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For the same atoms in a three-dimensional volume $V$, the one-atom translational partition function is $q_{\mathrm{trans}}^{(3)}=V/{\Lambda }^3$. Comparing the same $N$ atoms at the same temperature cancels the common $N!$ in their canonical partition functions. Therefore the molar adsorption confinement free energy, with the sign defined as adsorbed minus free, isThe logarithm is dimensionless: $V/\sigma$ and $\Lambda$ are both lengths. For a macroscopic gas depth $V/\sigma \gg \Lambda$, confinement raises the translational Helmholtz free energy. Equivalently, the translational state count is reduced by a factor $\sigma \Lambda /V$ per atom. The question supplies neither a volume nor an adsorption binding energy, so this answer is symbolic and describes the confinement cost, not a numerical net adsorption free energy.

For example, a binding energy $D>0$ per atom would supply a separate $-N_{\mathrm{A}}D$ to $\Delta F_m$, together with any normal-motion internal contribution retained by a more detailed model. Adsorption can then be favourable despite the loss of free translational phase space. Merely calling the change a loss must not reverse the sign of the explicitly defined $F_{\mathrm{ads}}-F_{\mathrm{gas}}$.

The two protons are identical spin-one-half fermions, so the total state is antisymmetric under exchange. In the electronic and vibrational ground state, an even rotational $J$ has a symmetric spatial wavefunction and therefore uses the antisymmetric nuclear-spin singlet state; odd $J$ uses the symmetric triplet state. Thus parahydrogen has one nuclear-spin state, and orthohydrogen has three. Each rotational level also has spatial state degeneracy $2J+1$.

The translational partition functions and all common electronic/vibrational factors cancel in the ratio of equilibrium populations. Using the same zero of energy for the two spin species givesSubtracting the $J=1$ energy separately from all odd levels without restoring that offset in the equilibrium ratio would incorrectly remove the low-temperature preference for parahydrogen. The nuclear spin statistics account for the factor three, independently of the spatial factor $2J+1$.

Use ground-state references and the supplied ideal-gas, rigid-rotor and harmonic-oscillator approximations. The two ions have closed electronic shells; take their electronic ground-state degeneracies and the molecular electronic ground-state degeneracy as one, and neglect higher electronic levels. Consistently treated nuclear-spin factors cancel between the species. Since the molecule is heteronuclear, its rotational symmetry number is one.

Let ${\mu }_r=m_{\mathrm{Cs}}{m}_{\mathrm{I}}/m_{\mathrm{CsI}}$ be the reduced mass, neglecting the negligible mass correction from electron transfer. The ratio of translational partition functions per volume reduces toThe molecule has additional rotational and vibrational states, whereas the monatomic ions do not. At $3000\mathrm{K}$,Hence the dimensional equilibrium quotient isThe last form makes the unassigned standard concentration explicit. For the common choice of $1\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$, the corresponding microscopic reference is $c^{\circ }=6.02214\times 10^{26}{\mathrm{m}}^{-3}$, givingEquivalently, the dimensional molar quotient is $8.27\times 10^{-7}\mathrm{mol}\mathrm{d}{\mathrm{m}}^{-3}$. A pressure-based gas standard is a different convention: taking $c^{\circ }=p^{\circ }/(k_{B}T)$ with $p^{\circ }=1\mathrm{bar}$ would give $2.06\times 10^{-4}$, rather than the same numerical constant.

The well-minimum convention gives the identical answer becauseConstants used here are $h=6.62607015\times 10^{-34}\mathrm{J}\mathrm{s}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, $m_u\simeq 1.66053907\times 10^{-27}\mathrm{kg}$ and $N_A=6.02214076\times 10^{23}\mathrm{mo}{\mathrm{l}}^{-1}$.^[ref]

Use the same standard pressure and temperature, the same bond length, and the same electronic and vibrational-ground-state approximations. The translational partition function scales as mass to the power $3/2$, giving the factor $(47/44{)}^{3/2}$. For the original isotopologue, $I=32m_u{R}^2$ and the rotational symmetry number is two. For the isotope-substituted molecule, the two oxygen ends are distinguishable, so the symmetry number is one. HenceThe change in standard molar entropy issoThis uses a convention omitting a separate nuclear-spin entropy. If nuclear-spin degeneracies are instead included explicitly, the spin-$1/2$ carbon-13 nucleus adds a further $R\log 2$ relative to the spin-zero original isotopologue; that convention must be stated rather than silently mixed with the quoted partition functions.

An ideal gas whose particles move in an area $A$ with two independent translational momentum components. In its dilute classical limit the one-particle translational partition function is $q=A/{\Lambda }^2$, where $\Lambda$ is the thermal de Broglie wavelength, and $Z_N=q^N/N!$. The translational internal energy is $N{k}_{\mathrm{B}}T$, and the surface pressure conjugate to area is $N{k}_{\mathrm{B}}T/A$. A mobile dilute adsorbed gas is an example; a fully occupied array of fixed adsorption sites has different state counting.