Maxwell Relations

Some thermodynamic quantities are easy to measure and some are nearly impossible. You can measure how much a gas expands when you warm it at constant pressure — that is just thermal expansion, an afternoon in the lab. But how does the entropy of that gas change when you squeeze it at constant temperature? Entropy has no meter; there is no probe you can dip into the gas to read it off. It would seem that such a quantity is beyond direct reach. The Maxwell relations are the astonishing statement that it is not: the unmeasurable entropy change equals a perfectly measurable expansion coefficient, and the equality is exact, forced by nothing more than calculus. The mechanism is disarmingly simple. Every thermodynamic potential — $U$ , $F$ , $H$ , $G$ — is a state function, so its differential is exact, which means its mixed second partial derivatives commute: $\partial^2\Phi/\partial x\,\partial y = \partial^2\Phi/\partial y\,\partial x$ . Apply this elementary theorem of calculus to each of the four potentials and out fall four identities relating derivatives of $S$ , $T$ , $P$ , $V$ to one another. That is the entire content: Maxwell relations are the commutativity of mixed partials, dressed in thermodynamic clothing. Their power lies in what they connect. Two of the four relations express derivatives of the entropy — which no instrument reads — in terms of derivatives of the pressure or volume, which any equation of state supplies. This is what makes thermodynamics predictive rather than merely descriptive: measure a few response functions, and the Maxwell relations let you compute quantities you could never measure directly, including the internal pressure that distinguishes a real gas from an ideal one and drives refrigeration by throttling. Because the four potentials are Legendre transforms of one another, the four Maxwell relations are not four independent facts about nature — they are four faces of the same fundamental relation. They also serve as a consistency check: any proposed equation of state, heat capacity, or model must satisfy them, and a violation is a guaranteed sign of a modeling error. They are the exact scaffolding that holds the whole structure of equilibrium thermodynamics rigid.

Key takeaways

Maxwell relations equate cross-derivatives of $S$ , $T$ , $P$ , $V$ ; they follow purely from the exactness of the potentials' differentials (mixed partials commute).
There are four, one from each potential: $U$ gives $(\partial T/\partial V)_S = -(\partial P/\partial S)_V$ ; $F$ gives $(\partial S/\partial V)_T = (\partial P/\partial T)_V$ ; $H$ gives $(\partial T/\partial P)_S = (\partial V/\partial S)_P$ ; $G$ gives $(\partial S/\partial P)_T = -(\partial V/\partial T)_P$ .
Two of them convert unmeasurable entropy derivatives into measurable equation-of-state derivatives.
A key consequence is the internal pressure $(\partial U/\partial V)_T = T(\partial P/\partial T)_V - P$ : zero for an ideal gas, $aN^2/V^2$ for a van der Waals gas.
They are a consistency test: any equation of state must satisfy them.
Key idea: the four Maxwell relations are the four mixed-partial identities of the four thermodynamic potentials, and they convert unmeasurable entropy changes into measurable response functions.

The core insight

Because each thermodynamic potential ( $U, F, H, G$ ) is a state function with an exact differential, its mixed second partial derivatives commute. The four resulting Maxwell relations equate cross-derivatives of $S, T, P, V$ — crucially expressing unmeasurable entropy derivatives in terms of measurable equation-of-state derivatives, and giving the internal pressure $(\partial U/\partial V)_T = T(\partial P/\partial T)_V - P$ .

