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At chemical equilibrium, ${\mu }_{\mathrm{AB}}={\mu }_{\mathrm{A}}+{\mu }_{\mathrm{B}}$. Using the number-concentration form givesHere $\Delta {\epsilon }_0$ is the reaction energy between the chosen molecular reference states. Rearranging,Define the dimensionless concentration equilibrium constant byThusThe standard concentration $c^{\circ }$ has the same units as $c_i$. It converts concentration ratios to ideal thermodynamic activities; changing it changes the numerical dimensionless $K_c$ for this dissociation, which creates an extra particle. The unnormalized quotient $c_{\mathrm{A}}{c}_{\mathrm{B}}/c_{\mathrm{AB}}$ instead has concentration units.^[ref]

At low temperature, $x=\epsilon /(k_{B}T)\gg 1$, so the denominator in the exact expression approaches one:The exponential suppression dominates the $T^{-2}$ factor. Almost all particles remain in the ground state, so very little volume change is thermally accessible.

At high temperature, $e^{-x}\to 1$, and thereforeThe excited fraction has already approached $3/4$, so further heating barely changes it. Both limits are consistent with a bounded two-level system; the high-temperature vanishing does not imply that ordinary materials cease expanding when additional levels or other mechanisms become accessible.

The definition uses a fixed reference volume, so it is a reference-volume thermal expansion coefficient:Let $y=3e^{-\epsilon /(k_{B}T)}$. Sincewe obtainIt is positive and has units of inverse temperature. Equivalently, $\alpha =[\epsilon /(k_B{T}^2)]P_1(1-P_1)$: the response reflects changes in the thermally occupied excited population. The usual instantaneous fractional thermal expansion coefficient $V^{-1}dV/dT$ would instead equal $\alpha V_0/V$; substituting that different definition would change the answer.

Use the additive-volume two-level system implicit in the model: each excited particle increases the volume by the same increment. The difference between the two endpoint volumes is $V_0$, so that increment is $v=V_0/N$.

If $N_1$ particles are excited, $V=V_0+(V_0/N)N_1$. Linearity of the expectation gives $⟨N_1⟩=N{P}_1$, henceAs $T\to 0$, $V\to V_0$. As $T\to \infty$, only three quarters of the particles are excited on average, giving $V\to 7V_0/4$, rather than the all-excited endpoint $2V_0$. These limits also check the inclusion of state degeneracy.

Choose the ground-state energy as zero and write $x=\epsilon /(k_{B}T)$. The molecular partition function must sum individual states, so the triply degenerate excited level contributes three Boltzmann factors:The probabilities of occupying the two levels areEach of the three individual excited states has probability $e^{-x}/q$. At high temperature the four states become equiprobable, so the two levels have probabilities $1/4$ and $3/4$, not $1/2$ each. This is the distinction between a level and its state degeneracy.

For the uniform-binding multilayer adsorption model, ${\theta }_j$ describes sites of exactly height $j$, not the fraction whose height is at least $j$. At equilibrium the upward and downward fluxes on each adjacent pair of heights balance:For completeness, stationarity alone gives a height-independent probability current; the reflecting boundary at height zero sets that current to zero. This justifies pairwise balance for a normalizable equilibrium distribution.

Writing $x=K_{\mathrm{ads}}p$ gives the recurrence ${\theta }_j=x{\theta }_{j-1}$. Successive substitution then yieldsAn equilibrium geometric distribution of unbounded heights requires $0\le x<1$. The first-layer-only Langmuir adsorption model instead truncates the allowed heights at one, which changes the normalization and removes this convergence restriction.

At chemical equilibrium, adsorption and desorption balance:Canceling $M$ and rearranging givesThus $K_{\mathrm{ads}}=b$ when the kinetic and thermodynamic descriptions refer to the same ideal surface and energy convention. In particular,The ratio fixes the adsorption isotherm, while the individual rate constants also determine relaxation toward equilibrium. Here $k_a$ has units of inverse pressure per time and $k_d$ of inverse time, so $K_{\mathrm{ads}}$ is dimensional. This equivalent-site arrival/departure argument is the simple case of the Langmuir adsorption model.^[ref]

Let ${\epsilon }_s$ be the energy of one adsorbed atom, using the same energy reference as the gas-phase value ${\epsilon }_0$. For equivalent, noninteracting adsorption sites, choosing the occupied positions givesEvery configuration has energy $N_1{\epsilon }_s$, so the canonical lattice-gas partition function isThere is no further $1/N_1!$ factor: the combinatorial choice already counts indistinguishable adsorbates correctly. The Helmholtz free energy, using Stirling's approximation for a macroscopic number of occupied and empty sites, isHence the adsorbed-phase chemical potential at fixed $T,M$ isThe logarithm comes from configurational entropy: adding an atom is less favorable when vacancies are scarce. In the gas, the stipulated expression can be writtenwhere $\Lambda$ is the thermal de Broglie wavelength. Equality of chemical potentials at chemical equilibrium givesThereforeThe binding energy $E_b={\epsilon }_0-{\epsilon }_s$ is positive for favorable binding. The units of $b$ are inverse pressure, making $bp$ dimensionless. Choosing the adsorbed energy as zero simply sets ${\epsilon }_s=0$ and leaves the same physics; it must not remove the relative binding energy. This is the Langmuir isotherm, with low-pressure linear uptake and high-pressure monolayer saturation.