Statistical Mechanics - Physics at Oregon State University

238 APPENDIX A. SOLUTIONS TO SELECTED PROBLEMS.
(b) Because CV =
∂U
∂T we start at zero for low temperature, go through a
V
maximum near T0 and approach zero again for high temperature.
(c) Like in the previous problem we can write the partition function in the
following manner, by defining a variable si = ±1 for each particle by ɛi
1
2
=
(ɛ1 + ɛ2) + si 1
2 (ɛ2 − ɛ1) = A + Bsi. This gives
Z(T ) = 1 −
e kT i (A+Bsi)
which is
Using
we get
or
which leads to
or
with
s1,···,sN
NA −
Z(T ) = e kT (Z1(T )) N
Z1(T ) = e B
B
kT −
+ e kT
2 ∂
U = kT log(Z(T ))
∂T
2 ∂
U = kT −
∂T
NA
+ N log(Z1)
kT
B
B
2 e kT − − e kT
U = NA + NkT
e B
B
kT + e− kT
U = N(A − Bf(T ))
−B
kT 2
U = N( 1
2 (ɛ1 + ɛ2) − 1
2 (ɛ2 − ɛ1)f(T ))
f(T ) =
B
B
e kT − − e kT
e B
B
kT + e− kT
This gives the correct limits indeed. f(0) = 1 which gives U(0) = Nɛ1. f(∞) =
0 which gives U(∞) = 1
2 (ɛ1 + ɛ2).
Also
and using
CV = − N
2 (ɛ2 − ɛ1)f ′ (T )