9.13600

=

11-16.Khi lng ring ca khng kh trong mt ci bong bng di yca mt h nc su 6m ln gp 5 ln khi lng ring ca khngkh kh quyn ( nhit bng nhit y h). Xc nh bnknh bong bng.

Gii

Gi bn knh ca bong bng l R, p sut bn trong bong bng l:

2op p gh

r

= + + (1)


52/54


T phng trnh Menleep Claprn suy ra4 khi lng ring ca kh bn

trong v bn ngoi bong bng ln lt lRT

p= v

RT

poo

=

T oo p

pn ==

(2)

Thay (1) vo (2)

( ) ghp1n

2r

p

gh

rp

21n

ooo

=++=

Thay s:( )

m4,06.8,9.100010.15

073,0.2r

5=

=

11-17.Trn mt nc ngi ta mt ci kim c bi mt lp m mng (cho khi b nc lm t). Kim c ng knh ln nht l bao nhiu

n c th c gi trn mt nc m khng b chm xung di?Cho bit khi lng ring ca thp lm kim l =7,7k.103kg/m3.

Gii

kim khng b chm th p sut do trong lng ca kim ti mt tip xcgia kim v nc phi nh hn p sut gy ra bi mt cong ca nc v psut do lc y Acsimet tc dng ln kim.

( )

rl

gmmg

S

Fmg

r

rlrlS

A

2

2/2

(1)

trong r, l,r, l th t l bn knh, chiu di, khi lng ring ca kim v

ca nc. Cn khi lng ca kim: rlrm 2=

(2)

Thay (2) vo (1)

( ) ( )mm6,1

10.8,9.17,7.2.14,3

073,0.16

g2

16dr2

3

lr

=

=

Ht

4RT

p

V

mRT

mpV

===


53/54


Ph lcA. Qu trnh on nhitTrong qu trnh on nhit

0dTnCpdV0Q V =+= (P1)H s on nhit

+

==V

V

V

p

C

RC

C

C

1

RCV

=

;1

RCp

=

(P2)

Th vo (1), v lu phng trnh Mendeleev CrapayronnRdTVdppdVnRTpV =+=

Cho ta

( ) 0VdppdV1

pdV =+

1+

( ) 0pVd0VdppdV ==+ Hay

constpV = y l phng trnh cho qu trnh on nhitHay cng c th vit cch khc

( ) ( )== 11 TVnRVpVpV constTV 1 = (P3)

( )( )

==

11 pnR

p

pVpV

constTp1 = (P4)

Cng trong qu trnh on nhit( )

TnR1

1TnCUA V

2

===

Do( ) 112212 VpVpTTnRTnR ==

Nn

1

VpVpA 2211

= (P5)

Cng c th vit( )

=

=1

2

1113

1

21

V

V1

1

Vp

T

T1nRT

1

1A

(P6)

B. Cc hng s Van de Walls xc nh qua cc thng s ti hn p k,

Tk, VkPhng trnh Van de Walls cho mt mol kh thc

( ) RTbVV

a

p 2 =

+ 2V

a

bV

RT

p = Ti im ba

2

kk

kk

V

a

bV

RTp

= (P7)


54/54

( ) ( ) 3k2

k

k

3

k

2

k

k

0V V

a2

bV

RT0

V

a2

bV

RT0

V

p

k

=

=+

=

=

(P8)

( ) ( ) 4k3

k

k

4

k

3

k

k

0V

2

2

V

a3

bV

RT0

V

a6

bV

RT20

V

p

k

=

=

=

=

(P9)

Chia (P9) cho (P8) ta nhn cb3Vk = (P10)

Th (P10) vo (P7) ta nhn c

( )( )

2k2

k

k2

k

kk

b27

ap

V

abV

bV

RTp =

= (P11)

T (P10) v (P8) ta nhn c

bR27

a8Tk = (P12)

Cc hng s Van de WallsT (P7) v (P8) ta c

k

kk

3

kk

kk

p8

RTb

2

V

V

a2

bV

RTp =

= (P13)

Hng s a c tm t (P11) v (P12)

k

2

k

2

k

2

p64

TR27pb27a == (P14)

***Ht***

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