Thermodynamic Potentials
Roll a ball into a bowl and it settles at the bottom: nature, left alone, slides downhill in
energy. That single instinct — equilibrium sits at a minimum — is the backbone
of thermodynamics too. But it hides a trap. A chemist mixing reagents in an open beaker on the bench,
a gas sealed in a rigid flask, a star's core held at fixed entropy — each reaches equilibrium, yet
they are not all minimising the same energy. Which "energy" a system minimises
depends entirely on what you are holding fixed while it settles.
A gas free to soak up heat from the room and to push its piston outward does not
minimise its internal energy — it happily borrows energy from the surroundings if that lets it spread
out. What it actually minimises is a different bookkeeping quantity, one that already has the heat
traffic and the pushing-back of the atmosphere folded in. These tailored quantities are the
thermodynamic potentials, and this page is about the family of four —
internal energy, enthalpy, Helmholtz free energy, and Gibbs free energy — how they
are built from one another, and the beautiful web of identities (the Maxwell relations)
that falls out for free once you have them.
The one idea to carry the whole way: for every set of experimental constraints there is a
potential that is minimised at equilibrium, and the four potentials are just the same
physics re-expressed in whichever variables you actually control in the lab.
Which knobs are you holding? — the beaker on the bench
Picture a reaction fizzing in an open beaker. The lab's thermostat pins the temperature
T; the atmosphere pins the pressure p.
Those are the two knobs held fixed — not the entropy, not the volume. Heat flows freely across the
glass and the liquid level rises or falls against the air. Under those constraints the
quantity that runs downhill to a minimum is the Gibbs free energy
G. That is why chemists live in "G-land": constant T
and p is the natural habitat of bench chemistry, and a reaction is
spontaneous exactly when it lowers G.
Seal that same reaction in a rigid, insulated bomb instead — fix the volume
V and let no heat escape (fix entropy S) — and
now the potential that is minimised is the plain internal energy
U. Put it in a rigid flask in a heat bath (fix T
and V) and it is the Helmholtz free energy
F that bottoms out. Same molecules, same chemistry — three different
"energies" minimised, because you clamped three different pairs of knobs.
- Fixed S, V, N: internal energy U is minimised.
- Fixed S, p, N: enthalpy H is minimised.
- Fixed T, V, N: Helmholtz free energy F is minimised.
- Fixed T, p, N: Gibbs free energy G is minimised.
The four potentials and their fundamental differentials
Each potential is a function of its own natural variables — the very knobs held
fixed when it is the one being minimised. Written in those variables, the total differential
of each potential is exact and clean. Start from the first law wrapped up in the combined statement
for a simple substance:
dU = T\,dS - p\,dV + \mu\,dN,
so the natural variables of U are (S, V, N).
Reading the coefficients straight off gives the intensive quantities:
T = \left(\partial U/\partial S\right)_{V,N},
-p = \left(\partial U/\partial V\right)_{S,N}, and the
chemical potential \mu = \left(\partial U/\partial N\right)_{S,V}.
The other three potentials are built by adding or subtracting a product of a conjugate pair:
-
Internal energy U(S,V,N):
dU = T\,dS - p\,dV + \mu\,dN.
-
Enthalpy H = U + pV, with
H(S,p,N):
dH = T\,dS + V\,dp + \mu\,dN.
The heat absorbed at constant pressure.
-
Helmholtz free energy F = U - TS, with
F(T,V,N):
dF = -S\,dT - p\,dV + \mu\,dN.
The maximum useful work extractable at constant temperature.
-
Gibbs free energy G = U - TS + pV = H - TS = F + pV,
with G(T,p,N):
dG = -S\,dT + V\,dp + \mu\,dN.
The chemist's potential; for a single component G = \mu N.
Notice the pattern in the differentials. Every time you swap a variable — say entropy
S for temperature T — the term
+T\,dS turns into -S\,dT: the coefficient and
the differential trade places and pick up a minus sign. Swap volume V for
pressure p and -p\,dV becomes
+V\,dp. That is no accident — it is a Legendre transform,
and it is the engine that generates the whole family.
Legendre transforms: trading a variable for its slope
The internal energy U(S,V,N) knows everything, but it is written in
S — a quantity you cannot dial on a bench. What you can control is
T, the slope of U with respect to
S. A Legendre transform rewrites a function in terms of
its slope instead of its original variable, without losing any information.
