A planar rigid rotor has angular eigenfunctions $e^{im\varphi }$ and single-valuedness imposes $m=0,\pm 1,\pm 2,\dots$. Its energy depends on $m^2$, so if the level is labelled by $J=|m|$, thenThe two states at $J>0$ represent opposite directions of angular momentum. If the signed quantum number itself is called $J$, each signed state is counted once instead; one must not both sum over both signs and multiply by two again.

For an unrestricted three-dimensional linear rigid rotor, $J=0,1,2,\dots$ has $M=-J,\dots ,J$ and state degeneracy $2J+1$, with $E_J={\hslash }^{2}J(J+1)/(2I)$. Planar confinement removes one rotational coordinate and changes both the spectrum and degeneracy. These are the spatial counts before nuclear spin statistics. For homonuclear hydrogen, the allowed even/odd spatial states must also be paired with the appropriate parahydrogen or orthohydrogen nuclear states, as considered explicitly in A8.

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 $F_{\mathrm{ads}}-F_{\mathrm{gas}}$.

Let $\Lambda =h/\sqrt{2\pi m{k}_{\mathrm{B}}T}$ be the thermal de Broglie wavelength. The two free momentum components of one adsorbed atom give the phase-space integralEach momentum integral is a Gaussian integral, soThe second expression is the classical ideal-gas partition function for mobile, indistinguishable surface atoms. Its validity requires a dilute nondegenerate gas, $N{\Lambda }^2/\sigma \ll 1$, and neglects interactions and excited motion normal to the surface. The $N!$ appears here because particles move between positions, unlike the fully occupied fixed-site model of A6.

Only the two lower orientations remain populated, giving $2^N$ equiprobable microstates. ThusThis residual entropy follows from the stipulated independent-site ground-state degeneracy. The zero-entropy version of the third law of thermodynamics presupposes a unique ordered ground state; that assumption does not hold here. Intermolecular interactions or a small splitting of the two in-plane states could remove this degeneracy, but they are excluded from the given model.

Using $dx/dT=-x/T$ in the internal energy givesThis is a Schottky anomaly with unequal degeneracies, with lower-level state degeneracy two and upper-level state degeneracy one. Its limiting forms areIt vanishes at low temperature because exciting the upper orientation is exponentially unlikely. It also vanishes at high temperature because the populations have nearly saturated and further heating hardly changes the internal energy. Between these limits it has one smooth maximum. The original plot shows its shape; numerical maximization gives $k_{\mathrm{B}}{T}_{\max }/\delta \simeq 0.449$ and $C_{\max }/(N{k}_{\mathrm{B}})\simeq 0.241$.

For $k_{\mathrm{B}}T\gg \delta$, the Boltzmann factors of the orientations become almost equal, giving $3^N$ nearly equiprobable microstates. The Boltzmann entropy of that distribution approachesThe exact limiting value is reached only as $T\to \infty$ for a nonzero splitting. The useful crossover scale is $T\sim \delta /k_{\mathrm{B}}$; one needs $T\gg \delta /k_{\mathrm{B}}$ for a good saturation approximation. The binding offset $\epsilon$ is not an additional population scale here because all sites remain occupied and it shifts every orientation equally.

At low temperature, the upper orientation has negligible Boltzmann weight, whereas at high temperature all three orientations become equally probable. ConsequentlyThe high-temperature limit is a finite arithmetic mean of three available one-site energy levels. It is not an unbounded translational or rotational equipartition contribution: the prescribed site model has only three states per molecule. Their common binding offset can keep this limiting internal energy negative without any inconsistency.

Write $\beta =(k_{\mathrm{B}}T{)}^{-1}$ and $x=\beta \delta$. The one-site partition function includes both degenerate lower states,The occupied adsorption sites are distinguishable positions, so the canonical partition function is a product, without an additional $N!$ divisor:Differentiating $\ln Z_N$ gives the internal energy,The final factor is the population of the perpendicular orientation. The shared binding offset $-\epsilon$ shifts all energy levels equally and does not determine their relative populations.

Both in-plane orientations minimize the one-site energy. Since every adsorption site is occupied, the sites label independent choices even though the molecules are chemically identical. The ground state of the whole system therefore hasThe ground-state degeneracy is extensive: independently choosing either in-plane orientation at each site costs no energy. There is no extra vacancy or free-translational contribution in this model.