Maxwell relations

You cannot put a probe into a beaker and read off its entropy. There is no "entropy-meter" the way there is a thermometer or a pressure gauge. Yet chemists routinely quote $\Delta S$ for a compression, a reaction, or a phase change. How? Because thermodynamic potentials such as $U, H, A, G$ are state functions with exact differentials, and exactness forces a hidden symmetry between their mixed partial derivatives. That symmetry — the Maxwell relations — is a translation dictionary: it converts an unmeasurable entropy derivative like $(\partial S/\partial V)_T$ into an ordinary, measurable $P$ – $V$ – $T$ derivative like $(\partial P/\partial T)_V$ , obtainable from an equation of state or a simple experiment. This article derives all four Maxwell relations from a single mathematical fact, shows how to generate any of them on demand rather than memorising a table, and uses them to derive the general thermodynamic equation of state and to compute a real $\Delta S$ for an isothermal gas compression — setting up the real-gas (van der Waals) treatment that follows.

The core insight

Exactness of the differentials of $U, H, A, G$ forces equalities between mixed partial derivatives, and these four Maxwell relations let you replace an unmeasurable entropy derivative with a measurable $P$ – $V$ – $T$ derivative.

The core idea

Exactness is the engine. If $f(x,y)$ is a state function with $df = M\,dx + N\,dy$ , path-independence of $f$ requires the mixed second partials to match:

\left(\frac{\partial M}{\partial y}\right)_x = \left(\frac{\partial N}{\partial x}\right)_y

where $M$ and $N$ are the coefficients of $dx$ and $dy$ in the exact differential. Apply this once to each of the four thermodynamic potentials, in their own natural variables, and you get four relations for free.

The four relations. From $dU = TdS - PdV$ , $dH = TdS + VdP$ , $dA = -SdT - PdV$ , $dG = -SdT + VdP$ :

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

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

where $S$ = entropy, $V$ = volume, $P$ = pressure, $T$ = temperature. These are the complete set; no fifth independent relation exists among $U,H,A,G$ built this way.

Generate, don't memorise. CY11001 expects you to reconstruct any relation from scratch: write the potential's differential in its natural variables, read off $M$ and $N$ as the coefficients of the two differentials, and equate the cross partials. This is faster and safer than recalling a table under exam pressure, and it is the only reliable way to avoid sign errors.

Why they matter. The left-hand sides above involve entropy, which has no direct instrument. The right-hand sides are ordinary $P$ – $V$ – $T$ derivatives — the thermal expansion coefficient $(\partial V/\partial T)_P$ or its pressure analogue $(\partial P/\partial T)_V$ — read straight off an equation of state or a simple dilatometry/manometry experiment. Maxwell relations are the bridge that lets $\Delta S$ , and through it $\Delta U$ , $\Delta H$ , $\Delta G$ , be computed for real, non-ideal systems purely from measurable mechanical data.

Figure

Four boxes show each potential differential feeding its Maxwell relation through the exactness condition

Derivation AFrom dA to the thermodynamic equation of state

Derive the Maxwell relation hidden in $dA = -SdT - PdV$ from the exactness condition alone, then combine it with the fundamental relation $dU = TdS - PdV$ to obtain the general thermodynamic equation of state $(\partial U/\partial V)_T = T(\partial P/\partial T)_V - P$ , and finally show that this expression collapses to exactly zero for an ideal gas, recovering Joule's law as a derived result rather than an assumed postulate.

Step 1. Write the Helmholtz differential as $dA = -SdT - PdV = M\,dT + N\,dV$ , so $M = -S$ and $N = -P$ , with $T,V$ as the natural variable pair.

Step 2. Apply exactness, $(\partial M/\partial V)_T = (\partial N/\partial T)_V$ :

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

This is Maxwell relation 3, obtained purely from the exact-differential structure of $A$ , with no physical assumption about the substance.

Step 3. Take $dU = TdS - PdV$ and divide through by $dV$ at constant $T$ to isolate the "internal pressure":

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

Step 4. Substitute the Maxwell relation from Step 2 for $(\partial S/\partial V)_T$ :

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

This is the general thermodynamic equation of state, valid for any substance, ideal or real.

