Can I derive the Maxwell-Boltzmann Statistics with rotational energy using this method?

Question

I understand that the derivation of the Maxwell-Boltzmann statistics is based on particles only having translational kinetic energy and that it is derived by maximizing the number of ways $\Omega_T$ a microstate can be achieved. $$\Omega_T = \frac{N!}{n_{T1}!n_{T2}!n_{T3}!...n_{Tn}!}$$ I also understand that rotational kinetic energy would only be present in the case of very high temperatures (especially for one-atomic particles). Thinking of that scenario, I would deduce that the number of ways a microstate can be achieved would be by multiplying $\Omega_T$ by the number of ways that a microstate can have in the rotational energy distribution $\Omega_R$. Thus: $$\Omega=\Omega_T \cdot \Omega_R = \frac{N!}{n_{T1}!n_{T2}!n_{T3}!...n_{Tn}!}\cdot \frac{N!}{n_{R1}!n_{R2}!n_{R3}!...n_{Rn}!}$$ However, when maximizing this by taking the derivative (after rewriting the formula in terms of basenumber $e$ and applying Stirling's Approximation), I'd have to deal with two variables, $n_T$ and $n_R$.

If this formula is correct, is it actually possible to derive the MB Statistics formula using this approach?

Answer 1

Yes! Rotational kinetic energy can absolutely be accounted-for with Maxwell-Boltzmann statistics.

For monoatomic gases, there are no rotational degrees of freedom, but for diatomic gases ($H_2$, $O_2$) and more complicated molecules, rotational and molecular vibrational modes contribute to the heat capacity. These modes are frozen out at low temperatures, but 'turn on' as you raise the temperature. See Hyperphysics for molecular hydrogen gas.

I am not sure that it is easy to do this calculation in your current approach, however. I think you will want to use the Canonical ensemble partition function

For low temperatures (just translational energy):

$Z = \frac{1}{N!} \left( \frac{1}{h^3} \int \int \exp \left[-\beta \frac{p^2}{2m} \right] d^3p d^3x \right)^N$

When you add rotational kinetic energy in, you just add those to the exponent:

$Z = \frac{1}{N!} \left( \frac{1}{h^3} \int \int \exp \left[ -\beta \left( \frac{\vec p^2}{2m} + \frac{I\vec \omega^2}{2} \right) \right] d^3p d^3x d^3\omega \right)^N$

Disclaimer: I have not double checked these equations.