9.9 An Introduction to Entropy
341
Q H /T H − Q C /T C ≤ 0
for each cycle of any engine. From this, Clausius saw that if one takes heat going
into the engine as positive and the heat coming out of the engine as negative, and
includes in the thermodynamic cycle variations in temperature by using a sequence
of Carnot engines running between the system and the environment, the relation
above becomes:
δQ
T
≤ 0 .
(9.19)
The equality will apply only for a reversible process.
Now suppose we divide the full cycle in the process into two: Start at some state
of the system, called A and proceed to state B. Then go from state B back to A.
In (9.19), we can move the second part of the cyclic integral to the right-hand-side,
to get
B
A
δQ
T
≤ −
A
B
δQ
T
.
(9.20)
Now if one selects a sequence of changes, called a path, from B to A to be a
reversible process, then
B
A
δQ
T
≤
B
A rev
δQ
T
.
(9.21)
The equality holds if the process used for the integral on the left-hand side is also
reversible. But even then, the relation is not trivial, because the thermodynamic
‘path’ that one takes going from A to B may be different.
In performing the integral, the temperature T is that of a series of reservoirs
in thermal contact with the system, and the system must be allowed to reach the
same temperature as the reservoir during each infinitesimal step of the sum. The
reservoirs, by definition, must be capable of supplying and absorbing heat without
significant changes in their own temperature. Even with heat transfer, the process
can be reversible, because the state of any one reservoir will be restored by reversing
the direction of the heat transfer at fixed temperature.
For the reversible process cases, we see that the integral in Eq. (9.21) does not
depend on the path from A to B, just on the endpoints:
S B − S A =
B
A rev
δQ
T
.
(9.22)
The state A can be selected as a standard state, so the quantity S at any other
thermodynamic state of the system can be defined by integrating δQ/T from the
standard state to the given one:
341
Q H /T H − Q C /T C ≤ 0
for each cycle of any engine. From this, Clausius saw that if one takes heat going
into the engine as positive and the heat coming out of the engine as negative, and
includes in the thermodynamic cycle variations in temperature by using a sequence
of Carnot engines running between the system and the environment, the relation
above becomes:
δQ
T
≤ 0 .
(9.19)
The equality will apply only for a reversible process.
Now suppose we divide the full cycle in the process into two: Start at some state
of the system, called A and proceed to state B. Then go from state B back to A.
In (9.19), we can move the second part of the cyclic integral to the right-hand-side,
to get
B
A
δQ
T
≤ −
A
B
δQ
T
.
(9.20)
Now if one selects a sequence of changes, called a path, from B to A to be a
reversible process, then
B
A
δQ
T
≤
B
A rev
δQ
T
.
(9.21)
The equality holds if the process used for the integral on the left-hand side is also
reversible. But even then, the relation is not trivial, because the thermodynamic
‘path’ that one takes going from A to B may be different.
In performing the integral, the temperature T is that of a series of reservoirs
in thermal contact with the system, and the system must be allowed to reach the
same temperature as the reservoir during each infinitesimal step of the sum. The
reservoirs, by definition, must be capable of supplying and absorbing heat without
significant changes in their own temperature. Even with heat transfer, the process
can be reversible, because the state of any one reservoir will be restored by reversing
the direction of the heat transfer at fixed temperature.
For the reversible process cases, we see that the integral in Eq. (9.21) does not
depend on the path from A to B, just on the endpoints:
S B − S A =
B
A rev
δQ
T
.
(9.22)
The state A can be selected as a standard state, so the quantity S at any other
thermodynamic state of the system can be defined by integrating δQ/T from the
standard state to the given one: