Thermodynamic potentials — U, H, F, G (preview): Worked examples

This page is the "drill ground" for the potentials note. We will not re-derive the four differentials — we assume you have seen $d U = T d S - p d V, d H = T d S + V d p, d F = - S d T - p d V, d G = - S d T + V d p .$ Our goal here is different: hit every kind of situation these equations can be dropped into, including the weird corners (a sign flip, a zero, an ideal gas, a phase change, an exam trap). Each example tells you which cell of the matrix it fills.

The scenario matrix

Every cell below is covered by at least one worked example (its label is in brackets). The map figure below shows how the potentials and their slopes hang together — refer back to it as you work.

Figure — Thermodynamic potentials — U, H, F, G (preview)

Case class	Specific scenario	Example
Read a slope	positive slope (get $T$ or $V$ )	Ex 1
Sign trap	negative slope (get $- S$ , $- p$ )	Ex 2
Hold-fixed + integrate	$d p = 0$ : enthalpy = heat	Ex 3
Second derivative	Maxwell relation, both signs	Ex 4
Real-world number	ideal-gas $Δ H$ heating	Ex 5
Zero / degenerate input	$Δ (p V) = 0$ so $Δ H = Δ U$	Ex 6
Limiting behaviour	$T \to 0$ : free energy $\to$ internal energy	Ex 7
Spontaneity call	sign of $Δ G$ , all three sign cases	Ex 8
Exam-style twist	which potential? wrong-constraint trap	Ex 9

Ex 1 — Read a positive slope (cell: read a slope)

Forecast: guess before reading — is $T$ the $S$ -slope or the $V$ -slope of $U$ , and can it be negative?

Match coefficients. Write the chain rule $d U = (\frac{\partial U}{\partial S})_{V} d S + (\frac{\partial U}{\partial V})_{S} d V$ next to $d U = T d S - p d V$ . Why this step? Two expressions for the same $d U$ must have equal coefficients in front of $d S$ and $d V$ — that is the whole engine of the topic.
Pick off the $d S$ coefficient: $(\frac{\partial U}{\partial S})_{V} = T$ . Why this step? The coefficient of $d S$ on the physics side is exactly $+ T$ .
Sign meaning: $T$ is absolute temperature, so $T > 0$ always. Hence internal energy always rises as you add entropy at fixed volume — the $U$ -vs- $S$ curve has a strictly positive slope. Why this step? A sign check catches nonsense: a negative $T$ here would be unphysical.

Verify: For an ideal monatomic gas $U = \frac{3}{2} N k_{B} T$ and (up to constants) $S = N k_{B} ln (V T^{3/2}) + const$ , so at fixed $V$ , $T \propto e^{2 S /3 N k_{B}} > 0$ — always positive, slope always up. ✓

Ex 2 — The negative-slope trap (cell: sign trap)

Forecast: does the entropy come out with a plus or a minus in front?

Match the $d T$ coefficient: on the physics side it is $- S$ , so $(\frac{\partial F}{\partial T})_{V} = - S$ . Why this step? Same coefficient-matching, but now the coefficient itself carries a minus sign.
Solve for $S$ : $S = - (\frac{\partial F}{\partial T})_{V}$ . The student dropped the minus. Why this step? $S$ must be positive (third law), while $F$ decreases with rising $T$ , so $\partial F / \partial T < 0$ ; the extra minus turns it positive. Sign and physics agree only with the minus.
Same for pressure: $(\frac{\partial F}{\partial V})_{T} = - p$ , so $p = - (\frac{\partial F}{\partial V})_{T}$ . Why this step? Squeezing the box ( $V ↓$ ) raises $F$ , so $\partial F / \partial V < 0$ , and the minus makes $p > 0$ . ✓

Verify: Take the standard ideal-gas Helmholtz energy $F = - N k_{B} T ln (V) + f (T)$ , where $f (T)$ is the volume-independent piece (it collects all the purely temperature-dependent terms — the thermal/kinetic and constant parts — which do not matter for the $V$ -derivative). Then $- \partial F / \partial V = N k_{B} T / V = p$ , the ideal-gas law — positive, as required. ✓

Ex 3 — Hold pressure fixed, integrate (cell: hold-fixed + integrate)

Forecast: will $Q$ depend on how the volume changed, or only on the endpoints?

