Equilibrium as Free-Energy Minimization

Derivation

Equilibrium as Free-Energy Minimization

D-152 Home PU-203 Threads energy · matter Depends on The Clausius Inequality and Entropy as a State Function, Thermodynamic Potentials via Legendre Transforms

Statement

For a system exchanging heat with a reservoir at fixed temperature T₀ and doing work only against a fixed external pressure p₀, the Clausius inequality forces the availability A = U − T₀S + p₀V to decrease in every spontaneous process, dA ≤ 0, so that at equilibrium the thermodynamic potential appropriate to the imposed constraints is stationary and minimal: entropy S is maximised at fixed (U,V,N); internal energy U at fixed (S,V,N); Helmholtz F = U − TS at fixed (T,V,N); and Gibbs G = U − TS + pV at fixed (T,p,N).

Why it matters

The second law in the form dS_univ ≥ 0 governs the isolated whole. But experiments are almost never run on isolated systems — they are run in thermostats and at atmospheric pressure. This derivation converts the global entropy criterion into a variational principle expressed entirely in the system's own variables, so that "which way does it go, and where does it stop?" becomes "minimise the right potential."

Every quantitative equilibrium condition in chemistry, materials science, and soft matter — phase coexistence, the law of mass action, equality of chemical potentials, the common-tangent construction — is a corollary of one of these minimisations. It is also the bridge to statistical mechanics, where the same F emerges as −k_BT ln Z.

Assumptions

The reservoir is arbitrarily large.Its temperature T₀ and pressure p₀ stay constant as the system exchanges heat and volume with it. If dropped, T₀ and p₀ drift, A ceases to be a function of the system alone, and one must minimise the total entropy of system+reservoir explicitly.

Work is purely p–V against the external pressure.đW = p₀ dV. If other work modes act (magnetic −m dB, surface γ d𝒜, electrical φ dq), each adds a term and the minimised potential is the correspondingly transformed one.

The reservoir temperature, not the instantaneous system temperature, enters the Clausius inequality.The bound is dS ≥ đQ/T₀, where T₀ is the temperature of the body supplying the heat. If dropped and one writes T of a non-uniform system, the inequality is unjustified during irreversible stages.

The potential F, G uses the system's own intensive variable, which equals the reservoir's only at the endpoints.Identifying T₀S with d(TS) needs the system to be internally equilibrated so that a single T exists and equals T₀. Along an irreversible interior path F(T) is not even defined; only the availability A is monotone throughout.

The relevant constraints are actually held fixed by the environment.Minimising G presumes the walls conduct heat and transmit pressure but block the constrained variable (e.g. N). Choosing the wrong potential for the true boundary conditions gives the wrong equilibrium.

Derivation

dS ≥ đQ / T₀

Clausius inequality for a system receiving heat đQ from a reservoir at temperature T₀; equality only for a reversible exchange. Prior result: clausius-inequality-and-entropy. A

đQ = dU + đW = dU + p₀ dV

First law with the only work being expansion against the constant external pressure p₀. A

T₀ dS ≥ dU + p₀ dV

Multiply step 1 by T₀ > 0 (inequality preserved) and substitute step 2 for đQ. A

dU + p₀ dV − T₀ dS ≤ 0

Collect on one side; the sign flips relative to step 3 because a term is moved across. This is the master inequality. B

dA ≡ d(U − T₀S + p₀V) = dU − T₀ dS + p₀ dV

Because T₀ and p₀ are constants, they may be moved inside the differential; the right-hand side is exactly the master inequality. Hence dA ≤ 0. The availability is a monotonically non-increasing Lyapunov function for the approach to equilibrium. B

fixed T,V: dV=0, T=T₀ ⇒ dF = d(U−TS) = dU − T dS ≤ 0

Specialise the master inequality: rigid diathermal walls set dV=0 and pin the system temperature to T₀, so A reduces to the Helmholtz free energy up to the constant p₀V. Legendre-transform structure from thermodynamic-potentials-legendre-transforms. C

fixed T,p: T=T₀, p=p₀ ⇒ dG = d(U−TS+pV) = dU − T dS + p dV ≤ 0

Now the diathermal wall also transmits pressure, so A is precisely the Gibbs free energy. Both F and G therefore fall until they can fall no further. C

at equilibrium: dΦ = 0 (extremum), d²Φ > 0 (minimum)

