= Solution
\b[Gibbs free energy and its minimum.] For fixed atom number $N$, there are $N+n$ sites and
$$
W(N,n)=\binom{N+n}{n}
$$
configurations 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> $n\epsilon$ and pressure-<volume> work $npv$. Put $A=\epsilon+pv>0$. Then
$$
G(n)=G_0+nA-k_BT\ln\frac{(N+n)!}{N!\,n!}.
$$
At leading thermodynamic order, <Stirling's approximation> gives
$$
G(n)\simeq G_0+nA-k_BT\big[(N+n)\ln(N+n)-N\ln N-n\ln n\big].
$$
The minimum obtained from the stationary point below is
$$
\boxed{G_{\min}\simeq G_0+Nk_BT\ln\left(1-e^{-A/(k_BT)}\right).}
$$
If 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.
\b[Equilibrium vacancy population.] Differentiating the extensive <free energy> at fixed $N,T,p$ gives
$$
\frac{\partial G}{\partial n}=A+k_BT\ln\frac n{N+n},\qquad
\frac{\partial^2G}{\partial n^2}=\frac{k_BTN}{n(N+n)}>0.
$$
The stationary point is consequently the unique interior minimum. Write $q=e^{-A/(k_BT)}$; its condition is $n/(N+n)=q$, giving the requested thermodynamic result for <equilibrium vacancies with variable lattice size>,
$$
\boxed{n_*\simeq\frac{Nq}{1-q}
=\frac{N}{e^{(\epsilon+pv)/(k_BT)}-1}.}
$$
Substitution back into $G$ cancels the terms containing $n_*$ and yields the boxed minimum. In the dilute-defect limit $A\gg k_BT$, $n_*\simeq Ne^{-A/(k_BT)}$. 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 sum
$$
\mathcal Y=\sum_{n=0}^\infty\binom{N+n}{n}q^n=(1-q)^{-N-1},
$$
so
$$
\boxed{\langle n\rangle_{\rm exact}=q\partial_q\ln\mathcal Y
=\frac{(N+1)q}{1-q}.}
$$
Its relative equilibrium <free energy> is $-k_BT\ln\mathcal Y=(N+1)k_BT\ln(1-q)$. 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 $N\to\infty$.
\b[Thermal volume.] Each atom or vacancy occupies <volume> $v$, so at the same thermodynamic accuracy
$$
\boxed{\langle V(T)\rangle\simeq v(N+n_*)
=\frac{Nv}{1-e^{-(\epsilon+pv)/(k_BT)}}.}
$$
For the exact counting model, replace $n_*$ by $(N+1)q/(1-q)$ in $v(N+\langle n\rangle)$. The vacancy-free limit is $Nv$ as $T\to0$. This model isolates expansion from vacancies, taking $v$ and $\epsilon$ 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.
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