Physical Chemistry , 1st ed.

The temperature units cancel to give us, after rearranging,

p



(

1.

1

6

0

3

)

(22

1

.

0

0



4

3

J)

L

We have to use the given conversion factor to get a recognizable unit of

pressure:

p



(

1.

1

6

0

3

)

(22

1

.

0

0



4

3

J)

L

1

1

L

0



0

ba

J

r

The units of J and L cancel, leaving units of bar, which is the standard unit

of pressure. Solving:

p1.35 

103 bar

Since 1 bar equals 0.987 atm, it takes about 1330 atm to lower the melting

point of water to  10 °C. This is an estimate, since Vand Swould be

slightly different at  10 °than at 0°C (the normal melting point of ice) or at

25 °C (the common thermodynamic temperature). However, it is a very good

estimate, since both Vand Sdo not vary much over such a small tem-

perature range.

The Clapeyron equation can be applied to substances under extreme con-

ditions of temperature and pressure, since it can estimate the conditions of

phase transitions—and therefore the stable phase of a compound—at other

than standard conditions. Such conditions might exist, say, at the center of a

gas giant planet like Saturn or Jupiter. Or, extreme conditions might be applied

in various industrial or synthetic processes. Consider the synthesis of dia-

monds, which normally occurs deep within the earth (or so it is thought). The

phase transition from the stable phase of carbon, graphite, to the “unstable”

phase, diamond, is a viable target for the Clapeyron equation, even though the

two phases are solids.

Example 6.5

Estimate the pressure necessary to make diamond from graphite at a tem-

perature of 2298 K, that is, with T(2298 298) K 2000 K. (This

conversion was first achieved industrially by General Electric in 1955.) Use

the following information:

C (s, graphite) C (s, diamond)

S(J/K) 5.69 2.43

V(L) 4.41 

10 ^3 3.41 

10 ^3

Solution

Using the Clapeyron equation, we find that

20



00

p

K



1

1

L

0



0

ba

J

r

where we have included the conversion factor from J to Lbar. Solving for p,

we get

p65,200 bar

(2.43 5.69)^ KJ^

(3.41

10 ^3 4.41
10 ^3 ) L

QP

150 CHAPTER 6 Equilibria in Single-Component Systems