A monotone-decreasing bounded-below function halts where its first variation vanishes; the second law's inequality guarantees the stationary point is a minimum (a maximum would be unstable to fluctuations). Φ denotes whichever potential the constraints select. B

Result

dA = dU − T₀ dS + p₀ dV ≤ 0 ⇒ δΦ = 0, δ²Φ > 0 at equilibrium

Reading. The one master inequality dA ≤ 0 contains every equilibrium rule. Whichever pair of variables the environment clamps, the availability collapses onto the matching potential — S (max), U, F, H, or G — and that potential is driven to a minimum (or S to a maximum). Equilibrium is not a special force balance; it is the exhaustion of the system's capacity to lower its free energy.

Units check. Each of U, T₀S, and p₀V is an energy: [T][S] = K · J K⁻¹ = J, and [p][V] = Pa · m³ = (N m⁻²)(m³) = J. So A, F, G are all in joules and dA ≤ 0 compares like with like.

Limiting cases

Isolated system (đQ=0, dV=0): step 4 gives −T₀ dS ≤ 0, i.e. dS ≥ 0. Entropy maximisation is recovered as the special case.
Fixed S,V: dS=0, dV=0 ⇒ dU ≤ 0. Internal energy is minimised — the mechanical intuition "systems seek lowest energy" holds only at fixed entropy.
Fixed S,p: dS=0 ⇒ dU+p₀dV = dH ≤ 0. Enthalpy is minimised (the natural potential for insulated, pressure-controlled flow).
Reversible limit: the Clausius equality holds, dA=0 exactly; the potential is constant along the path — the system is at every instant infinitesimally close to equilibrium.
High-temperature limit of F=U−TS: the −TS term dominates, so minimising F becomes maximising S; at low T it becomes minimising U. Free energy interpolates the energy–entropy competition.

Breaks when

The reservoir is finite. If the surroundings are comparable in heat capacity to the system, T₀ and p₀ change during equilibration; A is no longer a state function of the system and one must return to maximising the total entropy S_sys+S_res.
The system is far from equilibrium with no well-defined temperature. A strongly driven or steeply non-uniform body has no single T, so F and G are undefined and the Clausius bound in step 1 loses its reservoir reference.
Driven, open, or dissipative steady states. A system fed by external fluxes (a laser above threshold, a Bénard cell, living matter) settles into a steady state that does not minimise any equilibrium free energy; minimum-entropy-production or other criteria govern it instead.
Mesoscopic systems. When thermal fluctuations are comparable to the potential's curvature (ΔΦ ∼ k_BT), the minimum is only a most-probable value; the state fluctuates and fluctuation theorems, not a sharp minimum, describe it.
Long-range or non-additive interactions. Gravitating systems or unscreened Coulomb systems have non-extensive energy; ensembles become inequivalent, F need not be convex, and the canonical minimisation can disagree with the microcanonical result.

Failure modes

Wrong temperature in Clausius. Using the instantaneous (possibly ill-defined) system temperature instead of the reservoir T₀ in step 1, then wondering why the bound seems to be violated.
Potential/constraint mismatch. Claiming G is minimised at fixed T,V, or F at fixed T,p. Each potential is minimal only when its own natural variables are the ones held constant.
Sign inversion. Writing dF ≥ 0 or "free energy increases toward equilibrium," usually from mishandling the sign flip in step 4.
đQ as an exact differential. Treating heat as dQ with a state function Q; only đQ/T₀ is bounded by the exact dS.
Keeping the pV term when volume is fixed. Carrying G into a rigid-container problem where F is the correct potential, so the constant p₀V is mistaken for physics.
Confusing entropy max with free-energy min. Believing the two criteria conflict; they are the same second law seen through different constraints.

