Maxwell relations — derivation from each potential: Question bank

True or false — justify

True or false: A Maxwell relation is a new law of physics beyond the first two laws.

False — it is pure calculus (equality of mixed partials, see Equality of mixed partial derivatives (Schwarz theorem)) applied to a state function whose exact differential already came from the first two laws.

True or false:

(\frac{\partial S}{\partial V})_{T} = (\frac{\partial P}{\partial T})_{V}

has no minus sign, so all Maxwell relations are sign-free.

False — this one (from

F

) happens to be positive because

- S d T

and

- P d V

each carry a minus that cancels; the relations from

U

and

G

genuinely carry a minus.

True or false: Maxwell relations require the substance to be an ideal gas.

False — they hold for any simple compressible closed system because they follow only from the potentials being state functions, not from any equation of state.

True or false: You can write

d U = T d S - P d V

only for a reversible process, so Maxwell relations only apply reversibly.

False — the differential relates state variables, and since

U, S, V

are all state functions the equation holds between equilibrium states regardless of the path taken.

True or false: The enthalpy relation

(\frac{\partial T}{\partial P})_{S} = (\frac{\partial V}{\partial S})_{P}

tells you how temperature changes with pressure at constant entropy.

True — constant

S

is exactly the adiabatic-reversible condition, so this quantity is the reversible-adiabatic temperature response to squeezing.

True or false: Because

U

and

F

both have

- P d V

, their Maxwell relations look identical.

False —

U

uses natural variables

(S, V)

giving

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

, while

F

uses

(T, V)

giving

(\partial S / \partial V)_{T} = (\partial P / \partial T)_{V}

; different independent variables, different relations.

True or false: For an ideal gas

(\partial U / \partial V)_{T} = 0

is an assumption we plug in by hand.

False — it is derived from the energy equation plus the ideal-gas law, showing

T (\partial P / \partial T)_{V} = P

exactly (see Joule expansion and internal energy).

Spot the error

"Since

U = U (T, V)

is perfectly valid, I can read off

d U = T d S - P d V

using

(T, V)

as variables." — find the error.

The clean coefficients

T

and

- P

only appear when

U

is expressed in its natural variables

(S, V)

; in

(T, V)

the differential becomes

d U = C_{V} d T + [T (\partial P / \partial T)_{V} - P] d V

, which is not the Maxwell-generating form.

"From

d G = - S d T + V d P

I get

(\frac{\partial S}{\partial P})_{T} = (\frac{\partial V}{\partial T})_{P}

." — spot the mistake.

Missing minus:

M = - S

, so

(\partial M / \partial P)_{T} = - (\partial S / \partial P)_{T}

, giving

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

(\frac{\partial T}{\partial V})_{S} = (\frac{\partial T}{\partial V})_{P}

because it's the same partial." — what's wrong?

The subscript is part of the quantity; holding

S

fixed versus

P

fixed gives genuinely different numbers, so these are two different derivatives.

H = U + P V

, so

d H = d U + P d V = T d S - P d V + P d V = T d S

." — find the dropped term.

The product rule gives

d (P V) = P d V + V d P

, so the correct result is

d H = T d S + V d P

; forgetting

V d P

loses enthalpy's natural pressure dependence.

"The Maxwell relation from

F

says

(\partial S / \partial V)_{T} = (\partial P / \partial T)_{V}

, so entropy always rises when I increase volume." — where's the flaw?

The sign of

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

equals the sign of

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

, which is positive for normal substances but can be negative (e.g. water near

4^{\circ}

C), so the claim isn't universal.

"Since mixed partials are equal,

(\frac{\partial P}{\partial S})_{V} = (\frac{\partial T}{\partial V})_{S}

with a plus sign." — check it.

The

U

-relation carries a minus:

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

, because

N = - P

inherits the minus from the

- P d V

term.

"I'll memorize the four signs as

-, +, +, -

and be done." — why is this risky?

Memorized signs are easily scrambled under exam pressure; re-deriving from the differential (read off

M

N

with their signs, cross-differentiate) is foolproof and takes seconds.

Why questions

Why does the Legendre transform

H = U + P V

swap

V

for

P

as the natural variable?

Adding

P V

injects

+ P d V + V d P

; the

+ P d V

annihilates the existing

- P d V

, leaving

V d P

so that

d P

(not

d V

) becomes the independent differential (see Thermodynamic potentials & Legendre transforms).

Why are the relations from

F

and

G

called the "measurable" ones?

They put

S

(hard to measure directly) on the left equal to derivatives of

P, V, T

on the right, letting you infer entropy changes from ordinary lab quantities like thermal expansion.

Why must the potential be a state function for the Maxwell relation to hold?

Only state functions have exact differentials with path-independent, smooth second partials, which is the precise condition for the equality of mixed partials to apply.

Why is

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

physically satisfying?

The right side is (up to

- V

) the thermal expansion coefficient

α = \frac{1}{V} (\partial V / \partial T)_{P}

, so a substance that expands on heating loses entropy when compressed — expansion and entropy-pressure response are locked together.

Why does combining

d U = T d S - P d V

with a Maxwell relation give

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

Dividing by

d V

at fixed

T

yields

(\partial U / \partial V)_{T} = T (\partial S / \partial V)_{T} - P

, then the

F

-relation replaces the unmeasurable

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

with the measurable

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

Why can't we build a fifth independent Maxwell relation from these variables?

There are only four ways to pair the two conjugate couples

(T, S)

and

(P, V)

into a potential, so exactly four distinct relations exist for a simple system.

Edge cases

Edge case: For an ideal gas, what does the

F

-relation give for

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

(\partial P / \partial T)_{V} = n R / V

, so

(\partial S / \partial V)_{T} = n R / V > 0

— entropy always increases with volume at fixed

T

, consistent with a free (Joule) expansion increasing disorder.

Edge case: What happens to the

G

-relation for a substance with zero thermal expansion at some temperature?

(\partial V / \partial T)_{P} = 0

there, then

(\partial S / \partial P)_{T} = 0

too — squeezing at fixed

T

changes no entropy at that special point (as near water's density maximum).

Edge case: Do Maxwell relations still hold when phase transitions occur?

Within a single phase they hold, but at a first-order transition the potentials have kinks (discontinuous first derivatives), so the mixed-partial identity fails exactly at the coexistence line where derivatives aren't smooth.

Edge case: At absolute zero, what does the third law imply for

(\partial S / \partial P)_{T}

and

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

Since

S \to 0

independent of

P

and

V

T \to 0

, both

(\partial S / \partial P)_{T}

and

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

vanish, which via Maxwell forces

(\partial V / \partial T)_{P} \to 0

and

(\partial P / \partial T)_{V} \to 0

— thermal expansion dies at absolute zero.

Edge case: What if the system is not closed (matter can enter)?

A chemical-potential term

μ d N

appears in each differential, so the simple four-corner Maxwell relations are supplemented by new ones involving

μ

and

N

; the four given here assume fixed

N

Edge case: Is

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

ever trivially zero on both sides?

For a system where entropy and volume are effectively independent (e.g. an ideal incompressible/no-work idealization), both sides can vanish, but for any real gas the reversible-adiabatic temperature-volume coupling is nonzero.

Connections

Maxwell relations — derivation from each potential — the parent derivation
Equality of mixed partial derivatives (Schwarz theorem) — the calculus engine behind every relation
Thermodynamic potentials & Legendre transforms — why four potentials, why natural variables
Joule expansion and internal energy — the $(\partial U / \partial V)_{T}$ application
Heat capacities $C_P - C_V$ and Thermal expansion coefficient and isothermal compressibility — where these relations get cashed out