Concise Physical Chemistry

c03 JWBS043-Rogers September 13, 2010 11:24 Printer Name: Yet to Come

38 THE THERMODYNAMICS OF SIMPLE SYSTEMS

First

Path

Second

Path

T

p

1 10

290

310

A

B

FIGURE 3.1 Different path transformations from A to B.

The work done on the system in the first compression (lower horizontal) is

w=−

∫V 2

V 1

pdV=−

∫V 2

V 1

RT

V

dV=−

∫V 2

V 1

RT

dV

V

=−RTln

V 2

V 1

=−RTln 0. 100 =− 8 .314(290) ln 0. 100 =5552 J

The work done in the heating step (rightmost vertical) is isobaric withp=10.0 bar:

w=p(V 2 −V 1 )= 10 .0 (266. 0 − 244 .1)=219 J

so the total over the first path is 5771 J.

Over the second path, the leftmost vertical givesw= 1 .00(2660−2441)=219 J;

thus the two isobaric steps require the same work into the system. The isothermal

compressions are not the same. The topmost horizontal is

−RTln 0. 100 =− 8 .314(320) ln 0. 100 =6126 J

so the total work done over the second path is obviously not the same as by the first

path. The difference is 574 J.

3.2.1 Hey, Let’s Make a Perpetual Motion Machine!

Let’s run the thermodynamic system in Fig. 3.1 around a cycle such that the transfor-

mation over the second path runs backwards; that is, the top horizontal is an expansion

instead of a compression and the leftmost step is a temperature decrease. The first

path remains as before.