Physical Chemistry Third Edition

5.4 The Gibbs Energy and Phase Transitions 215

5.23At− 78. 5 ◦C, the vapor pressure of solid carbon dioxide is
equal to 760 torr. The triple point is at 216.55 K and
5 .112 atm. Find the average enthalpy change of sublimation.

5.24 The normal boiling point of oxygen is 90.18 K. The vapor
pressure at 100.0 K is equal to 2.509 atm. Find the enthalpy
change of vaporization.

5.25 The following data give the vapor pressure of liquid
aluminum as a function of temperature:

T/K 1557 1760 1908 2022 2220 2329

P/torr 1 10 40 100 400 760

Using a linear least-squares procedure, find the enthalpy

change of vaporization of aluminum.

5.26 Find the pressure necessary to lower the freezing
temperature of water to− 10. 00 ◦C.
5.27 a.Derive a modified version of the Clausius–Clapeyron
equation using the relation

∆Hm(T)∆Hm(T 1 )+∆CP, m[T−T 1 ]

where∆CP, mis assumed to be constant.

b.Using data from Table A.8 and the vapor pressure of

water at 25. 0 ◦C, equal to 23.756 torr, calculate the

vapor pressure of water at 50. 0 ◦C using the modified

Clausius–Clapeyron equation of part a and using the

unmodified form of Eq. (5.3-15). Compare both results

with the experimental value, 92.5 torr.

5.28 Find the pressure on the liquid phase that can raise the vapor

pressure of water by 1.00% at 25◦C.

5.29 Find the vapor pressure of water at 60. 00 ◦C, assuming that

the heat capacities of liquid and gaseous water are constant

between 25◦C and 60◦C and given that at 25◦C the vapor

pressure is 23.756 torr. Compare your answer with the

experimental value of 149.38 torr.

5.30 The average enthalpy change of vaporization of benzene

between room temperature and the normal boiling

temperature is 34.085 kJ mol−^1. The vapor pressure at

26. 1 ◦Cis100.0 torr.

a. Find the normal boiling temperature. Compare with the

correct value, 80. 1 ◦C. State any assumptions.

b. Find the entropy change of vaporization at the normal

boiling temperature. Compare with the prediction of

Trouton’s rule. State any assumptions.

5.4 The Gibbs Energy and Phase Transitions

We ask the following question: Why is water a liquid with a molar volume of 18 mL

mol−^1 at 1.000 atm and 373.14 K, but a vapor with a molar volume of 30 L mol−^1

at 1.000 atm and 373.16 K? Why should such a small change in temperature make

such a large change in structure when we generally expect a small cause to produce a

small effect? The thermodynamic answer to this question comes from the fact that at

equilibrium at constantTandP, the Gibbs energy of the system must be at a minimum.

Figure 5.5 shows schematically the molar Gibbs energy (chemical potential) of

liquid and gaseous water as a function of temperature at 1.000 atm pressure. In order

to construct Figure 5.5, we have used Eq. (4.2-20):

(

∂Gm

∂T

)

P

−Sm (5.4-1)

Liquid

Liquid

Vapor

Vapor

373.15 K

G

m

T

Figure 5.5 The Molar Gibbs Energy

of Water as a Function of Temper-

ature Near the Liquid–Vapor Phase

Transition (Schematic).

The molar entropy of the water vapor is greater than the molar entropy of the liquid

water, so that the tangent to the vapor curve in Figure 5.5 has a more negative slope

than that of the liquid curve. The temperature at which the curves intersect is the

temperature of phase coexistence at this pressure, since this is the temperature at which

the values of the molar Gibbs energy in the two phases are equal. We have continued

the curves past the coexistence point with broken curves to represent metastable states.

If one phase has a more negative value of the molar Gibbs energy than the other phase,