The recipe is mechanical. To swap a natural variable for its conjugate
(S\leftrightarrow T are a conjugate pair, as are
V\leftrightarrow p), subtract or add the product of that conjugate
pair:
-
Trade S for T: form
F = U - TS. Then
dF = dU - T\,dS - S\,dT = -S\,dT - p\,dV + \mu\,dN — the
T\,dS cancels and S is gone, replaced by
T.
-
Trade V for p: form
H = U + pV, giving
dH = T\,dS + V\,dp + \mu\,dN.
-
Do both trades: G = U - TS + pV = H - TS = F + pV, whose
natural variables are the two bench knobs (T, p, N).
So the family is a little lattice of transforms — two independent trades, giving
2\times 2 = 4 potentials:
U \;\xrightarrow{\,+pV\,}\; H \;\xrightarrow{\,-TS\,}\; G, \qquad U \;\xrightarrow{\,-TS\,}\; F \;\xrightarrow{\,+pV\,}\; G.
Both routes from U arrive at the same G — the
transforms commute. This is exactly the same operation that turns a Lagrangian into a Hamiltonian in
mechanics: give up a variable, gain its slope.
A picture that remembers all of it: the thermodynamic square
There is a lovely mnemonic — the thermodynamic square (Guggenheim's square) — that
packs all four potentials, their natural variables, their differentials, and the Maxwell
relations into one little diagram. Each edge carries a potential; the two
corners flanking that edge are its natural variables. The corners on each diagonal
are conjugate pairs: S opposite
T, and V opposite p.
Reveal it step by step.
To read off a differential — say for F on the right edge — its flanking
corners are V and T, so
dF is built from dT and
dV. Each differential is multiplied by the variable
diagonally opposite its corner (T opposite
S, V opposite p),
with signs read from the arrows, recovering
dF = -S\,dT - p\,dV. The mnemonic sentence for the border letters,
starting at bottom-middle and going round, is "Good Physicists Have Studied Under Very Fine
Teachers": G, p, H, S, U, V, F, T.
Maxwell relations: entropy on the cheap
Here is where the potentials earn their keep. Each differential
d\Phi is an exact differential (a genuine change in a
state function), and for an exact differential the mixed second partial derivatives are
equal — the order of differentiation does not matter. Apply that to
dF = -S\,dT - p\,dV: since
-S = (\partial F/\partial T)_V and
-p = (\partial F/\partial V)_T, equating
\partial^2 F/\partial V\,\partial T = \partial^2 F/\partial T\,\partial V
gives
\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial p}{\partial T}\right)_V.
Do the same for each potential and you get the four Maxwell relations:
- From U:
\left(\dfrac{\partial T}{\partial V}\right)_S = -\left(\dfrac{\partial p}{\partial S}\right)_V.
- From H:
\left(\dfrac{\partial T}{\partial p}\right)_S = \left(\dfrac{\partial V}{\partial S}\right)_p.
- From F:
\left(\dfrac{\partial S}{\partial V}\right)_T = \left(\dfrac{\partial p}{\partial T}\right)_V.
- From G:
\left(\dfrac{\partial S}{\partial p}\right)_T = -\left(\dfrac{\partial V}{\partial T}\right)_p.
Why is this a big deal? The left-hand sides all involve entropy — a quantity you
cannot stick a meter on. The right-hand sides involve only pressure, volume and
temperature — the easy, mechanical, everyday measurables. Maxwell relations let you
trade a derivative of entropy you can never measure directly for a derivative of the equation of
state you can. For an ideal gas pV = nRT, for instance,
(\partial p/\partial T)_V = nR/V, so immediately
\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial p}{\partial T}\right)_V = \frac{nR}{V},
telling you exactly how the entropy of a gas grows as it expands isothermally — deduced without ever
measuring entropy at all.
