新版物理化学第二章热力学第二定律练习题及答案课件 doc - 图文

3 3

He (0.04649 m ,283.2 K )

3,293.2 K )

H2 (0.02406 m

T

2

( 0.07055 m ,287.7 K ) ( 0.07055 m3,287.7 K )

V

2

S He nC V

,m

He ln

T

1

nR ln

V

1

所以:

287 .7

2 1 .5

0 .07055

-1

8 .314 ln

283 .2

-1

2 8. 314 ln

0 .04649

7 .328 J K

同理: ∴ΔS =

SH

2

8. 550 J K

SHe

+

S H

2

= 7.328 + 8.550 = 15.88 J · K

10. 解:

ΔH = ΔH1 + ΔH2 + ΔH3 = (75.4 × 10) - 6032 - (37.7 × 10) = - 5648.4 J

ΔS = ΔS1 + ΔSS3 = 75.4 2 + Δ

×

273 ln

263

6032

263 ln 273

- 1

- 273 + 37.7 ×

= - 20.66 J K·

ΔG = ΔH- TΔS = - 5648.4 + 263

p

l

(×- 20.66) = - 214.82 J

ln

p

ΔG≈ ΔG2 = - RT ×

s

p ln l

p

s

G RT

214 .82

p

l

= = 263 8.314

p

= 0.09824,

s

= 1.1032

11. 解: 用公式 ΔS = - R∑ni lnxi

- 1

= - 8.314 × (0.2 × ln0.2 + 0.5 × ln0.5 + 0.3 × ln0.3) = 8.561 J K ·

12.解:

- 1

ΔCp = 65.69 - 2 ×26.78 - 0.5 ×31.38 = - 3.56 J K· ΔHT = ∫ ΔCpdT + Const = - 3.65T + Const ,∵ T = 298K时, ΔH298 = - 30585J

代入,求得: Const = - 29524,ΔHT = - 3.65T - 29524,代入吉 -赫公式,

G

积分,

2

G 298

823

1

3 .65 29524

298

2

dT

66 .91

823 T

G

2

10836 .6

G

66 .91 30 .55

2

25143 J

823 298

恒温恒压下, ΔG > 0 ,反应不能自发进行,因此不是形成 13.解:

Ag2O 所致。

RT ln

p p

2

= 8.314 ×373 ×ln0.5 = - 2149.5 J

ΔG1 = 0 ;ΔG2 = S

m

1

373

ln

S 298 C

p 373 - 1 -1 R ln 196 .24 R ln 2 202 .0 J K mol 298 p

473 p -1 -1

R ln 209 .98 J K mol 298 p

S

m

p , m

473 ln

S 298 C

p ,m

ΔH3 = Cp,m × (473 - 373) = 33.58 × 100 = 3358 J ΔG3 = ΔH3 - (S2T2 - S1T1) = 3358 - (209.98 × 473 –202.00× 373) = -20616.5 9 J ΔG = ΔG2 + ΔG3 = - 20616.5 - 2149.5 = - 22766 J = - 22.77 kJ

1

7 =

p

1

2 7

576 K

14.解:

5

, T

2

T

298

1

.1

0

p

2

5

Q = 0 , W = ΔU = n CV,m( T2- T1) = 10 × 2 R × (576 - 298) = 57.8 kJ

7

ΔU = 57.8 kJ ,ΔH = n Cpm(T2- T1) = 10 × 2

R × (576 - 298) = 80.8 kJ

- -3

= - 282.1kJ

ΔS = 0 ,ΔG = ΔH - Sm,298ΔT = 80.9 - 10 × 130.59 × (576 - 298) × 10

ΔA = ΔU - SΔT = - 305.2 kJ 15.解:这类题目非常典型,计算时可把热力学量分为两类:一类是状态函数的变化, 包括 ΔU、ΔH、ΔS、ΔA、ΔG,计算时无需考虑实际过程;另一类是过程量,包括 Q、W,不同的过程有不同的数值。 先求状态函数的变化,状态变化为 :(p1,V1,T1) ( p2,V2,T2)

U

T

dU = V TdS - pdV ,

p 对状态

方程

T

S V

T

p

p T p T V

a V

2

V

RT

p p RT 2a R , T V V V T V V

2

3

而言:

U

V

T

R V

p

a

2

T

V

V

2

U

dV

V

2

a dV

2

1 a V

1

U

V

1

V

V

V

1

V

1

所以:

T

2

H a 1 V

2

U 1

1

pV

U

p2 V 2

p1V1

a

2

a

1

V S

V V

V

2a

1 V

2

V

1

1

V R

V

2

S

V

2

p

dV T

2 2

dV

1

dV V

1

R ln

V

1

V

1

V

T

V

V

V

A

U T S a

1 V

2

1

1

V

V

2 RT ln

V

1

G H T S

1

2a 1

2

V RT ln 2

V

1

V

1

V

再求过程量,此时考虑实际过程恒温可逆:

W

V

V

2

2

RT

a

dV

2

V

1

1 a V

1

RT ln

V

V

2

pdV

V1

V

V

V

1

2 1

V

2

Q T S RT ln

V

1

对于恒温可逆过程:

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