m m m - pku.edu.cn...- 4 - = 37.61 kJ mol-1...

- 1 -

V

V 2

m m m

m

2

第一章习题答案

1．解（1）W = nRT ln(V2 /V1 )

p V = nRT = nRT = W

1 1 1

ln(V2 /V1 )

V = W

= 41850J = 8.97 10−2 m3 1

p ln(V /V ) (2.0265105Pa)ln10 1 2 1

（2） T = W

= 41850J = 1093 nR ln(V /V ) (2mol)(8.314J K-1 mol-1 ) ln10

2 1

2．解（1）V1 = nRT

p1

(1mol)(8.314J K -1 mol -1 )(423) = = 0.03517m3

100000Pa

W = nRT ln V

2

V1

= (1mol)(8.314J K-1 mol-1 )(423) ln 0.01m3

= −4423J 0.03517m3

（2） ( p + a

m

)(Vm − b) = RT

即 (100000Pa+ 0.417Pa m

m

6 mol-2 ) (Vm − 3.7110

− m5 m3 ol-1 )

= (8.314J K-1 mol-1 )(423)

整理得 V 3 − 3.520 10-2V 2 + 4.17 10−6V −1.547 10−10 = 0

V = 0.0353m3

V2

W = pdV = 1

V2 (

nRT −

an

1 V − nb V

)dV

V V

= nRT ln V

2 − nb

+ an2 ( 1

V1 − nb V2

2

− 1

)

V1

= 1mol(8.314J K-1 mol-1 )(423) ln 0.01m

3 − (1mol)(3.7110−5 m3 mol-1 )

0.0353m3 − (1mol)(3.7110−5 m3 mol-1 )

2

- 2 -

p p p

+(0.417Pa m6

mol-2

)(1mol)2

(

= −4415J

1

0.01m3

−

1 )

0.0353m3

3. 解（1）W = pV = (10000Pa)(10−3 m3 ) = 100J

（2）W = p(V −V ) = p ( nRT

− nRT

) = nRT (1− p

2 ) 2 1 2

2 1 1

= (10mol)(8.314J K-1

mol-1

)(300) (1− 100kPa

) 1013.25kPa

= 22.48kJ

（3）W = nRT ln p

1

p2

= (10mol)(8.314J K

-1 mol

-1 )(300) (ln

1013.25 )

100

= 57.76kJ

5．解（1）W = p(V g − Vl )

= (100000Pa)(1.677 −1.04310−3 )m3 kg-1 (18.0 10−3 kg)

= 3.017kJ

（2）W = pVg = (100000Pa)(18.010−3 kg) (1.677m3 kg-1 ) 10−3

= 3.019kJ

x = [(3019 − 3017) / 3017]100% = 0.066%

（3）W = pVg = nRT

= (1mol)(8.314J K-1 mol-1 )(373)

= 3.101kJ

(4)

∆𝑣𝑎𝑝𝑈𝑚 = 𝑄𝑝−𝑊

𝑛 = 40.63kJ mol-1 − 3.017kJ mol-1

- 4 -

= 37.61 kJ mol-1

（5）水在蒸发过程中吸收的热量一部分用于胀大自身体积对外作功。另一

部用于克服分子间引力，增加分子间距离，提高分子内能。（因为

∆U 0, ∆U = Q −W 0, 所以 Q W ）

6. 解状态 1 V = 0.0224m3 ,T = 273, p = 101.3kPa 1 1 1

状态 2 V = 0.0224m3 ,T = 546, p = 202.6kPa 1 1 1

状态 3 V = 0.0448m3 ,T = 546, p = 101.3kPa 1 1 1

44.8

22.4

O 273 546

T / K

3

C B

1 A

2

V /

dm

3m

ol-

1

- 5 -

步骤过程的名称 Q / J W / J ∆U / J

A 等容 3405 0 3405

B 等温 3147 3147 0

C 等压 -5674 -2269 -3405

7. 解

𝑛 =𝑝1𝑉1

𝑅𝑇1 =

(100000𝑃𝑎)(0.003𝑚3)

(8.314𝐽∙𝐾−1∙𝑚𝑜𝑙−1)(293𝐾) = 0.123mol

𝑉2 = 𝑛𝑅𝑇2

𝑃2=

(0.123𝑚𝑜𝑙)(8.314𝐽 ∙ 𝐾−1 ∙ 𝑚𝑜𝑙−1)(353𝐾)

100000𝑃𝑎= 3.61 × 10−3𝑚3

𝑊 = 𝑝(𝑉2 − 𝑉1) = 100000𝑃𝑎 (3.61 × 10−3

𝑚3 − 0.003𝑚3) = 61 𝐽

= 0.123𝑚𝑜𝑙 ∫ [(27.28 + 3.26 × 10−3𝑇/𝐾)𝐽 ∙ 𝐾−1 ∙ 𝑚𝑜𝑙−1]353𝐾

293𝐾𝑑𝑇

= 0.123 mol {27.28(353K-293K)+1

2× 3.26 × 10−3[(353𝐾)2 −

(293𝐾)2]}𝐽 ∙ 𝐾−1 ∙ 𝑚𝑜𝑙−1

= 209.1J

- 6 -

∆𝑈 = 𝑄 − 𝑊 = 209.1𝐽 − 61𝐽 = 148.1𝐽

8.