Start from $d H = T d S + V d p$ . Set $d p = 0$ : $d H = T d S$ . Why this step? "Constant pressure" is literally the instruction $d p = 0$ , which deletes the second term.
Recognise the heat. For a reversible path $δ Q = T d S$ , so $δ Q_{p} = d H$ . Why this step? This is the definition of entropy tying $T d S$ to reversible heat (see First Law of Thermodynamics).
Bring in $C_{p}$ . By definition $C_{p} = (\frac{\partial H}{\partial T})_{p}$ , so at constant $p$ , $d H = C_{p} d T$ . Why this step? We want an answer in measurable temperatures, and $C_{p}$ is exactly the $T$ -slope of $H$ (see Heat Capacities Cp and Cv).
Integrate: $Q_{p} = Δ H = \int_{T_{1}}^{T_{2}} C_{p} d T = C_{p} (T_{2} - T_{1})$ for constant $C_{p}$ . Why this step? Constant $C_{p}$ pulls out of the integral, leaving a plain difference.

Verify (numbers): $C_{p} = 29.1 J K^{- 1} mol^{- 1}$ (diatomic gas), heating $1$ mol from $T_{1} = 300$ K to $T_{2} = 350$ K: $Q_{p} = 29.1 \times (350 - 300) = 1455 J .$ Units: $J K^{- 1} \times K = J$ . ✓ Depends only on endpoints — good, $H$ is a state function.

Ex 4 — Both signs of a Maxwell relation (cell: second derivative)

Forecast: will the two mixed partials come out with the same sign or opposite signs?

The figure below is the Maxwell square for $G$ : the two arrows are the two routes to the same mixed second derivative $\partial^{2} G / \partial T \partial p$ , and following them forces the relation at the bottom.

Read both first slopes. From $d G$ : $(\frac{\partial G}{\partial T})_{p} = - S$ and $(\frac{\partial G}{\partial p})_{T} = + V$ . Why this step? $S$ follows $T$ (minus), $V$ follows $p$ (plus) — exactly the signs in $d G$ ; these are the two labels on the top of the square.
Cross-differentiate. Because $G$ is a state function, its mixed second derivative is path-independent: $\frac{\partial}{\partial p} (\frac{\partial G}{\partial T})_{p} = \frac{\partial}{\partial T} (\frac{\partial G}{\partial p})_{T} .$ Why this step? Equality of mixed partials (an exact differential) is the mathematical fact that manufactures every Maxwell relation; in the figure it is the statement that both arrows reach the same corner value.
Substitute the slopes from step 1: left side is $\frac{\partial ( - S )}{\partial p} = - (\frac{\partial S}{\partial p})_{T}$ ; right side is $\frac{\partial V}{\partial T} = (\frac{\partial V}{\partial T})_{p}$ . Why this step? Look at the amber arrow in the figure: differentiating the $- S$ label by $p$ must equal differentiating the $+ V$ label by $T$ .
Result: $(\frac{\partial S}{\partial p})_{T} = - (\frac{\partial V}{\partial T})_{p}$ . The minus survives here (contrast the $F$ -relation which had no leftover minus). Why this step? The lone minus came from the $- S$ but not from the $+ V$ — asymmetric signs give an asymmetric result.

Verify: Ideal gas $V = n R T / p$ . Then $(\frac{\partial V}{\partial T})_{p} = n R / p$ , so the relation predicts $(\frac{\partial S}{\partial p})_{T} = - n R / p$ . From $S = n R ln (V) + g (T) = n R ln (n R T / p) + g (T)$ , where $g (T)$ is the volume-independent (purely temperature-dependent) part of the entropy that drops out under $\partial / \partial p$ , we indeed get $\partial S / \partial p = - n R / p$ . ✓

Ex 5 — A real-world number: heating a gas (cell: real-world number)

Forecast: which is bigger, and by how much — the gap should equal $Δ (p V)$ .