Where this lives today

The Maxwell relations are among the most heavily used tools in practical thermal science, precisely because they let one build a full thermodynamic description from a handful of measurable quantities. Every real-fluid equation-of-state package — the ones inside process simulators, refrigeration-cycle design software, and the reference formulations for water (IAPWS-95) and industrial refrigerants — relies on Maxwell relations to compute entropy and enthalpy tables from measured $P$ – $V$ – $T$ data, since entropy is never measured directly. The design of every refrigerator and gas-liquefaction plant rests on the internal pressure $(\partial U/\partial V)_T$ and the Joule–Thomson coefficient it feeds, both accessible only through these relations. Beyond gases, the same reciprocity structure generalizes to any conjugate work pair. Replacing $-P\,\mathrm{d}V$ with $B\,\mathrm{d}M$ (magnetic) yields magnetic Maxwell relations that underpin adiabatic-demagnetization cooling to millikelvin temperatures; with $\mathcal{E}\,\mathrm{d}\Pi$ (electric) they describe pyroelectric and electrocaloric materials now being explored for solid-state cooling. In condensed-matter experiments, magnetocaloric entropy changes $\Delta S(T)$ — impossible to measure directly — are routinely extracted from magnetization data $M(T,B)$ via the Maxwell relation $(\partial S/\partial B)_T = (\partial M/\partial T)_B$ , a technique central to the search for room-temperature magnetic refrigeration. Wherever entropy needs to be known but cannot be measured, a Maxwell relation is doing the work.

The physical picture

The idea behind every Maxwell relation is a single fact from multivariable calculus: for a smooth function $\Phi(x,y)$ , it does not matter whether you differentiate first by $x$ then by $y$ , or the other way around — the mixed second derivatives are equal. Thermodynamic potentials are smooth functions of state, so this applies to all of them. The "magic" of relating entropy to expansion is really just this commutativity, applied to $\Phi = F(T,V)$ or $\Phi = G(T,P)$ . To see it concretely: from $\mathrm{d}F = -S\,\mathrm{d}T - P\,\mathrm{d}V$ (at fixed $N$ ) we read off $-S = (\partial F/\partial T)_V$ and $-P = (\partial F/\partial V)_T$ . Differentiate the first by $V$ and the second by $T$ ; commutativity of the mixed partial forces $(\partial S/\partial V)_T = (\partial P/\partial T)_V$ . The left side is how entropy changes when you expand at fixed temperature — unmeasurable. The right side is how pressure changes when you warm at fixed volume — trivially measurable. The equality is exact.

Figure

The thermodynamic square: read off potentials, natural variables, and Maxwell relations

Once you internalize this, the Maxwell relations stop being four formulas to memorize and become one idea with four instances. The thermodynamic square above is the classic mnemonic: place $S, V, T, P$ at the corners and the potentials on the edges, and each Maxwell relation is read off directly, including its sign. The punchline plot below shows why this matters physically — the internal pressure $(\partial U/\partial V)_T$ , computable only through a Maxwell relation, cleanly separates the ideal gas (a flat zero line) from real gases (a nonzero curve), and that separation is exactly what makes real gases cool on expansion.

Figure

Punchline: the internal pressure a/v² separates real gases from the ideal-gas zero line

The core equations

Every Maxwell relation is the commutativity of mixed partials of a potential. Stating the master principle first: for any $C^2$ potential $\Phi(x,y)$ ,

\frac{\partial^2 \Phi}{\partial x\,\partial y} = \frac{\partial^2 \Phi}{\partial y\,\partial x} \qquad (1)

where: $\Phi$ — any thermodynamic potential; $x, y$ — its two natural variables (at fixed $N$ ). This is the Euler reciprocity condition for an exact differential $\mathrm{d}\Phi = M\,\mathrm{d}x + N\,\mathrm{d}y$ : $(\partial M/\partial y)_x = (\partial N/\partial x)_y$ . Applying (1) to each of the four potentials gives the four Maxwell relations. From $U$ ( $\mathrm{d}U = T\,\mathrm{d}S - P\,\mathrm{d}V$ ):

\left(\frac{\partial T}{\partial V}\right)_S = -\left(\frac{\partial P}{\partial S}\right)_V \qquad (2)

From the Helmholtz free energy $F$ ( $\mathrm{d}F = -S\,\mathrm{d}T - P\,\mathrm{d}V$ ) — the workhorse relation that turns entropy into pressure:

\left(\frac{\partial S}{\partial V}\right)_T = \left(\frac{\partial P}{\partial T}\right)_V \qquad (3)

where: the left side is an entropy derivative (no meter), the right side an equation-of-state derivative (easy). This is the relation used most in derivations. From the enthalpy $H$ ( $\mathrm{d}H = T\,\mathrm{d}S + V\,\mathrm{d}P$ ):