9.

- 7 -

10.

- 9 -

11.

= 1

1.4−1[(100000Pa)(0.020𝑚3) − (2 × 100000𝑃𝑎)](0.01219𝑚3)

= -1095J

= (2×100000𝑃𝑎)[0.02𝑚3+(0.02𝑚3−0.01219𝑚3)]

(0.8067𝑚𝑜𝑙)(8.314𝐽∙𝐾−1𝑚𝑜𝑙−1)=829.3K

左方膨胀气体耗用的热量一部分用来升高左室温度，另一部分用来推动活塞向

- 10 -

右移动。设左室理想气体为体系。

∆𝑈左 = 𝑛𝐶𝑣,𝑚(𝑇2 − 𝑇1)

=(0.8067mol)(20.92J∙ 𝐾−1 ∙ 𝑚𝑜𝑙−1) × (829.3𝐾 − 298.2𝐾) = 8963𝐽

𝑊左 = −𝑊右 = 1095𝐽

𝑄左 = ∆𝑈左 + 𝑊左 = 10058J

-13 32

22

(1mol)(8.314J K mol)(136.5 ) 2.8 10 m(4 101325Pa)

nRTVp

−⋅ ⋅ Κ= = = ××

（2）理想气体任何过程2

1

T

VTU C dΔ = ∫ T T

2

1

T

pTH C dΔ = ∫

2 1( )VU C T TΔ = −

-1 -13(1mol) (8.314J K mol ) (136.5 273 )2

= × ⋅ ⋅ × Κ − Κ

1702J= −

2 1( )pH C T TΔ = −

-1 -15(1mol) (8.314J K mol ) (136.5 273 )2

= × ⋅ ⋅ × Κ − Κ

2837J= −

（3）W pdV= ∫

pT C= 2nRT nRTV

p C= =

2nRTdV dTC

=

2 2 21 1 1

2 12 2 2 (

V T T

V T T

C nRTW pdV dT nRdT nR T TT C

= = = = −∫ ∫ ∫ )

Κ-1 -12(1mol)(8.314J K mol ) (136.5 273 )= ⋅ ⋅ × Κ −

2270J= −

13. 解（1）U H pV= −

( ) ( ) ( ) ( )p p p pU H V Vp C pT T T T

∂ ∂ ∂ ∂= − = −

∂ ∂ ∂ ∂ p

（2）理想气体的内能和焓只是温度的函数，若温度不变焓也不变。

令 ( , )H f T V=

( ) ( )T VH HdH dV dTV T∂ ∂

= +∂ ∂

若则0dT = ( )THdH dVV∂

= =∂

0

因为所以0dV ≠ ( )THV∂ 0=∂

- 9 -

- 10 -

（3）(𝜕𝐶𝑉

𝜕𝑉)

𝑇= [

𝜕

𝜕𝑉(

𝜕𝑈

𝜕𝑇)

𝑉]

𝑇= [

𝜕

𝜕𝑇(

𝜕𝑈

𝜕𝑇)

𝑇]

𝑉

对于理想气体(𝜕𝑈

𝜕𝑉)𝑇 = 0 所以(

𝜕𝐶𝑉

𝜕𝑉)𝑇 = 0

14. 解（1） 𝑈 = 𝐻 − 𝑝𝑉

(𝜕𝑈

𝜕𝑉)𝑃 = (

𝜕𝐻

𝜕𝑉)𝑃 − 𝑝 = (

𝜕𝐻

𝜕𝑇)

𝑃(

𝜕𝑇

𝜕𝑉)

𝑃− 𝑝 = 𝐶𝑝 (

𝜕𝑇

𝜕𝑉)

𝑃− 𝑝

（2）𝐶𝑝 − 𝐶𝑉 = (𝜕𝐻

𝜕𝑇)

𝑝− (

𝜕𝑈

𝜕𝑇)

𝑉= (

𝜕𝐻

𝜕𝑇)

𝑃− [(

𝜕(𝐻−𝑝𝑉)

𝜕𝑇)

𝑉]

= (𝜕𝐻

𝜕𝑇)𝑃 − (

𝜕𝐻

𝜕𝑇)

𝑉+ 𝑉(

𝜕𝑝

𝜕𝑇)𝑉

（1）

令𝐻 = 𝑓(𝑇, 𝑝)

dH = (𝜕𝐻

𝜕𝑇)

