Physical Chemistry Third Edition

128 3 The Second and Third Laws of Thermodynamics: Entropy

∆S 2 

∆H 2

T



− 1. 202 × 104 J

273 .15 K

− 44 .00JK−^1

∆S 3 

∫ 258 .15 K

273 .15 K

CP(s)

T

dTCP(s) ln

(

258 .15 K

273 .15 K

)

(2.000 mol)(37.15JK−^1 mol−^1 )ln(0.9451)− 4 .20JK−^1

∆S∆S 1 +∆S 2 +∆S 3 − 39 .68JK−^1

From Example 2.28,∆H−10870 J.

∆Ssurr

qsurr

Tsun

−

q

Tsurr

−

∆H

Tsurr



10870 J

258 .15 K

 42 .09JK−^1

∆Suniv∆S+∆Ssurr− 39 .69JK−^1 + 42 .09JK−^1  2 .40JK−^1

Nonequilibrium Steady States

In some cases a system undergoes an irreversible process in such a way that the nonequi-

librium state of the system does not change during the process. We say that the system

is in asteady state.We cannot calculate the entropy of the system since it is not in

an equilibrium state, but we can conclude that its entropy does not change during the

process. Any entropy changes occur in the surroundings.

The system depicted in Figure 3.9a provides an example of such a situation. The

surroundings consist of one large object at temperatureT 1 and another large object at

a higher temperatureT 2. These objects are insulated from the rest of the universe and

from each other except for a thin bar (our system) connecting the objects. If the objects

are very large compared with the system and if they have large thermal conductivities,

they will have nearly uniform and constant temperatures, and we can treat them as

though they operate reversibly. After an induction period the system will come to a

steady state with a temperature that depends on position but not on time, as depicted

in Figure 3.9b. Consider a period of time∆tduring which a quantity of heatqpasses

through the bar. Since the bar is in a steady state the amount of heat entering one end

of the bar is equal to the amount of heat leaving the other end.

The entropy change of object 1 is

∆Ssurr,1

∫

dq 1

T 1



q 1

T 1



q

T 1

(3.3-9)

The entropy change of object 2 is

∆S 2 

q 2

T 2

−

q

T 2

(3.3-10)

The entropy change of the surroundings is

∆Sq 1

(

1

T 1

−

1

T 2

)

(3.3-11)