\left(\frac{\partial T}{\partial P}\right)_S = \left(\frac{\partial V}{\partial S}\right)_P \qquad (4)

From the Gibbs free energy $G$ ( $\mathrm{d}G = -S\,\mathrm{d}T + V\,\mathrm{d}P$ ) — the relation used most experimentally, since $T$ and $P$ are directly controllable:

\left(\frac{\partial S}{\partial P}\right)_T = -\left(\frac{\partial V}{\partial T}\right)_P \qquad (5)

where: the right side is $-V\alpha$ , with $\alpha = \tfrac1V(\partial V/\partial T)_P$ the measurable thermal expansion coefficient. So the entropy's pressure dependence is just the expansivity. Combining relation (3) with $\mathrm{d}U = T\,\mathrm{d}S - P\,\mathrm{d}V$ gives the internal pressure, the flagship consequence:

\left(\frac{\partial U}{\partial V}\right)_T = T\left(\frac{\partial P}{\partial T}\right)_V - P \qquad (6)

where: for an ideal gas $(\partial P/\partial T)_V = P/T$ makes this vanish; for a van der Waals gas it equals $aN^2/V^2 > 0$ , the energy stored in intermolecular attraction.

Derivation AAll four Maxwell relations from the four potentials

The goal is to derive all four Maxwell relations at once, exhibiting them as the four instances of one calculus theorem — the commutativity of mixed second partials of a state function — rather than four unrelated formulas. For each potential we read off its two first derivatives from its differential, then equate the two ways of taking the mixed second derivative. Doing all four together reveals the pattern (and its sign structure, captured by the thermodynamic square) and makes clear that these are four faces of one fundamental relation, not four independent statements about nature.

Step 1. From $\mathrm{d}U = T\,\mathrm{d}S - P\,\mathrm{d}V$ : $T = (\partial U/\partial S)_V$ and $-P = (\partial U/\partial V)_S$ . Equating $\partial^2 U/\partial V\,\partial S = \partial^2 U/\partial S\,\partial V$ gives $(\partial T/\partial V)_S = -(\partial P/\partial S)_V$ — relation (2). Why: differentiating $T$ by $V$ and $-P$ by $S$ produces the same mixed partial of $U$ .

Step 2. From $\mathrm{d}F = -S\,\mathrm{d}T - P\,\mathrm{d}V$ : $-S = (\partial F/\partial T)_V$ and $-P = (\partial F/\partial V)_T$ . Commuting gives $(\partial S/\partial V)_T = (\partial P/\partial T)_V$ — relation (3), the entropy-to-pressure workhorse. Why: the same mixed partial of $F$ computed two ways, with the two minus signs cancelling to leave a plus.

Step 3. From $\mathrm{d}H = T\,\mathrm{d}S + V\,\mathrm{d}P$ : $T = (\partial H/\partial S)_P$ , $V = (\partial H/\partial P)_S$ , giving $(\partial T/\partial P)_S = (\partial V/\partial S)_P$ — relation (4). And from $\mathrm{d}G = -S\,\mathrm{d}T + V\,\mathrm{d}P$ : $-S = (\partial G/\partial T)_P$ , $V = (\partial G/\partial P)_T$ , giving $(\partial S/\partial P)_T = -(\partial V/\partial T)_P$ — relation (5). Why: each potential's two first derivatives, cross-differentiated, yield one relation; the signs track which variable carries the minus in the differential.

Step 4. The sign pattern ( $+,-,+,-$ across $F, U/G$ ) is exactly encoded by the thermodynamic (Born) square: place $S, V$ on one diagonal and $T, P$ on the other, and each relation reads off with its sign from the arrows. The four are not independent — since $U, F, H, G$ are Legendre transforms of one another, they are four projections of the single fundamental relation. Why: the square is a mnemonic for the shared origin, and the shared origin is why any equation of state must satisfy all four simultaneously. *Verified: all four Maxwell relations are checked symbolically for a general $C^2$ potential and explicitly for the ideal-gas potentials (matching cross-derivatives on both sides) in verify_aq-sm-maxwell-relations.py.*

Go deeper · Derivation B · 1 minThe internal pressure and what distinguishes a real gas

Derivation BThe internal pressure and what distinguishes a real gas

The goal is to derive the "thermodynamic equation of state" $(\partial U/\partial V)_T = T(\partial P/\partial T)_V - P$ — the internal pressure — using the Helmholtz Maxwell relation, and to show it is the single quantity that operationally separates an ideal gas from a real one. This result is not a curiosity: the internal pressure is the energy stored in intermolecular forces, and it directly determines whether a gas cools or warms on throttled expansion (the Joule–Thomson effect that runs every refrigerator).

