Solution
= Solution
At the critical point, $(\partial p/\partial v)_T=(\partial^2p/\partial v^2)_T=0$. The first logarithmic <derivative> gives
$$
\frac{a}{k_BT v^2}=\frac1{v-b}.
$$
At a stationary point the second-derivative condition reduces to
$$
\frac1{(v-b)^2}-\frac{2a}{k_BT v^3}=0.
$$
Combining the equations gives $v_c=2b$, and substitution then gives
$$
\boxed{T_c=\frac{a}{4bk_B}.}
$$