Spring-gas equilibrium derivation

Setup: Spring (force constant $k$ ) + gas in heat bath. Piston at displacement $x$ from equilibrium; gas volume $V = V_{1} + A x$ where $A$ is piston area.

Equilibrium condition: $\frac{\partial F}{\partial x} = 0$

\frac{\partial F}{\partial x} = \frac{\partial E}{\partial x} - T \frac{\partial S}{\partial x}

Energy term — gas is at constant $T$ , so its energy doesn't change with $x$ . Only the spring contributes:

\frac{\partial E}{\partial x} = F_{piston} = - k x

Entropy term — piston has no entropy; only the gas contributes:

\frac{\partial S}{\partial x} = \frac{\partial S}{\partial V} \frac{\partial V}{\partial x} = \frac{P}{T} \cdot A = \frac{F_{gas}}{T}

Substituting:

\frac{\partial F}{\partial x} = F_{piston} - F_{gas} = 0 ⟹ F_{piston} = F_{gas}

Result: Minimising free energy is equivalent to balancing forces.