Discussion

The deep content is that a single global principle — total entropy never decreases — wears different clothes depending on what the laboratory holds fixed. The availability A = U − T₀S + p₀V is really the negative of the total entropy change of system+reservoir multiplied by T₀: dA = −T₀ dS_univ. So "minimise A" and "maximise S_univ" are literally the same statement, re-expressed in variables the experimenter can read off a gauge. The Legendre transforms that build F, H, G from U are exactly the bookkeeping that trades a controlled extensive variable for its conjugate intensive one.

The structure F = U − TS makes the physics vivid: equilibrium is a treaty between energy, which wants to be low, and entropy, which wants to be high, brokered at a rate set by temperature. Ice melts not because water is lower in energy — it is higher — but because above 273 K the entropic gain TΔS outweighs the enthalpic cost. Every phase diagram is a map of which competitor wins where.

Setting the first variation to zero yields the working equilibrium conditions. For a Gibbs system open to particle exchange, ∂G/∂N_i = μ_i = 0 at fixed constraints reproduces equality of chemical potentials across phases and the law of mass action; the common-tangent construction is d²G ≥ 0 (convexity) made geometric. The second-variation condition d²Φ > 0 is the origin of the stability criteria C_V > 0 and κ_T > 0.

The identification dA = −T₀ dS_univ also exposes the availability (exergy) as the maximum useful work extractable before the system equilibrates with its environment: W_useful,max = −ΔA. In statistical mechanics the same F reappears non-phenomenologically as F = −k_BT ln Z, and its minimisation is the large-deviation statement that the equilibrium macrostate is the one of overwhelmingly greatest microstate weight; the Bogoliubov–Gibbs inequality F ≤ F₀ + ⟨H−H₀⟩₀ is the variational form used throughout mean-field theory.

Common misconceptions. (i) "Systems always seek minimum energy" — true only at fixed entropy; at fixed T they seek minimum free energy, which can mean higher U. (ii) "Free energy minimisation contradicts the second law's entropy increase" — it is the second law, for a subsystem in contact with a reservoir. (iii) "ΔG < 0 means the reaction is fast" — it fixes direction and extent, never rate; kinetics is a separate question governed by barriers.

Worked examples

Two-level system: free-energy minimisation gives the Boltzmann factor. One particle with states of energy 0 and ε, upper-state probability p, at fixed T. Minimise F = ⟨U⟩ − TS.

Constraints are fixed T and (implicitly) volume, so the Helmholtz potential is the one to minimise (step 6). A

⟨U⟩ = pε, S = −k_B[p ln p + (1−p) ln(1−p)]

Mean energy and Gibbs entropy of a two-outcome distribution; symbols first. A

dF/dp = ε + k_BT ln[ p/(1−p) ] = 0 ⇒ p/(1−p) = e^−ε/k_BT

Set the first variation to zero (step 8); d²F/dp² = k_BT/[p(1−p)] > 0 confirms a minimum. B

ε = 0.025 eV, T = 300 K ⇒ k_BT = (8.617×10⁻⁵ eV K⁻¹)(300 K) = 0.02585 eV

Numbers substituted only now; ε/k_BT = 0.967. A

p/(1−p) = e^−0.967 = 0.380 ⇒ p = 0.380/1.380 = 0.275

Solve the ratio for the probability. A

p ≈ 0.28

Reading. Minimising the Helmholtz free energy of a single two-level particle reproduces the Boltzmann distribution from pure thermodynamics; about 28% upper-state occupation at room temperature for a 25-meV gap. Units check. The exponent ε/k_BT = eV/eV is dimensionless, as it must be; p is a pure number.

Melting of ice: Gibbs minimisation selects the stable phase. Compare G_solid and G_liquid for water at p = 1 atm using ΔG_fus = ΔH_fus − TΔS_fus (solid→liquid).