Step 5. For an ideal gas, $PV = nRT \Rightarrow P = nRT/V \Rightarrow (\partial P/\partial T)_V = nR/V$ . Substituting into Step 4:

\left(\frac{\partial U}{\partial V}\right)_T = T\cdot\frac{nR}{V} - P = \frac{nRT}{V} - P = P - P = 0

So an ideal gas's internal energy depends on $T$ only, never on $V$ — Joule's law, here derived rather than postulated. For a real gas (treated in the next article via the van der Waals equation) this quantity is generally nonzero, reflecting intermolecular attraction or repulsion.

Worked example

Use Maxwell relation 4, $(\partial S/\partial P)_T = -(\partial V/\partial T)_P$ , to find $\Delta S$ when 1 mol of an ideal gas is compressed isothermally from 1 bar to 5 bar at 298 K.

Step 1. For an ideal gas, $V = nRT/P$ , so $(\partial V/\partial T)_P = nR/P$ . By Maxwell relation 4:

\left(\frac{\partial S}{\partial P}\right)_T = -\frac{nR}{P}

Step 2. Integrate at constant $T = 298$ K from $P_1 = 1$ bar to $P_2 = 5$ bar, with $n = 1$ mol:

\Delta S = \int_{P_1}^{P_2}\left(\frac{\partial S}{\partial P}\right)_T dP = -nR\int_{1}^{5}\frac{dP}{P} = -nR\ln\frac{P_2}{P_1} = -(1)(8.314)\ln(5)

Step 3. Evaluate: $\ln 5 = 1.6094$ , so

\Delta S = -8.314 \times 1.6094 = \mathbf{-13.38\ J\,K^{-1}}

This matches the direct Clausius-integral formula $\Delta S = -nR\ln(P_2/P_1)$ exactly, as it must — both routes describe the same state function. The negative sign is physically expected: isothermal compression to higher pressure and lower volume reduces the number of accessible microstates.

Competitive stretch

If instead the gas expanded isothermally from 5 bar to 1 bar, the same formula gives $\Delta S = -(1)(8.314)\ln(1/5) = +13.38\ \text{J K}^{-1}$ — the identical magnitude with the sign flipped, since expansion increases accessible microstates. Recognising this symmetry avoids re-deriving the integral for the reverse process.

Common traps

Mixing up which relation carries a minus sign — relations 1 and 4 do, relations 2 and 3 do not; re-deriving from the potential's own differential (rather than recalling a table) avoids this.
Pulling a derivative from the wrong potential's natural variables — e.g. trying to get $(\partial S/\partial V)_T$ from $dG = -SdT + VdP$ (natural variables $T,P$ ) instead of the correct source $dA = -SdT - PdV$ (natural variables $T,V$ ).
Forgetting that Maxwell relations are exact identities for any substance; the ideal-gas equation of state is plugged in only at the last step to evaluate the measurable right-hand side, not baked into the relation itself.
Treating $(\partial U/\partial V)_T = 0$ as a universal law rather than an ideal-gas-specific consequence of the thermodynamic equation of state — real gases give a nonzero value here.

Ask the tutor: If the four relations keep blurring together, ask the AI tutor to quiz you by naming a random potential and having you derive its Maxwell relation from scratch until it becomes automatic.

Why can't we just measure entropy directly with an instrument

Entropy is not a mechanical quantity like pressure or volume that a gauge can sense; it is a statistical measure of accessible microstates, defined through heat exchanged reversibly divided by temperature. There is no "entropy probe." Maxwell relations sidestep this by rewriting any entropy derivative as an equivalent $P$ – $V$ – $T$ derivative, which an equation of state or a simple experiment can supply directly.

Do I need to memorise all four Maxwell relations for exams

No — CY11001 rewards the ability to regenerate any relation on demand rather than recall. Write the differential of the relevant potential in its natural variables, label the coefficients $M$ and $N$ , and apply the exactness condition. This method is more exam-safe than memorising a table, since a single misremembered sign in a memorised relation is a common, hard-to-catch error.

Why does the same Maxwell relation appear in both the derivation and the worked example

It does not — the derivation block uses relation 3 (from $dA$ ) to build the general thermodynamic equation of state, while the worked example uses relation 4 (from $dG$ ) to compute an entropy change with pressure as the swept variable. Seeing both in one article shows how the same exactness idea produces different, complementary tools depending on which variable you need to eliminate.

Maxwell relations

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads