Thermodynamic potentials — U (internal), H (enthalpy), F (Helmholtz), G (Gibbs): Question bank

True or false — justify

Each line: decide true or false, then give the why. Reveal only after committing.

d U = T d S - p d V

holds only for reversible processes.

False.

U

is a state function, so

d U

depends only on endpoints; we derive it reversibly but it holds for any process between the same two states. Only the individual pieces

T d S = δ Q

and

p d V = δ W

are reversible-path statements.

F

and

G

are just two names for "the free energy" and are interchangeable.

False. They differ by

p V

(recall

G = F + p V

) and have different natural variables —

F = F (T, V)

G = G (T, p)

. Which one is minimized at equilibrium depends on what you clamp: volume (

F

) or pressure (

G

At equilibrium every thermodynamic potential is minimized.

False. Each is minimized only when its own natural variables are held fixed:

U

at fixed

(S, V)

H

at fixed

(S, p)

F

at fixed

(T, V)

G

at fixed

(T, p)

. Minimize the wrong one for your constraints and the criterion is simply false.

H = U + p V

, so enthalpy is always larger than internal energy.

False in general.

p V

is positive for a gas so

H > U

there, but the definition is just a Legendre transform for convenience; nothing in physics requires

p V

to dominate any comparison, and for other systems (e.g. negative pressures) it can flip.

The Legendre transform to get

H

adds

p V

while the one to get

F

subtracts

T S

, so the rule is inconsistent.

False — it is one consistent rule. You always subtract the conjugate-pair product to swap a variable for its slope; the sign is inherited from

d U = T d S - p d V

. Since

V

carries the minus (

- p d V

), swapping it adds

p V

; since

S

carries the plus, swapping it subtracts

T S

- Δ F

at constant

T

equals the maximum work the system can do.

True. From

d F = - S d T - p d V

, at constant

T

we get

d F = - p d V = - δ W

, so

- Δ F

is the total (maximum, reversible) work extractable when temperature is held fixed.

(Δ H)_{p}

equals the heat exchanged at constant pressure.

True. At constant

p

d H = T d S + V d p = T d S = δ Q

, so enthalpy change is the constant-pressure heat — this is why chemists tabulate reaction heats as

Δ H

For an ideal gas,

G

depends on pressure but

H

does not.

True.

(\partial H / \partial p)_{T} = 0

for an ideal gas so

H = H (T)

only; but

(\partial G / \partial p)_{T} = V = n R T / p \neq = 0

, giving the famous

Δ G = n R T ln (p / p_{0})

A Maxwell relation lets you replace an unmeasurable derivative like

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

with a measurable one.

True. Exactness of

d F

forces

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

; the right side is read off the equation of state, so you never need to measure entropy's volume-dependence directly.

Spot the error

Each line states a flawed claim/derivation. Name the mistake and correct it.

d G = - S d T + V d p

, so at constant temperature

Δ G = V Δ p

Error: treating

V

as constant.

V

depends on

p

, so you must integrate

Δ G = \int V d p

; for an ideal gas

V = n R T / p

gives

n R T ln (p / p_{0})

, not

V Δ p

H = U + p V

therefore

d H = d U + p V

Error: forgot the product rule.

d (p V) = p d V + V d p

, so

d H = d U + p d V + V d p = T d S + V d p

. Writing

d H = d U + p V

isn't even dimensionally a differential.

"From

d U = T d S - p d V

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

Error: wrong partial.

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

(coefficient of

d S

); the

V

-derivative gives

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

. Match each coefficient to its own differential.

"The Maxwell relation from

G

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

Error: sign. From

d G = - S d T + V d p

, equating mixed partials of

- S

and

V

gives

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

— the minus is forced by the

- S d T

term, not optional.

"Since

F = U - T S

and

U

is minimized at equilibrium (fixed

S, V

F

is also minimized at fixed

S, V

Error: mixing constraint sets.

F

is minimized at fixed $(T, V)$ , not

(S, V)

. Under fixed

(S, V)

it's

U

that governs; each potential owns exactly one constraint pair.

"For an ideal gas

C_{p} - C_{V} = n R

, and this is a special coincidence of ideal gases."

Half-wrong. The general result

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

holds for any substance; substituting the ideal-gas equation of state happens to collapse it to

n R

. The formula is universal, the value

n R

is not.

d F = - S d T - p d V

shows

F

decreases whenever temperature rises, since

- S d T < 0

Error: this only holds at constant

V

, and

S > 0

makes

(\partial F / \partial T)_{V} = - S < 0

, so

F

does fall as

T

rises at fixed

V

— but you cannot conclude it from

d F

alone if

V

also changes; the

- p d V

term contributes too.

