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core26 min studyLesson 317 in Physics

Free energies and natural variables

Use Legendre transforms to match equilibrium potentials to imposed constraints and derive useful-work bounds without confusing a minimum principle with a coordinate derivative.

Before you begin
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The question for today

What you’ll understand

  • Derive the natural differentials of Helmholtz, enthalpy, and Gibbs potentials.
  • Choose a potential for fixed reservoir conditions.
  • Calculate the appropriate upper bound on useful work.

Entropy is a natural energy coordinate, but a laboratory often controls temperature through a bath instead. A pressure reservoir can similarly be easier to specify than a volume. Thermodynamic potentials reorganize the same physical information around those controlled intensive variables. Their definitions are Legendre transforms, not arbitrary combinations chosen because their names sound useful. We will derive the cancellation behind each transform and connect the resulting potential to the entropy balance of a system and the reservoirs that maintain its conditions.

Swap a coordinate for its conjugate

At fixed amount, the energy differential is dU=T dS−p dVdU=T\,dS-p\,dV. Define Helmholtz free energy F=U−TSF=U-TS. Applying the product rule cancels the T dST\,dS term and leaves dF=−S dT−p dVdF=-S\,dT-p\,dV. Its natural coordinates are therefore temperature and volume. The entropy is minus the temperature derivative of F at fixed volume, and pressure is minus its volume derivative at fixed temperature. Re-expressing U merely as U of temperature and volume would not produce these same simple derivative coefficients.

Enthalpy H=U+pVH=U+pV exchanges volume for pressure, while Gibbs free energy G=U−TS+pVG=U-TS+pV exchanges both entropy and volume. The corresponding differentials identify their natural coordinates directly. Amount remains an additional natural variable, with chemical potential as its coefficient. Every potential contains the same equilibrium content when the transforms and coordinate changes are well defined. Which one is convenient depends on the constraints, rather than on one potential being universally more fundamental or always minimized.

F=U−TS,dF=−S dT−p dV+μ dν;H=U+pV,dH=T dS+V dp+μ dν;G=U−TS+pV,dG=−S dT+V dp+μ dνF=U-TS,\quad dF=-S\,dT-p\,dV+\mu\,d\nu;\qquad H=U+pV,\quad dH=T\,dS+V\,dp+\mu\,d\nu;\qquad G=U-TS+pV,\quad dG=-S\,dT+V\,dp+\mu\,d\nu
F naturally uses (T,V,nu), H uses (S,p,nu), and G uses (T,p,nu). Each potential is in joules; the displayed identities connect neighboring equilibrium states of a simple compressible substance.↔ Scroll wide equations horizontally.

A bath converts entropy maximization into a free-energy test

Place a closed system in a fixed-volume container in contact with a bath at constant temperature T. If there is no work exchange, the system’s energy gain is the bath’s energy loss. The bath entropy change is minus that heat divided by T. The combined entropy condition therefore becomes a nonpositive change of U−TSU-TS. Equilibrium minimizes F over accessible internal changes at fixed temperature, volume, and amount, such as a reaction coordinate or an internal phase distribution.

At fixed bath temperature and external pressure, account for boundary pressure-volume work and the analogous quantity is G. A negative Gibbs change can drive an allowed process under those constraints. These minimum statements compare possible internal configurations while the imposed coordinates are fixed; they do not assert that F has positive curvature in its temperature argument or that every nonequilibrium system has a single equilibrium temperature at all times. Natural-variable derivatives and constrained relaxation are distinct uses of the same potential.

ΔF≤Won(T,V,ν fixed),Wby≤−ΔF;WnonPV,by≤−ΔG(T,p,ν fixed)\Delta F\le W_{\mathrm{on}}\quad(T,V,\nu\ \text{fixed}),\qquad W_{\mathrm{by}}\le-\Delta F;\qquad W_{\mathrm{nonPV,by}}\le-\Delta G\quad(T,p,\nu\ \text{fixed})
The first bound allows non-volume work at fixed volume and uses a single thermal bath. The Gibbs bound subtracts the unavoidable pressure-volume work at constant external pressure. Equality requires a reversible complete interaction.↔ Scroll wide equations horizontally.