Fixed T,p, so the stable phase is the one of lower G (step 7). A

ΔH_fus = 6010 J mol⁻¹, ΔS_fus = ΔH_fus/T_m = 6010/273.15 = 22.0 J mol⁻¹ K⁻¹

At the melting point the two phases coexist, so ΔG=0 there fixes ΔS_fus. B

ΔG_fus(T) = ΔH_fus − TΔS_fus = 6010 − 22.0 T [J mol⁻¹]

Treat ΔH, ΔS as constant over a narrow range; symbolic form before numbers. A

T = 263 K: ΔG_fus = 6010 − 22.0×263 = 6010 − 5786 = +224 J mol⁻¹

Positive ΔG for melting means liquid has the higher G; the solid is favoured. A

T = 283 K: ΔG_fus = 6010 − 22.0×283 = 6010 − 6226 = −216 J mol⁻¹

Negative ΔG for melting means liquid has the lower G; the liquid is favoured. A

ΔG_fus(−10°C) = +224 J mol⁻¹ (solid wins) | ΔG_fus(+10°C) = −216 J mol⁻¹ (liquid wins)

Reading. The system always slides to the phase of lower Gibbs energy; the sign of ΔG flips through zero exactly at T_m = 273.15 K, which is the coexistence line. Units check. TΔS = K · (J mol⁻¹ K⁻¹) = J mol⁻¹, matching ΔH; ΔG is a molar energy.

Problems

(A) Fixed S,V potential. Starting from the master inequality dU + p₀dV − T₀dS ≤ 0, show which potential is minimised when entropy and volume are held fixed.
Solution
Set dS = 0 and dV = 0. The inequality collapses to dU ≤ 0. So the internal energy itself is the relevant potential and it is minimised at equilibrium. This is the mechanical-energy intuition, valid only under the adiabatic (fixed S), rigid (fixed V) constraints.
(B) Gibbs from availability. Show explicitly that for a diathermal, pressure-transmitting boundary (fixed T,p) the availability equals the Gibbs energy up to a constant, and hence dG ≤ 0.
Solution
With the system equilibrated to the reservoir, T = T₀ and p = p₀. Then A = U − T₀S + p₀V = U − TS + pV = G. Differentiating with T,p constant, dA = dU − TdS + pdV = dG. Since dA ≤ 0 (step 5), dG ≤ 0, minimised at equilibrium.
(B) Two-level occupation at higher temperature. For the two-level system of Worked Example 1 (ε = 0.025 eV), find the upper-state probability at T = 600 K, and comment on the trend.
Solution
k_BT = (8.617×10⁻⁵)(600) = 0.0517 eV, so ε/k_BT = 0.025/0.0517 = 0.484. Ratio p/(1−p) = e^−0.484 = 0.616, giving p = 0.616/1.616 = 0.381 ≈ 0.38. Doubling T raises occupation from 0.28 toward the infinite-temperature limit p = 0.5, because the entropic (−TS) term in F increasingly favours the disordered equal-population state.
(B) Reaction direction and equilibrium constant. A reaction A ⇄ B has ΔG° = −5.0 kJ mol⁻¹ at T = 298 K. Find the equilibrium constant K = exp(−ΔG°/RT) and the equilibrium ratio [B]/[A].
Solution
RT = (8.314)(298) = 2477 J mol⁻¹ = 2.477 kJ mol⁻¹. Exponent −ΔG°/RT = +5.0/2.477 = 2.018. K = e^2.018 = 7.5. For the ideal A⇄B, [B]/[A] = K = 7.5, so about 88% B at equilibrium. The negative ΔG° drives the reaction toward B until the reaction Gibbs energy ∂G/∂ξ reaches zero.
(C) Exergy of a hot block. A block of heat capacity C = 500 J K⁻¹ starts at T₁ = 400 K in an environment at T₀ = 300 K (constant pressure, negligible volume change). Using dA = −T₀dS_univ, find the maximum useful work −ΔA extractable as it cools to T₀.
Solution
The block's energy change: ΔU = C(T₀ − T₁) = 500(300−400) = −5.00×10⁴ J. Its entropy change: ΔS = C ln(T₀/T₁) = 500 ln(300/400) = 500(−0.2877) = −143.8 J K⁻¹. With negligible pΔV, ΔA = ΔU − T₀ΔS = −5.00×10⁴ − 300(−143.8) = −5.00×10⁴ + 4.31×10⁴ = −6.9×10³ J. The maximum useful (reversible) work is −ΔA = 6.9 kJ. Note this is far less than the 50 kJ of heat released: the environment "taxes" the extraction through the T₀ΔS term, which is the availability/exergy statement of the second law.