 Rotational partition functions of hydrogen

Write the linear rigid rotor energy eigenvalues as $E_J=k_{\mathrm{B}}{\theta }_{r}J(J+1)$, where ${\theta }_r={\hslash }^2/(2I{k}_{\mathrm{B}})=hB/k_{\mathrm{B}}$ is the rotational temperature scale, $I$ the moment of inertia, $B$ the frequency-valued rotational constant, and $k_{\mathrm{B}}$ the Boltzmann constant. Put $t=T/{\theta }_r$ and ${ϵ}_J=J(J+1)$. Define sums that include rotational multiplicity but exclude nuclear-spin multiplicity:The canonical partition function for one equilibrating internal rotor is $q_{\mathrm{eq}}=q_{\mathrm{p}}+3q_{\mathrm{o}}$, so its equilibrium orthohydrogen fraction isFor $t\ll 1$, the lowest allowed levels give $q_{\mathrm{p}}\simeq 1$ and $q_{\mathrm{o}}\simeq 3e^{-2/t}$, hence $x_{\mathrm{o}}^{\mathrm{eq}}\simeq 9e^{-2/t}/(1+9e^{-2/t})\to 0$. For $t\gg 1$, the densely populated even and odd ladders have equal leading sums, $q_{\mathrm{p}}\sim q_{\mathrm{o}}\sim t/2$, hence $x_{\mathrm{o}}^{\mathrm{eq}}\to 3/4$. The factor nine belongs to the low-temperature two-level approximation, and the factor three to the high-temperature spin multiplicities.

A frozen mixture requires two separately normalized probability distributions,Multiply these by the prescribed fractions $1-x$ and $x$, respectively. The nuclear-spin factor three cancels from the conditional orthohydrogen distribution. For fixed counts $N_{\mathrm{o}}$ and $N_{\mathrm{p}}$, the internal part of the many-molecule partition function is proportional to $q_{\mathrm{p}}^{N_{\mathrm{p}}}(3q_{\mathrm{o}}{)}^{N_{\mathrm{o}}}$, with combinatorial factors independent of temperature. Its logarithm yields the two conditional means weighted by the fixed counts. Using $q_{\mathrm{p}}+3q_{\mathrm{o}}$ instead would silently permit conversion and change the physical constraint.