Why questions

Give a one-or-two-sentence mechanistic reason, not a restatement.

Why do we need four potentials instead of just using

U

for everything?

Because you control in the lab what you can clamp — usually

T

(thermostat) and

p

(atmosphere), not

S

and

V

. Each potential re-expresses the same physics in the variables you can actually hold fixed, so its minimum directly signals equilibrium.

Why must every

d Φ

be an exact differential, and why does that matter?

Because each

Φ

is a state function — its value depends only on the state, so its differential closes on any loop. Exactness forces equality of mixed second partials, which is the source of every Maxwell relation.

Why is

G

, not

F

, the chemist's workhorse for reactions?

Reactions in a beaker happen at fixed

T

and fixed

p

(open to the room), and those are exactly

G

's natural variables — so

G

being minimized is the correct equilibrium criterion for that setup.

Why does the Legendre transform subtract the conjugate product rather than substitute a formula?

Subtracting

x y

where

y

is the slope of

Φ

x

makes the

d x

term cancel and a

d y

term appear, cleanly swapping the natural variable for its conjugate — no equation-of-state input needed.

Why is

(\partial H / \partial p)_{T} = 0

special about the ideal gas rather than universal?

It requires

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

, which the ideal-gas law satisfies exactly; real gases with intermolecular forces violate it, giving nonzero

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

(the basis of Joule–Thomson cooling).

Why do the four Maxwell relations carry different signs?

The sign of each is copied straight from the sign of the conjugate term in that potential's differential — e.g.

- S d T

injects a minus. They are never guessed; they are read off

d Φ

Why is

- Δ F

the maximum work and not just the work?

Only a reversible path saturates

δ W = p d V

; irreversibility dissipates some of the available energy, so real processes deliver less than

- Δ F

, making it the ceiling.

Edge cases

Boundary and degenerate scenarios — reason carefully.

What happens to

G = U - T S + p V

at absolute zero,

T \to 0

The

T S

term vanishes (and by the third law

S \to

const or

0

), so

G \to U + p V = H

; at

T = 0

the "messiness tax" disappears and Gibbs energy equals enthalpy.

If a process holds both

S

and

T

fixed, which potential governs equilibrium?

This over-constrains a simple one-component system —

S

and

T

are conjugate, so fixing both generically pins the whole state. There is no free minimization; the "which potential" question is ill-posed for that pair.

For an incompressible solid (

d V = 0

always), how do

F

and

G

, and

U

and

H

, relate?

With

V

effectively constant,

H = U + p V

and

G = F + p V

differ from

U

and

F

by an additive constant per unit pressure; changes

Δ H \approx Δ U

and

Δ G \approx Δ F

, so the volume-work distinction nearly collapses.

At a first-order phase transition (e.g. boiling), which potential is continuous across the boundary and why?

G

(per mole) is continuous — the two phases coexist precisely because they share the same

T

p

, and molar

G

. Its derivatives (

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

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

) jump, which is what makes it "first-order".

What is

Δ F

for a free (irreversible) isothermal expansion into vacuum for an ideal gas?

Since

F = F (T, V)

and

U

(hence

T

) is unchanged for ideal-gas free expansion,

Δ F = - T Δ S < 0

because

S

increases — even though no work was extracted, showing

- Δ F

is only the potential work, not the realized work.

If entropy

S = 0

(a perfectly ordered

T = 0

crystal), does

F = U

exactly?

Yes:

F = U - T S

and

T S = 0

when either factor is zero, so

F = U

. The "free" energy equals the total energy because there is no unusable, entropy-locked portion.

Can

Δ H

be negative while

Δ G

is positive for a process at fixed

T, p

Yes.

Δ G = Δ H - T Δ S

; a strongly negative

Δ S

(large ordering) can make

- T Δ S

outweigh a negative

Δ H

, giving

Δ G > 0

— the reaction is exothermic yet non-spontaneous.

Recall One-line takeaways

A potential is minimized only at fixed values of its own natural variables. ::: $U : (S, V)$ , $H : (S, p)$ , $F : (T, V)$ , $G : (T, p)$ .
Maxwell-relation signs come from ::: the signs in the matching $d Φ$ — never guessed.
$- Δ F$ (const $T$ ) and $- Δ G$ (const $T, p$ ) are ::: maximum obtainable work, saturated only reversibly.