The Maxwell Relations

No entropy meter exists. Yet you can measure how volume changes with temperature at constant pressure. You only need a ruler and a thermometer. The Maxwell relations let you trade the unmeasurable for the measurable. These four equations are not new physics. They are bookkeeping, guaranteed by the mathematics of exact differentials. This mathematics applies to the four thermodynamic potentials. Because $U$ , $H$ , $A$ , and $G$ are state functions, their mixed second partial derivatives must agree. That single fact of calculus generates four exact, testable relations. These relations connect $P$ , $V$ , $T$ , and $S$ . This article derives all four relations from the fundamental equations. You will see the general recipe for building a Maxwell relation from any potential. Then you will apply the relations to real numerical problems. Finally you will see where they lead next. They point toward equations of state and pressure dependence of the Gibbs energy.

Key takeaways

The Maxwell relations come from equating mixed partial derivatives of $U$ , $H$ , $A$ , $G$ .
Each relation trades an entropy derivative for a mechanical one you can measure directly.
They require only that the potential be a state function with an exact differential.
Applications include finding $(\partial S/\partial V)_T$ and $(\partial S/\partial P)_T$ without a calorimeter.
They feed directly into deriving thermodynamic equations of state and the pressure dependence of $G$ .
Key idea: if a quantity is a true state function, the order of differentiation cannot matter, and that constraint is a working formula.

The core insight

Because $dU$ , $dH$ , $dA$ , and $dG$ are exact differentials, the equality of mixed second partial derivatives, $\partial^2 U/\partial S\partial V = \partial^2 U/\partial V\partial S$ , converts abstract exactness into four concrete, experimentally checkable relations linking $P$ , $V$ , $T$ , $S$ .

The physical picture

Every thermodynamic potential is a function of its own natural variables. Internal energy $U(S,V)$ , enthalpy $H(S,P)$ , Helmholtz energy $A(T,V)$ , and Gibbs energy $G(T,P)$ each generate a fundamental equation from the first and second laws. Each fundamental equation is an exact differential: a tiny change in the potential can be split into two independent pieces, one for each natural variable, with well-defined coefficients. The deep fact of multivariable calculus is that for any well-behaved function $f(x,y)$ , differentiating first with respect to $x$ then $y$ gives the same answer as differentiating first with respect to $y$ then $x$ . Thermodynamic potentials, built from state functions, obey this automatically. Applying it to $U$ , $H$ , $A$ , $G$ produces the four Maxwell relations. Each one links an entropy derivative, hard to measure, to a $P$ – $V$ – $T$ derivative, easy to measure with standard lab equipment. The picture below shows the "thermodynamic square," a classic mnemonic. Each potential sits between the two natural variables it depends on, and each side of the square encodes one Maxwell relation.

Figure

The thermodynamic square: potentials, natural variables, and the four Maxwell relations they generate

The core relations

\left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V

where: derived from $U(S,V)$ ; both sides equal $\partial^2 U/\partial S\,\partial V$ up to sign.

\left(\frac{\partial T}{\partial P}\right)_S = \left(\frac{\partial V}{\partial S}\right)_P

where: derived from $H(S,P)$ ; links adiabatic temperature change to entropy-volume coupling.

\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V

where: derived from $A(T,V)$ ; the most-used relation, giving entropy change from an equation of state.

\left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P

where: derived from $G(T,P)$ ; connects isothermal entropy change to thermal expansion.

Derivation ABuilding a Maxwell relation from a fundamental equation

Goal. Derive the Maxwell relation hidden inside the Helmholtz energy, $A(T,V)$ , and show the general method that produces all four.

Step 1. Write the fundamental equation. Starting from $U = TS - PV$ for a closed system doing only $PV$ work, and combining the first law $dU = \delta q_{rev} - PdV$ with the second law $\delta q_{rev} = TdS$ , we get:

dU = TdS - PdV

Define $A \equiv U - TS$ . Differentiating,

dA = dU - TdS - SdT = TdS - PdV - TdS - SdT

which collapses to the fundamental equation for the Helmholtz energy:

dA = -SdT - PdV

Step 2. Identify the natural variables and read off the coefficients. Because $A = A(T,V)$ , its total differential is by definition

dA = \left(\frac{\partial A}{\partial T}\right)_V dT + \left(\frac{\partial A}{\partial V}\right)_T dV

Comparing term by term with Step 1 gives two identifications:

