Statistical Mechanics - Physics at Oregon State University

4.2. BOLTZMANN GAS AGAIN. 75
Although the distribution functions depend on µ, it is often possible to use
this approach
to obtain functions of N,T,V. Also note that for the ideal gas
∂U
∂V = 0 and hence that the pressure in an ideal gas is completely due to
T,N
the derivative of the entropy, as we have stated before. This is due to the fact
that
∂F
∂U
∂S
p = −
= −
+ T
(4.30)
∂V T,N ∂V T,N ∂V T,N
In a real gas system, of course, a change in volume will change the average
distance between particles and hence the inter-particle interactions. This will
give a small correction to the formula for U, and hence a small contribution to
the pressure. In solids, on the other hand, the pressure is mainly due to the
change in interaction energy.
Summation of the chemical potential.
We have transformed our independent variables to the set T, V, N and therefore
need to find the corresponding free energy. The Helmholtz free energy is
the important quantity and can be calculated as soon as µ(T, N, V ) is found, because
µ =
∂F
∂N . The Helmholtz free energy for a system with zero particles
T,V
is zero, and hence we have
N
0
F (N, T, V ) =
N
0
µdN ′ =
′ N
kBT log
dN
V nQ(T )
′
= NkBT
n
log( ) − 1
nQ(T )
(4.31)
where we used N
0 log(cN ′ )dN ′ = N log(c) + N log(N) − N.
Formally, there is an objection against using an integral over the number of
particles when this number is small. Remember that the chemical potential is
the free energy needed to add one particle to the system, and hence a correct
expression for F is
F (N, T, V ) =
N
µ( ˆ N, T, V ) =
ˆN=1
N
ˆN
kBT log
= kBT log(N!) − N log(nQ(T )V ) (4.32)
V nQ(T )
ˆN=1
Here we used the fact that N ˆN=1 log(c ˆ N) = N log(C) + log(N!). If N is very
large we can again use Stirling’s formula to show that this form really reduces