 Solution

The fixed common temperature, rigid external container and fixed particle numbers on either side make the total Helmholtz free energy the appropriate thermodynamic potential. Let the barrier have cross-sectional area $A$, and let increasing $x$ expand the left compartment. Then $d{V}_1=Adx$ and $d{V}_2=-Adx$. At fixed temperature and particle number, $d{F}_j=-p_{j}d{V}_j$, soThus mechanical equilibrium, $p_1=p_2$, makes $F$ stationary. The physical force on the barrier towards increasing $x$ is $f=A(p_1-p_2)=-F^{\prime }(x)$. If the barrier is displaced slightly from $x_{\ast }$, strict mechanical stability requires a restoring force:ConsequentlyThe Taylor expansion $F(x_{\ast }+\delta x)=F(x_{\ast })+\frac{1}{2}{F}^{\prime \prime }(x_{\ast })\delta x^2+O(\delta x^3)$ proves that thermodynamic equilibrium is a strict local minimum of the Helmholtz free energy, rather than merely a stationary point.

For individually stable gases, the positive isothermal compressibilities ${\kappa }_{T,j}=-(1/V_j)(\partial V_j/\partial p_j{)}_{T,N_j}$ make the sign especially transparent:If these inequalities hold throughout the accessible interval, $F$ is a strictly convex function and an interior stationary point is the unique global minimum of the Helmholtz free energy. Stability at one point alone proves only the local statement. A vanishing restoring derivative describes marginal stability and requires examining higher-order terms.