Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 4 / 34D (Statistical Physics)

The Gibbs free energy is $G=U-TS+pV$. For an extensive thermodynamic system, scaling the amount of matter at fixed pressure, temperature and composition scales $G$ by the same factor, hence $G(T,p,N)=N\mu (T,p)$. Euler's relation $U=TS-pV+\mu N$ gives the same conclusion. This uses the ordinary thermodynamic-limit extensivity assumption, rather than a universal claim about finite-size or long-range-interacting systems.

The fundamental thermodynamic identity gives $dG=-SdT+Vdp+\mu dN$. Comparing with $d(N\mu )$ at fixed $N$ yieldsThus the chemical potential changes with pressure at a rate equal to the volume per particle.

In the Van der Waals equation, $b>0$ accounts for excluded volume per particle, while $a>0$ accounts for attractive interactions that reduce the pressure relative to a hard-core ideal gas. Writing $v=V/N$ gives $p=k_{B}T/(v-b)-a/v^2$. At the critical point the first two volume derivatives vanish:Dividing gives $v_c=3b$ and thenAbove $T_c$ the isotherms decrease monotonically; at $T_c$ there is a horizontal inflection; below $T_c$ the formal curve develops a local minimum and maximum. The physically equilibrated isotherm replaces part of that loop by a constant-pressure coexistence segment.

At a fixed subcritical temperature, coexistence requires equal pressure and equal chemical potential of the liquid and gas. The Maxwell equal-area construction chooses $p_0$ and the two outer intersection volumes $V_l,V_g$ so thatIndeed $d\mu =(V/N)dp$, and integration by parts gives $N({\mu }_g-{\mu }_l)=p_0(V_g-V_l)-{\int }_{V_l}^{V_g}p(V)dV$, so the equal-area condition is exactly equality of chemical potentials.

Let $V_{s,l}<V_{s,g}$ be the two spinodal volumes where $p^{\prime }(V)=0$. A homogeneous phase is mechanically unstable on $V_{s,l}<V<V_{s,g}$, since $p^{\prime }>0$ gives negative compressibility. The negative-slope branches $V_l<V<V_{s,l}$ and $V_{s,g}<V<V_g$ are metastable homogeneous liquid and gas respectively: they resist small perturbations but a two-phase state has lower free energy. The single-phase stable branches are $V<V_l$ and $V>V_g$. Throughout the coexistence interval $[V_l,V_g]$, the equilibrium state is a mixture at pressure $p_0$ with its phase fractions chosen to produce the total volume. The preceding figure labels both the formal homogeneous branches and the equilibrium coexistence segment.

Use reduced variables $t=T/T_c$, $v=V/V_c$ and $P=p/p_c$, so $P=8t/(3v-1)-3/v^2$. Write $t=1-\tau$, $v=1+s$. Near the critical point,Set $s=\sqrt{\tau }z$. The leading departure from the coexistence pressure is ${\tau }^{3/2}(6z-3z^3/2)$. Equal pressure at the two outer endpoints together with the Maxwell equal-area condition selects the symmetric pair $z_l=-2$, $z_g=2$, with coexistence pressure $1-4\tau +O({\tau }^2)$. To see why, a horizontal line through the odd cubic has zero signed area between its outer equal-height roots only at zero height: the signed area is strictly decreasing with line height, and symmetry makes it zero there. ThusThis proves the coexistence exponent one half.

Along the critical isotherm $T=T_c$ itself, the expansion instead givesThe cubic scaling therefore has exponent three, with a negative coefficient. A fixed noncritical isotherm does not pass through the critical point, so that qualification is essential.