= Solution
\b[Ensemble and applications.] A <grand canonical ensemble> describes a system exchanging both <energy> and particles with a large reservoir. Its fixed control parameters are $T,V,\mu$, while microstates with different $N$ have <probabilities>
$$
P_r=\Xi^{-1}e^{-\beta(E_r-\mu N_r)},\qquad
\Xi=\sum_r e^{-\beta(E_r-\mu N_r)}.
$$
Weak coupling allows the system's <energy> levels to remain meaningful; equilibrium fixes the reservoir <temperature> and <chemical potential>. The <grand potential> is $\Phi=-k_BT\ln\Xi=U-TS-\mu N$, with $N$ now the mean population. It generates <entropy>, <pressure> and mean <particle number> by differentiation. This ensemble is convenient for gases in contact with particle reservoirs, independently occupied quantum levels and surface <adsorption>; its particle-number fluctuations are part of the description, not an error in counting.
\b[Classical single-level limit.] For the <single-level grand partition function in the dilute limit>, put $x_k=e^{-\beta(\epsilon_k-\mu)}$. Empty and singly occupied levels contribute $1$ and $x_k$. In the dilute classical limit $x_k\ll1$, higher occupancies contribute only at order $x_k^2$ or beyond:
$$
\boxed{\Xi_k=1+x_k+O(x_k^2)
\simeq1+\exp[-(\epsilon_k-\mu)/(k_BT)]}.
$$
This truncation does not impose the Pauli principle on a classical gas. A <fermion> state has exactly $1+x_k$, a <boson> state has $(1-x_k)^{-1}$, and Maxwell-Boltzmann occupancy weights give $e^{x_k}$; they all agree to this order. In particular, $\ln\Xi_k=x_k+O(x_k^2)$, with the classical Maxwell-Boltzmann result exactly $x_k$.
\b[<grand potential> and <chemical potential>.] With $z=e^{\beta\mu}$, summing the single-level <logarithms> and using the stated <density of states> gives
$$
\ln\Xi\simeq z\int_0^\infty g(\epsilon)e^{-\beta\epsilon}\,d\epsilon.
$$
Substitute $\epsilon=k_BTx^2$, for which
$$
\int_0^\infty\epsilon^{1/2}e^{-\beta\epsilon}\,d\epsilon
=2(k_BT)^{3/2}\int_0^\infty x^2e^{-x^2}dx
=\frac{\sqrt\pi}{2}(k_BT)^{3/2}.
$$
Combining this with the density-of-states prefactor gives $\ln\Xi=zV/\lambda^3$. Thus
$$
\boxed{\Phi=-k_BT\frac V{\lambda^3}e^{\mu/(k_BT)}}.
$$
Differentiating at fixed $T,V$,
$$
N=-\left(\frac{\partial\Phi}{\partial\mu}\right)_{T,V}
=\frac V{\lambda^3}e^{\mu/(k_BT)},
\qquad
\boxed{\mu=k_BT\ln\frac{N\lambda^3}{V}}.
$$
Similarly $p=-(\partial_V\Phi)_{T,\mu}=k_BTz/\lambda^3$, so $pV=Nk_BT$. This is the <grand-canonical chemical potential of a classical ideal gas>. The condition $N\lambda^3/V\ll1$ states the classical regime. The supplied <density of states> counts one internal state; extra internal degeneracies would have to be included consistently rather than inserted at only one stage.
\b[One <adsorption> site.] Take the free-vapour <energy> reference as zero and the <binding energy> as $-\epsilon$. A site is either empty or singly occupied; the latter has vibrational <partition function> $z_s(T)$. Its <adsorption-site grand partition function> and mean occupation are
$$
\Xi_{\rm site}=1+z_s e^{\beta(\mu_s+\epsilon)},\qquad
\theta=\frac{z_s e^{\beta(\mu_s+\epsilon)}}
{1+z_s e^{\beta(\mu_s+\epsilon)}}.
$$
The <chemical potential of an adsorbed species on equivalent sites> follows by inverting the occupation ratio
$$
\boxed{\mu_s=-\epsilon+k_BT\ln\frac{\theta}{(1-\theta)z_s}}.
$$
The factor $1-\theta$ is the vacant-site probability. The binding reference and the <energies> included in $z_s$ must be chosen consistently, so that vibrational <zero-point energy> is not counted twice.
\b[Isotherm.] The <vibrational factor in a Langmuir equilibrium constant> is fixed by <adsorption> equilibrium, $\mu_s=\mu_{\rm vap}$. Neglecting vapour internal degrees of freedom and writing its <number density> as $p/(k_BT)$ gives $e^{\beta\mu_{\rm vap}}=p\lambda^3/(k_BT)$. Therefore
$$
b_s(T)=\frac{z_s(T)\lambda^3}{k_BT}e^{\epsilon/(k_BT)},\qquad
\boxed{\theta(p,T)=\frac{b_s(T)p}{1+b_s(T)p}}.
$$
This is the <Langmuir isotherm>: initially linear, increasing monotonically and concave down, approaching unit coverage. Half coverage occurs at $p_{1/2}=1/b_s(T)$. The coefficient has inverse-pressure units. It assumes equivalent independent sites, at most one <molecule> per site and no lateral interaction or multilayer <adsorption>. The molecular mass appropriate to the vapour determines $\lambda$.
\Image[/past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2015/paper-1-adsorption-isotherm.png]
{title=Original dimensionless Langmuir coverage curve with half-coverage and limiting behavior}
A real uptake measurement must separate adsorbed <molecules> from free gas in the apparatus. Langmuir's original mica experiment compared a mica-containing bulb with a similar blank and inferred uptake from calibrated gas <volumes> and <pressures>. His stated blank contribution and bath-temperature drift limited the inferred <adsorption>; a saturating curve alone does not establish ideal equivalent sites.<The Adsorption of Gases on Plane Surfaces of Glass, Mica and Platinum by Irving Langmuir>[pp. 1376–1378: mica apparatus and blank; p. 1385: limitations]{ref}
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