\left(\frac{\partial A}{\partial T}\right)_V = -S, \qquad \left(\frac{\partial A}{\partial V}\right)_T = -P

Step 3. Impose exactness. Since $A$ is a state function, its differential is exact, so the mixed second partials must agree regardless of order:

\frac{\partial}{\partial V}\left(\frac{\partial A}{\partial T}\right)_V\bigg|_T = \frac{\partial}{\partial T}\left(\frac{\partial A}{\partial V}\right)_T\bigg|_V

Step 4. Substitute the Step 2 results. The left side becomes $-\left(\partial S/\partial V\right)_T$ and the right side becomes $-\left(\partial P/\partial T\right)_V$ :

-\left(\frac{\partial S}{\partial V}\right)_T = -\left(\frac{\partial P}{\partial T}\right)_V

Step 5. Cancel the sign and box the result.

\boxed{\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V}

The identical four-step method, applied to $U(S,V)$ , $H(S,P)$ , and $G(T,P)$ , produces the other three relations listed above. Only the starting fundamental equation and the sign bookkeeping change.

Worked examples

Example 1 — Entropy change of an ideal gas on isothermal expansion. Find $(\partial S/\partial V)_T$ for one mole of an ideal gas, then compute $\Delta S$ for isothermal expansion from 10.0 L to 20.0 L at 298 K. Using the Maxwell relation from Helmholtz energy,

\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V

For an ideal gas, $P = RT/V$ , so at constant $V$ ,

\left(\frac{\partial P}{\partial T}\right)_V = \frac{R}{V}

Therefore $(\partial S/\partial V)_T = R/V$ , independent of temperature — a genuinely non-obvious result obtained without ever measuring entropy directly. Integrating at constant $T$ from $V_1$ to $V_2$ :

\Delta S = \int_{V_1}^{V_2} \frac{R}{V}\,dV = R\ln\frac{V_2}{V_1}

Plugging in numbers, with $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ :

\Delta S = 8.314\ \text{J K}^{-1}\text{mol}^{-1} \times \ln\left(\frac{20.0}{10.0}\right) = 8.314 \times 0.6931 = 5.76\ \text{J K}^{-1}\text{mol}^{-1}

Sanity check: expansion increases accessible microstates, so $\Delta S > 0$ , consistent with the sign obtained. The value also matches the standard ideal-gas formula $\Delta S = nR\ln(V_2/V_1)$ derived independently from statistical mechanics, confirming the Maxwell-relation route.

Example 2 — Isothermal entropy and internal-energy change of a real gas from its equation of state. A gas obeys the truncated virial-type equation of state

P = \frac{RT}{V_m} + \frac{a}{V_m^2}

with $a = -0.220\ \text{Pa m}^6\text{mol}^{-2}$ (a small attractive correction, similar in spirit to a van der Waals term). Find $(\partial S/\partial V_m)_T$ and use it to estimate $\Delta S_m$ for isothermal expansion of the gas at $T = 350$ K from $V_{m,1} = 5.00\times10^{-4}\ \text{m}^3\text{mol}^{-1}$ to $V_{m,2} = 1.00\times10^{-3}\ \text{m}^3\text{mol}^{-1}$ . Then find the corresponding $(\partial U_m/\partial V_m)_T$ , the "internal pressure."

Step 1: entropy derivative. By the Maxwell relation,

\left(\frac{\partial S_m}{\partial V_m}\right)_T = \left(\frac{\partial P}{\partial T}\right)_{V_m} = \frac{R}{V_m}

because the $a/V_m^2$ term is temperature-independent and drops out upon differentiating with respect to $T$ . This is the same functional form as the ideal gas, since the correction term has no explicit $T$ dependence.

Step 2: integrate for $\Delta S_m$ .

\Delta S_m = R\ln\frac{V_{m,2}}{V_{m,1}} = 8.314\ \text{J K}^{-1}\text{mol}^{-1}\times\ln(2.00) = 8.314\times0.6931 = 5.76\ \text{J K}^{-1}\text{mol}^{-1}

Interestingly, because the correction term is $T$ -independent, the entropy change is identical to the ideal-gas value from Example 1, even though the gas is non-ideal.

