Thermodynamic Potentials via Legendre Transforms

Derivation

Thermodynamic Potentials via Legendre Transforms

D-151 Home PU-203 Threads energy Depends on The Clausius Inequality and Entropy as a State Function

Statement

Starting from the fundamental relation for the internal energy in its natural variables, dU = TdS − PdV, we perform Legendre transformations that exchange one or both of the extensive variables (S, V) for their conjugate intensive variables (T, P). This generates the enthalpy H = U + PV, the Helmholtz free energy F = U − TS, and the Gibbs free energy G = U − TS + PV, each with its exact differential and its own set of natural variables.

Why it matters

The four potentials are not four independent physical facts; they are one function U(S,V) viewed through four coordinate choices. A Legendre transform re-expresses the same information so that the variables you actually control in the laboratory become the independent ones: constant (T,V) selects F, constant (T,P) selects G, constant (S,P) selects H.

Because each transform preserves the full thermodynamic content, the minimum of the appropriate potential at fixed control variables is the criterion for equilibrium, and the second cross-derivatives of each potential give the Maxwell relations. Almost every practical calculation in chemical thermodynamics, phase equilibria, and materials science begins by choosing the potential whose natural variables match the constraints.

Assumptions

Simple compressible system with only PdV work.If other work modes exist (chemical μdN, magnetic −mdB, surface γdA), each adds a term to dU and the transforms multiply into further potentials; the three listed here are then incomplete.

States are equilibrium states and the fundamental relation is exact.The identification dU = TdS − PdV uses the reversible limit of the Clausius result TdS ≥ δQ; for irreversible paths TdS exceeds the heat and the intensive variables are not even defined.

Amount of matter is fixed.We hold N constant so it is a silent parameter. If N varies, every differential gains a +μdN term and G acquires its central role G = μN.

U is a smooth, jointly convex function of its extensive variables.The Legendre transform is single-valued and invertible only where U is convex; at a phase transition convexity fails, the transform develops flat regions or corners, and (T,P) no longer index states uniquely.

Derivation

δQ_rev = TdS, δW_by = PdV

Reversible limit of the Clausius inequality supplies the heat; quasi-static compression supplies the work. B

dU = δQ_rev − δW_by = TdS − PdV

First law with the two substitutions from step 1; U is a state function so this exact differential holds for any path. A

T = (∂U/∂S)_V, P = −(∂U/∂V)_S

Read the coefficients of the exact differential; (S,V) are the natural variables of U and (T,P) their conjugates. B

for f(x) with p ≡ df/dx, g ≡ f − px ⇒ dg = −xdp

General Legendre transform: subtracting the product of a variable and its conjugate swaps which of the pair is independent, since d(px) = pdx + xdp cancels the pdx term. C

H ≡ U + PV

Transform on the (V, −P) pair: subtract V·(∂U/∂V)_S = −PV, i.e. add PV, to trade V for P. B

dH = dU + PdV + VdP = TdS + VdP

Substitute step 2; the ∓PdV terms cancel, leaving natural variables (S,P). A

F ≡ U − TS

Transform on the (S, T) pair: subtract S·(∂U/∂S)_V = TS to trade S for T. B

dF = dU − TdS − SdT = −SdT − PdV

Substitute step 2; the ±TdS terms cancel, leaving natural variables (T,V). A

G ≡ U − TS + PV = H − TS = F + PV

Apply both transforms at once, trading S→T and V→P; equivalently transform H on (S,T) or F on (V,−P). B

dG = dU − TdS − SdT + PdV + VdP = −SdT + VdP

Substitute step 2; both TdS and PdV cancel, leaving natural variables (T,P). A

e.g. S = −(∂F/∂T)_V, P = −(∂F/∂V)_T; ∂²F/∂V∂T ⇒ (∂S/∂V)_T = (∂P/∂T)_V

Reading coefficients of each new differential gives the conjugate relations; equality of mixed second partials of each potential yields the four Maxwell relations. C

Result

dU = TdS − PdV  (S,V)
dH = TdS + VdP  (S,P),  H = U + PV
dF = −SdT − PdV  (T,V),  F = U − TS
dG = −SdT + VdP  (T,P),  G = U − TS + PV

Reading. Each free energy is the internal energy with a "stored" contribution removed so that an intensive variable you can clamp becomes independent. Subtracting TS removes the reversible heat reservoir bookkeeping and hands control to temperature; adding PV removes the atmospheric flow work and hands control to pressure. The sign of each differential coefficient tells you the conjugate variable: e.g. −(∂G/∂T)_P = S and (∂G/∂P)_T = V.

