 Solution

Gibbs free energy and its minimum. For fixed atom number $N$, there are $N+n$ sites andconfigurations with $n$ indistinguishable vacancies. Their configurational entropy is $k_B\ln W$. Relative to the vacancy-free Gibbs free energy $G_0$, creating $n$ vacancies costs energy $nϵ$ and pressure-volume work $npv$. Put $A=ϵ+pv>0$. ThenAt leading thermodynamic order, Stirling's approximation givesThe minimum obtained from the stationary point below isIf the perfect crystal's internal energy and other entropy are chosen as zero, its pressure term is $G_0=pNv$. More generally the information given fixes the vacancy contribution rather than the absolute bulk free energy. The pressure-volume term must be retained because the number of sites, and hence the volume, changes.

Equilibrium vacancy population. Differentiating the extensive free energy at fixed $N,T,p$ givesThe stationary point is consequently the unique interior minimum. Write $q=e^{-A/(k_{B}T)}$; its condition is $n/(N+n)=q$, giving the requested thermodynamic result for equilibrium vacancies with variable lattice size,Substitution back into $G$ cancels the terms containing $n_{\ast }$ and yields the boxed minimum. In the dilute-defect limit $A\gg k_{B}T$, $n_{\ast }\simeq N{e}^{-A/(k_{B}T)}$. The Bose-like denominator is a consequence of the changing number of lattice sites and combinatorial entropy; it does not assert that vacancies here are an ideal gas of bosonic particles.

There is a finite-size correction to a variable-site vacancy model if “average” is interpreted exactly. The microscopic count written above gives the relative isothermal-isobaric sumsoIts relative equilibrium free energy is $-k_{B}T\ln \mathcal{Y}=(N+1)k_{B}T\ln (1-q)$. Thus the source's $N$ numerator is correct to leading extensive order, but is not the exact finite-$N$ mean for that stated configuration count. A free-energy minimum is likewise not generally identical to the mean of a finite distribution. These distinctions vanish relatively as $N\to \infty$.

Thermal volume. Each atom or vacancy occupies volume $v$, so at the same thermodynamic accuracyFor the exact counting model, replace $n_{\ast }$ by $(N+1)q/(1-q)$ in $v(N+⟨n⟩)$. The vacancy-free limit is $Nv$ as $T\to 0$. This model isolates expansion from vacancies, taking $v$ and $ϵ$ constant; it omits ordinary phonon thermal expansion. Its growth at high temperature should not be extrapolated to arbitrarily large vacancy fractions in a stable crystal, and $A>0$ is essential for a normalizable variable-site ensemble.