= Van der Waals critical point
{c}
{title2=$(v_c,T_c,p_c)$}
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 $(\partial p/\partial v)_T=(\partial^2p/\partial v^2)_T=0$. The short calculation is
$$
-\frac{k_BT_c}{(v_c-b)^2}+\frac{2a}{v_c^3}=0,\qquad
\frac{2k_BT_c}{(v_c-b)^3}-\frac{6a}{v_c^4}=0.
$$
Dividing these relations gives $2/(v_c-b)=3/v_c$, hence
$$
v_c=3b,\qquad T_c=\frac{8a}{27bk_B},\qquad p_c=\frac{a}{27b^2}.
$$
Substitution gives the <Van der Waals law of corresponding states> and the parameter-independent ratio $p_cv_c/(k_BT_c)=3/8$. That ratio belongs to this <equation of state>; reducing a real <fluid>'s variables does not enforce it.<Statistical Physics by David Tong>[§5.1.3, equations 5.7–5.8: critical values, reduced equation and compressibility ratio]{ref}
An exact rearrangement of the reduced critical <Van der Waals isotherm> makes its flatness visible:
$$
q(x,1)-1=-\frac{3(x-1)^3}{x^2(3x-1)}.
$$
The first nonzero departure near $x=1$ is $-3(x-1)^3/2$. On the critical particle <volume (thermodynamics)> approached from $t>1$, $(\partial q/\partial x)_{x=1}=-6(t-1)$ and
$$
\kappa_T=\frac{1}{6p_c(t-1)}.
$$
The 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>.
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