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.