Helmholtz work inequality derivation

Setup: Isolated system at constant $T$ (in contact with heat bath) and constant $V$ . System starts and ends at same $V, T$ . Only interaction with surroundings is heat exchange.

Energy conservation:

W + Δ E_{sys} - Q = 0 ⟹ W = Q - Δ E_{sys}

Entropy change of bath: $Δ S_{bath} = - Q / T$

By the 2nd law ( $Δ S_{e} x t t o t a l \geq 0$ ):

Δ S_{sys} \geq \frac{Q}{T}

equality iff reversible.

Free energy change:

Δ F_{sys} = Δ E_{sys} - T Δ S_{sys} \leq Δ E_{sys} - T \cdot \frac{Q}{T} = Δ E_{sys} - Q = - W

Result:

W \leq - Δ F

equality iff reversible.

Corollary ( $W = 0$ ): $Δ F \leq 0$ — at constant $T, V$ with no work, free energy never increases.