Step 3: internal pressure from the thermodynamic equation of state. A standard result (itself derived from the same Maxwell relation, foreshadowing the thermodynamic-equation-of-state topic) states:

\left(\frac{\partial U_m}{\partial V_m}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_{V_m} - P

Compute each piece at, say, $V_{m,1} = 5.00\times10^{-4}\ \text{m}^3\text{mol}^{-1}$ , $T=350$ K:

T\left(\frac{\partial P}{\partial T}\right)_{V_m} = T\cdot\frac{R}{V_m} = 350 \times \frac{8.314}{5.00\times10^{-4}} = 350\times16{,}628 = 5.820\times10^{6}\ \text{Pa}

P = \frac{RT}{V_m} + \frac{a}{V_m^2} = \frac{8.314\times350}{5.00\times10^{-4}} + \frac{-0.220}{(5.00\times10^{-4})^2}

P = 5.820\times10^{6} + (-8.80\times10^{5}) = 4.940\times10^{6}\ \text{Pa}

\left(\frac{\partial U_m}{\partial V_m}\right)_T = 5.820\times10^{6} - 4.940\times10^{6} = 8.80\times10^{5}\ \text{Pa} = 0.880\ \text{MPa}

Sanity check: this equals exactly $-a/V_m^2$ , as it must, since for this equation of state the ideal term cancels identically, leaving only the correction term. A negative $a$ (net attraction) should give a positive internal-pressure derivative reflecting extra cohesive energy cost on expansion, matching the algebra above.

Limiting cases & common traps

Situation	What happens	Why
Ideal gas	$(\partial S/\partial V)_T = R/V$ exactly	$P=nRT/V$ has no other $T$ -dependent term
Incompressible solid or liquid	$(\partial V/\partial T)_P \approx 0$	thermal expansion coefficient is small, so $(\partial S/\partial P)_T \approx 0$
Any equation of state with $T$ -independent attractive term	entropy relation reduces to ideal-gas form	derivative with respect to $T$ kills the term
Using the wrong potential's natural variables	relation becomes meaningless or wrong	each relation only holds when the derivatives use that potential's own natural variable pair

The chemist behind it

James Clerk Maxwell is remembered mostly as a physicist for electromagnetism, but his 1871 textbook "Theory of Heat" introduced these thermodynamic identities using an elegant geometric argument based on the properties of a thermodynamic surface. Maxwell recognized that the reciprocity of partial derivatives, ordinary multivariable calculus, was hiding genuine experimental content: quantities nobody could measure directly, like $(\partial S/\partial V)_T$ , could be replaced by quantities anyone could measure with a thermometer, a manometer, and patience. This insight became foundational for chemical thermodynamics because entropy and Gibbs energy calculations for real reacting systems depend on exactly these substitutions.

Origins · 1 minJames Clerk Maxwell and how this was discovered

James Clerk Maxwell

1831–1879 · Scottish physicist who derived these relations from the geometry of the thermodynamic surface, better known for electromagnetic theory

Ask the tutor: Want to see how the internal-pressure formula itself is derived, and how it leads into the thermodynamic equation of state?

Do the Maxwell relations require reversibility?

The fundamental equations like $dU = TdS - PdV$ are written using $\delta q_{rev} = TdS$ , but $U$ , $H$ , $A$ , $G$ are state functions. Their differentials connect initial and final states regardless of the actual path taken. So the relations apply to any process, reversible or not, as long as you are comparing equilibrium states. Reversibility enters only in how the fundamental equation was originally derived, not in the domain where the resulting relation applies.

Why are there exactly four Maxwell relations and not more?

Each relation comes from one thermodynamic potential expressed in its own natural variables: $U(S,V)$ , $H(S,P)$ , $A(T,V)$ , $G(T,P)$ . These are the four potentials commonly built from Legendre transforms of $U$ for a simple closed system with $PV$ work. Additional potentials exist for open systems or systems with extra work terms, like chemical potential terms, and each of those generates its own additional Maxwell-type relation.

How do I remember all four relations quickly?

Use the thermodynamic square mnemonic shown in the figure. Write $A$ , $G$ on top, arranged with $T$ shared between them, and $U$ , $H$ on the sides, arranged with $S$ shared between them. Each potential sits between its two natural variables. Reading around the square with a consistent sign convention regenerates all four relations without memorizing them individually.

Are the Maxwell relations useful for real, non-ideal substances?

Yes, arguably they matter most there. For a real gas or liquid, you rarely have direct entropy data, but you often have precise $P$ – $V$ – $T$ data, from an equation of state or from measured thermal expansion and compressibility coefficients. The Maxwell relations turn that mechanical data into entropy and internal-energy information, which is exactly how the thermodynamic equation of state and pressure-dependence of Gibbs energy are derived next.