Past exam of the mathematics course of the University of Cambridge / 2017 / ii / Paper 4 / 34D (Statistical Physics)

For a simple homogeneous thermodynamic system, internal energy is extensive: $E(\lambda S,\lambda V,\lambda N)=\lambda E(S,V,N)$. Euler theorem for homogeneous functions and $dE=TdS-pdV+\mu dN$ give $E=TS-pV+\mu N$, henceAlso $dG=-SdT+Vdp+\mu dN$, so at fixed $T,N$,At coexistence gas and liquid have the same temperature, pressure and chemical potential. Integrating $d\mu =vdp$ along the intervening isotherm gives $\int vdp=0$. Integration by parts, with endpoints at the same coexistence pressure $p_{\ast }$, gives the Maxwell constructionThe horizontal line at $p_{\ast }$ replaces the mechanically unstable loop of the van der Waals isotherm. The extensivity argument presupposes the ordinary thermodynamic limit and neglects macroscopic interface contributions.

The liquid-gas critical point is the endpoint of the gas-liquid coexistence curve where the distinct phases merge. For this equation of state it is the horizontal inflection of an isotherm, so $p_v=p_{vv}=0$:Dividing these conditions gives $v_c=3b$, then substitution givesAll lie in the physical domain $v>b$.

Substitute $v=3b\bar{v}$, $T=(8a/(27bk))\bar{T}$, and $p=(a/(27b^2))\bar{p}$ into the equation of state. After cancellation of the material parameters,This is the Van der Waals law of corresponding states: fluids described by this model share the same reduced equation, although real fluids need not obey it exactly.

The expansion about the liquid-gas critical point isDifferentiating at fixed $t$ gives the requested $-9{\varphi }^2/2+O({\varphi }^3)+t[-6+O(\varphi )]$. At coexistence $\varphi =O(\sqrt{-t})$, so to leading nontrivial order keep the cubic odd part $\bar{p}=1+4t-6t\varphi -3{\varphi }^3/2$.

Let $a={\varphi }_l$, $b={\varphi }_g$ for this leading cubic model. Equality of endpoint pressures gives $a^2+ab+b^2=-4t$. The Maxwell condition in pressure coordinates is ${\int }_a^b(1+\varphi ){\bar{p}}^{\prime }(\varphi )d\varphi =0$. Since the endpoint pressures agree, its first term vanishes, leavingIf $a+b\ne 0$, these two equations would imply $(a-b{)}^2=0$, contrary to distinct phases. Therefore $a=-b$, and the pressure equality gives $a^2=b^2=-4t$. HenceThus the coexistence gap has exponent one half. The printed equality ${\varphi }_l(t)=-{\varphi }_g(t)$ is a leading-order statement, not an exact identity for the full Van der Waals equation. Discarded asymmetric terms shift both phase volumes at higher order. This distinction is needed even though the leading cubic calculation has exact antisymmetry within its truncation.