Step 1. Write the internal energy in natural variables $(T,V)$ : $\mathrm{d}U = (\partial U/\partial T)_V\,\mathrm{d}T + (\partial U/\partial V)_T\,\mathrm{d}V$ . We want the second coefficient, the internal pressure. Why: it measures how energy changes with volume at fixed temperature — zero only if energy is volume-independent.

Step 2. From $\mathrm{d}U = T\,\mathrm{d}S - P\,\mathrm{d}V$ at fixed $T$ : $(\partial U/\partial V)_T = T(\partial S/\partial V)_T - P$ . The entropy derivative is unmeasurable — but the Helmholtz Maxwell relation (3) rescues us: $(\partial S/\partial V)_T = (\partial P/\partial T)_V$ . Why: this is the whole point of the Maxwell relation — it swaps the unmeasurable $(\partial S/\partial V)_T$ for the measurable $(\partial P/\partial T)_V$ .

Step 3. Substitute to get $(\partial U/\partial V)_T = T(\partial P/\partial T)_V - P$ , Equation (6). For an ideal gas $P = Nk_BT/V$ , so $(\partial P/\partial T)_V = Nk_B/V = P/T$ , and the internal pressure is $T\cdot P/T - P = 0$ : ideal-gas energy depends only on temperature. Why: with no intermolecular forces there is no energy stored in the spacing of molecules, so expanding at fixed $T$ costs no energy.

Step 4. For a van der Waals gas $P = \tfrac{Nk_BT}{V - Nb} - \tfrac{aN^2}{V^2}$ : $(\partial P/\partial T)_V = \tfrac{Nk_B}{V - Nb}$ , and $(\partial U/\partial V)_T = \tfrac{Nk_BT}{V-Nb} - P = \tfrac{aN^2}{V^2} > 0$ . Attractive interactions store energy as the gas expands, so pulling the molecules apart requires energy drawn from the thermal motion — the gas cools. Why: this positive internal pressure, computed purely from a Maxwell relation, is the microscopic origin of Joule–Thomson cooling and hence of gas liquefaction. *Verified: the van der Waals internal pressure $(\partial U/\partial V)_T = aN^2/V^2$ and the ideal-gas result of zero are checked symbolically in the verify script.*

Limiting cases

Where the Maxwell relations bite hardest:

Case	Result	Meaning
Ideal gas, relation (3)	$(\partial S/\partial V)_T = (\partial P/\partial T)_V = Nk_B/V$	Both sides $Nk_B/V$ ; entropy rises on expansion.
Internal pressure, ideal	$(\partial U/\partial V)_T = 0$	Energy depends only on $T$ — Joule's free-expansion result.
Internal pressure, van der Waals	$(\partial U/\partial V)_T = aN^2/V^2$	Attraction stores energy; drives Joule–Thomson cooling.
Relation (5) via expansivity	$(\partial S/\partial P)_T = -V\alpha$	Entropy's pressure dependence = thermal expansion; usually negative.
Incompressible solid ( $\alpha \to 0$ )	$(\partial S/\partial P)_T \to 0$	Squeezing a rigid solid barely changes its entropy.
Negative expansion (water below 4°C)	$(\partial S/\partial P)_T > 0$	Anomalous: entropy rises with pressure where $\alpha < 0$ .

