60.

(a) The ideal gas is diatomic, so f = 5 (see Table 20-3). Since this is an isobaric (constant pressure)

process, with no change in the number ofmoles, then the ideal gas in ratio f

orm (see Sample

Problem 20-1) leads to

V

f

V

i

=

T

f

T

i

=

8

3

.

With C

V

=

f

2

R, Eq. 21-4 gives

∆S

gas

= nR ln

8

3



+ n

5

2

R



ln

8

3



where n is the number ofmoles (25 mol), not to be confused with the number ofreservoirs (also
denoted “n” in the later parts ofthis problem). Consequently, we obtain

∆S

gas

=

7

2

(25 mol)

8.31

J

mol

·K



ln

8

3



= 713 J/K .

Since Q = nC

p

∆T for this process, the entropy change of the reservoir (which transfers energy Q

to the gas, so it (the heat) is negative-valued in this context) is (using Eq. 21-2)

∆S

res

=

−Q

T

=

−

n



7
2

R



(800 K

− 300 K)

800 K

=

−454 J/K .

Therefore, ∆S

system

= ∆S

gas

+ ∆S

res

= 259 J/K.

(b) The change in entropy ofthe gas is the same regardless ofthe number ofintermediate reservoirs,

so long as the beginning state and final state ofthe gas is unchanged. The difference (relative to
part (a)) is that the sum ofthese two reservoirs’ entropy changes is not equivalent to that ofthe
one reservoir in the previous part:

∆S

res1

+ ∆S

res2

=

−Q

1

T

1

+

−Q

2

T

2

=

−

(25 mol)



7
2

R



(550 K

− 300 K)

550 K

−

(25 mol)



7
2

R



(800 K

− 550 K)

800 K

=

−(25 mol)

7

2

R



(250 K)

1

550 K

+

1

800 K



which yields

−558 J/K for the total loss of entropy from the reservoirs. The entire system change

in entropy is therefore 713

− 558 = 155 J/K.

(c) Towards the end ofthe calculation in part (b), a pattern emerges in the computation ofthe total

entropy loss from the original high-temperature reservoir plus the n intermediate reservoirs:

∆S

res total

=

−(25 mol)

7

2

R



500 K

n + 1

 

n+1



reservoirs

1

T



where the temperature ofa particular reservoir (the j

th

reservoir, where 1

≤ j ≤ n + 1) is T =

300+



500

n+1

(in Kelvins). For n = 10, this leads to ∆S

res total

=

−680 J/K and therefore 713−680 =

33 J/K for the entire system (including the gas) entropy change.

(d) For n = 50, this leads to ∆S

res total

=

−705.82 J/K and therefore 713.19 − 705.82 = 7.37 J/K for

the entire system (including the gas) entropy change.

(e) For n = 100, this leads to ∆S

res total

=

−709.45 J/K and therefore 713.19 − 709.45 = 3.74 J/K for

the entire system (including the gas) entropy change.