The Helmholtz Energy

Consider a closed system held at fixed volume, in thermal contact with a huge surroundings at temperature $T$. No work of expansion is possible because the volume cannot change, but the internal energy $U$ can still change as heat flows across the boundary. The second law says entropy is what decides whether a change happens spontaneously, but entropy is a property of the universe — system plus surroundings — not of the system alone. That makes it awkward to use directly for predicting the behavior of a beaker sitting on a bench. The trick is to rewrite the second law entirely in terms of system properties. If the system exchanges heat $dq$ with surroundings at temperature $T$, the surroundings' entropy changes by $-dq/T$. The Clausius inequality states that for any process,

At constant volume with no non-expansion work, the first law gives $dq = dU$. Substituting and rearranging turns a statement about total entropy into a statement about $U$ and $S$ of the system alone: $dU - T\,dS \le 0$. Define $A \equiv U - TS$. At constant $T$, $dA = dU - T\,dS$, so the condition becomes $dA \le 0$. This is the entire logical content of the Helmholtz energy: it is a bookkeeping device that folds the surroundings' entropy change into a single system-only function. Physically, $TS$ is the portion of the internal energy that is "entropy-locked" — energy the system must release as heat merely to keep its entropy consistent with contact at temperature $T$. Subtracting $TS$ from $U$ leaves the fraction of the energy that is, in principle, convertible to work under isothermal conditions. That is why $A$ has historically been called the work function (Helmholtz's own term, "Arbeit" — German for "work" — is the source of the symbol $A$).

Now picture the approach to equilibrium itself. At fixed $T$ and $V$, a system evolves so that $A$ falls continuously and flattens out at a minimum — the equilibrium state. Any further fluctuation away from that point would raise $A$, which the second law forbids, so the system sits there.

where: $U$ is internal energy, $T$ is absolute temperature, $S$ is entropy — all properties of the system.

where: this is the natural differential form for a closed system doing only $pV$ work, showing $A$ is naturally a function of $T$ and $V$.

where: these partial derivatives recover entropy and pressure directly from $A(T,V)$.

where: equality holds at equilibrium; strict inequality signals a spontaneous, irreversible change.

where: $w_{\max}$ is the maximum total work (expansion plus any other form) obtainable from a process at constant temperature; the system does work when $\Delta A$ is negative. ##

Example 1 — Computing ΔA for an isothermal ideal-gas expansion. One mole of an ideal gas expands isothermally and reversibly at $T = 298\text{ K}$ from $V_1 = 10.0\text{ L}$ to $V_2 = 20.0\text{ L}$. Find $\Delta U$, $\Delta S$, $\Delta A$, and the work done. For an ideal gas, $U$ depends only on $T$. Since $T$ is constant,

The reversible isothermal entropy change of an ideal gas is

With $n = 1\text{ mol}$, $R = 8.314\text{ J K}^{-1}\text{mol}^{-1}$, and $V_2/V_1 = 2.00$:

Now

Check via work: reversible isothermal work for an ideal gas is

The magnitudes match exactly, since for an ideal gas at constant $T$, $\Delta A = w_{\text{rev}}$ (heat $q = T\Delta S = +1717\text{ J}$ flows in, and $\Delta U = q + w = 1717 - 1717 = 0$, consistent). The negative sign means the gas does 1.72 kJ of work on the surroundings — exactly the maximum work available, confirming $w_{\max} = \Delta A$.

Example 2 — Spontaneity and A for a phase transition away from the normal transition temperature. Supercooled liquid water sits at $-5.0\,^\circ\text{C}$ ($T = 268.15\text{ K}$) at constant volume and constant temperature (imagine a rigid, thermostatted vessel). Freezing releases $\Delta H_{\text{fus}} = -6.01\text{ kJ mol}^{-1}$ at the normal melting point $T_m = 273.15\text{ K}$, with $\Delta S_{\text{fus}} = \Delta H_{\text{fus}}/T_m = -22.0\text{ J K}^{-1}\text{mol}^{-1}$ for freezing. Treating $\Delta H$ and $\Delta S$ as roughly temperature-independent over this small range, determine whether freezing is spontaneous at $-5.0\,^\circ\text{C}$ under constant $T,V$ conditions, using $\Delta A$. At constant volume the appropriate work-content criterion is $\Delta A$, and since $pV$ changes are negligible for a nearly incompressible liquid/solid pair, $\Delta A \approx \Delta U - T\Delta S \approx \Delta H - T\Delta S$ (ignoring the small $\Delta(pV)$ term, valid to a few J for condensed phases).

Since $\Delta A < 0$, freezing is spontaneous at $-5.0\,^\circ\text{C}$ under these constant $T,V$ conditions — consistent with everyday experience that supercooled water freezes readily once nucleated. As a sanity check, at exactly $T_m = 273.15\text{ K}$,

confirming equilibrium (melting point) is correctly recovered as the point where $\Delta A \to 0$, and that a few degrees of supercooling is enough to make freezing clearly favorable.

Hermann von Helmholtz was a physician-turned-physicist whose 1847 memoir on the conservation of energy helped crystallize the first law of thermodynamics into rigorous mathematical form. Decades later, in the 1880s, he introduced the free energy function that bears his name, showing how to separate the "free" (work-available) part of a system's energy from the part inevitably dissipated as heat at a given temperature. His broader scientific range — from nerve conduction speed to acoustics to the physiology of vision — reflects a 19th-century conviction that energy conservation was a truly universal principle, applicable equally to muscles, batteries, and chemical reactions.

Ask the tutor: How does the Helmholtz energy connect to the Gibbs energy when a reaction produces or consumes gas, and which one should you actually use in the lab?