Past exam of the thermal and statistical physics course of the University of Cambridge / 2014 / Paper 1 / 4

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,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.

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: $p_v=p_{vv}=0$. Thuswhich giveFor fluctuations about a stable equilibrium value $x_0$, the system plus reservoir has probability proportional to $e^{-A(x)/(k_{B}T)}$. Expanding availability,gives a Gaussian distribution and, to this order, $⟨x⟩=x_0$ withThis 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 $A(V)=F_s(V,T)+p_{\mathrm{bath}}V$, soOn the critical isochore, approached from $T>T_c$,ConsequentlyEquivalently, $⟨\delta V^2⟩=k_{B}T{\kappa }_T\bar{V}$, with ${\kappa }_T=4b/[3R(T-T_c)]$. 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.