PHYSICAL CHEMISTRY IN BRIEF

CHAP. 3: FUNDAMENTALS OF THERMODYNAMICS [CONTENTS] 105

on the path from the initial to the final state. In calculations, the irreversible path from the
initial to the final state is therefore substituted by an appropriately chosen reversible path. We
present two examples of this procedure.

Example

Determine the entropy change during an irreversible adiabatic transition of a system from the

state (T 1 ,p 1 ) to the state (T 2 ,p 2 ).

Solution

If the adiabatic process were reversible, we would have∆S = 0 because equation (3.6) would

apply and ̄dQwould be equal to 0. We substitute the irreversible adiabatic process by a reversible

isobaric and a reversible isothermal process. Then

∆S=

∫T 2

T 1

Cp(T, p 1 )

T

dT−

∫p 2

p 1

(

∂V

∂T

)

p,T=T 2

dp.

Note:Note that when calculating ∆Sin the above example we did not use the information

that the process was adiabatic. To determine changes of thermodynamic quantities, it is

enough to know the initial and final state of the system (T 1 ,p 1 ) and (T 2 ,p 2 ), respectively.

Example

Calculate the changes ∆H, ∆S, and ∆G during the solidification of a supercooled one-

component liquid atT < Tfus andp =pfus, whereTfus is the temperature of solidification

andpfusis the pressure of solidification. (Note that the temperature of melting and solidifica-

tion are the same for pure substances. However, enthalpy of solidification is minus enthalpy of

melting).