Physical Chemistry Third Edition

5.3 Phase Equilibria in One-Component Systems 211

Exercise 5.6

Compare the molar volumes of liquid water at 100◦C and water vapor at 100◦C and 1.00 atm.

Assume that the water vapor is an ideal gas.

For a liquid–vapor transition with our approximations

∆VmV

(gas)

m −V

(liq)

m ≈V

(gas)

m ≈

RT

P

(5.3-12)

The same approximation holds for a solid–vapor transition. From Eqs. (5.3-8) and

(5.3-12) we obtain the derivative form of theClausius–Clapeyron equation. For a

liquid–vapor transition

dP

dT



P∆vapHm

RT^2

(5.3-13)

where∆vapHmis the molar enthalpy change of vaporization. For sublimation (a solid–

vapor transition),∆vapHmis replaced by∆subHm, the molar enthalpy change of sub-

limation. We omit the subscript and apply the equation to either case. To obtain a

representation ofPas a function ofT, we need to integrate Eq. (5.3-13). We multiply

bydTand divide byP:

1

P

dP

dT

dT

1

P

dP

∆Hm

RT^2

dT (5.3-14)

Carrying out a definite integration with the assumption that∆Hmis constant gives the

integral form of the Clausius–Clapeyron equation:

ln

(

P 2

P 1

)

−

∆Hm

R

(

1

T 2

−

1

T 1

)

(Clausius–Clapeyron equation) (5.3-15)

Exercise 5.7

Carry out the steps to obtain Eq. (5.3-15).

EXAMPLE 5.5

Using the vapor pressure values for water at 25◦C and 100◦C, find the enthalpy change of

vaporization of water.

Solution

∆vapHm

−RT 2 T 1

T 1 −T 2

ln

(

P 2

P 1

)



(8.3145 J K−^1 mol−^1 )(298.15 K)(373.15 K)

75 .00 K

ln

(

760 .0 torr

23 .756 torr

)

 4. 274 × 104 J mol−^1  42 .74 kJ mol−^1

This value is an average value for the temperature range. As we expect, it is intermediate

between the experimental values: 44.0kJmol−^1 at 25◦C and 40.7 kJ mol−^1 at 100◦C.