Past exam of the chemistry A course of the University of Cambridge / 2022 / Paper 2 / A8 / a

Assume a dilute classical ideal gas of $N$ indistinguishable, noninteracting particles, with translational states dense enough to replace their sum by a continuum integral. Separate translation from all internal degrees of freedom. ThenUsing Stirling's approximation in the Helmholtz free energy,Differentiate at fixed $N,V$. Since ${\Lambda }^{-3}\propto T^{3/2}$,Only after this derivative set $N=N_A$ and $V=RT/p^{\circ }$ for one mole at standard pressure:This is the translational Sackur-Tetrode equation. The $N!$ factor is essential for the correct extensive entropy; replacing $V$ by a temperature-dependent standard molar volume before taking the fixed-volume derivative would give the wrong constant term.

For independent internal rotational states, the rotational factor in the many-molecule partition function is $q_{\mathrm{rot}}^N$. The indistinguishability factor has already been included in the translational part and is not applied a second time to the internal factor. ThusA nonlinear rigid rotor has three rotational quadratic degrees of freedom; in the high-temperature limit, $q_{\mathrm{rot}}\propto T^{3/2}$. ThereforeThis is the rotational entropy contribution, with the appropriate rotational symmetry number already inside $q_{\mathrm{rot}}$. It has no ideal-gas pressure dependence.