Maxwell relations — derivation from each potential: Worked examples

This page is the drill floor. The parent note built the four relations; here we throw every kind of situation at them until none can surprise you. Read the scenario matrix first, then watch each cell get solved.

Everything below uses only tools already earned in the parent: an exact differential $d z = M d x + N d y$ , the equality of mixed partials $(\partial M / \partial y)_{x} = (\partial N / \partial x)_{y}$ (the Schwarz theorem), and the four boxed relations. If a new symbol appears, it is defined the moment it lands.

Definition Symbols used on this page (all defined here before use)

$R$ = the universal gas constant, $R = 8.314 J K^{- 1} mol^{- 1}$ — the fixed proportionality in the ideal-gas law $P V = n R T$ , where $n$ is the number of moles.
$α \equiv \frac{1}{V} (\frac{\partial V}{\partial T})_{P}$ = thermal expansion coefficient (fractional volume rise per kelvin at fixed pressure), units $K^{- 1}$ .
$κ_{T} \equiv - \frac{1}{V} (\frac{\partial V}{\partial P})_{T}$ = isothermal compressibility (fractional volume shrink per pascal at fixed $T$ ), units $Pa^{- 1}$ . Used in Ex 8.
$C_{V} \equiv (\frac{\partial U}{\partial T})_{V}$ = heat capacity at constant volume (energy needed to warm the system by one kelvin while its volume is held fixed), units $J K^{- 1}$ . For a monatomic ideal gas $C_{V} = \frac{3}{2} n R$ .
In the van der Waals equation, $a$ measures molecular attraction ( $Pa\cdotm^{6} mol^{- 2}$ ) and $b$ is the excluded volume per mole ( $m^{3} mol^{- 1}$ ), the finite size of the molecules.

Formula Two workhorse identities we re-derive once, then reuse

(i) The energy equation. Start from the master differential $d U = T d S - P d V$ (from the parent). Divide by $d V$ holding $T$ fixed, so $d S \to (\partial S / \partial V)_{T} d V$ : $(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial S}{\partial V})_{T} - P .$ Now insert the Helmholtz Maxwell relation $(\partial S / \partial V)_{T} = (\partial P / \partial T)_{V}$ : $(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial P}{\partial T})_{V} - P .$ (ii) The triple-product (cyclic) identity. For any three quantities linked by one equation of state, $x, y, z$ with any one held fixed in turn, $(\frac{\partial x}{\partial y})_{z} (\frac{\partial y}{\partial z})_{x} (\frac{\partial z}{\partial x})_{y} = - 1 \Rightarrow (\frac{\partial x}{\partial y})_{z} = - \frac{( \partial x / \partial z ) _{y}}{( \partial y / \partial z ) _{x}} .$ Why the $- 1$ ? Writing $d z = (\partial z / \partial x)_{y} d x + (\partial z / \partial y)_{x} d y$ and then holding $z$ fixed ( $d z = 0$ ) forces $d x$ and $d y$ into a fixed ratio; solving that ratio and closing the loop through all three variables leaves exactly one leftover minus sign. We use this whenever we must swap which variable is held constant.

The scenario matrix

Every problem this topic can throw is one of the cells below. The rightmost column names the worked example that lands on it.

Cell	What makes it distinct	Covered by
A. Pick-the-right-potential	You are handed a target derivative; choose which of $U, H, F, G$ gives it	Ex 1
B. Sign bookkeeping (+ case)	A relation with no minus sign ( $H$ )	Ex 2
C. Sign bookkeeping (− case)	A relation with a minus sign ( $G$ )	Ex 3
D. Numeric plug-in, ideal gas	Put real numbers in, get a number with units	Ex 4
E. Zero / degenerate input	A derivative that comes out exactly zero, and why	Ex 5
F. Limiting behaviour	What happens as $T \to 0$ or a coefficient $\to 0$	Ex 6
G. Real-world word problem	Rubber band / adiabatic cooling, physical story	Ex 7
H. Exam twist — chain the relations	Combine two Maxwell relations + a triple product	Ex 8

Ex 1 — Cell A · choosing the potential

Worked example Statement

You need $(\frac{\partial S}{\partial P})_{T}$ but your lab can only measure how volume changes with temperature. Which Maxwell relation delivers it, and what do you measure?

Forecast: Guess first — which potential has $T$ and $P$ as its natural variables?

Step 1. List the natural-variable pairs. $U : (S, V)$ , $H : (S, P)$ , $F : (T, V)$ , $G : (T, P)$ . Why this step? The target derivative holds $T$ constant and differentiates by $P$ . The potential whose natural variables are exactly $(T, P)$ is the one whose Maxwell relation is expressed purely in $T, P$ (and their partners $S, V$ ). That is $G$ .

