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The chemical potential of an ideal single-occupancy lattice gas with occupied-site internal partition function $z$ and surface coverage $\theta$. The ratio $\theta /(1-\theta )$ accounts for the entropy of distributing particles among occupied and vacant adsorption sites.

A lattice gas describes particle positions by occupations of fixed cells. In the single-occupancy model, $n_{\alpha }\in \{0,1\}$ and $N={\sum }_{\alpha }{n}_{\alpha }$. Its grand canonical partition function sums configurations with weight $\exp [-\beta (U-\tilde{\mu }N)]$. Discretizing a classical position integral into cells of volume $b$ gives $\tilde{\mu }=\mu +k_{B}T\ln (b/{\lambda }^3)$, with thermal de Broglie wavelength $\lambda$.

For energy $U=-4J{\sum }_{⟨\alpha {\alpha }^{\prime }⟩}{n}_{\alpha }{n}_{{\alpha }^{\prime }}$ on a regular lattice of coordination $q$, the substitution $n_{\alpha }=(1+{\sigma }_{\alpha })/2$ maps the model to the Ising model with coupling $J$ and effective field $h=qJ+\tilde{\mu }/2$. The mean Ising spin is a density imbalance, $⟨\sigma ⟩=2bn-1$. For $J>0$ below the critical temperature, the field-driven Ising transition becomes gas-liquid phase coexistence.

For the adsorbed lattice gas, the Helmholtz free energy isUse Stirling's approximation with $M,N,M-N$ large, then differentiate at fixed $M,T,B$. Writing surface coverage as $\theta =N/M$ gives the adsorption chemical potentialThe occupation ratio comes from site-mixing entropy; it is essential near saturation.

Let $N_g$ be the number of free atoms and $n_g=N_g/V$. In the field-free gas, both spin states have zero energy, so the one-atom partition function is $q_g=2V/{\Lambda }^3$, where $\Lambda$ is the thermal de Broglie wavelength. For a dilute ideal gas,ThereforeThe last equality uses $p=n_g{k}_{\mathrm{B}}T$. The bulk spin degeneracy is two even though the bulk magnetic field vanishes: degeneracy is not the absence of spin states.

Write $M=\mathcal{N}=V/b$ for the number of cells; this is fixed even though the particle number $N$ fluctuates. Approximate each position integral by the cell sum $\int d^3{r}_i\simeq b{\sum }_{{\alpha }_i=1}^M$, evaluating the slowly varying intercell pair potential at cell centres. With the assumed single-occupancy restriction, the fixed-$N$ contribution becomesAn occupation configuration $\{n_{\alpha }\}$ with $N={\sum }_{\alpha }{n}_{\alpha }$ selects an unordered set of $N$ occupied cells. There are exactly $N!$ ways to assign the temporary particle labels to that set. They all have the same interaction energy, cancelling the prefactor $1/N!$.

Summing over all particle numbers is therefore equivalent to summing each cell occupation independently over $0,1$:The activity per cell, $e^{\beta \mu }b/{\lambda }^3$, is dimensionless. This is a lattice gas, with each cell either empty or occupied.

The exclusion of multiple occupation is an additional physical modeling assumption, explicitly imposed by the question. Making a cell small does not by itself forbid two point particles from entering it; a repulsive core or a dilute-occupancy approximation would be needed to justify that restriction starting from a general continuum pair potential. In the remaining parts it is part of the defined cell model.