Common trap“Maxwell relations are new physical laws you have to postulate”

They are not independent laws at all — they are pure calculus consequences of the potentials being state functions with exact differentials. Nothing about specific materials, forces, or the second law beyond "entropy is a state function" goes into them. This has two implications. First, you never need to memorize or postulate them separately: derive any one in three lines from the relevant potential's differential (or read it off the thermodynamic square). Second, they are universal constraints — every real substance, every valid equation of state, every correct model must satisfy all four exactly. If a proposed model violates a Maxwell relation, the model is wrong, not the relation. They are the rigidity of thermodynamics, not extra input.

Worked examples

1 · Turning an unmeasurable entropy change into a measured one (confidence builder)

One mole of an ideal gas at $T = 300\ \mathrm{K}$ , $V = 22.4\ \mathrm{L}$ is expanded isothermally to $2V$ . Use the Maxwell relation (3) to compute the entropy change without ever "measuring" entropy — then check against the known $\Delta S = R\ln 2$ .

Step 1. The Maxwell relation (3) gives $(\partial S/\partial V)_T = (\partial P/\partial T)_V$ . For an ideal gas $P = RT/V$ (per mole), so $(\partial P/\partial T)_V = R/V = \mathbf{371\ Pa/K}$ at $V = 22.4\ \mathrm{L}$ (since $R/V = 8.314/0.0224$ ). The unmeasurable entropy slope equals this measured pressure slope.

Step 2. Integrate at fixed $T$ : $\Delta S = \int_V^{2V}(\partial S/\partial V)_T\,\mathrm{d}V' = \int_V^{2V}\frac{R}{V'}\,\mathrm{d}V' = R\ln\frac{2V}{V} = R\ln 2$ .

Step 3. Evaluate: $\Delta S = (8.314)(0.693) = \mathbf{5.76\ J/K}$ . Positive, as expected — the gas gains entropy on expansion.

Step 4. Cross-check: the direct statistical result for isothermal expansion is exactly $\Delta S = Nk_B\ln(V_f/V_i) = R\ln 2 = 5.76\ \mathrm{J/K}$ . The Maxwell route, using only the measurable equation-of-state derivative $(\partial P/\partial T)_V$ , reproduces the entropy change — the promise of the relations made concrete.

2 · Internal pressure and energy of expansion for CO₂ (exam level)

Using relation (6), find the internal pressure of a van der Waals gas and compute the energy required to isothermally expand $1\ \mathrm{mol}$ of CO $_2$ ( $a = 0.364\ \mathrm{Pa\,m^6/mol^2}$ ) from $1\ \mathrm{L}$ to $\infty$ (fully separating the molecules).

Step 1. Internal pressure: $(\partial U/\partial V)_T = T(\partial P/\partial T)_V - P$ . For van der Waals, $(\partial P/\partial T)_V = \tfrac{R}{V - b}$ (per mole), so $(\partial U/\partial V)_T = \tfrac{RT}{V-b} - \big(\tfrac{RT}{V-b} - \tfrac{a}{V^2}\big) = \mathbf{\tfrac{a}{V^2}}$ . Purely the attraction term.

Step 2. Energy to expand isothermally: $\Delta U = \int_{V_0}^{\infty}(\partial U/\partial V)_T\,\mathrm{d}V = \int_{V_0}^{\infty}\frac{a}{V^2}\,\mathrm{d}V = \frac{a}{V_0}$ .

Step 3. Evaluate for CO $_2$ at $V_0 = 1\times10^{-3}\ \mathrm{m^3/mol}$ : $\Delta U = 0.364/10^{-3} = \mathbf{364\ J/mol}$ . This is the energy stored in intermolecular attraction, released as the gas is compressed and required back to pull it apart.

Step 4. Physical consequence: because expanding CO $_2$ against its own attraction costs energy, an adiabatic throttled expansion (no heat in) draws that energy from thermal motion, so the gas cools by roughly $\Delta U/C_V \sim 364/28 \approx 13\ \mathrm{K}$ per full expansion — the Joule–Thomson effect. A single Maxwell relation predicted the cooling that liquefies gases; without it, $(\partial U/\partial V)_T$ would be inaccessible.

