Partial Derivatives — Maths for Thermodynamics & Kinetics

Varying one thing at a time

The calculus of thermodynamic surfaces

When a quantity depends on several variables, a partial derivative asks how it changes with one of them while the rest are held fixed. The curly $\partial$ replaces the straight $d$ to signal this, and the subscript names what is held constant:

(\frac{\partial f}{\partial x})_{y} means: differentiate f w.r.t. x keeping y fixed

Computing a partial derivative is easy: treat every other variable as a plain number and differentiate as usual. For the ideal gas $p = n RT / V$ , holding $T$ and $n$ constant and differentiating with respect to $V$ just means $n RT$ acts as a constant: $(\partial p / \partial V)_{T} = - n RT / V^{2}$ .

The real power comes from combining partial derivatives. A total differential says how a function changes when all its variables change together:

d G = (\frac{\partial G}{\partial T})_{p} d T + (\frac{\partial G}{\partial p})_{T} d p

Because the order of mixed partial derivatives doesn't matter (Schwarz's theorem), setting $\partial^{2} G / \partial T \partial p = \partial^{2} G / \partial p \partial T$ generates the Maxwell relations — identities that connect measurable quantities to derivatives of thermodynamic potentials.

(\partial G / \partial T)_{p} = - S

Free energy falls with temperature; the faster the fall, the higher the entropy.

(\partial G / \partial p)_{T} = V

Free energy rises with pressure; the steeper the rise, the larger the molar volume.

(\partial U / \partial V)_{T}

The internal pressure — zero for an ideal gas, positive for a real gas (molecules attract).

Total differential

d p = (\frac{\partial p}{\partial V})_{T} d V + (\frac{\partial p}{\partial T})_{V} d T

Thermodynamics runs on these

Free energy, enthalpy and entropy all depend on several variables, so thermodynamics is written in partial derivatives:

(\partial G / \partial T)_{p} = - S

(\partial G / \partial p)_{T} = V

, and the Maxwell relations that connect them. Open the slice explorer.

The systematic framework for Maxwell relations comes from the four fundamental thermodynamic potentials and their exact differentials. For an exact differential $d Z = M d x + N d y$ , Euler's reciprocity (the Schwarz theorem) states $(\partial M / \partial y)_{x} = (\partial N / \partial x)_{y}$ . This is the exactness test — a quick way to verify whether a differential is exact, and to extract Maxwell relations without memorising them. The four potentials give four Maxwell relations:

d U : (\frac{\partial T}{\partial V})_{S} = - (\frac{\partial p}{\partial S})_{V}

d G : (\frac{\partial S}{\partial p})_{T} = - (\frac{\partial V}{\partial T})_{p}

An important partial derivative in calorimetry is $(\partial H / \partial p)_{T}$ , known as the isothermal pressure coefficient of enthalpy. From the fundamental relation $d H = T d S + V d p$ , differentiating with respect to $p$ at constant $T$ and using the Maxwell relation $(\partial S / \partial p)_{T} = - (\partial V / \partial T)_{p}$ :

(\frac{\partial H}{\partial p})_{T} = V - T (\frac{\partial V}{\partial T})_{p}

For an ideal gas this is zero (enthalpy is independent of pressure — a direct consequence of no intermolecular forces); for real gases it is the basis of the Joule–Thomson coefficient.

Euler reciprocity (exactness test)

For

d Z = M d x + N d y

: exact iff

(\partial M / \partial y)_{x} = (\partial N / \partial x)_{y}

(\partial H / \partial p)_{T}

= V - T (\partial V / \partial T)_{p}

. Zero for ideal gas; negative for most real gases below the inversion temperature (Joule–Thomson cooling).

Total differential

d V = (\frac{\partial V}{\partial T})_{p} d T + (\frac{\partial V}{\partial p})_{T} d p

— mixes thermal expansion and compressibility.

Cross-partials equal

\frac{\partial ^{2} f}{\partial x \partial y} = \frac{\partial ^{2} f}{\partial y \partial x}

(Schwarz theorem) — the foundation of every Maxwell relation.

