Thermal Physics II – Solutions for Problem Sheet 1
1. Consequences of the Third Law of Thermodynamics
• Start:
cp − cV =
∂U
∂V
+p
T
∂V
∂T
p
• Take (∂U/∂V )T from first law: dU = T dS − pdV
∂U
∂V
=T
T
∂S
∂V
−p
T
Note that (∂U/∂V )S = −p ONLY holds if the entropy is being kept
constant, NOT if the volume is kept constant!
•
cp − cV = T
∂S
∂V
T
∂V
∂T
p
• In the limit T → 0, the entropy S is only a function of the temperature.
Hence, we have
lim
T →0
∂S
∂V
=0
T
• This means that
lim (cp − cV ) = T × 0
T →0
or
cp − cV
=0
T →0
T
lim
This means that the difference of the heat capacities at constant volume and
constant pressure decays faster than the temperature if T → 0.
Of course, it follows directly from the relation above that
lim (cp − cV ) = 0 ,
T →0
which is a much weaker restriction. As cp approaches zero in the limit T → 0,
the heat capacity at constant volume, CV , must do the same.
2. Maxwell Relation
• Start from first law: dU = T dS − pdV + µdN
• Then the first derivatives are
∂U
∂S
=T,
V,N
∂U
∂V
= −p ,
S,N
∂U
∂N
=µ
S,V
• The mixed derivatives of U must be equal as the internal energy is a total
differential. Hence, we have
∂
∂N
∂U
∂S
V,N
!
∂
∂S
=
S,V
∂U
∂N
S,V
!
V,N
• Now we insert the first derivatives from above (the first on the left side,
the last on the right side) and obtain as required
∂T
∂N
S,V
=
∂µ
∂S
V,N
.
3. Stability of the Equilibrium
• We start from the fundamental law of thermodynamics:
dU = T dS − pdV
or
dS =
i
1h
dU + pdV .
T
• Now we take into account that entropy, internal energy and volume are
extensive quantities (T and p can be different in both subsystems):
dS = dS1 + dS2 =
i
i
1h
1h
dU1 + p1 dV1 +
dU2 + p2 dV2 .
T1
T2
• Total internal energy and total volume are conserve (we only allow energy
transfer between the two subsystems and replacement of one gas with the
other). This means dU1 = −dU2 and dV1 = −dV2 . We then have
dS = dS1 + dS2 =
h 1
T1
−
hp
1i
p2 i
2
−
dU1 +
dV1
T2
T1 T2
• In equilibrium, the entropy reaches its maximum and we find
∂S
∂U1
=
V1
h1
T1
−
1i
=0
T2
and
∂S
∂V1
U1
=
hp
2
T1
−
p2 i
=0
T2
These conditions can only be fulfilled if T1 = T2 and p1 = p2 as required.