Worked examples
Example 1 — squeezing the equation of state out of F. For
an ideal gas one can show (via the partition function — see
the partition function)
that the Helmholtz free energy has the form
F(T,V,N) with
\left(\partial F/\partial V\right)_T = -nRT/V. Because
dF = -S\,dT - p\,dV, the pressure is just the (negative) volume slope:
p = -\left(\frac{\partial F}{\partial V}\right)_T = \frac{nRT}{V},
which is the ideal-gas law, recovered by differentiating a single potential. Likewise the entropy is
minus the temperature slope, S = -\left(\partial F/\partial T\right)_V. One
potential, and every thermodynamic property tumbles out.
Example 2 — a Maxwell relation in anger. Find how the entropy of one mole of ideal
gas changes with volume at fixed temperature. Direct measurement is hopeless, so use the
F-Maxwell relation:
\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial p}{\partial T}\right)_V.
From p = nRT/V with n = 1, the right side is
R/V. Integrating at fixed T from
V_1 to V_2 gives
\Delta S = R\ln(V_2/V_1) — the familiar entropy of isothermal expansion,
pulled straight from the equation of state.
Example 3 — a Legendre mini-exercise. You are handed
U(S,V,N) and asked for the potential natural to constant pressure. You want
to trade V for p, so add the conjugate product
pV:
H = U + pV, \qquad dH = dU + p\,dV + V\,dp = T\,dS + V\,dp + \mu\,dN.
The -p\,dV and +p\,dV cancel, leaving
dH written in dS and dp.
So the enthalpy's natural variables are (S, p, N), and at constant pressure
(dp = 0) with no particle change, dH = T\,dS = \delta Q
— enthalpy change is the heat absorbed at constant pressure, which is why reaction enthalpies
are what a bench calorimeter reads.
Two phases, one winner: minimising G foreshadows melting
Because a system at fixed T and p minimises
G, whichever phase of a substance has the lower Gibbs free energy
is the stable one. Since G = H - TS and
(\partial G/\partial T)_p = -S, a plot of
G against temperature is a line whose slope is
-S: the higher-entropy phase (the liquid) falls off more steeply. The two
lines cross at the melting temperature T_m, and below it
the solid wins, above it the liquid wins.
This is the seed of the whole theory of phase transitions: the substance always
occupies the lower-G branch, jumping from one to the other where the lines
cross. The kink in the stable-phase curve at T_m — where the slope
(-S) jumps — is precisely the entropy of melting, the signature of a
first-order transition.
Three classic exam-wreckers hide here. First, the constraint mix-up. Helmholtz
F is the potential minimised at constant T and
V; Gibbs G is the one for constant
T and p. Reaching for F
in an open beaker (constant pressure) — or G in a sealed rigid vessel — is
the single most common slip. Match the potential to the knobs you are holding.
Second, the signs and the subscripts. In
dG = -S\,dT + V\,dp the entropy term is negative and the volume
term positive; flip a sign and every Maxwell relation you derive comes out wrong. And the
subscript on a partial derivative is not decoration:
(\partial S/\partial V)_T (temperature fixed) and
(\partial S/\partial V)_p (pressure fixed) are genuinely different
numbers. Always name what is held constant.
Third, "free energy decreases" is not "energy is conserved". Energy is always
conserved (first law). The statement that G or F
decreases toward equilibrium is a statement about spontaneity (the second law) — it
is entropy of the universe increasing, viewed through the system's variables. A falling free energy
does not mean energy is vanishing; it means the combined system-plus-surroundings entropy is rising.
Chemistry happens in open flasks on benches at room temperature and atmospheric pressure — constant
T, constant p. That is precisely the domain
where G is minimised, so the sign of
\Delta G tells a chemist whether a reaction will go, and
\Delta G = \Delta H - T\Delta S neatly splits "does it release heat?"
from "does it increase disorder?". Split the reaction into products and reactants and the condition
for equilibrium is that the chemical potentials balance — and for a pure substance
\mu = G/N, so G and
\mu are almost the same idea.
Josiah Willard Gibbs (1839–1903) was a famously quiet Yale professor who worked out
essentially this entire framework — potentials, chemical potential, phase rule, the statistical
ensembles — in two long papers buried in the Transactions of the Connecticut Academy, a
journal almost nobody read. Maxwell in Britain grasped their importance immediately and even built a
plaster thermodynamic-surface model to send him. Gibbs published little, taught patiently, drew no
salary for his first decade at Yale, and quietly became one of the greatest scientists America ever
produced. The G you minimise on every bench is his.