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## 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?