= Solution
For the \b[ideal dilute Maxwell-Boltzmann gas] implicit in a single-particle <partition function>, indistinguishability gives $Z_N=Z_1^N/N!$. With <Stirling's approximation>,
$$
F=-k_BT\ln Z_N\simeq-Nk_BT\ln Z_1+k_BT(N\ln N-N).
$$
Differentiate at fixed $T,V$:
$$
\boxed{\mu_{\rm ch}=\left(\frac{\partial F}{\partial N}\right)_{T,V}=k_BT\ln\frac N{Z_1}=-k_BT[\ln Z_1-\ln N].}
$$
This is a thermodynamic-limit ideal-gas result; arbitrary interacting classical gases do not have $Z_N=Z_1^N/N!$.
\b[Equilibrium constraints and the Gibbs minimum.] The <open-system equilibrium contacts> allow exchange of heat, <volume> and matter. For unconstrained contacts without external-field or interface corrections, equilibrium requires equal temperatures, pressures and each exchangeable species' <chemical potentials> across the contacts. These conditions follow by maximizing total <entropy> under conserved total <energy>, <volume> and species amounts.
The next minimization statement fixes the system's particle number. For a system in a reservoir at $T_0,p_0$, with $N$ fixed and only pressure-volume work, the reservoir <entropy> change gives
$$
\Delta S_{\rm tot}=\Delta S-\frac{\Delta U+p_0\Delta V}{T_0}=-\frac{\Delta(U+p_0V-T_0S)}{T_0}\geq0.
$$
For candidate states at the imposed $T=T_0$, $p=p_0$, the expression in parentheses is $G$. Thus \b[the <Fixed-particle-number Gibbs minimum> selects stable equilibrium at fixed $T,p,N$], with stationary first variation and nonnegative allowed second variation. If particle exchange remains unconstrained, $N$ is not fixed and this minimization statement must be replaced by one incorporating the particle reservoir; it is not a justification for minimizing $G$ over arbitrary $N$.
The simple-compressible-system first law gives
$$
dG=d(U-TS+pV)=\boxed{-S\,dT+V\,dp+\mu_{\rm ch}\,dN}.
$$
For a homogeneous extensive single-component phase, the <Single-component Gibbs Euler relation> follows from $U=TS-pV+\mu_{\rm ch}N$, yielding \b[$G=\mu_{\rm ch}N$]. For a mixture, the corresponding identities are $G=\sum_i\mu_iN_i$ and $dG=-S\,dT+V\,dp+\sum_i\mu_i\,dN_i$. Surface effects or other nonextensive contributions require additional terms.<Classical Thermodynamics by David Tong>[Free energies, chemical potential and equilibrium with imposed reservoir variables]{ref}
\b[Reaction equilibrium.] Let $d\xi$ be an allowed reaction extent for the forward process, so $dN_A=-d\xi$ and $dN_B=2d\xi$. At fixed $T,p$,
$$
dG=(2\mu_B-\mu_A)d\xi.
$$
For an interior equilibrium allowing both reaction directions, the coefficient vanishes:
$$
\boxed{\mu_A=2\mu_B.}
$$
At fixed $T,V$, minimizing $F$ gives the same condition. The <reaction formula-unit conservation> gives $M=N_A+N_B/2$, not the total number $N_A+N_B$ of gas particles. This <conservation law> replaces fixed total particle number for the reacting mixture.
Using the stipulated single-particle functions,
$$
\mu_i=k_BT\ln\frac{N_i}{C_iVT^{3/2}},\qquad\frac{N_A}{C_AVT^{3/2}}=\left(\frac{N_B}{C_BVT^{3/2}}\right)^2.
$$
The <number-based dissociation equilibrium ratio> is therefore
$$
\boxed{K_N(V,T)=\frac{N_B^2}{N_A}=\frac{C_B^2}{C_A}VT^{3/2}.}
$$
This number-based quantity has a factor of <volume>; the corresponding concentration ratio $n_B^2/n_A=K_N/V$ depends on <temperature> alone in the stated ideal model. A standard activity-based equilibrium constant would additionally specify reference states. No bond-energy factor is added to the functions given in the paper.
Define the dissociated fraction of original A formula units by $\alpha=N_B/(2M)$. Then $N_A=M(1-\alpha)$ and $N_B=2M\alpha$, so
$$
\frac{\alpha^2}{1-\alpha}=\frac{K_N}{4M}\equiv q,\qquad\boxed{\alpha=\frac{\sqrt{q^2+4q}-q}{2}.}
$$
The <dissociation fraction versus conserved density> follows because the left side increases monotonically from zero to infinity on $0<\alpha<1$. At fixed $V,T$, increasing $M$ decreases $q$, hence \b[decreases dissociation]. Equivalently, at fixed $M,T$ compression raises $M/V$ and favours A. Association produces fewer independent translating particles; at high <density> the translational-entropy advantage of the two-particle side is smaller, so equilibrium shifts toward the one-particle side.
\Image[/past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2016/paper-1-dissociation-density.png]
{title=At fixed temperature the dissociated fraction falls as the conserved formula-unit density increases}
{source=paper-1-dissociation-density.py}
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