Pick capacities. Monatomic ideal gas: $C_{v} = \frac{3}{2} R$ , $C_{p} = \frac{5}{2} R$ per mole, with $R = 8.314 J K^{- 1} mol^{- 1}$ . Why this step? $Δ H$ uses $C_{p}$ , $Δ U$ uses $C_{v}$ ; the whole point of $H$ vs $U$ is this pair.
Compute $Δ H = n C_{p} Δ T$ : $2.0 \times \frac{5}{2} (8.314) \times 50 = 2078.5 J$ . Why this step? Ex 3 gave $Δ H = \int C_{p} d T$ ; constant $C_{p}$ makes it $n C_{p} Δ T$ .
Compute $Δ U = n C_{v} Δ T$ : $2.0 \times \frac{3}{2} (8.314) \times 50 = 1247.1 J$ . Why this step? $U$ 's natural heat measure at fixed volume is $C_{v}$ ; for an ideal gas $Δ U$ depends only on $T$ .
Check the gap: $Δ H - Δ U = 831.4 J$ , which should equal $Δ (p V) = n R Δ T = 2.0 \times 8.314 \times 50 = 831.4 J$ . Why this step? $H = U + p V$ , so $Δ H - Δ U = Δ (p V)$ ; matching confirms no arithmetic slip.

Verify: $2078.5 - 1247.1 = 831.4$ and $n R Δ T = 831.4$ . Units all joules. ✓ $Δ H > Δ U$ because expanding gas does work on the atmosphere. ✓

Ex 6 — A degenerate case: $Δ (p V) = 0$ (cell: zero / degenerate input)

Forecast: if $p V$ returns to its old value, does the enthalpy "extra term" vanish?

Write the definition change. $Δ H = Δ U + Δ (p V)$ . Why this step? Enthalpy is defined as $U + p V$ , so its change splits cleanly.
Insert the degenerate condition. $Δ (p V) = p_{2} V_{2} - p_{1} V_{1} = 0$ by hypothesis. Why this step? The "zero input" here is $Δ (p V) = 0$ ; this is the corner case where the added $p V$ term contributes nothing.
Conclude: $Δ H = Δ U + 0 = Δ U$ . Why this step? With the correction term gone, the two potentials track each other exactly.

Verify (numbers): ideal gas isothermal at $T = 300$ K, $n = 1$ : $p V = n R T = 2494.2$ J at both endpoints, so $Δ (p V) = 0$ and $Δ U = 0$ (ideal gas, constant $T$ ) — hence $Δ H = 0$ too. ✓ Consistent.

Ex 7 — Limiting behaviour as $T \to 0$ (cell: limiting behaviour)

Forecast: what happens to the "tax" term $T S$ when $T$ hits zero?

Take the definition to the limit. $F = U - T S$ . As $T \to 0$ , the entropy of a perfect crystal tends to zero (third law), and even for a general system $S$ stays bounded, so the product $T S \to 0$ . Hence $F \to U$ . Why this step? The entropic tax $T S$ scales with $T$ ; with $T \to 0$ and $S$ finite (indeed $S \to 0$ for a perfect crystal), the tax is squeezed to nothing.
Compare differentials. $d F = - S d T - p d V$ . Consider a change made at (or infinitesimally near) $T = 0$ , where $S \to 0$ : the term $- S d T$ vanishes because its coefficient $S$ goes to zero, leaving $d F \to - p d V$ . Why this step? We want to see which term of $d F$ survives at absolute zero; since $S \to 0$ kills the coefficient of $d T$ , only the mechanical $- p d V$ term remains — this is what makes $F$ behave like $U$ there.
Recover $U$ 's slope. At fixed $S$ , $d U = - p d V$ too, so the $- p d V$ mechanical part is shared: $(\frac{\partial F}{\partial V})_{T} \to - p = (\frac{\partial U}{\partial V})_{S}$ in this limit. Why this step? Both potentials reduce to pure mechanical work when the thermal terms die.