Units check. All four potentials are energies (J). TS: K × (J K⁻¹) = J. PV: Pa × m³ = (N m⁻²)(m³) = N m = J. In each differential, TdS = K × J K⁻¹ = J, PdV = Pa × m³ = J, SdT = J K⁻¹ × K = J, VdP = m³ × Pa = J. Dimensionally consistent throughout.

Limiting cases

Constant S and V: dU = 0 at equilibrium; U is minimized — the isolated-system criterion.
Constant T,V: dF = 0; the Helmholtz free energy is minimized and −ΔF bounds the extractable work.
Constant T,P: dG = 0; Gibbs free energy is minimized — the standard bench-chemistry condition.
Constant S,P (adiabatic, isobaric): dH = 0; enthalpy is minimized and ΔH is the heat at constant pressure.
Incompressible / condensed phase: PV is nearly constant, so H ≈ U and G ≈ F up to a fixed offset.

Breaks when

First-order phase transitions. U(S,V) loses strict convexity across coexistence; the Legendre transform flattens onto a tie-line and (T,P) map to a whole range of volumes, so G is continuous but its first derivatives jump. The transform is then the convex-hull (Maxwell) construction rather than a smooth swap.
Irreversible or non-equilibrium processes. When the path is not quasi-static, TdS > δQ and PdV is not the actual work; the intensive variables T,P are undefined for the system as a whole, so the differentials above no longer identify measurable heat and work — only the endpoint state-function changes survive.
Non-extensive or long-range-interacting systems. For self-gravitating or small systems the entropy is not concave and can be non-additive; the Legendre transform is not invertible and different ensembles cease to be equivalent, so the potentials do not carry the same information.

Failure modes

Writing G = U − TS − PV (wrong PV sign): the sign follows from P = −(∂U/∂V)_S, so the transform adds PV.
Reporting the wrong natural variables, e.g. treating F as a function of (T,P): differentiating a potential with respect to a non-natural variable loses information and gives wrong Maxwell relations.
Confusing −(∂F/∂T)_V = S with +S: the minus sign is intrinsic to the Legendre transform dg = −xdp.
Using dU = TdS − PdV to claim heat = TdS along an irreversible leg; it is exact for the state change but only equals δQ when reversible.
Forgetting the μdN term in open systems, then being surprised that G ≠ Σμ_iN_i.

Discussion

The deep point is geometric. A convex function is completely encoded either by its graph (value versus variable) or by the envelope of its tangent lines (intercept versus slope). The Legendre transform switches between these two descriptions, and thermodynamics exploits both: the extensive description U(S,V) is natural for isolated systems, while the tangent-line description is natural when a reservoir fixes a slope — a heat bath fixes T = (∂U/∂S), a pressure reservoir fixes P = −(∂U/∂V). Choosing the potential is choosing which reservoir you have coupled to.

Each potential also has a variational meaning. If the system exchanges heat and volume with reservoirs at fixed T₀,P₀, the total entropy of system-plus-reservoir is maximized exactly when the system's G = U − T₀S + P₀V is minimized. So the "minimum free energy" rules are not new postulates — they are the second law rewritten from the system's viewpoint after the reservoir degrees of freedom have been eliminated, which is precisely what the Legendre transform accomplishes.

Because each transform leaves a well-defined exact differential, equality of mixed second partials immediately yields the Maxwell relations, one per potential: from F, (∂S/∂V)_T = (∂P/∂T)_V; from G, (∂S/∂P)_T = −(∂V/∂T)_P; and so on. These convert hard-to-measure entropy derivatives into mechanical (P,V,T) measurements — the practical payoff of the whole construction.

There is a statistical-mechanical mirror. The transform between U(S) and F(T) corresponds exactly to the passage from the microcanonical entropy S(E) = k_B ln Ω(E) to the canonical free energy F = −k_BT ln Z via a Laplace transform, Z = ∫ Ω(E) e^−E/k_BT dE. In the thermodynamic limit the Laplace integral is dominated by its saddle point, and the saddle-point (steepest-descent) evaluation reduces the Laplace transform precisely to a Legendre transform. Ensemble equivalence is therefore exactly the statement that S(E) is concave; where it is not, the ensembles inequivalence is the same failure that breaks the thermodynamic Legendre transform.

Common misconceptions. The free energies are not "the energy available to do work" in any universal sense — that identity holds only under specific constraints (−ΔF bounds isothermal work; −ΔG bounds isothermal-isobaric non-expansion work). And subtracting TS does not "remove energy that is unavailable"; it changes which variable is independent. The numerical value of any potential also carries an arbitrary additive constant (and, for F and G, an arbitrary linear-in-T piece from the entropy reference); only differences and derivatives are physical.

