Statistical Mechanics - Physics at Oregon State University

2.4. HELMHOLTZ FREE ENERGY. 31
Suppose that in the previous model we replace the energy relation by ɛs(V ) =
ɛs(V0) α V0
V . Then we would find pV0 = αU0. This is positive when α is
positive. For an ideal gas we have this form of relation between pressure and
. This value of α is indeed consistent with the form of the
energy, with α = 2
3
energy eigenvalues for free particles. The energy relation is of the form ɛ = 2k 2
and k is proportional to inverse length.
More about pressure.
In summary, in the previous discussion we used the traditional definition of
temperature to evaluate the work done on the system in a volume change. We
linked this work to the change in eigenvalues in a particular process, in which
the entropy remaind the same, by using conservation of energy. In this way
the pressure could be directly related to the thermodynamic functions we have
available.
In order to find the pressure from the energy, we need U(S,V). This form is
usually not available. Counting states we derived S(U,V) and hence we would
like to relate the pressure to this function. A standard trick will do. A change
in the entropy due to changes in the energy and volume is given by
dS =
∂S
∂U
V
dU +
∂S
∂V
U
2m
dV (2.31)
It is possible to change both the energy and the volume at the same time in
such a way that the entropy does not change:
∂S
∂S
0 = (∆U)S + (∆V )S
(2.32)
∂U
∂V
V
After dividing by ∆V and taking the limit ∆V → 0 we get
∂S ∂U ∂S
0 =
+
(2.33)
∂U V ∂V S ∂V U
or
∂S
p = T
(2.34)
∂V U
which is a more useful expression. In the next section we will also give the
expression for the pressure if the temperature and the volume are specified. Of
course, these tricks have been discussed extensively in thermodynamics, where
changes of variables are the bread and butter of the game.
2.4 Helmholtz free energy.
The formula relating the differentials of the entropy, energy, and pressure is
simplified by introducing the pressure and temperature. We find, as expected,
U