Verify: For a perfect crystal $S \to 0$ as $T \to 0$ (third law), so with $U = U_{0}$ finite, at $T = 0$ : $F = U_{0} - 0 \cdot 0 = U_{0} = U$ . Numerically, take $U_{0} = 100$ J and residual $S = 0$ J/K: $F = 100 - 0 \times 0 = 100$ J $= U$ . ✓ (Even a hypothetical finite $S$ still gives $T S \to 0$ , so the conclusion is robust.) See Entropy and the Second Law for the third-law statement $S \to 0$ .

Ex 8 — Spontaneity: all three signs of $Δ G$ (cell: spontaneity call)

Forecast: guess the sign of $Δ G$ for each before computing.

Use the right potential. At constant $T, p$ the boss is Gibbs: $Δ G = Δ H - T Δ S$ , and $Δ G < 0$ means spontaneous. Why this step? The matching-constraint rule: $T, p$ fixed ⇒ minimise $G$ (see Gibbs Free Energy and Chemical Equilibrium).
Case (a): $Δ G = - 40000 - 298 (50) = - 40000 - 14900 = - 54900 J = - 54.9 kJ < 0$ . Spontaneous (both terms help). Why this step? Exothermic and entropy-increasing — the easy always-go case.
Case (b): $Δ G = + 30000 - 298 (- 20) = 30000 + 5960 = + 35960 J = + 35.96 kJ > 0$ . Not spontaneous (both terms oppose). Why this step? Endothermic and entropy-decreasing — the always-stuck case.
Case (c): $Δ G = + 50000 - 298 (100) = 50000 - 29800 = + 20200 J = + 20.2 kJ > 0$ at 298 K — not yet spontaneous, but becomes spontaneous above $T^{*} = Δ H /Δ S = 50000/100 = 500 K$ . Why this step? When $Δ H > 0$ and $Δ S > 0$ , the entropy term wins only once $T$ is large enough — a temperature-triggered case.

Verify: (a) $- 54900$ J, (b) $+ 35960$ J, (c) $+ 20200$ J, crossover $T^{*} = 500$ K. ✓ Signs: (a) −, (b) +, (c) + below 500 K. ✓

Ex 9 — Exam twist: the wrong-constraint trap (cell: exam-style twist)

Forecast: is $F$ even the right potential to judge this?

Identify the constraints. The lab holds $T$ and $p$ fixed (open flask + thermostat). Why this step? The very first move in any potential problem: what is held constant decides the boss.
Pick the matching potential. Constant $T, p$ ⇒ spontaneity is governed by $Δ G$ , not $Δ F$ . $F$ is minimised only at constant $T, V$ . Why this step? Using $F$ here is the classic wrong-constraint error; $F$ need not decrease when $V$ is free to change.
Show they can disagree. $G = F + p V$ , so $Δ G = Δ F + Δ (p V)$ . If the volume shrinks so that $Δ (p V) = - 8$ kJ, then $Δ G = + 5 - 8 = - 3 kJ < 0$ — spontaneous despite $Δ F > 0$ . Why this step? A concrete number proves the student's logic is unsound: the wrong potential gave the wrong verdict.

Verify: $Δ G = Δ F + Δ (p V) = 5 + (- 8) = - 3$ kJ $< 0$ . ✓ Correct verdict: it can proceed, because at constant $T, p$ only the sign of $Δ G$ matters.

Recall Quick self-test

A gas heats at constant $p$ ; is $Q = Δ H$ or $Δ U$ ? ::: $Q = Δ H$ (constant pressure, Ex 3) Sign of $S$ from $F$ ? ::: $S = - (\partial F / \partial T)_{V}$ , the minus is essential (Ex 2) When does $Δ H = Δ U$ ? ::: When $Δ (p V) = 0$ (Ex 6) Reaction with $Δ H > 0, Δ S > 0$ turns spontaneous above which $T$ ? ::: $T^{*} = Δ H /Δ S$ (Ex 8c, 500 K) Which potential judges constant- $T, p$ spontaneity? ::: Gibbs $G$ , not $F$ (Ex 9)

Related builders: Legendre Transform (how the potentials are made), Maxwell Relations (Ex 4), Heat Capacities Cp and Cv (Ex 3, 5).