Physical Chemistry , 1st ed.

Solution

First, we calculate the change in volume. For 1.00 mole of water that has a mass

of 18.01 g, the volume of the liquid is 18.01/0.958 18.8 mL. For 1.00 mole

of steam, the volume is 18.01/0.5983 30.10 L.Vis 30.10 L 18.8 mL 

30.08 L per mole of water. (Notice the units on the volumes.) Using equation

6.12, we find

p

4

3

0

0

,

7

0

0

8

0

L

J

ln

3

3

7

7

0

3

K

K

Notice that we have converted Hinto units of J. The temperature units can-

cel; we get

p1353 J/L (0.00808)

p10.9 J/L

At this point, we invoke our conversion factor between J and Latm:

p10.9

L

J

1

1

0

L

1



a

3

t

2

m

J

The J and L units cancel, leaving units of atm, which are units of pressure:

p0.108 atm

This is the change in pressure from the original pressure of 1.000 atm; the actual

pressure at which the boiling point is 97°C is therefore 1.000 0.108 atm 

0.892 atm. This would be the pressure about 1000 meters above sea level, or

about 3300 feet. Since many people live at that altitude or higher around the

world, substantial populations experience water with a boiling point of 97°C.

6.5 The Clausius-Clapeyron Equation

If a gas is involved in the phase transition, we can make a simple approxima-

tion. The volume of the gaseous phase is so much larger than the volume of

the condensed phase (as Example 6.6 showed) that we introduce only a tiny bit

of error if we simply neglect the volume of the condensed phase. We simply

use Vgasin equation 6.11 and get

d

d

T

p



T



V

H

gas

If we also assume that the gas obeys the ideal gas law, we can substitute RT/p

for the molar volume of the gas:

d

d

T

p





T

H

R



T

p



H

R



T



2

p

Rearranging, we get

d

p

p





R

H



d

T

T

(^2)
Recognizing that dp/pis equal to d(ln p), we have
d(ln p) 



R

H



d

T

T

2 (6.13)

152 CHAPTER 6 Equilibria in Single-Component Systems