Origins · 1 minmaxwell relations and how this was discovered

maxwell relations

concept · exact differentials of the potentials

The relations are named for James Clerk Maxwell, who presented them in his 1871 textbook Theory of Heat, deriving them from the geometry of the thermodynamic surface that Gibbs would soon formalize. What Maxwell recognized — and what makes the relations timeless — is that they require no assumptions about the substance: they are consequences of a thermodynamic potential being a well-behaved function of state, so that its differential is exact and its mixed partial derivatives commute. The compact mnemonic now taught (the "thermodynamic square" or "Born square") was popularized by Max Born in 1929 and by the physicist Koenig; it arranges the natural variables $S, V, T, P$ at the corners of a square with the potentials on the edges, letting one read off every Maxwell relation, including its sign, by inspection. The relations remain one of the most-used tools in all of thermal physics, precisely because they convert quantities you cannot measure into ones you can.

History

1859

— Maxwell derives the velocity distribution, sharpening the statistical foundation of thermodynamics.

1865

— Clausius states dU = T dS − P dV, the exact differential from which the relations flow.

1871

— Maxwell presents the reciprocity relations in Theory of Heat.

1873

— Gibbs's thermodynamic surface makes the "exact differential ⇒ commuting partials" structure geometric.

1875

— Gibbs's potentials give the four differentials that generate all four relations.

1929

— Born popularizes the thermodynamic-square mnemonic for reading off the relations and signs.

Going further

References. Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed., ch. 7 (Maxwell relations and the thermodynamic square) · Pathria & Beale, Statistical Mechanics, 3rd ed., ch. 1 · Reif, Fundamentals of Statistical and Thermal Physics, §5.6 · Kardar, Statistical Physics of Particles, ch. 1 · Landau & Lifshitz, Statistical Physics Part 1, §16 · Maxwell, Theory of Heat (1871), the original derivation.

Provenance. The four Maxwell relations (2)–(5) follow from equating mixed partials of $U, F, H, G$ (Callen ch. 7; Reif §5.6). The internal pressure $(\partial U/\partial V)_T = T(\partial P/\partial T)_V - P$ gives $0$ for an ideal gas and $aN^2/V^2$ for van der Waals — both checked symbolically. Numerical instances: $(\partial P/\partial T)_V = R/V = 371\ \mathrm{Pa/K}$ and $\Delta S = R\ln 2 = 5.76\ \mathrm{J/K}$ for isothermal doubling; CO $_2$ internal energy of expansion $\Delta U = a/V_0 = 364\ \mathrm{J/mol}$ ( $a = 0.364\ \mathrm{Pa\,m^6/mol^2}$ ). All in verify_aq-sm-maxwell-relations.py. $R = 8.314\ \mathrm{J\,mol^{-1}K^{-1}}$ .

Builds on: Thermodynamic Potentials and the Fundamental Relation · Legendre Transforms · Work, Heat, and Inexact Differentials · Second Law of Thermodynamics and Entropy. Leads to: Helmholtz Free Energy, Enthalpy, and Gibbs Free Energy · Euler Relation and Gibbs–Duhem Equation · Temperature, Pressure, and Chemical Potential as Entropy Derivatives.

How do I remember the four Maxwell relations without memorizing them?

Use the thermodynamic (Born) square, or just re-derive them — each takes three lines. To re-derive: pick the potential whose two natural variables you want ( $F$ for $(T,V)$ , $G$ for $(T,P)$ , etc.), write its differential (e.g. $\mathrm{d}F = -S\,\mathrm{d}T - P\,\mathrm{d}V$ ), read off the two first derivatives, and set the mixed second partials equal. The signs come from the signs in the differential. For the square: draw a box with $V, T$ on top corners and $S, P$ on bottom, label the edges $F$ (top), $G$ (right), $H$ (bottom), $U$ (left); each Maxwell relation connects the two variables not on a given edge, with a sign set by whether the arrows point together. Deriving beats memorizing because you will never mis-recall a sign.

Why can two of the Maxwell relations "measure" entropy but the other two can't?

