Obtain the critical exponent $\beta$ for the van der Waals model, as follows. To find the liquid-vapour density difference we need to find the locations of the points $A$ and $B$ indicated in the Maxwell construction shown on Figure 15.6. Let the two volumes be $V_A$ and $V_B$, then since they are at the same pressure on the same isotherm, we may write
$$
p=\frac{8 T}{3 V_A-1}-\frac{3}{V_A^2}=\frac{8 T}{3 V_B-1}-\frac{3}{V_B^2} .
$$
Solve this equation for $T$, to obtain
$$
T=\frac{\left(3 V_A-1\right)\left(3 V_B-1\right)}{8 V_A^2 V_B^2}\left(V_B+V_A\right)
$$
At the critical point we have $V_A=V_B=1$ and then the equation gives $T=1$ as expected. In general we would need the full Maxwell construction to find $V_A$ and $V_B$, but now argue that near the critical point the two volumes must fall equally either side of 1 , i.e. $V_A=1-x$ and $V_B=1+x$ for some $x$ (this can also be shown by expanding the equation of state to third order in powers of $V-1$ and looking at the limit when $T \rightarrow 1$ ). Expand the equation for small $x$ and thus find $T \sim 1-x^2 / 4$, which implies $\left(V_v-V_l\right) \propto\left(T-T_c\right)^{1 / 2}$. Obtain the density difference, and hence the exponent $\beta$ in equation (25.9).