PHYSICAL CHEMISTRY IN BRIEF

CHAP. 7: PHASE EQUILIBRIA [CONTENTS] 194

Solution

We use equation (7.19) in which

∆fusV = 18× 10 −^6 − 19. 8 × 10 −^6 =− 1. 8 × 10 −^6 m^3 mol−^1.

Given the definition of the normal melting point [see7.1.9],p 1 = 101 325 Pa. We obtain

p 2 = 101325 +

6008

− 1. 8 × 10 −^6

ln

272. 15

273. 15

= 12. 2 × 106 Pa= 12. 2 MPa.

7.5.6 Solid-solid equilibrium

This section deals with the temperature-pressure dependence during transformations of one
crystalline form into another. When the process is a first-order phase transition, we use the
Clapeyron equation (7.13) and its integral forms (7.18), (7.19), where ∆Hmand ∆Vmrepresent
changes in molar enthalpy and volume during transition from one solid phase into another.

7.5.7 Equilibrium between three phases

This equilibrium occurs at triple points [see7.1.11]. The criterion of phase equilibrium applies

G(1)m =G(2)m =G(3)m , (7.20)

whereG(mj)is the molar Gibbs energy in thejthphase. The criterion written using the fugacities
is
f(1)=f(2)=f(3). (7.21)

At the triple point, at which the solid, liquid and gaseous phases are in equilibrium, the following
relation applies between the enthalpy of sublimation, melting and vaporization:

∆subH= ∆fusH+ ∆vapH. (7.22)