Thermodynamic potentials — U (internal), H (enthalpy), F (Helmholtz), G (Gibbs): Worked examples

This page is a drill room. The parent note built the four potentials, their differentials, and the Maxwell relations. Here we throw every kind of problem at those tools — every constraint you can clamp, every degenerate case, and the exam-style traps — and work each one from the ground up.

The scenario matrix

Every problem this topic can pose lives in one of these cells. The worked examples below are tagged with the cell(s) they cover, so by the end no cell is left dark.

#	Cell class	What is clamped	Right tool	Example
C1	Constant $V$ (rigid box)	$T, V$	$F$ , $C_{V}$	Ex. 1
C2	Constant $p$ (open beaker)	$T, p$	$G$ , $H$ , $C_{p}$	Ex. 2
C3	Constant $S$ (adiabatic + reversible)	$S, V$ or $S, p$	$U$ , $H$	Ex. 3
C4	Isothermal work / free energy	$T$	$Δ F = - W_{m a x}$	Ex. 4
C5	Sign / direction of spontaneity	$T, p$	$Δ G$ sign	Ex. 5
C6	Degenerate input: ideal gas ( $H, U$ depend on $T$ only)	—	Maxwell + EoS	Ex. 6
C7	Limiting case: $p \to 0$ , $T \to 0$ , $V \to \infty$	limits	asymptotics	Ex. 7
C8	Real-world word problem (phase change)	$T, p$	$G$ equality	Ex. 8
C9	Exam twist: which potential? wrong-tool trap	mixed	pick $Φ$	Ex. 9

Throughout, for a monatomic ideal gas we use $C_{V} = \frac{3}{2} n R$ , $C_{p} = \frac{5}{2} n R$ , and the gas constant $R = 8.314 J mol^{- 1} K^{- 1}$ . See Heat capacities Cp and Cv.

Example 1 — Cell C1: heating a gas in a rigid sealed box

Worked example Constant volume, constant

$N$ $2.0$ mol of monatomic ideal gas sits in a rigid sealed tank (volume cannot change). We warm it from $T_{1} = 300 K$ to $T_{2} = 400 K$ . Find the heat absorbed $Q$ and the internal-energy change $Δ U$ .

Forecast: Guess first — will $Q$ equal $Δ U$ here, or will some energy leak off as work?

Step 1. Name the clamp. Volume is fixed $\Rightarrow$ $d V = 0$ . Why this step? Rigid walls mean the gas can push on nothing that moves, so $p V$ -work $= p d V = 0$ .

Step 2. Use the master equation $d U = T d S - p d V$ with $d V = 0$ : $d U = T d S = δ Q .$ Why this step? With no work, the first law $d U = δ Q - δ W$ collapses to $d U = δ Q$ . So at constant volume, heat is the internal-energy change.

Step 3. Integrate using $C_{V} = (\partial U / \partial T)_{V} = \frac{3}{2} n R$ : $Δ U = Q = \int_{T_{1}}^{T_{2}} C_{V} d T = \frac{3}{2} n R (T_{2} - T_{1}) .$ Why this step? $C_{V}$ is precisely the amount of heat needed per degree at fixed volume — that's its definition.

Step 4. Numbers: $Q = \frac{3}{2} (2.0) (8.314) (400 - 300) = 2494.2 J \approx 2.49 kJ .$

Verify: Units: $mol \cdot J mol^{- 1} K^{- 1} \cdot K = J$ ✓. Since no work was done, $Q = Δ U$ exactly — matches the forecast if you predicted "all heat becomes internal energy." Positive $Q$ for heating ✓.

Example 2 — Cell C2: heating the same gas in an open beaker

Worked example Constant pressure, constant

$N$ Same $2.0$ mol of monatomic ideal gas, same $300 \to 400 K$ , but now in a cylinder with a frictionless piston open to the atmosphere (pressure fixed). Find the heat $Q$ and compare to Example 1.

Forecast: Will the heat needed be more or less than the $2.49 kJ$ of Example 1?

Step 1. Name the clamp: pressure fixed $\Rightarrow d p = 0$ . Why this step? At constant $p$ the natural potential is enthalpy, whose differential is $d H = T d S + V d p$ .

