 Solution

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.