Past exam of the chemistry A course of the University of Cambridge / 2025 / Paper 2 / A6

The internal energy is the population-weighted sum of the molecular energy levels:At fixed volume, assume the energy levels themselves are temperature-independent. Differentiating the molecular partition function givesChoosing the zero of energy changes $U$ by the corresponding constant per molecule, so the ground-level convention must be consistent throughout.

Relative to the ground state, the quantum harmonic oscillator levels are $vhc\tilde{\omega }$. Put ${\theta }_v=hc\tilde{\omega }/k$. Summing the geometric series gives the vibrational partition function $q=(1-e^{-{\theta }_v/T}{)}^{-1}$. Therefore the vibrational internal energy isThe omitted zero-point energy would add $Nhc\tilde{\omega }/2$ if absolute oscillator energies were used.

“Low” and “high” compare $kT$ with the vibrational spacing $hc\tilde{\omega }$, not temperature with an arbitrary number of kelvin. For $T\ll {\theta }_v$, occupation of excited levels is exponentially suppressed andThis vanishing result uses the ground-relative energy zero. For $T\gg {\theta }_v$, expand $e^{{\theta }_v/T}-1\sim {\theta }_v/T$ to obtainThis agrees with equipartition: the oscillator has one quadratic kinetic and one quadratic potential term, contributing $kT/2$ each. The corresponding limiting vibrational heat capacity is zero at low temperature and $Nk$ at high temperature.

Subtracting the $v=0$ energy in the Morse oscillator givesThe first three relative wavenumbers are $0$, $213$, and $424{\mathrm{cm}}^{-1}$. Thus, with ${\theta }_1=213/0.695\mathrm{K}$ and ${\theta }_2=424/0.695\mathrm{K}$,This is a three-level partition function, as stipulated, rather than an unrestricted sum of the anharmonic formula to arbitrarily large $v$.

Differentiating each Boltzmann factor yieldsThe derivative is positive: warming increases occupation of both excited energy levels. At $100\mathrm{K}$,

For one mole, $Nk=R$, so the three-level Morse oscillator result isFor the full quantum harmonic oscillator with the given wavenumber,The specified anharmonic calculation is about $1.7\%$ higher. Anharmonicity lowers the first excitation spacings and increases excited-state occupation. Both results remain small because $100\mathrm{K}$ is well below the vibrational temperature. The comparison also mixes a three-level anharmonic truncation with an infinite harmonic sum; omitted higher Morse levels would make a further small positive correction at this temperature.