PHYSICAL CHEMISTRY IN BRIEF

CHAP. 4: APPLICATION OF THERMODYNAMICS [CONTENTS] 113

Solution

It follows from equation (4.24) that

T 2 =T 1

(

P 1

p 2

)(1−κ)/κ

We calculateκfrom the Mayer relation (3.67) and from relation (4.25)

κ=

Cpm

Cpm−R

=

5

3

and after substituting into the relation forT 2 we calculate

T 2 = 300

(

106

105

)(1− 5 /3)/(5/3)

= 119.3 K.

Example

Prove that a reversible adiabatic process is identical with an isentropic process.

Solution

During an adiabatic process, ̄dQ= 0, and relation (3.6) holds for reversible processes. HencedS

= 0, and consequently entropy does not change during this process, (∆S= 0). The process is

thus isentropic.

4.2.2 Irreversible adiabatic process.

A typical example is an irreversible adiabatic expansion against a constant external pressure
pexcompleted at pressure equalization (p=pex). For the volume work, relation (4.8) applies

Wvol=−pex(V 2 −V 1 ). (4.26)

If the heat capacityCV is constant, we also have [see (4.6)]

Wvol=CV(T 2 −T 1 ). (4.27)