CHAP. 3: FUNDAMENTALS OF THERMODYNAMICS [CONTENTS] 106
Solution
For an irreversible phase transition, we have the inequalities (3.90) and (3.99) from which the
changes ofS andGcannot be calculated. Hence we substitute the irreversible process with a
sequence of three reversible processes: the heating of the liquid fromT toTfus, the equilibrium
(i.e. reversible) change of the state of matter, and the cooling of the solid from the temperature
of solidification to temperatureT. We thus obtain:
∆H =
∫Tfus
T
Cp(l)dT−∆fusH+
∫T
Tfus
Cp(s)dT ,
∆S =
∫Tfus
T
C(l)p
T
dT−
∆fusH
Tfus
+
∫T
Tfus
Cp(s)
T
dT ,
∆G = ∆H−T∆S
= −∆fusH
[
1 −
T
Tfus
]
+
∫T
Tfus
(Cp(s)−Cp(l)) dT
−T
∫T
Tfus
Cp(s)−C(l)p
T
dT.
