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

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.

The integer molecular mass of the stated isotopologue isTake $p^{\circ }=1\mathrm{bar}=10^5\mathrm{Pa}$, the recommended standard pressure.^[ref]

At $298\mathrm{K}$, the thermal de Broglie wavelength is $\Lambda \simeq 9.3100\times 10^{-12}\mathrm{m}$. The result of part (a) givesIf $1\mathrm{atm}$ is used instead, the result decreases by $R\ln (1.01325)\simeq 0.1094\mathrm{J}{\mathrm{K}}^{-1}\mathrm{mo}{\mathrm{l}}^{-1}$, giving $168.23$ in the same units. Specifying the pressure removes that small convention dependence.

Chloroform has the C3v molecular point group. Its three identical ${}^{35}\mathrm{Cl}$ atoms are interchanged by three proper rotations, so $\sigma =3$. The three mirror operations do not increase the rotational symmetry number to six. Isotopic identity matters for this count.

Using the supplied wavenumber form of $k_{\mathrm{B}}$,All are far below $298\mathrm{K}$, justifying the high-temperature nonlinear rigid rotor expression. ThenandThe calculation treats the molecule as rigid and separates electronic, vibrational and nuclear-spin contributions from the requested rotational term.

For each coordinate, the horizontal graph position iswith $\tilde{\omega }$ in inverse centimetres. Read $S/R$ from the supplied curve at these positions. Each $A_1$ frequency contributes one coordinate and each $E$ frequency contributes two. The graphical estimates areThus the listed frequency groups contribute approximately $0$, $1.4$, $4.7$, $0.33$, $2.0$ and $13.8\mathrm{J}{\mathrm{K}}^{-1}\mathrm{mo}{\mathrm{l}}^{-1}$, respectively, andThe nine coordinates are consistent with $3N-6=9$ for a five-atom nonlinear molecule. The lowest-frequency doubly degenerate bend dominates the vibrational entropy; the C–H stretch is effectively frozen at this temperature.

As a numerical check on the graphical reading, the harmonic expression $S/R=x/(e^x-1)-\ln (1-e^{-x})$, with $x=1/r$, gives a total $22.10\mathrm{J}{\mathrm{K}}^{-1}\mathrm{mo}{\mathrm{l}}^{-1}$. This agrees with the precision of the supplied graph.