Step 2. Set $d p = 0$ : $d H = T d S = δ Q .$ Why this step? At constant pressure, heat equals the enthalpy change — this is why chemists call $Δ H$ the "heat of reaction."

Step 3. Integrate with $C_{p} = (\partial H / \partial T)_{p} = \frac{5}{2} n R$ : $Q = Δ H = \frac{5}{2} n R (T_{2} - T_{1}) = \frac{5}{2} (2.0) (8.314) (100) = 4157.0 J \approx 4.16 kJ .$

Step 4. The extra heat over Example 1 is $Q_{p} - Q_{V} = (C_{p} - C_{V}) Δ T = n R Δ T = (2.0) (8.314) (100) = 1662.8 J .$ Why this step? At constant $p$ the gas expands and spends energy pushing the atmosphere back ( $p Δ V$ ); that pushed-out work is the extra heat you must supply. This is exactly the $C_{p} - C_{V} = n R$ from the parent note's Example (c).

Verify: $Q_{p} > Q_{V}$ ✓ (constant- $p$ heating always costs more). $p Δ V = n R Δ T = 1662.8 J$ matches $Q_{p} - Q_{V}$ ✓. Units all Joules ✓.

Example 3 — Cell C3: reversible adiabatic compression (constant entropy)

Worked example Constant

$S$ (adiabatic + reversible) $1.0$ mol of monatomic ideal gas is compressed reversibly and adiabatically from $V_{1} = 22.4 L$ to $V_{2} = 11.2 L$ , starting at $T_{1} = 273 K$ . Find $T_{2}$ and $Δ U$ .

Forecast: Adiabatic means no heat in or out. Does the temperature rise, fall, or stay put when you squeeze it?

Step 1. Reversible + adiabatic $\Rightarrow$ $δ Q = 0$ and reversible $\Rightarrow δ Q = T d S$ , so $d S = 0$ : entropy is constant. Why this step? Constant $S$ makes $U (S, V)$ the natural potential; $d U = T d S - p d V = - p d V$ .

Step 2. For an ideal gas at constant $S$ , $T V^{γ - 1} = const$ with $γ = C_{p} / C_{V} = 5/3$ . Why this step? Setting $d S = 0$ in $d S = \frac{C _{V}}{T} d T + \frac{n R}{V} d V$ and integrating gives exactly this relation. See Entropy and the Second Law.

Step 3. Solve for $T_{2}$ : $T_{2} = T_{1} (\frac{V _{1}}{V _{2}})^{γ - 1} = 273 (\frac{22.4}{11.2})^{2/3} = 273 \cdot 2^{2/3} = 433.4 K .$

Step 4. Since $δ Q = 0$ , all internal-energy change came from work: $Δ U = \int C_{V} d T = \frac{3}{2} n R (T_{2} - T_{1}) = \frac{3}{2} (1.0) (8.314) (433.4 - 273) = 2000.4 J .$

Verify: Temperature rose (compression heats a gas) — matches the forecast if you predicted "rise" ✓. $Δ U > 0$ and equals the work done on the gas since $Q = 0$ ✓. $2^{2/3} \approx 1.587$ , so $273 \times 1.587 = 433.4$ ✓.

Example 4 — Cell C4: maximum work from an isothermal expansion

Worked example Constant

$T$ , free-energy = available work $1.0$ mol of ideal gas expands isothermally at $T = 300 K$ from $V_{1} = 10 L$ to $V_{2} = 30 L$ . Find $Δ F$ and the maximum work obtainable.

Forecast: The parent note says $- Δ F$ at constant $T$ is the maximum work. Do you expect $Δ F$ positive or negative when the gas expands?

Step 1. Constant $T$ $\Rightarrow d T = 0$ . Use $d F = - S d T - p d V = - p d V$ . Why this step? Helmholtz $F (T, V)$ has $T, V$ as natural variables — exactly what an isothermal process controls.

Step 2. Integrate with $p = n R T / V$ : $Δ F = - \int_{V_{1}}^{V_{2}} p d V = - n R T ln \frac{V _{2}}{V _{1}} .$ Why this step? Substitute the equation of state and integrate $1/ V$ .

