Physical Chemistry Third Edition

5.3 Phase Equilibria in One-Component Systems 209

Exercise 5.3

For most substances a solid–liquid coexistence curve has a positive slope, but in the water phase

diagram the ice I–liquid curve has a negative slope. Explain this phenomenon. In the^3 He phase

diagram a horizontal region occurs in the solid–normal liquid curve and a region with a negative

slope also occurs. Interpret these two phenomena given that the explanation of the negative slope

is not the same as with the liquid water–ice I curve.

In order to have a representation of the functionPP(T) corresponding to the

equilibrium of two phases, we integrate the Clapeyron equation written in the form

dP

∆Sm

∆Vm

dT (5.3-9)

For a solid–liquid or a solid–solid phase transition, we assume that the phases have

nearly constant volume so that∆Vmis nearly constant. For a sufficiently small range

of temperature we can assume that∆Smis nearly constant, so that

P 2 −P 1 ≈

∆Sm

∆Vm

(T 2 −T 1 ) (5.3-10a)

whereP 1 is the pressure corresponding to temperatureT 1 andP 2 is the pressure corre-

sponding to temperatureT 2. We can also write this equation in the form

PP(T)P 1 +

∆Sm

∆Vm

(T−T 1 )P 1 +

∆Hm

T 1 ∆Vm

(T−T 1 ) (5.3-10b)

Equation (5.3-10) represents a linear function. In the water phase diagram several of

the solid–solid equilibrium curves are nearly linear, and this equation should be an

adequate approximation for these curves.

EXAMPLE 5.3

Estimate the pressure on a system of liquid and solid water if the equilibrium melting temper-

ature is equal to− 0. 100 ◦C. The density of ice is 0.917 g cm−^3 , the density of liquid water is

1.000 g cm−^3 , and the molar enthalpy change of fusion is 6008 J mol−^1.

Solution

∆Vm(18.01 g mol−^1 )

[

1

1 .00 g cm−^3

−

1

0 .917 g cm−^3

]

1m^3

106 cm^3

− 1. 63 × 10 −^6 m^3 mol−^1

P 2 −P 1 ≈

6008 J mol−^1

(273.15 K)

(

− 1. 63 × 10 −^6 m^3 mol−^1

)− 0 .100 K

≈ 1. 35 × 106 Jm−^3  1. 35 × 106 Nm−^2  1. 35 × 106 Pa 13 .3 atm

P 2 ≈ 14 .3 atm