Step 2. Read off the Gibbs relation: $(\frac{\partial S}{\partial P})_{T} = - (\frac{\partial V}{\partial T})_{P} .$ Why this step? $G$ 's differential is $d G = - S d T + V d P$ , so $M = - S$ , $N = V$ , and $(\partial M / \partial P)_{T} = (\partial N / \partial T)_{P}$ gives exactly this.

Step 3. Name the measurable. Using the thermal expansion coefficient $α$ defined above, $(\frac{\partial S}{\partial P})_{T} = - V α .$

Verify: RHS units — $V$ is $m^{3}$ , $α$ is $K^{- 1}$ , product $m^{3} K^{- 1}$ . LHS is entropy $/$ pressure $= (J/K) / (Pa) = J K^{- 1} Pa^{- 1} = m^{3} K^{- 1}$ (since $J = Pa\cdotm^{3}$ ). ✓ Units match. See Thermal expansion coefficient and isothermal compressibility.

Ex 2 — Cell B · the plus-sign relation (enthalpy)

Ex 3 — Cell C · the minus-sign relation, numerically (Gibbs)

Worked example Statement

For water near $25^{\circ} C$ , $α = 2.57 \times 1 0^{- 4} K^{- 1}$ and molar volume $V_{m} = 1.81 \times 1 0^{- 5} m^{3} / mol$ . Find $(\frac{\partial S}{\partial P})_{T}$ per mole. Does entropy rise or fall when you squeeze water isothermally?

Forecast: Sign guess — squeezing usually orders molecules, so entropy should drop. Let's see.

Step 1. Use the Gibbs relation with $α$ : $(\frac{\partial S}{\partial P})_{T} = - V_{m} α .$ Why this step? Ex 1 already established this exact link; now we insert numbers.

Step 2. Plug in: $(\frac{\partial S}{\partial P})_{T} = - (1.81 \times 1 0^{- 5}) (2.57 \times 1 0^{- 4}) = - 4.6517 \times 1 0^{- 9} J K^{- 1} Pa^{- 1} mol^{- 1} .$ Why this step? Direct multiply; keep units $m^{3} K^{- 1}$ which equals $J K^{- 1} Pa^{- 1}$ .

Step 3. Interpret the sign. Negative ⇒ raising $P$ at fixed $T$ lowers entropy. Why this step? The Maxwell relation converts a hard entropy measurement into an easy expansion measurement; the sign of $α$ (positive for water at $25^{\circ} C$ ) fixes the sign of the answer.

Verify: Our forecast (entropy drops on compression) matches the negative sign. Magnitude $\sim 5 \times 1 0^{- 9}$ is tiny because liquids barely change — physically sensible. ✓

Ex 4 — Cell D · numeric plug-in for an ideal gas (energy equation)

Worked example Statement

Using the energy equation re-derived at the top of this page, $(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial P}{\partial T})_{V} - P$ , evaluate it for $n = 2 mol$ of ideal gas at $T = 300 K$ , $V = 0.05 m^{3}$ . Then repeat for a van der Waals gas to see a non-zero result.

Forecast: For an ideal gas the parent proved this is $0$ . Guess the van der Waals value's sign.

Step 1 (ideal). With gas constant $R = 8.314 J K^{- 1} mol^{- 1}$ , $P = \frac{n R T}{V} \Rightarrow (\frac{\partial P}{\partial T})_{V} = \frac{n R}{V}$ . Then the energy equation gives $T \cdot \frac{n R}{V} - P = \frac{n R T}{V} - \frac{n R T}{V} = 0.$ Why this step? $T (\partial P / \partial T)_{V}$ reproduces $P$ exactly, so they cancel — Joule's law falls out.

Step 2 (van der Waals). Equation of state $P = \frac{n R T}{V - nb} - \frac{a n ^{2}}{V ^{2}}$ , where (as defined above) $a$ is the molecular-attraction parameter and $b$ is the excluded volume per mole. Since only the first term contains $T$ , $(\frac{\partial P}{\partial T})_{V} = \frac{n R}{V - nb} .$ Why this step? The $- a n^{2} / V^{2}$ term is temperature-independent, so it drops out on differentiating by $T$ . The $b$ term survives inside the first fraction but is likewise $T$ -independent as a constant offset to $V$ .

Step 3. Assemble via the energy equation: $(\frac{\partial U}{\partial V})_{T} = \frac{n R T}{V - nb} - (\frac{n R T}{V - nb} - \frac{a n ^{2}}{V ^{2}}) = \frac{a n ^{2}}{V ^{2}} .$ Notice the whole $b$ -dependence cancels. With $a = 0.14 Pa\cdotm^{6} mol^{- 2}$ , $n = 2$ , $V = 0.05 m^{3}$ : $\frac{( 0.14 ) ( 2 ) ^{2}}{( 0.05 ) ^{2}} = \frac{0.56}{0.0025} = 224 Pa = 224 J/m^{3} .$ Why this step? The leftover $a n^{2} / V^{2}$ is the internal pressure from molecular attraction — real gases do store energy in volume.