Step 3. Numbers: $Δ F = - (1.0) (8.314) (300) ln \frac{30}{10} = - (2494.2) (1.0986) = - 2740.2 J .$

Step 4. Maximum work $W_{m a x} = - Δ F = 2740.2 J$ . Why this step? At constant $T$ , $d F = - δ W$ , so the drop in $F$ is the work the gas can deliver.

Verify: $Δ F < 0$ for expansion (gas can do work), matching the forecast ✓. Compare to reversible isothermal work $W = n R T ln (V_{2} / V_{1}) = + 2740.2 J = - Δ F$ ✓ — the two computations agree. Units: J ✓.

Example 5 — Cell C5: sign of $Δ G$ decides spontaneity

Worked example Which direction does it go?

At $T = 298 K$ and constant pressure, a reaction has $Δ H = - 50.0 kJ$ and $Δ S = - 100.0 J K^{- 1}$ . Is it spontaneous? At what temperature does it flip?

Forecast: Exothermic ( $Δ H < 0$ ) but entropy-lowering ( $Δ S < 0$ ) — a tug of war. Which wins at room temperature?

Step 1. At constant $T, p$ , spontaneity is governed by $G$ : spontaneous $⟺ Δ G < 0$ . Why this step? The parent note: at fixed $T, p$ , $G$ is minimized at equilibrium, so systems slide toward lower $G$ .

Step 2. Use $Δ G = Δ H - T Δ S$ (valid at constant $T$ ): $Δ G = - 50000 - (298) (- 100) = - 50000 + 29800 = - 20200 J = - 20.2 kJ .$ Why this step? $G = H - T S$ , so at fixed $T$ , $Δ G = Δ H - T Δ S$ .

Step 3. $Δ G < 0 \Rightarrow$ spontaneous at $298 K$ . The enthalpy term wins.

Step 4. Flip temperature: set $Δ G = 0$ : $T^{*} = \frac{Δ H}{Δ S} = \frac{- 50000}{- 100} = 500 K .$ Why this step? Above $T^{*}$ the $- T Δ S = + 100 T$ term (positive, since $Δ S < 0$ ) overwhelms $Δ H$ , making $Δ G > 0$ .

Verify: Check $T = 500$ : $Δ G = - 50000 - 500 (- 100) = 0$ ✓. At $T = 600 > T^{*}$ : $Δ G = - 50000 + 60000 = + 10000 > 0$ — non-spontaneous ✓, confirming the flip.

Example 6 — Cell C6: degenerate input, ideal-gas special structure

Worked example Show

$(\partial U / \partial V)_{T} = 0$ for an ideal gas (Joule's law) Prove the internal energy of an ideal gas does not change when volume changes at fixed temperature.

Forecast: Squeeze a gas at constant $T$ . Does its internal energy shift?

Step 1. Start from $d U = T d S - p d V$ and take $(\partial / \partial V)_{T}$ : $(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial S}{\partial V})_{T} - p .$ Why this step? We want a measurable expression; $(\partial S / \partial V)_{T}$ is not measurable directly, so we'll swap it via Maxwell.

Step 2. Apply the Maxwell relation from $F$ : $(\partial S / \partial V)_{T} = (\partial p / \partial T)_{V}$ : $(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial p}{\partial T})_{V} - p .$ Why this step? This is the parent note's boxed Maxwell relation — it trades an entropy derivative for a pressure derivative we can compute from the equation of state.

Step 3. Ideal gas $p = n R T / V \Rightarrow (\partial p / \partial T)_{V} = n R / V = p / T$ : $(\frac{\partial U}{\partial V})_{T} = T \cdot \frac{p}{T} - p = 0.$

Verify: This mirrors the parent's Example (a) which showed $(\partial H / \partial p)_{T} = 0$ ; here $(\partial U / \partial V)_{T} = 0$ is its twin. Both say ideal-gas $U$ and $H$ depend on $T$ only — the degenerate structure of the ideal gas ✓.

Example 7 — Cell C7: the three limiting cases $p \to 0$ , $T \to 0$ , $V \to \infty$

Worked example Isothermal Gibbs energy at its edges

Use $Δ G = n R T ln (p / p_{0})$ (parent Example b), where $p_{0}$ is a freely chosen reference pressure — a fixed baseline (typically the standard $p_{0} = 1 bar = 1 0^{5} Pa$ ) against which we measure the pressure $p$ . Because only the ratio $p / p_{0}$ enters the logarithm, $p_{0}$ merely sets the zero of $G$ ; the physics lives in how $p$ compares to it. Examine the three limits $p \to 0^{+}$ , $T \to 0^{+}$ , and $V \to \infty$ .

Forecast: As you let a gas expand at fixed $T$ , does its Gibbs energy blow up, vanish, or head to $- \infty$ ? And what happens when you cool the whole thing toward absolute zero?

Step 1. Degenerate anchor $p = p_{0}$ : $ln (1) = 0 \Rightarrow Δ G = 0$ . Why this step? Sanity check — the reference state must give zero change, confirming $p_{0}$ only fixes the baseline.

Step 2. (Limit $p \to 0^{+}$ .) $ln (p / p_{0}) \to - \infty$ , so $Δ G \to - \infty$ . Why this step? $ln$ of a shrinking positive argument diverges to $- \infty$ ; a gas always lowers its Gibbs energy by spreading into more volume (lower pressure) at fixed $T$ . Concrete point, $n = 1$ , $T = 300 K$ , $p / p_{0} = 0.01$ : $Δ G = (1) (8.314) (300) ln (0.01) = 2494.2 \times (- 4.6052) = - 11487.5 J .$

Step 3. (Limit $V \to \infty$ .) At fixed $T$ , ideal gas $p = n R T / V$ , so $p \to 0$ as $V \to \infty$ . Substituting $p / p_{0} = n R T / (p_{0} V)$ : $Δ G = n R T ln \frac{n R T}{p _{0} V} V \to \infty - \infty.$ Why this step? $V \to \infty$ is the same physical limit as $p \to 0$ read through the equation of state — infinite dilution drives $G$ to $- \infty$ , which is why a gas never spontaneously re-compresses itself.

Step 4. (Limit $T \to 0^{+}$ .) The prefactor $n R T \to 0$ . Even though $ln (p / p_{0})$ may be large, the product $T ln (p / p_{0}) \to 0$ for any fixed finite $p / p_{0}$ : $lim_{T \to 0^{+}} n R T ln (p / p_{0}) = 0.$ Why this step? The $T$ outside beats the (bounded, fixed) logarithm, so the isothermal pressure-dependence of $G$ switches off at absolute zero — consistent with the Third Law that entropy (and $S = - (\partial G / \partial T)_{p}$ -type responses) freeze out as $T \to 0$ . See Entropy and the Second Law.

Verify: $Δ G (p = p_{0}) = 0$ ✓. Sign negative for $p < p_{0}$ ✓. The $p \to 0$ point $- 11487.5 J$ is already large and negative, consistent with divergence ✓. $V \to \infty$ reproduces the $p \to 0$ limit ✓. $T \to 0$ gives $0$ ✓.

Example 8 — Cell C8: real-world phase change (Clausius–Clapeyron)

Worked example Water–vapour equilibrium at the boiling point

At a phase boundary, the two phases coexist. Water boils at $T = 373.15 K$ with latent heat $L = 40.66 kJ mol^{- 1}$ and molar volume change $Δ V = 3.03 \times 1 0^{- 2} m^{3} mol^{- 1}$ (vapour minus liquid). Estimate the slope $d p / d T$ of the boiling curve.

Forecast: Guess the order of magnitude of $d p / d T$ near boiling — thousands of Pa per K? tens?

Step 1. At the phase boundary the molar Gibbs energies are equal: $G_{liq} = G_{vap}$ , and staying on the boundary means $d G_{liq} = d G_{vap}$ . Why this step? Two coexisting phases at the same $T, p$ must have equal $G$ (else all matter flows to the lower- $G$ phase). See Phase equilibria & Clausius–Clapeyron.

Step 2. Insert $d G = - S d T + V d p$ for each phase and set them equal: $- S_{liq} d T + V_{liq} d p = - S_{vap} d T + V_{vap} d p \Rightarrow \frac{d p}{d T} = \frac{S _{vap} - S _{liq}}{V _{vap} - V _{liq}} = \frac{Δ S}{Δ V} .$ Why this step? The natural differential of $G$ carries $S$ and $V$ ; equating the two phases isolates the slope.

