= Solution
Use $v$ for molar <volume> to distinguish it from the fluctuating parcel <volume> below. The equation is $p=RT/(v-b)-a/v^2$. The parameter $b>0$ models excluded <volume>, while $a>0$ models the reduction in pressure from attractive interactions.
Below the <critical temperature>, a raw isotherm has a loop. The positive-slope segment is mechanically unstable and surrounding metastable portions do not give the stable equilibrium mixture. Replace the loop by a constant coexistence pressure $p_s$ satisfying the Maxwell equal-area condition,
$$
\int_{v_l}^{v_g}[p(v,T)-p_s],dv=0.
$$
The endpoints are liquid and vapour molar <volumes>; intermediate mean <volumes> correspond to <phase> mixtures with changing proportions. Above the <critical temperature>, liquid-like and gas-like states connect continuously without a coexistence plateau.
\Image[/past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2014/paper-1-tsp2014-van-der-waals.png]
{title=Original reduced van der Waals isotherms with computed Maxwell coexistence plateaus and the pressure–temperature coexistence curve ending at the critical point}
In the pressure-temperature diagram, the liquid-vapour coexistence line ends at the <critical point>. Below it, the low-pressure side is vapour and the high-pressure side liquid. “Gas” conventionally denotes the region above the <critical temperature>, where compression alone cannot cross a liquid-vapour transition. The model has no solid <phase>.
The critical isotherm has an inflection: $p_v=p_{vv}=0$. Thus
$$
-\frac{RT_c}{(v_c-b)^2}+\frac{2a}{v_c^3}=0,\qquad
\frac{2RT_c}{(v_c-b)^3}-\frac{6a}{v_c^4}=0,
$$
which give
$$
\boxed{v_c=3b,\qquad T_c=\frac{8a}{27Rb},\qquad p_c=\frac{a}{27b^2}.}
$$
For fluctuations about a stable equilibrium value $x_0$, the system plus reservoir has probability proportional to $e^{-A(x)/(k_BT)}$. Expanding <availability>,
$$
A(x)=A(x_0)+\frac12 A''(x_0)(x-x_0)^2+\cdots,
$$
gives a <Gaussian distribution> and, to this order, $\langle x\rangle=x_0$ with
$$
\boxed{\langle(x-\langle x\rangle)^2\rangle\simeq\frac{k_BT}{A''(x_0)}.}
$$
This requires positive curvature and sufficiently small fluctuations; it is not exact when higher-order terms matter.
The whole cylinder's <volume> is fixed. For the <volume fluctuations of a small fluid parcel>, interpret the small quantity as a material parcel containing $s$ <moles>, whose <volume> can fluctuate against the much larger remainder. Its mean <volume> at the critical <density> is $\bar V=s v_c=3sb$. At fixed <temperature> its <availability> is $A(V)=F_s(V,T)+p_{\rm bath}V$, so
$$
A_{VV}=-\left(\frac{\partial p}{\partial V}\right)_{T,s}
=-\frac1s\left(\frac{\partial p}{\partial v}\right)_T.
$$
On the critical isochore, approached from $T>T_c$,
$$
\left(\frac{\partial p}{\partial v}\right)_{v_c,T}
=-\frac{R(T-T_c)}{4b^2}.
$$
Consequently
$$
\boxed{\langle\delta V^2\rangle
\simeq\frac{4k_Bs b^2T}{R(T-T_c)}
=\frac{4k_B b\bar V T}{3R(T-T_c)}.}
$$
Equivalently, $\langle\delta V^2\rangle=k_BT\kappa_T\bar V$, with $\kappa_T=4b/[3R(T-T_c)]$. \b[The mean-field <variance> diverges as $(T-T_c)^{-1}$, so its <critical exponent> is $1$]; the root-mean-square fluctuation has exponent $1/2$. A fixed spatial observation box instead has number fluctuations, not fluctuating geometric <volume>. Finite size and non-Gaussian terms cut off the literal divergence; the approximation fails exactly at the <critical point>.
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