1. Specific Heat We’ll concentrate first on CV.
The specific heat is defined as variation with temperature of the average energy <E>, keeping the other two independent variables (N and V) or (N and P) constant.
We’ll concentrate first on CV.
High temperature limit: At high temperature, were kT is large compared to the spacing between energy levels, the average energy becomes ½ kT for each degree of freedom. This is the case for translational motion and rotational motion. In this case, the specific heat is just ½k for each degree of freedom.
example: a monatomic gas has CV = ³⁄₂ R, whereas for a diatomic gas (N2, say)
CV = ⁵⁄₂ R (³⁄₂ for translation and 1 for rotation). The larger the number of internal degrees of freedom of the molecule, the more energy is needed to raise the temperature by one degree, and, consequently, the greater the value of Cv.
Low temperature: At temperatures low enough that kT is much less than the spacing between the ground state and the first excited state, then adding a small amount of energy will be ineffective in raising the energy of the system. Thus, at low temperature the specific heat will be zero.

2. Example: two-level system
Consider first a two-level system, with energies ±ε. Letting x=exp(-εβ), we have
q= 1/x + x = (x²+1)/x
<E> = -∂lnq/dβ = -(1/q)∂q/∂x × ∂x/∂β = (1-x²)/[x(1+x²)] × ∂x/∂β (using the chain rule)
we know that ∂x/∂β = -εexp(-εβ) = - ε x, so
<E> = -ε(1-x²)/(1+x²)
Differentiating again will give Cv. We can again use the chain rule
Cv=∂<E>/dT = ∂<E>/∂x × ∂x/∂β × ∂β/∂T
This is
Cv= 4εx/(1+x²)² × (-ε x) × (-1/kT²) = 4ε²x²/(1+x²)² × 1/kT² =k (2ε/kT)² x²/(1+x²)²
If we let y = kT/2ε, then x² = exp(-εβ)² = exp(-2εβ)= exp(-2ε/kT)=exp(-1/y), so
Cv = k/y² exp(-1/y)/[1+exp(-1/y)]²
The variable y goes from 0 (at T=0) to ∞
Here is a plot of the specific heat. It is zero
at low T, and also zero at high T. Why? Because
at high T, both levels are equally populated and
there is no way the system can be heated up more!
1/V
The probability that the atoms will have energy Ej is just the number that have energy j divided by the total number: pj = Nj/N

3. Example: harmonic oscillator
Consider now a harmonic oscillator. Letting x=exp(-hνβ) (note that this is a different definition of x than in the case of the two-level system) we have (we did this in class)
q= x½/(1-x)
<E> = -∂lnq/dβ = -(1/q)∂q/∂x × ∂x/∂β = -(1+x)/[2x(1-x)] × ∂x/∂β (using the chain rule)
we know that ∂x/∂β = -hνexp(-hνβ) = - hν x, so
<E> = (hν/2)×(1+x)/(1-x)
Differentiating again will give Cv. We can again use the chain rule
Cv=∂<E>/dT = ∂<E>/∂x × ∂x/∂β × ∂β/∂T
This is
Cv= hν/(1-x)² × (- hν x) × (-1/kT²) = k(hν/kT)²x/(1-x)²
If we let y = kT/hν, then x = exp(-hνβ) = exp(-hν/kT)=exp(-1/y), so
Cv = (k/y²) exp(-1/y)/[1-exp(-1/y)]²
The variable y goes from 0 (at T=0) to ∞
The book defines a vibrational “temperature”
Θv = hν/k, so that y=T/Θv
In terms of Θv, the vibrational specific heat is
Cv = k(Θv/T)² exp(-Θv/T)/[1-exp(-Θv/T)]²
Here is a plot of the specific heat. It is zero
at low T (as expected). It goes to k at high T.