Physical Chemistry , 1st ed.

as the pressure needed to promote the conversion from graphite to diamond.

This is over 65,000 times atmospheric pressure. In reality, much higher pres-

sures, on the order of 100,000 bar, are used to produce synthetic diamonds

at these temperatures.

The Clapeyron equation also works for liquid-gas and solid-gas phase tran-
sitions, but as we will see shortly, some approximations can be made that allow
us to use other equations with minimal error.
Recall that for phase equilibria,G0, so that
0 transHTtransS

This rearranges to

transS

tr

T

ans H

If we assume molar amounts, we can substitute for Sin equation 6.9. The
Clapeyron equation becomes

d

d

T

p



T





H

V

(6.11)

where again we have dropped the “trans”label from H. Equation 6.11 is par-
ticularly useful because we can bring dTover to the other side of the equation
where temperature is a variable:

dp

T





H

V

dT

Rearranging, we get

dp





H

V





d

T

T

We can now take the definite integral of both sides, one with respect to pres-
sure and one with respect to temperature. Assuming Hand Vare inde-
pendent of temperature, we get



pi

pf

dp





H

V







Ti

Tf

d

T

T

The integral on the pressure side is the change in pressure,p. The integral on
the temperature side is the natural logarithm of the temperature, evaluated at
the temperature limits. We get

p





H

V





ln

T

T

f

i

(6.12)

This expression relates changes in phase-change conditions, but in terms of the
molar quantities transHand transV.

Example 6.6

What pressure is necessary to change the boiling point of water from its

1.000-atm value of 100°C (373 K) to 97°C (370 K)? The heat of vaporization

of water is 40.7 kJ/mol. The density of liquid water at 100°C is 0.958 g/mL

and the density of steam is 0.5983 g/L. You will have to use the relationship

101.32 J 1 Latm.

6.4 The Clapeyron Equation 151