𝑝𝑑𝑇 + (

𝜕𝐻

𝜕𝑝)

𝑇𝑑𝑝

(𝜕𝐻

𝜕𝑇)𝑉 = (

𝜕𝐻

𝜕𝑇)𝑝 + (

𝜕𝐻

𝜕𝑝)𝑇(

𝜕𝑝

𝜕𝑇)𝑉

（2）

将（2）式代入（1）式得：

𝐶𝑝 − 𝐶𝑉 = (𝜕𝐻

𝜕𝑇)

𝑝− [(

𝜕𝑈

𝜕𝑉)

𝑝+ (

𝜕𝐻

𝜕𝑝)

𝑇(

𝜕𝑝

𝜕𝑇)

𝑉 ] + 𝑉(

𝜕𝑝

𝜕𝑇)𝑉

= −(𝜕𝑝

𝜕𝑇)𝑉[(

𝜕𝐻

𝜕𝑝)

𝑇− 𝑉]

15. 解（1）设 H2 为理想气体

𝑉2 =𝑛𝑅𝑇1

𝑝=

(1𝑚𝑜𝑙)(8.314𝐽 ∙ 𝑘−1 ∙ 𝑚𝑜𝑙−1)(298.2𝐾)

100000𝑃𝑎= 0.02479𝑚3

γ =𝐶𝑝,𝑚

𝐶𝑉,𝑚=

(5/2)𝑅+𝑅

(5/2)𝑅=

7

5= 1.4

𝑇1𝑉1𝛾−1

= 𝑇2𝑉2𝛾−1

𝑇2 = 𝑇1(𝑉1𝑉2

)𝛾−1 = 298.2𝐾 (0.02249𝑚3

0.005𝑚3)

1.4−1

= 565.8K

（2）𝑝2 =𝑛𝑅𝑇2

𝑉2=

(1𝑚𝑜𝑙)(8.314𝐽∙𝐾−1∙𝑚𝑜𝑙−1)(565.8𝐾)

0.005𝑚3= 940.8kPa

- 11 -

（3）W = 𝐶𝑉(𝑇1 − 𝑇2) =5

2𝑅(𝑇1 − 𝑇2)

=5

2(8.314𝐽 ∙ 𝐾−1 ∙ 𝑚𝑜𝑙−1)(298.2 − 565.8)

= −5562J

16. 解（1）等温可逆过程：∆U = 0, ∆H = 0

n =𝑝1𝑉1𝑅𝑇1

=(20 × 101325𝑃𝑎)(0.020𝑚3)

(8.314𝐽 ∙ 𝐾−1 ∙ 𝑚𝑜𝑙−1)(298𝐾)= 16.36𝑚𝑜𝑙

Q = W = nRTln𝑝1/𝑝2

= (16.36mol)(8.314𝐽 ∙ 𝐾−1 ∙ 𝑚𝑜𝑙−1)(298K)ln20 × 𝑝𝜃

𝑝𝜃

= 1.214 × 105𝐽 = 121.4𝑘𝐽

（2）绝热可逆过程

Q = 0

γ =𝐶𝑝,𝑚𝐶𝑉,𝑚

=(5/2)𝑅

(3/2)𝑅= 1.667

𝑝11−𝛾

𝑇1𝛾

= 𝑝21−𝛾

𝑇2𝛾

𝑇2 = （𝑝1𝑝2

）

1−𝛾𝛾

𝑇1 = (20 × 𝑝𝜃

𝑝𝜃)

1−1.6671.667

(298K) = 89.9K

∆𝑈 = n𝐶𝑉,𝑚(𝑇2 − 𝑇1) = 𝑛3

2𝑅(𝑇2 − 𝑇1)

= (16.36mol)3

2(8.314𝐽 ∙ 𝐾−1 ∙ 𝑚𝑜𝑙−1) × (89.9𝐾 − 298𝐾)

= −4.245 × 104𝐽 = −42.45𝑘𝐽

W = −∆U = 42.45kJ

∆H = n𝐶𝑝,𝑚(𝑇2 − 𝑇1) = 𝑛5

2𝑅(𝑇2 − 𝑇1)

= (16.36mol)5

2(8.314𝐽 ∙ 𝐾−1 ∙ 𝑚𝑜𝑙−1) × (89.9𝐾 − 298𝐾)

= −7.076 × 104𝐽 = −70.76𝑘𝐽

17. 解 (1) ×1

2+ (2) ×

1

2+ (3) − (4) ×

1

2得：

Ag(s) + 12

Cl2(g) = AgCl(s)