Look at what each relates. Relations (3) and (5) — from $F$ and $G$ — have an entropy derivative on one side and a pressure/volume derivative on the other: $(\partial S/\partial V)_T = (\partial P/\partial T)_V$ and $(\partial S/\partial P)_T = -(\partial V/\partial T)_P$ . Since $P$ – $V$ – $T$ derivatives are directly measurable from an equation of state, these two let you compute entropy changes you could never measure with an instrument. Relations (2) and (4) — from $U$ and $H$ — hold entropy fixed (they are derivatives at constant $S$ ), so they relate quantities on adiabats and are less immediately useful for extracting entropy, though they are essential in adiabatic processes and for the speed of sound. The "useful for entropy" pair is exactly the pair whose natural variables are $T$ (not $S$ ), because those are the potentials you use when temperature is controlled.

Do Maxwell relations only apply to gases?

Not at all — they apply to any system with an exact-differential potential, which is essentially every equilibrium system. The general structure is one Maxwell relation per pair of conjugate variables. Replace the mechanical work term $-P\,\mathrm{d}V$ with a magnetic term $B\,\mathrm{d}M$ , an electric term $\mathcal{E}\,\mathrm{d}\Pi$ , a surface term $\gamma\,\mathrm{d}A$ , or an elastic-stress term, and you get magnetic, electric, surface, or elastic Maxwell relations. The magnetic one, $(\partial S/\partial B)_T = (\partial M/\partial T)_B$ , is the basis of magnetocaloric cooling and is how experimentalists extract entropy changes from magnetization data. So the relations are a completely general feature of thermodynamics, not a property of gases; gases are just the simplest place to meet them.

Is the internal pressure the same thing as the ordinary pressure?

No — they are different quantities with different meanings. The ordinary pressure $P$ is the force per area the gas exerts on its walls (mechanical). The internal pressure $\pi_T = (\partial U/\partial V)_T$ is how the internal energy changes with volume at fixed temperature, and it measures the energy stored in intermolecular forces. For an ideal gas $\pi_T = 0$ even though $P \ne 0$ : the molecules push on the walls but store no energy in their mutual spacing. For a real gas $\pi_T = aN^2/V^2 > 0$ because attractive forces do store energy as the molecules separate. The two are related through the thermodynamic equation of state $\pi_T = T(\partial P/\partial T)_V - P$ — a Maxwell-relation consequence — but conceptually one is a wall force and the other is an interaction-energy density.

What role do Maxwell relations play in deriving other thermodynamic identities?

They are the universal joints of thermodynamic algebra. Almost every nontrivial identity connecting response functions — the difference $C_P - C_V = TV\alpha^2/\kappa_T$ , the internal pressure, the Joule–Thomson coefficient, the ratio $\gamma = C_P/C_V = \kappa_T/\kappa_S$ — is derived by taking a derivative you cannot measure, using a Maxwell relation to swap it for one you can, and simplifying. The general strategy is: express the target quantity in terms of derivatives of a potential, hit any entropy derivative with the appropriate Maxwell relation to convert it to a $P$ – $V$ – $T$ derivative, and reduce to measurable response functions ( $C_P$ , $\alpha$ , $\kappa_T$ ). Learning to deploy the four relations fluently is what turns thermodynamics from a set of formulas into a calculational method.

Why does the sign differ between the four relations — two have a minus sign and two don't?

The sign is inherited directly from the differential of the potential. In $\mathrm{d}U = T\,\mathrm{d}S - P\,\mathrm{d}V$ the pressure term carries a minus, so the $U$ -relation $(\partial T/\partial V)_S = -(\partial P/\partial S)_V$ picks it up; the same happens for $G$ (where $-S\,\mathrm{d}T$ contributes the minus), giving $(\partial S/\partial P)_T = -(\partial V/\partial T)_P$ . In $\mathrm{d}F = -S\,\mathrm{d}T - P\,\mathrm{d}V$ both first derivatives ( $-S$ and $-P$ ) carry a minus, and the two minus signs cancel when you commute the mixed partial, leaving the plus relation $(\partial S/\partial V)_T = (\partial P/\partial T)_V$ ; likewise for $H$ . The thermodynamic square encodes exactly this by the direction of its arrows. So the signs are not arbitrary — they are bookkeeping of the minus signs in each potential's differential, which is why re-deriving is safer than memorizing.