Thermodynamic stability conditions

Remarks

By the principle of minimum energy, a stable equilibrium requires $U$ to be convex in $(S, V)$, and local stability in the strict sense is positive definiteness of the Hessian. The proof computes the Hessian entries. First, $U_{SS} = (\partial T / \partial S)_V = T / C_V$, which is positive if and only if $C_V > 0$. Second, the Helmholtz free energy $F(T, V)$ is the Legendre transform of $U$ in $S$, and $F_{VV} = U_{VV} - U_{SV}^2 / U_{SS} = \det(\text{Hessian}) / U_{SS}$. Since $F_{VV} = -(\partial p / \partial V)_T = 1/(V \kappa_T)$, the determinant is positive if and only if $\kappa_T > 0$, given $U_{SS} > 0$.

Consequences are $C_p \ge C_V > 0$ and $\kappa_T \ge \kappa_S > 0$, through the identities $C_p - C_V = T V \alpha^2 / \kappa_T$ and $\kappa_T - \kappa_S = T V \alpha^2 / C_p$, where $\alpha$ is the thermal expansion coefficient. Example: an ideal gas has $C_V = \tfrac{3}{2} nR > 0$ (monatomic case) and $\kappa_T = 1/p > 0$, so all its states are stable. The van der Waals isotherms below the critical temperature contain a stretch with $(\partial p / \partial V)_T > 0$, that is $\kappa_T < 0$; these states are unstable, and the fluid separates into liquid and vapour instead.

Going deeper

For a student

The conditions are the second-order part of the extremum principles of thermodynamics: entropy is concave and energy is convex in the extensive variables. Each Legendre transform turns convexity in one variable into concavity in its conjugate, which is why $F$ is concave in $T$ and convex in $V$. Where the conditions fail, the system is unstable against splitting into phases, and the boundary $(\partial p / \partial V)_T = 0$ is the spinodal; the Maxwell construction then gives the actual coexisting phases. With more variables, such as several species, the same statement holds with the full Hessian of $U$ in all extensive variables but one.