= Van der Waals isotherm
{c}
{title2=$p(v,T)$}
A <Van der Waals isotherm> is the <pressure>–particle <volume (thermodynamics)> curve at a fixed <temperature>. Put $x=v/v_c$, $t=T/T_c$, and $q=p/p_c$, using the <Van der Waals critical point> values below. Then the existing <Van der Waals law of corresponding states> becomes
$$
q(x,t)=\frac{8t}{3x-1}-\frac{3}{x^2},\qquad x>\frac13.
$$
All curves in the figure are calculations in this reduced model.
\Image[/van-der-waals-reduced-isotherms.png]
{title=<Van der Waals isotherms> below, at and above the <Van der Waals critical point>}
{id=van-der-waals-reduced-isotherms}
{description=The dashed subcritical homogeneous continuation contains an unstable red segment. The green <Maxwell construction> plateau describes equilibrium <phase coexistence>; the two shaded areas are equal. Curves are model calculations, not <carbon dioxide> measurements.}
For a uniform <thermodynamic phase>, the <isothermal compressibility> is
$$
\kappa_T=-\frac1v\left(\frac{\partial v}{\partial p}\right)_T
=-\frac{1}{v(\partial p/\partial v)_T}.
$$
Thus the upward-sloping portion of a subcritical <Van der Waals isotherm> has $\kappa_T<0$: compression would reduce <pressure>, allowing a density disturbance to grow. It is mechanically unstable. The <spinodal points> bound that portion; setting the <derivative> $\partial q/\partial x$ to zero gives their curve $t=(3x-1)^2/(4x^3)$. Negative slope alone guarantees only local mechanical stability: portions between the <spinodal> and the <phase coexistence> endpoints have <metastability>. Equilibrium replaces the entire interval between those endpoints, including these metastable portions, by a <Maxwell construction> plateau.<Statistical Physics by David Tong>[§5.1, equation 5.3 and §5.1.1, Maxwell Construction and Meta-Stable States: unstable middle branch and coexistence versus spinodal]{ref}
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