Common pitfalls

Forgetting which variable is held constant. $\partial p / \partial T$ is ambiguous without specifying what is held fixed. Always write the subscript: $(\partial p / \partial T)_{V} \neq = (\partial p / \partial T)_{p}$ (the second is identically zero).
Treating $\partial$ like $d$ in a chain rule. The chain rule $df / d x = (\partial f / \partial u) (d u / d x)$ works for functions of one variable, but for a function of two variables you must include all partial contributions: $df = (\partial f / \partial x) d x + (\partial f / \partial y) d y$ .
Sign errors in Maxwell relations. The sign depends on the potential. From $d G$ the relation carries a minus sign: $(\partial S / \partial p)_{T} = - (\partial V / \partial T)_{p}$ . From $d A$ the relation also carries a minus: $(\partial S / \partial V)_{T} = + (\partial p / \partial T)_{V}$ — check the sign from the fundamental differential each time.
Assuming that $(\partial U / \partial V)_{T} = 0$ for all gases. This is true only for ideal gases. For real (van der Waals) gases the internal pressure $(\partial U / \partial V)_{T} = a / V_{m}^{2}$ is positive and non-negligible at high densities.

Partial-slice explorer

Freeze one variable and slide the other — the tangent line gives the partial derivative at each point.

Ideal gas p = RT/V. Vary:

(\frac{\partial p}{\partial V})_{T} = - \frac{RT}{V ^{2}}

held T300 K

evaluate at V4.0 L

p at this point6.15 atm

partial slope-1.5386

Freezing one variable turns the 3-D surface into a single curve — and the partial derivative is just the ordinary slope of that slice. At constant T, p falls steeply as V grows (slope −RT/V²).

Worked example 1Evaluating (∂p/∂V)_T for an ideal gas

1.00 mol of an ideal gas is in a 4.00 L container at 300 K. Evaluate the partial derivative $(\partial p / \partial V)_{T}$ at these conditions ( $R = 0.08206 L atm mo l^{- 1} K^{- 1}$ ).

Worked example 2Fundamental thermodynamic relation: (∂G/∂T)_p = −S

The molar Gibbs free energy of a substance is $G = H - TS$ . By differentiating with respect to $T$ at constant $p$ , and using the Gibbs–Helmholtz relation $(\partial H / \partial T)_{p} = C_{p}$ , show that $(\partial G / \partial T)_{p} = - S$ .

Worked examples

Partial derivatives across thermodynamics, kinetics and molecular properties.

Worked example 3Total differential of the ideal gas pressure

Write the total differential of $p = n RT / V$ in terms of $d T$ and $d V$ (holding $n$ constant). Use it to estimate the change in pressure when temperature increases by 5 K and volume increases by 0.20 L, starting from $n = 1 mol$ , $T = 300 K$ , $V = 5.00 L$ .

Worked example 4Internal pressure of a van der Waals gas

Show that for a van der Waals gas $(p + \frac{a}{V _{m}^{2}}) (V_{m} - b) = RT$ , the internal pressure $(\partial U / \partial V_{m})_{T} = a / V_{m}^{2}$ . Use the thermodynamic identity $(\partial U / \partial V)_{T} = T (\partial p / \partial T)_{V} - p$ .

Worked example 5A Maxwell relation: entropy and volume

Starting from $d A = - S d T - p d V$ (the Helmholtz free energy), derive the Maxwell relation $(\partial S / \partial V)_{T} = (\partial p / \partial T)_{V}$ .

Worked example 6Euler reciprocity / exactness test for a thermodynamic differential

Test whether the differential $d Z = \frac{p}{T} d V + \frac{V}{T} d p - \frac{p V}{T ^{2}} d T$ is exact (i.e. whether $Z$ is a state function). Use the Euler reciprocity condition applied to the $V$ and $T$ terms.

Worked example 7(∂H/∂p)_T for an ideal gas and a real gas

Using the relation $(\partial H / \partial p)_{T} = V - T (\partial V / \partial T)_{p}$ (derived from $d H = T d S + V d p$ ), evaluate this partial derivative for (a) an ideal gas and (b) a van der Waals gas at large molar volume.

ChallengeDerive (∂S/∂V)_T = (∂p/∂T)_V and evaluate for an ideal gas

Starting from the Helmholtz free energy $A$ with exact differential $d A = - S d T - p d V$ :

(a) Derive the Maxwell relation $(\partial S / \partial V)_{T} = (\partial p / \partial T)_{V}$ .

(b) Evaluate both sides for 1 mol of ideal gas at $T = 300 K$ , $V = 5.00 L$ and confirm they are equal.

Check yourself

Four quick questions on partial derivatives and thermodynamic relations.

Question 1 of 4 · Score 0

A partial derivative ∂f/∂x measures how f changes as you vary x while:

(\frac{\partial f}{\partial x})_{y}

Choose an answer.