Physical Chemistry Third Edition

210 5 Phase Equilibrium

For some solid–liquid or solid–solid phase transitions, it might be a better approx-

imation to assume that the quotient∆Hm/∆Vmis approximately constant. If so, one

can integrate

dP

1

∆Vm

∆Hm

T

dT

to obtain

P 2 −P 1 ≈

∆Hm

∆Vm

ln

(

T 2

T 1

)

(∆Hmconstant) (5.3-11a)

or

PP(T)P 1 +

∆Hm

∆Vm

ln

(

T

T 1

)

(∆Hmconstant) (5.3-11b)

The relation of Eq. (5.3-11) is probably a better approximation than that of Eq. (5.3-10)

for large temperature differences. An even better approximation can be obtained by

assuming that the heat capacities of the two phases are nearly constant.

Exercise 5.4

Estimate the pressure of the system of Example 5.3, using Eq. (5.3-11) instead of Eq. (5.3-10).

Compare the answer with that of Example 5.3 to see whether the assumption of constant∆H

gives different results from the assumption of constant∆S.

EXAMPLE 5.4

Integrate the Clapeyron equation for a solid–solid or liquid–solid phase transition under the

assumption that∆Vmis constant and that∆Hm(T)∆Hm(T 1 )+∆CP,m(T−T 1 ) where

∆CP, mis constant.

Solution

P 2 −P 1 

1

∆Vm

∫T 2

T 1

∆Hm(T 1 )+∆CP,m(T−T 1 )

T

dT



∆Hm(T 1 )

∆Vm

ln

(

T 2

T 1

)

+

1

∆Vm

∆CP,m(T 2 −T 1 )−

1

∆Vm

∆CP,mT 1 ln

(

T 2

T 1

)

Exercise 5.5

Estimate the pressure of the system of Example 5.3, assuming that∆CP, mis constant.

The Clausius–Clapeyron Equation

The Clausius–Clapeyron equation is obtained by integrating the Clapeyron equation in

the case that one of the two phases is a vapor (gas) and the other is acondensed phase

(liquid or solid). We make two approximations: (1) that the vapor is an ideal gas, and

(2) that the molar volume of the condensed phase is negligible compared with that of

the vapor (gas) phase. These are both good approximations.