Physical Chemistry , 1st ed.

and using equation 6.8, we get

Sphase1dTVphase1dpSphase2dTVphase2dp

Because the temperature change dTand pressure change dpare experienced by

both phases simultaneously, there is no need to put labels on them. However,

each phase will have its own characteristic molar entropy and molar volume,

so each Sand Vmust have a label to distinguish it. We can rearrange to col-

lect the dpterms and the dTterms on opposite sides:

(Vphase2Vphase1) dp(Sphase2Sphase1) dT

We write the differences inside the parentheses as Vand S, since they rep-

resent the changes in molar volume and entropy from phase 1 to phase 2.

Substituting,

VdpSdT

which is rearranged to get the following equation:

d

d

T

p







V

S



 (6.9)

This is called the Clapeyron equation,after Benoit P. E. Clapeyron, a French

engineer who worked out this relationship in 1834. (See Figure 6.2.) The

Clapeyron equation relates pressure and temperature changes for all phase

equilibria in terms of the changes in molar volumes and entropies of the

phases involved. It is applicable to any phase equilibrium. It is sometimes es-

timated as





T

p







V

S



 (6.10)

One very useful application of the Clapeyron equation is to estimate the pres-

sures necessary to shift phase equilibria to other temperatures. The following

example illustrates this.

Example 6.4

Estimate the pressure necessary to melt water at  10 °C if the molar volume

of liquid water is 18.01 mL and the molar volume of ice is 19.64 mL.Sfor

the process is 22.04 J/K and you can assume that these values remain rel-

atively constant with temperature. You will need this conversion factor:

1 Lbar 100 J.

Solution

The change in molar volume for the reaction

H 2 O (s) H 2 O ()

is 18.01 mL 19.64 mL 1.63 mL. In units of liters, this is 1.63

10 ^3 L.Tfor this process is  10 °C, which is also 10 K. (Recall that

changesin temperature have the same magnitude in kelvins as they have

in degrees Celsius.)Sis given, so we use the Clapeyron equation and get





10

p

K



22.04^ KJ^

1.63

10 ^3 L

JQPJ

6.4 The Clapeyron Equation 149

Figure 6.2 Benoit P. E. Clapeyron (1799–1864),
French thermodynamicist. Using principles laid
down by Carnot, Clapeyron deduced concepts of
entropy that eventually led to the second law of
thermodynamics.