= Solution
\b[<Pressure> and its physical terms.] Solving the molar <Van der Waals equation> for <pressure> gives
$$
\boxed{P=\frac{RT}{V-b}-\frac a{V^2},\qquad V>b.}
$$
The excluded-volume parameter $b$ reduces the <volume (thermodynamics)> 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 (thermodynamics)> with the molar amount already fixed; $b$ has units of that <volume (thermodynamics)> and $a$ of <pressure> times <volume (thermodynamics)> squared.
\b[<volume (thermodynamics)> dependence of <entropy>.] The <Helmholtz free energy> has differential $dF=-S\,dT-P\,dV$. Equality of mixed <derivatives> yields the <Maxwell relation>
$$
\left(\frac{\partial S}{\partial V}\right)_T
=\left(\frac{\partial P}{\partial T}\right)_V.
$$
For $a=0$ this <derivative> is $R/(V-b)$, giving
$$
\boxed{S(V,T)-S(V_*,T)=R\ln\frac{V-b}{V_*-b},\qquad V,V_*>b.}
$$
Equivalently, $S=R\ln(V-b)+f(T)$ with a common reference <volume (thermodynamics)> understood inside the <logarithm>. The <equation of state> fixes this <volume (thermodynamics)> dependence, not the otherwise unspecified function of <temperature>.
\b[Excess <entropy> at fixed <volume (thermodynamics)> and <temperature>.] The <ideal gas> has $(\partial S_{\mathrm{id}}/\partial V)_T=R/V$. Therefore
$$
\left(\frac{\partial\Delta S^{\mathrm{ex}}}{\partial V}\right)_T
=R\left(\frac1{V-b}-\frac1V\right).
$$
Integrate from infinite <volume (thermodynamics)>, where the <entropy> difference is specified to vanish:
$$
\Delta S^{\mathrm{ex}}(V,T)
=\int_\infty^V R\left(\frac1{v-b}-\frac1v\right)dv
=\boxed{R\ln\left(1-\frac bV\right).}
$$
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>.
\b[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 from
$$
F_{\mathrm{ex}}=-RT\ln(1-b/V)-\frac aV,
\qquad S_{\mathrm{ex}}=-\left(\frac{\partial F_{\mathrm{ex}}}{\partial T}\right)_V.
$$
The attraction affects <internal energy> and the equilibrium <volume (thermodynamics)>, 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.
\b[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 (thermodynamics)> $V_l$ and vapour <volume (thermodynamics)> $V_g$ by a constant saturation <pressure> $P_s$. The two <volume (thermodynamics)> are fixed by
$$
P(V_l,T)=P(V_g,T)=P_s,
\qquad
\boxed{\int_{V_l}^{V_g}[P(V,T)-P_s]dV=0.}
$$
The <Maxwell equal-area construction> expresses equality of the two phases' chemical potentials. Intermediate total <volume (thermodynamics)> are mixtures: changing the vapour fraction changes total <volume (thermodynamics)> 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.
\Image[/past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2019/paper-2-c10-coexistence.png]
{title=Original subcritical van der Waals isotherm showing the equal-area plateau, coexistence volumes and spinodals}
\b[Dense-phase <volume (thermodynamics)> 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 explicitly
$$
RT\ln\frac{V_g-b}{V_l-b}
+a\left(\frac1{V_g}-\frac1{V_l}\right)
=P_s(V_g-V_l).
$$
This supplies an exact implicit determination of the <liquid> <volume (thermodynamics)> even when a simple closed form is unavailable.
For the asymptotically low-temperature limit, put $\tau=RTb/a\ll1$. The <liquid> is dense, $V_l\simeq b$, while the vapour is dilute, $P_sV_g\simeq RT$ and $V_g\gg V_l$. The <area> condition then gives
$$
\ln\frac{V_g}{V_l-b}\simeq1+\frac{a}{RTV_l}.
$$
Thus the vapour <volume (thermodynamics)> 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 $\tau$:
$$
0\simeq\frac{RT}{V_l-b}-\frac a{V_l^2}
\quad\Longrightarrow\quad RTV_l^2-aV_l+ab\simeq0.
$$
Choose the root near $b$, not the large-volume zero-pressure root:
$$
\boxed{V_l\simeq\frac a{2RT}\left(1-\sqrt{1-\frac{4RTb}{a}}\right)
=\frac{2b}{1+\sqrt{1-4\tau}}.}
$$
Expanding gives the leading requested dense-phase behavior,
$$
\boxed{V_l=b+\frac{RTb^2}{a}+O(b\tau^2),\qquad T\to0\ \text{within the model}.}
$$
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 (thermodynamics)> 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.
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