Statistical Mechanics - Physics at Oregon State University

246 APPENDIX A. SOLUTIONS TO SELECTED PROBLEMS.
Problem 7.
2 ∂
U = kBT log(Z) = 3NkBT
∂T
T S − pV + µN = 3NkBT
n
µ = kBT log(
nQ(T ) )
We need to include both inside and outside the bottle!
∆F = (µin − µout)∆N
when ∆N molecules move from outside to inside the bottle. This leads to
∆F = kBT log( nin
)∆N
nout
This is only negative for ∆N > 0 when nin < nout, and hence when the outside
density is higher.
Problem 8.
This is Planck’s law for the distribution of photons. Assume that the volume
dependence is in the frequency ω, but that the frequency does not depend on
the number of photons present. This gives:
N(T, V, µ) =
ω(V ) −1
k e B T − 1
which is independent of µ! The grand potential is therefore
Ω(T, V, µ) = Ω(T, V, 0) −
µ
and therefore the Helmholtz free energy is
0
Ndµ ′ = Ω(T, V, 0) − Nµ
F (T, V, N) = Ω(T, V, µ(N)) + Nµ(N) = Ω(T, V, 0)
which does not depend on N!!
Hence
µ =
∂F
∂N T,V
which means that it does not cost energy to create and destroy photons, which
is, of course, what should happen in a cavity, where photons are always created
and destroyed.
= 0