1 1 1(AgCl,T)= (1) (2) (3) (4)2 2 2f m r m r m r m r m

H H H Hθ θ θ θΔ Δ + Δ + Δ − Δ H θ

-1 -1 -11 1( 324.9kJ mol ) ( 30.57kJ mol ) ( 92.31kJ mol )2 2

= − ⋅ + − ⋅ + − ⋅ −

-11 ( 285.84kJ mol )2− ⋅

-1127.14kJ mol= − ⋅

18. 解 0.500g 正庚烷燃烧后放出的恒容热效应为：

-1(8.177kJ K )( 2.94 ) 24.04kJVQ = ⋅ − Κ = −

1mol 正庚烷燃烧后放出的等容热效应为：

-1-124.04kJ 4818kJ mol

/ 0.500g/(100.2g mol )V

C mQU

W MΔ = = − = − ⋅

⋅

C7H16(l) + 11O2(g) = 7CO2(g) + 8H2O(l)

正庚烷的燃烧焓为：

7 16,(C H )C m C m BB

H T Uθ νΔ = Δ +∑ RT

-1 3 1 -14818kJ mol (7 11) (8.314 10 kJ mol )(298 )− −= − ⋅ + − × × ⋅Κ ⋅ Κ

3 6

Bθ

6 )}

-14828kJ mol= − ⋅

19. 解（1）（环丙烷） 23C(s) + 3H (g) C H→

该反应的反应热就是环丙烷的生成焓

3 6(C H )= ( ) ( )f m r m B C mB

H H T Hθ θ νΔ Δ = − Δ∑

2 33 {C(s)} 3 {H ( )} {C H (gC m C m C mH H g Hθ θ θ= Δ + Δ −Δ

-1 -1 -13( 393.8kJ mol ) 3( 285.84kJ mol ) ( 2092kJ mol )= − ⋅ + − ⋅ − − ⋅

=53.08 kJ.mol-1

（2） 3 6 3 2C H ( ) CH CH CH→ =环

( ) ( ) ( )r m f m f mH T H T H Tθ θΔ = Δ −Δ丙稀, 环丙烷,

-1 -120.5kJ mol 53.08kJ mol= ⋅ − ⋅

- 12 -

-132.58kJ mol= − ⋅

20. 解（1）C(s) = C(g) -1(1) 711.1kJ molr mHΔ = ⋅

（2）H2(g) = 2H(g) -1(2) 431.7kJ molr mHΔ = ⋅

（3） 2 4C(s) 2H (g) CH (g)+ →-1(3) 74.78kJ molr mHΔ = − ⋅

（3）－（1）－2（2）得：

C(g) + 4H(g) = CH4(g)

(3) (1) 2 (2)r m r m r m r mH H H HΔ = Δ −Δ − Δ

-1( 74.78 711.1 2 431.7)kJ mol= − − − × ⋅

-11649kJ mol= − ⋅

21. 解设计如下过程：

2H O(g)

800Κ2 2

1H (g) O (g)2

+

800 , pθΚ

298

, 2 , 2800

1(1) [ (H ) (O )]2r m p m p m

H C CΚ

ΚΔ = +∫ dT

298 -1 -1 4 -2 -1

800[47.15J K mol (4.14 10 J K mol ) ]T dT

Κ −

Κ= ⋅ ⋅ + × ⋅ ⋅∫

-1 -1 4 -2 -11(47.15J K mol )(298 800K) ( 4.14 10 J K mol )2

−= ⋅ ⋅ Κ − − × × ⋅ ⋅ ×

2 2[(298 ) (800 ) ]Κ − Κ

-123419.5J mol= − ⋅

2 21H (g) O (g)2

+

298 , pθΚ

2H O(l)

298 , pθΚ

2H O(l)

373 , pθΚ

2H O(g)

373 ,

(1) (5)

(2) (3) (4)

pθΚ

- 13 -

Qtonight标注此处为“-”

（题意）-1(2) 285.84kJ molr mHΔ = − ⋅

373

, 2298(3) {H O(l)}r m p mH C

Κ

ΚΔ = ∫ dT

dT

(5)

373 -1 -1

298(75.26J K mol )dT

Κ

Κ= ⋅ ⋅∫

-15644.4J mol= ⋅

(4)r mHΔ = 40650J mol⋅-1

800

, 2373(5) {H O(g)}r m p mH C

Κ

ΚΔ = ∫

800 -1 -1 3 -2 -1

373[30.00J K mol (10.7 10 J K mol ) ]T dT

Κ −

Κ= ⋅ ⋅ + × ⋅ ⋅∫

-115531.7J mol= ⋅

(1) (2) (3) (4)r m r m r m r m r m r mH H H H H HΔ = Δ + Δ + Δ + Δ + Δ

-1247.4kJ mol= − ⋅

- 14 -

1-1212-13第10页和第11页18-21

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m m m - pku.edu.cn...- 4 - = 37.61 kJ mol-1...

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