Step 3. Replace $Δ S$ by $L / T$ . Why this step? The phase change happens reversibly at fixed $T$ (both phases coexist at that one temperature), so the entropy jump is the reversible heat over temperature: $Δ S = δ Q_{rev} / T = L / T$ , where $L$ is the molar latent heat absorbed in going liquid $\to$ vapour. This is the crucial substitution that turns an un-measurable entropy difference into a measured latent heat. $\frac{d p}{d T} = \frac{Δ S}{Δ V} = \frac{L}{T Δ V} .$

Step 4. Numbers: $\frac{d p}{d T} = \frac{40660}{( 373.15 ) ( 3.03 \times 1 0 ^{- 2} )} = 3596.3 Pa K^{- 1} .$

Verify: Units: $\frac{J mol ^{- 1}}{K \cdot m ^{3} mol ^{- 1}} = \frac{J}{K m ^{3}} = \frac{Pa}{K}$ ✓ (since $1 J = 1 Pa \cdot m^{3}$ ). The result $\approx 3.6 kPa/K$ matches the known real slope of water's boiling curve near $10 0^{\circ} C$ (about $3.6 kPa/K$ ) ✓, and the positive sign ✓ says higher pressure raises the boiling point — exactly why a pressure cooker cooks hotter. See figure.

Example 9 — Cell C9: exam twist, "which potential?" trap

Worked example The wrong-tool trap

A gas is confined in a rigid insulated box and expands into a vacuum partition inside it (free expansion). A student writes " $Δ G = 0$ so nothing happens." Diagnose the error and find the correct invariant for an ideal gas.

Forecast: Is $G$ even the right potential when neither $T$ nor $p$ is controlled?

Step 1. Identify what is actually clamped. Rigid box $\Rightarrow$ total $V$ fixed. Insulated $\Rightarrow Q = 0$ . Free expansion does no external work $\Rightarrow W = 0$ , so $Δ U = Q - W = 0$ : internal energy is conserved. Why this step? The controlled/conserved quantity here is $U$ with $V$ fixed — not $T, p$ . So $G$ (natural vars $T, p$ ) is the wrong tool; neither $T$ nor $p$ is held fixed during free expansion.

Step 2. For an ideal gas, $U$ depends on $T$ only (Example 6), so $Δ U = 0 \Rightarrow Δ T = 0$ : temperature unchanged. Why this step? Constant $U$ + $U = U (T)$ forces constant $T$ for an ideal gas.

Step 3. Entropy, however, rises. With $T$ fixed and $V : V_{1} \to V_{2}$ , $Δ S = n R ln \frac{V _{2}}{V _{1}} > 0.$ Why this step? $(\partial S / \partial V)_{T} = (\partial p / \partial T)_{V} = n R / V$ ; integrate over the volume increase. The process is irreversible, so entropy increases even with $Q = 0$ .

Step 4. Numbers for $n = 1$ , doubling the volume ( $V_{2} / V_{1} = 2$ ): $Δ S = (1) (8.314) ln 2 = 5.763 J K^{- 1} .$

Verify: The student's " $Δ G = 0$ " is meaningless because $G$ 's natural clamp isn't active. Correct facts: $Δ U = 0$ , $Δ T = 0$ , $Δ S = + 5.763 J K^{- 1} > 0$ ✓ — consistent with the Second Law for an irreversible, isolated process ✓.

Active recall

Recall Match each scenario to its tool (cover the answers)

Constant volume, sealed tank — which heat capacity and potential? ::: $C_{V}$ and $F$ ; $Q = Δ U$ . Open beaker at fixed pressure — heat equals change in what? ::: $Δ H$ (constant- $p$ heat), via $C_{p}$ . Reversible adiabatic — which quantity is held constant? ::: entropy $S$ ( $d S = 0$ ). Isothermal — $- Δ F$ equals what physically? ::: the maximum work obtainable. Spontaneous at constant $T, p$ means the sign of what? ::: $Δ G < 0$ . Phase coexistence boundary condition? ::: equal molar $G$ of the two phases. Free expansion in an insulated rigid box conserves what? ::: internal energy $U$ (and for ideal gas, $T$ ).