Available work depends on the environment

At fixed temperature and volume, the entropy-production inequality gives Q≤TΔSQ\le T\Delta S. Combining it with ΔU=Q+Won\Delta U=Q+W_{\mathrm{on}} yields the displayed Helmholtz work bound. If no work is supplied or extracted, F cannot increase. If useful work is extracted, the maximum is the free-energy decrease, not necessarily the internal-energy decrease. The bath can provide or absorb heat while maintaining temperature, so the environment participates in determining what portion of energy change is available mechanically.

At fixed temperature and pressure, a gas or reacting system also does expansion work on the pressure reservoir. The Gibbs decrease bounds the remaining non-pressure-volume output, such as reversible electrical work. Other constraints require a different availability analysis. Reference choices must be applied consistently to all states and species, especially when composition changes. A quoted “free energy released” has a practical meaning only after the boundary, reservoir conditions, allowed exchanges, and the definition of useful work are stated.

Work it through

Useful work from a fixed-volume process

A closed system changes between equilibrium states at 300 K and the same volume. Its internal energy falls by 120 J and its entropy falls by 0.20 J K−10.20\,\mathrm{J\,K^{-1}}. A single 300 K bath is available. Find the Helmholtz change and maximum work output, and give the reversible heat exchange.

  1. The constant-temperature potential change is ΔF=ΔU−TΔS=−120−300(−0.20)=−60 J\Delta F=\Delta U-T\Delta S=-120-300(-0.20)=-60\,\mathrm J. The entropy decrease partly offsets the energy decrease.

  2. The maximum work by the system is −ΔF=60 J-\Delta F=60\,\mathrm J, equivalent to work on it of −60 J. Equality assumes reversible operation including its bath contact.

  3. Reversible heat is Q=TΔS=−60 JQ=T\Delta S=-60\,\mathrm J, so the system rejects heat. The first law checks as −120=−60+(−60)-120=-60+(-60), and the bath gains 0.20 J/K, canceling the system entropy change.

The result

The Helmholtz decrease permits at most 60 J of work output. The full 120 J energy decrease is not all available as work under the specified thermal constraint.

The living notebook

Explore the model

Interactive lab

Interactive lab · optimization

A minimum is defined over allowed changes

Use gradient descent on the fixed illustrative potential x²/2. Reset, set Step size η = 0.5 and Initial position x₀ = 2, then Step once to predict x = 1. For a fresh comparison Reset, set η = 1 and x₀ = 2, then Step once to predict x = 0.

A minimum is defined over allowed changesGradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Current readings: Position x0: 2; Objective x²/2: 2; Iteration behavior: Stable step. All parameters are adjustable with the labeled controls below.−4−202402.174.356.528.7Parameter xf(x) = x²/2n = 0
ObjectiveDiscrete iterate
Position x0
2
Objective x²/2
2
Iteration behavior
Stable step
n = 0 · t = 0 s

Gradient descent on f(x) = x²/2: xₙ = (1 − η)ⁿx₀. Playback advances two discrete iterations per second. Values outside |x| ≤ 4 are explicitly reported. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThis deterministic optimization model illustrates descent toward a minimum. Its step size is numerical, not a bath temperature, and it does not calculate a thermodynamic potential or relaxation rate.

Pause & reason

Check your understanding

Which potential bounds non-pressure-volume work output for a closed process at fixed bath temperature and external pressure?

Your turn

Try it yourself

A closed process at 400 K and constant external pressure has ΔH=−300 J\Delta H=-300\,\mathrm J and ΔS=−0.50 J K−1\Delta S=-0.50\,\mathrm{J\,K^{-1}}. Find ΔG\Delta G and the maximum non-pressure-volume work output.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Use G equals H minus T S, retaining the sign of the entropy change.

Compare with the solution

At constant temperature, ΔG=−300−400(−0.50)=−100 J\Delta G=-300-400(-0.50)=-100\,\mathrm J. The maximum non-PV work by the system is 100 J. For reversible operation the system rejects 200 J of heat to the bath; the remaining enthalpy decrease supplies the useful work after the pressure-volume contribution has been accounted for. Irreversibility reduces this maximum output.

Keep the idea

Take it with you

  • Legendre transforms change natural variables by canceling conjugate differential terms.
  • Minimum principles depend on the constraints and the accessible internal changes.
  • Helmholtz and Gibbs decreases bound different work outputs under different reservoir conditions.

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