4Classical thermodynamics
II Statistical Physics
4.6 Third law of thermodynamics
We end by a quick mention of the third law. The first three laws (starting from
0!) are “fundamental” in some sense, but the third one is rather different.
Law (Third law of thermodynamics). As T → 0, we have
lim
T →0
S = S
0
,
which is independent of other parameters (e.g.
V, B
etc). In particular, the limit
is finite.
Recall that classically,
S
is only defined up to a constant. So we can set
S
0
= 0, and fix this constant.
What does this imply for the heat capacities? We have
C
V
= T
∂S
∂T
V
.
Integrating this equation up, we find that
S(T
2
, V ) −S(T
1
, V ) =
Z
T
2
T
1
C
V
(T, V )
T
dT.
Let
T
1
→
0. Then the left hand side is finite. But on the right, as
T →
0, the
1
T
term diverges. For the integral to be finite, we must have
C
V
→
0 as
T →
0.
Similarly, doing this with C
p
, we find that
S(T
2
, p) − S(T
1
, p) =
Z
T
2
T
1
C
p
(T, p)
T
dT.
Saying the same words, we deduce that the third law says
C
p
(
T, V
)
→
0 as
T → 0.
But for an ideal gas, we know that
C
p
− C
V
must be constant! So it cannot,
in particular, tend to 0. So if we believe in the third law, then the approximation
of an ideal gas must break down at low temperatures.
If we go back and check, we find that quantum gases do satisfy the third
law. The statistical mechanical interpretation of this is just that as
T →
0, the
number of states goes to 1, as we are forced to the ground state, and so
S
= 0.
So this says the system has a unique ground state.