Past exam of the physics B course of the University of Cambridge / 2019 / Paper 2 / Section C / C10

Pressure and its physical terms. Solving the molar Van der Waals equation for pressure givesThe excluded-volume parameter $b$ reduces the volume available for translational motion, raising the kinetic pressure relative to $RT/V$. The attraction parameter $a$ lowers the pressure because attractive interactions reduce the momentum transferred to the boundary. This mean-field model treats $a,b$ as temperature-independent material constants. For one mole, $V$ is the sample volume with the molar amount already fixed; $b$ has units of that volume and $a$ of pressure times volume squared.

volume dependence of entropy. The Helmholtz free energy has differential $dF=-SdT-PdV$. Equality of mixed derivatives yields the Maxwell relationFor $a=0$ this derivative is $R/(V-b)$, givingEquivalently, $S=R\ln (V-b)+f(T)$ with a common reference volume understood inside the logarithm. The equation of state fixes this volume dependence, not the otherwise unspecified function of temperature.

Excess entropy at fixed volume and temperature. The ideal gas has $(\partial S_{\mathrm{id}}/\partial V{)}_T=R/V$. ThereforeIntegrate from infinite volume, where the entropy difference is specified to vanish:It is negative because excluding configurations reduces the accessible state count. It tends to zero as $V\to \infty$ and has leading dilute limit $-Rb/V$. This is the entropy excess of a van der Waals gas.

Effect of attraction. Restoring a temperature-independent $a$ changes the pressure by $-a/V^2$, but not its derivative with respect to temperature. With the same infinite-volume entropy reference, it therefore does not change $\Delta S^{\mathrm{ex}}$ at fixed $V,T$. This can also be seen fromThe attraction affects internal energy and the equilibrium volume, so it can affect entropy indirectly when different state points are compared. If $a$ were temperature dependent, its derivative would contribute directly to the entropy; that is outside the stipulated constant-parameter model.

Coexistence isotherm. Below the Van der Waals critical point, the formal homogeneous isotherm has a local minimum and maximum. Its middle branch has $\partial P/\partial V>0$, hence negative isothermal compressibility, and is mechanically unstable. Portions of the negative-slope branches can be metastable rather than globally stable. The equilibrium isotherm replaces the loop between liquid volume $V_l$ and vapour volume $V_g$ by a constant saturation pressure $P_s$. The two volume are fixed byThe Maxwell equal-area construction expresses equality of the two phases' chemical potentials. Intermediate total volume are mixtures: changing the vapour fraction changes total volume at the same pressure, rather than following the unstable homogeneous curve. The spinodals are the turning points of the formal isotherm; they are distinct from the coexistence endpoints.

Dense-phase volume at low temperature. At a general subcritical temperature, the two coexistence equations above must be solved together. For this equation of state the area condition is explicitlyThis supplies an exact implicit determination of the liquid volume even when a simple closed form is unavailable.

For the asymptotically low-temperature limit, put $\tau =RTb/a\ll 1$. The liquid is dense, $V_l\simeq b$, while the vapour is dilute, $P_s{V}_g\simeq RT$ and $V_g\gg V_l$. The area condition then givesThus the vapour volume grows exponentially in $1/\tau$, and the saturation pressure is exponentially small. It is consequently consistent to neglect $P_s$ in the liquid equation to algebraic order in $\tau$:Choose the root near $b$, not the large-volume zero-pressure root:Expanding gives the leading requested dense-phase behavior,This is the low-temperature liquid volume of a van der Waals gas. Setting $P_s=0$ is an asymptotic approximation, not an exact coexistence condition at every temperature below $T_c$. The figure illustrates generic subcritical coexistence; its finite-temperature liquid volume is determined by the full equal-area equations. The van der Waals liquid also ceases to be a realistic description at sufficiently low physical temperature, where other phases or quantum effects can matter.