Page 113 - Physical Chemistry

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Chapter 3 to state 3. From Eq. (3.23), S is zero for a reversible adiabatic process. Hence S
3
The Second Law of Thermodynamics
S . (As always, state functions refer to the system unless otherwise specified. Thus S
2 3
and S are the system’s entropies in states 3 and 2.) We next either add or withdraw
2
enough heat q isothermally and reversibly at temperature T to make the entropy
3→4 hr
of the system equal to S . This brings the system to state 4 with S S . (q is pos-
1 4 1 3→4
itive if heat flows into the system from the reservoir during the process 3 → 4 and neg-
ative if heat flows out of the system into the reservoir during 3 → 4.) We have
3
4
S S 4 dq rev 1 dq rev q 3S4
4
3 T T hr 3 T hr
Since states 4 and 1 have the same entropy, they lie on a line of constant S, an isen-
trop. What is an isentrop? For an isentrop, dS 0 dq /T, so dq 0; an isentrop
rev rev
is a reversible adiabat. Hence to go from 4 to 1, we carry out a reversible adiabatic
process (with the system doing work on the surroundings). Since S is a state function,
we have for the cycle 1 → 2 → 3 → 4 → 1

0 dS syst 1S S 2 1S S 2 1S S 2 1S S 2
3
2
2
1
4
1
4
3

dS syst 1S S 2 0 q 3S4 >T 0 0
hr
1
2
S S q 3S4 >T hr
2
1
The sign of S S is thus the same as the sign of q . We have for the cycle
2 1 3→4
dU 0 1dq dw2 q 3S4 w
The work done on the system in the cycle is thus w q . The work done by the
3→4
system on the surroundings is w q . Suppose q were positive. Then the work
3→4 3→4
w done on the surroundings would be positive, and we would have a cycle (1 → 2
→ 3 → 4 → 1) whose sole effect is extraction of heat q from a reservoir and its
3→4
complete conversion to work w q 0. Such a cycle is impossible, since it
3→4
violates the second law. Hence q cannot be positive: q 0. Therefore
3→4 3→4
S S q 3S4 >T 0 (3.36)
1
hr
2
We now strengthen this result by showing that S S 0 can be ruled out. To do
2 1
this, consider the nature of reversible and irreversible processes. In a reversible process,
we can make things go the other way by an infinitesimal change in circumstances.
When the process is reversed, both system and surroundings are restored to their orig-
inal states; that is, the universe is restored to its original state. In an irreversible process,
the universe cannot be restored to its original state. Now suppose that S S 0. Then
2 1
q , which equals T (S S ), would be zero. Also, w, which equals q , would
3→4 hr 2 1 3→4
be zero. (Points 3 and 4 would coincide.) After the irreversible process 1 → 2, the path
2 → 3 → 4 → 1 restores the system to state 1. Moreover, since q 0 w for the cycle
1 → 2 → 3 → 4 → 1, this cycle would have no net effect on the surroundings, and at
the end of the cycle, the surroundings would be restored to their original state. Thus we
would be able to restore the universe (system surroundings) to its original state. But
by hypothesis, the process 1 → 2 is irreversible, and so the universe cannot be restored
to its original state after this process has occurred. Therefore S S cannot be zero.
2 1
Equation (3.36) now tells us that S S must be positive.
2 1