2024年3月27日发(作者：沃诗柳)

Thermodynamics in Materials Science

Homework #4

Due Nov. 22, 2004

Chapter 7: Unary heterogeneous systems

1. From the vapor-temperature relationship for liquid silver, calculate the heat of

evaporation of liquid silver at the normal boiling temperature of 2147 K and

the heat capacity difference between liquid and gaseous silver.

33200

lnp(atm)0.85lnT20.31

T

对于这个反应过程有Q

L-G

=ΔH

L-G

; ΔV

L-G

=RT/P; ΔS

L-G

=ΔH

L-G

/T;

dPH

L-G

PH

LG



同时

LG2

dT

TVRT

33200

0.85lnT20.31

且有已知条件

lnp(atm)

T

dP332000.85dPP33200

dTdT;(0.85)

得：

pTdTT

T

2

T

2

P33200H

LG

PH

LG

PH

LG

PQ

LG

(0.85)

TT

TV

LG

RT

2

RT

2

RT

2

把 T=2147K,R=8.314J/(mol.K)带入 解方程：

蒸发热 ：

Q

LG

H

LG

260.1kg/mol;H

LG

33200R0.85RT

汽化热容变化：

C

P

C

P

(T)C

P

GL

dH

LG

(T)7.0669J/(mol.K)

dT

2. Calculate the approximate pressure required to distill mercury at 100 ˚C.

7610

lnp(atm)

LiquidHg

0.795lnT17.214

T

T373.15K

Pe0.795ln373.1517.21437.6Pa

3. Below the triple point (-56.2 ˚C) the vapor pressure of solid CO

2

is give as

3116

lnp(atm)16.01

T

The molar heat of melting of CO

2

is 8330 joules. Calculate the vapor pressure

exerted by liquid CO

2

at 25 ˚C, and illustrate why solid CO

2

, sitting on the

laboratory bench , evaporates rather than melts.

三相点 T

0

=216.95K

3116

16.01

得此时P

0

=5.19atm； 由

lnp(atm)

T

7610

T

Thermodynamics in Materials Science

在（298.15K,1atm）状态下，因为1atm <5.19atm,所以CO

2

此时是气态，

故标准状况下 ，应该是升华为主。

dP3116PPH

SG



对于S-G转变：

22

dT

TRT

S-G

得：ΔH=3116R=25906J

ΔH

L-G

=ΔH

S-G

-ΔH

S-L

=25906-8330=17576J

dPPH

LG



对于L-G转变：

2

dT

RT

得：

P

dPdPPH

LG

dTH

LG

dTPH

LG

11



2









Ln()

P

0

PPRP

0

RT

0

T

TRT

2

T=25℃带入的：P=73.74atm；故平衡气压为73.74atm。

Chapter 8: Solutions

1. Given that the volume change on mixing of a solution obeys the relation

cc

V

mix

2.7

x

1

x

2

()

mol

a. Derive expressions for the partial molal volumes of each of the

components as functions of composition.

b. Demonstrate that your result is correct by using it to

compute

V

mix

; demonstrating the equation above is recovered.

(a)

2组分x

1

,x

2

: x

1

+x

2

=1;dx

1

=-dx

2

设混合体积：

VV(x

1

,x

2

)V

1

x

1

V

2

x

2

VV

dVdx

1

dx

2

x

1

x

2

V

V

令

V

1



；

V

2





dVV

1

dx

1

V

2

dx

2

x

2

x

1

dV



x

1

V

1

x

1

V

2

x

1

dx

2

dV



V

1

x

1

x

1

V

2

x

1

dx

2

dVdV



Vx

1

V

2

x

1

V

2

x

2

V

2

x

1

dx

2

dx

2



V

2



V



x

1

dVdV



V

(1

x

2

)2.7

x

1

2.7

x

1

x

2

dx

2

dx

2

2024年3月27日发(作者：沃诗柳)

Thermodynamics in Materials Science

Homework #4

Due Nov. 22, 2004

Chapter 7: Unary heterogeneous systems

1. From the vapor-temperature relationship for liquid silver, calculate the heat of

evaporation of liquid silver at the normal boiling temperature of 2147 K and

the heat capacity difference between liquid and gaseous silver.

33200

lnp(atm)0.85lnT20.31

T

对于这个反应过程有Q

L-G

=ΔH

L-G

; ΔV

L-G

=RT/P; ΔS

L-G

=ΔH

L-G

/T;

dPH

L-G

PH

LG



同时

LG2

dT

TVRT

33200

0.85lnT20.31

且有已知条件

lnp(atm)

T

dP332000.85dPP33200

dTdT;(0.85)

得：

pTdTT

T

2

T

2

P33200H

LG

PH

LG

PH

LG

PQ

LG

(0.85)

TT

TV

LG

RT

2

RT

2

RT

2

把 T=2147K,R=8.314J/(mol.K)带入 解方程：

蒸发热 ：

Q

LG

H

LG

260.1kg/mol;H

LG

33200R0.85RT

汽化热容变化：

C

P

C

P

(T)C

P

GL

dH

LG

(T)7.0669J/(mol.K)

dT

2. Calculate the approximate pressure required to distill mercury at 100 ˚C.

7610

lnp(atm)

LiquidHg

0.795lnT17.214

T

T373.15K

Pe0.795ln373.1517.21437.6Pa

3. Below the triple point (-56.2 ˚C) the vapor pressure of solid CO

2

is give as

3116

lnp(atm)16.01

T

The molar heat of melting of CO

2

is 8330 joules. Calculate the vapor pressure

exerted by liquid CO

2

at 25 ˚C, and illustrate why solid CO

2

, sitting on the

laboratory bench , evaporates rather than melts.

三相点 T

0

=216.95K

3116

16.01

得此时P

0

=5.19atm； 由

lnp(atm)

T

7610

T

Thermodynamics in Materials Science

在（298.15K,1atm）状态下，因为1atm <5.19atm,所以CO

2

此时是气态，

故标准状况下 ，应该是升华为主。

dP3116PPH

SG



对于S-G转变：

22

dT

TRT

S-G

得：ΔH=3116R=25906J

ΔH

L-G

=ΔH

S-G

-ΔH

S-L

=25906-8330=17576J

dPPH

LG



对于L-G转变：

2

dT

RT

得：

P

dPdPPH

LG

dTH

LG

dTPH

LG

11



2









Ln()

P

0

PPRP

0

RT

0

T

TRT

2

T=25℃带入的：P=73.74atm；故平衡气压为73.74atm。

Chapter 8: Solutions

1. Given that the volume change on mixing of a solution obeys the relation

cc

V

mix

2.7

x

1

x

2

()

mol

a. Derive expressions for the partial molal volumes of each of the

components as functions of composition.

b. Demonstrate that your result is correct by using it to

compute

V

mix

; demonstrating the equation above is recovered.

(a)

2组分x

1

,x

2

: x

1

+x

2

=1;dx

1

=-dx

2

设混合体积：

VV(x

1

,x

2

)V

1

x

1

V

2

x

2

VV

dVdx

1

dx

2

x

1

x

2

V

V

令

V

1



；

V

2





dVV

1

dx

1

V

2

dx

2

x

2

x

1

dV



x

1

V

1

x

1

V

2

x

1

dx

2

dV



V

1

x

1

x

1

V

2

x

1

dx

2

dVdV



Vx

1

V

2

x

1

V

2

x

2

V

2

x

1

dx

2

dx

2



V

2



V



x

1

dVdV



V

(1

x

2

)2.7

x

1

2.7

x

1

x

2

dx

2

dx

2