Partition function, Ideal gas, Entropy

[Incand]

Homework Statement

For a diatomic gas near room temperature, the internal partion function is simply the rotational partition function multiplied by degeneracy $Z_e$ of the electronic ground state.
Show that the entropy in this case is
$S = Nk\left[ \ln \left( \frac{VZ_eZ_\text{rot}}{Nv_Q}\right) + \frac{7}{2}\right].$

Homework Equations

The entropy of an ideal gas is given by
$S = -\left( \frac{\partial F}{\partial T}\right)_{V,N} = Nk\left[ \ln \left( \frac{V}{Nv_q}\right) + \frac{5}{2} \right]-\frac{\partial F_\text{int}}{\partial T}.$

The rotational partion function should be
$Z_\text{rot} = \frac{kT}{2B}$
where $B$ is the rotational constant.

Helmholtz free energy
$F = -kT \ln Z$.

The Attempt at a Solution

The internal free energy is
$F_\text{int} = -kT\left[ \ln Z_\text{rot} + \ln Z_e\right].$
Which gives
$-\frac{\partial F_\text{int}}{\partial T} = k\left[\ln Z_\text{rot}Z_e+1\right].$
Apparantly this is for a single molecule and scaling this with $N$ give the correct answer.

However what I don't understand is if I want to derive it from the full partition function I should have $Z_\text{int} = (Z_eZ_\text{rot})^N$. But this is the partition function for distinguishable particles. Shouldn't I instead have $Z_\text{int} \approx \frac{1}{N!} (Z_eZ_\text{rot})^N$ for undistinguishable particles?

 

[Charles Link]

It's been quite a number of years since I worked through the details of some of these, but your last line looks correct for Maxwell-Boltzmann type statistics. You use Stirling's formula to evaluate $ln (N!) =N ln(N)-N$.

 

[Incand]

It's been quite a number of years since I worked through the details of some of these, but your last line looks correct. You use Stirling's formula to evaluate $ln (N!) =N ln(N)-N$.

Well that's pretty much my question. The second statement seems correct to me but that wouldn't give the correct answer. Instead the the partition function for distinguishable particles give the correct answer which is confusing.

When thinking about it now, I think I see the reason for that. The indistinguishable particle part is already accounted for in the original equation (the $N$ there is indeed from stirlings approximation). And then it doesn't matter that the internal states are the same since we already accounted for that.