Verify: Ideal answer $= 0$ matches parent. Van der Waals answer positive (attraction) as expected; units $Pa = J/m^{3}$ correct for $\partial U / \partial V$ . ✓

Ex 5 — Cell E · a derivative that is exactly zero (degenerate)

Worked example Statement

An ideal gas expands into vacuum (free / Joule expansion): no work, no heat, so $U$ is constant. Predict $(\frac{\partial T}{\partial V})_{U}$ and confirm it is zero.

Forecast: Does an ideal gas cool on free expansion? Guess yes or no.

Step 1. Apply the triple-product identity (re-derived at the top) to $T (U, V)$ : $(\frac{\partial T}{\partial V})_{U} = - \frac{( \partial U / \partial V ) _{T}}{( \partial U / \partial T ) _{V}} .$ Why this step? We want a derivative at constant $U$ , but we know derivatives at constant $T$ (energy equation). The triple product is exactly the tool for swapping which variable is held fixed.

Step 2. Numerator: from Ex 4, $(\partial U / \partial V)_{T} = 0$ for an ideal gas. Denominator: $(\partial U / \partial T)_{V} = C_{V} = \frac{3}{2} n R \neq = 0$ , the heat capacity at constant volume defined in the Symbols box. Why this step? A zero numerator over a finite denominator gives zero.

Step 3. $(\frac{\partial T}{\partial V})_{U} = - \frac{0}{C _{V}} = 0.$ Why this step? No temperature change on free expansion — the degenerate/zero case.

Verify: Forecast check — an ideal gas does not cool (real gases do, via the $a n^{2} / V^{2}$ term of Ex 4). The zero is genuine, not undefined, because $C_{V} > 0$ . ✓

Ex 6 — Cell F · limiting behaviour as $T \to 0$

Worked example Statement

The third law says entropy $S \to$ const as $T \to 0$ . Use a Maxwell relation to show the thermal expansion coefficient $α \to 0$ as $T \to 0$ .

Forecast: Guess whether materials keep expanding right down to absolute zero.

Step 1. Gibbs relation: $(\frac{\partial V}{\partial T})_{P} = - (\frac{\partial S}{\partial P})_{T} .$ Why this step? We want the temperature dependence of $V$ (i.e. $α$ ); the Gibbs relation ties it to how $S$ depends on $P$ , which the third law controls.

Step 2. Third law ⇒ as $T \to 0$ , $S$ approaches the same constant for all pressures, so $(\frac{\partial S}{\partial P})_{T} \to 0$ . Why this step? If $S (T = 0)$ is independent of $P$ , its $P$ -derivative vanishes at $T = 0$ .

Step 3. Therefore $(\frac{\partial V}{\partial T})_{P} \to 0$ , hence $α = \frac{1}{V} (\frac{\partial V}{\partial T})_{P} \to 0 (T \to 0) .$ Why this step? A finite $V$ times a vanishing derivative gives $α \to 0$ .

Verify: Experimentally, thermal expansion of solids does flatten to zero near absolute zero — matches. The Maxwell relation turned an entropy statement (third law) into a mechanical prediction (no expansion). ✓

Ex 7 — Cell G · real-world word problem (rubber band)

Figure — Maxwell relations — derivation from each potential

Reading the figure. The horizontal axis is the band's length $L$ (metres) — this is the mechanical variable, playing the role that volume $V$ plays for a gas. The lower black wavy line is the relaxed band; the upper red wavy line (the key object) is the same band stretched to a longer $L$ . The black arrows at each end are the tension $f$ (the inward pull, in newtons). The diagonal black arrow shows the stretching process. The pedagogical point the red curve makes visible: stretching straightens the tangled polymer chains, so the band becomes more ordered — its entropy $S$ falls as $L$ grows. Keep that red "ordered" state in mind through the steps below.

Worked example Statement

A stretched rubber band is an "elastic" system: its work term is $δ W = - f d L$ where $f$ is tension (force, N) and $L$ is length (m) — so $L$ plays the role of $V$ and $- f$ plays the role of $P$ . Experimentally, a rubber band warms when stretched adiabatically. Using the analogue of the Helmholtz relation, predict the sign of $(\partial f / \partial T)_{L}$ (does tension rise with temperature at fixed length?).

Forecast: From the figure, guess: at fixed stretch, does a hotter rubber band pull harder or softer?