Worked examples

Isothermal Gibbs-energy change of an ideal gas. Compress n = 1.00 mol of ideal gas reversibly and isothermally at T = 300 K from P₁ = 1.00 bar to P₂ = 5.00 bar.

Set-up: choose G because T,P are the controlled variables. A

dG = −SdT + VdP = VdP (T const)

The −SdT term vanishes at fixed temperature. A

ΔG = ∫_P₁^P₂ V dP = ∫ (nRT/P) dP = nRT ln(P₂/P₁)

Insert V = nRT/P and integrate; symbols rearranged before numbers. A

ΔG = (1.00)(8.314 J K⁻¹ mol⁻¹)(300 K) ln(5.00)

Substitute values; ln 5.00 = 1.609. A

ΔG = 2494 J × 1.609 ≈ +4.01 × 10³ J = +4.01 kJ

Reading. Compression raises the Gibbs free energy by about 4.0 kJ; the positive sign means work must be done on the gas. Units check. J K⁻¹ mol⁻¹ × mol × K = J, and the logarithm is dimensionless, so ΔG is in joules.

Enthalpy versus internal energy on heating. Heat n = 1.00 mol of a monatomic ideal gas at constant pressure from T₁ = 300 K to T₂ = 400 K, and verify H = U + PV numerically.

Set-up: constant P makes H the natural bookkeeper of heat. A

ΔU = nC_VΔT = n(3/2)R ΔT, ΔH = nC_PΔT = n(5/2)R ΔT

Ideal-gas heat capacities; symbols first. B

Δ(PV) = Δ(nRT) = nR ΔT ⇒ ΔH − ΔU = nR ΔT

For an ideal gas PV = nRT, so the Legendre offset PV changes by nRΔT; this is the identity H = U + PV in difference form. B

ΔU = (1.00)(1.5)(8.314)(100) = 1247 J, ΔH = (1.00)(2.5)(8.314)(100) = 2079 J

Substitute ΔT = 100 K. A

ΔH − ΔU = 2079 − 1247 = 832 J = nRΔT = (8.314)(100) = 831 J ✓

Reading. The 831 J gap is exactly the flow-work term PV that distinguishes enthalpy from internal energy; heating at constant pressure delivers more energy as heat than the internal energy rises, the excess going into pushing back the atmosphere. Units check. mol × J K⁻¹ mol⁻¹ × K = J for every quantity; the two routes agree to rounding.

Problems

Starting from H = U + PV and dU = TdS − PdV, derive dH and state the natural variables of H.
Solution
dH = dU + d(PV) = (TdS − PdV) + (PdV + VdP) = TdS + VdP. The −PdV and +PdV cancel, so H depends naturally on (S,P), with T = (∂H/∂S)_P and V = (∂H/∂P)_S.
From dU = TdS − PdV, derive the Maxwell relation associated with U.
Solution
Here T = (∂U/∂S)_V and −P = (∂U/∂V)_S. Because U is a state function, ∂²U/∂V∂S = ∂²U/∂S∂V, i.e. (∂T/∂V)_S = (∂(−P)/∂S)_V. Hence (∂T/∂V)_S = −(∂P/∂S)_V.
Compute ΔG for n = 2.00 mol of ideal gas compressed isothermally at T = 298 K from 1.00 bar to 10.0 bar.
Solution
At constant T, ΔG = nRT ln(P₂/P₁) = (2.00)(8.314)(298) ln(10.0). Now nRT = 4955 J and ln 10.0 = 2.303, so ΔG = 4955 × 2.303 ≈ +1.141 × 10⁴ J = +11.4 kJ. Positive, as expected for compression.
Show that G = U − TS + PV is the double Legendre transform of U, and hence find the conjugate relations for S and V from dG.
Solution
dG = dU − TdS − SdT + PdV + VdP. Substituting dU = TdS − PdV cancels both TdS and PdV, giving dG = −SdT + VdP. Both extensive variables have been traded for their intensive conjugates, so G = G(T,P). Reading coefficients: S = −(∂G/∂T)_P and V = +(∂G/∂P)_T.
Using G = H − TS and S = −(∂G/∂T)_P, derive the Gibbs–Helmholtz relation (∂(G/T)/∂T)_P = −H/T².
Solution
From G = H − TS and S = −(∂G/∂T)_P we get G = H + T(∂G/∂T)_P. Now differentiate G/T: (∂(G/T)/∂T)_P = (1/T)(∂G/∂T)_P − G/T². Substitute G = H + T(∂G/∂T)_P: the (1/T)(∂G/∂T) terms cancel, leaving −H/T². This lets equilibrium constants' temperature dependence be tied directly to reaction enthalpies.