Van der Waals critical point ($(v_c,T_c,p_c)$)

The liquid-gas critical point of the Van der Waals equation occurs where its two subcritical stationary points merge into a horizontal inflection point. The conditions are $(\partial p/\partial v{)}_T=({\partial }^{2}p/\partial v^2{)}_T=0$. The short calculation isDividing these relations gives $2/(v_c-b)=3/v_c$, henceSubstitution gives the Van der Waals law of corresponding states and the parameter-independent ratio $p_c{v}_c/(k_B{T}_c)=3/8$. That ratio belongs to this equation of state; reducing a real fluid's variables does not enforce it.^[ref]

An exact rearrangement of the reduced critical Van der Waals isotherm makes its flatness visible:The first nonzero departure near $x=1$ is $-3(x-1{)}^3/2$. On the critical particle volume approached from $t>1$, $(\partial q/\partial x{)}_{x=1}=-6(t-1)$ andThe divergent isothermal compressibility and cubic flattening explain why the critical region is mechanically soft. These are mean-field theory predictions; thermal fluctuations limit their use for real fluids very close to a thermodynamic critical point.

For molar equation $p=RT/(v-b)-a/v^2$, the critical values are $v_c=3b$ and $T_c=8a/(27Rb)$. At $v=v_c$, $(\partial p/\partial v{)}_T=-R(T-T_c)/(4b^2)$. Thus ${\kappa }_T=-v_c^{-1}(\partial v/\partial p{)}_T=4b/[3R(T-T_c)]$ on the stable branch above $T_c$. The mean-field compressibility exponent is one; the continuation below the critical temperature is unstable rather than an equilibrium negative compressibility.