Step 1. Build the elastic Helmholtz differential. Replace $- P d V \to + f d L$ in $d F = - S d T - P d V$ : $d F = - S d T + f d L .$ Why this step? For a rubber band you do work on it by stretching, so the mechanical term flips sign relative to a gas; $L$ is the natural mechanical variable (the red-curve axis in the figure).

Step 2. Read $M = - S$ , $N = f$ , apply Schwarz: $- (\frac{\partial S}{\partial L})_{T} = (\frac{\partial f}{\partial T})_{L} \Rightarrow (\frac{\partial f}{\partial T})_{L} = - (\frac{\partial S}{\partial L})_{T} .$ Why this step? Same cross-derivative logic as the gas, just with the elastic variables.

Step 3. Sign it. The figure showed stretching lines up the polymer chains, lowering disorder, so $(\partial S / \partial L)_{T} < 0$ . Hence $(\partial f / \partial T)_{L} = - (negative) > 0$ . Why this step? The minus flips a negative into a positive.

Verify: $(\partial f / \partial T)_{L} > 0$ means a hotter band pulls harder at fixed length — this is the famous rubber-band effect (a stretched band contracts when heated). And by the adiabatic route, warming-on-stretch follows from the same $S$ -decrease. Prediction matches everyday experiment. ✓

Ex 8 — Cell H · exam twist, chaining relations ( $C_{P} - C_{V}$ )

Worked example Statement

Derive the general relation $C_{P} - C_{V} = T V \frac{α ^{2}}{κ _{T}}$ , with $α$ and $κ_{T}$ as defined at the top. Then evaluate for $1 mol$ ideal gas at $T = 300 K$ to confirm it gives $R$ .

Forecast: For an ideal gas you already know $C_{P} - C_{V} = R$ . Does the messy formula collapse to it?

Step 1. Start from $C_{P} - C_{V} = T (\frac{\partial P}{\partial T})_{V} (\frac{\partial V}{\partial T})_{P}$ (standard result from $d S$ ; see Heat capacities $C_P - C_V$). Why this step? This already contains one awkward factor, $(\partial P / \partial T)_{V}$ , which we now rewrite with the triple product from the top of the page.

Step 2. Triple product: $(\frac{\partial P}{\partial T})_{V} = - \frac{( \partial V / \partial T ) _{P}}{( \partial V / \partial P ) _{T}} = \frac{α}{κ _{T}} .$ Why this step? It converts a constant- $V$ derivative (hard) into the two measurables $α$ and $κ_{T}$ . The minus in the triple product cancels the minus in the definition of $κ_{T}$ .

Step 3. Substitute and use $(\partial V / \partial T)_{P} = V α$ : $C_{P} - C_{V} = T \cdot \frac{α}{κ _{T}} \cdot V α = \frac{T V α ^{2}}{κ _{T}} .$ Why this step? Collecting the two $α$ 's gives $α^{2}$ ; done.

Step 4 (ideal-gas check). For $P V = n R T$ we get the two coefficients cleanly:

\kappa_T = -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T = -\frac{1}{V}\left(-\frac{nRT}{P^2}\right) = \frac{1}{P}.$$ Substitute into the boxed formula: $$C_P-C_V = \frac{TV\,(1/T)^2}{1/P} = \frac{TV\cdot P}{T^2} = \frac{PV}{T} = \frac{nRT}{T} = nR.$$ With $n=1$: $C_P-C_V = R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$. **Why this step?** Substituting the ideal-gas $\alpha=1/T$ and $\kappa_T=1/P$ collapses everything to $PV/T=nR$. **Verify:** The general formula reproduces the textbook $C_P-C_V=R$ for one mole. Also $C_P>C_V$ always because $\alpha^2\ge0$, $\kappa_T>0$, $T>0$ — a bonus universal fact the derivation hands us for free. ✓

Recall Quick self-test

Which relation gives $(\partial S / \partial V)_{T}$ and does it carry a minus? ::: Helmholtz $F$ : $(\partial S / \partial V)_{T} = (\partial P / \partial T)_{V}$ , no minus (two minuses cancel). Why is $(\partial U / \partial V)_{T} = 0$ for an ideal gas? ::: Because $T (\partial P / \partial T)_{V} = P$ exactly, so the energy equation gives $P - P = 0$ . As $T \to 0$ , why does $α \to 0$ ? ::: Third law ⇒ $(\partial S / \partial P)_{T} \to 0$ , and Gibbs relation ties that to $(\partial V / \partial T)_{P} \to 0$ .

Connections

Parent: Maxwell relations
Joule expansion and internal energy — Ex 4 & 5
Heat capacities $C_P - C_V$ — Ex 8
Thermal expansion coefficient and isothermal compressibility — $α, κ_{T}$
First and Second Laws of Thermodynamics — source of the master differential
Equality of mixed partial derivatives (Schwarz theorem) — the engine behind every step