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

L1 — Recognition

Recall Solution 1.1

WHAT we do: just read the table.

$U (S, V)$ , $H (S, p)$ , $F (T, V)$ , $G (T, p)$ .
Open beaker ⇒ pressure $p$ is fixed by the atmosphere. Hot-plate/thermostat ⇒ temperature $T$ is fixed.
The potential whose natural variables are exactly $(T, p)$ is $== G ==$ . So Gibbs free energy $G$ is minimized.

WHY this is the whole point: you pick the potential whose natural variables match what you physically hold constant. Match $(T, p) \to G$ .

Recall Solution 1.2

From $d G = - S d T + V d p$ , compare with the generic $d G = (\frac{\partial G}{\partial T})_{p} d T + (\frac{\partial G}{\partial p})_{T} d p$ . Matching the coefficient of $d T$ and of $d p$ : $- S = (\frac{\partial G}{\partial T})_{p}, V = (\frac{\partial G}{\partial p})_{T} .$ WHY: the differential is the dictionary; the coefficient sitting in front of $d T$ is the slope of $G$ in the $T$ direction at fixed $p$ .

L2 — Application

Recall Solution 2.1

Step 1 — pick the tool. We hold $T$ fixed, so use $d G = - S d T + V d p$ . With $d T = 0$ the first term dies and $d G = V d p$ . Step 2 — substitute the equation of state. $V = n R T / p$ , so $Δ G = \int_{p_{0}}^{p} \frac{n R T}{p} d p = n R T ln \frac{p}{p _{0}} .$ Step 3 — put numbers in. $n R T = 1 \times 8.314 \times 300 = 2494.2 J$ , and $ln (5/1) = 1.6094$ . $Δ G = 2494.2 \times 1.6094 \approx 4014 J .$ WHY positive: compressing a gas (raising $p$ ) at fixed $T$ raises $G$ , and this same $R T ln (p / p_{0})$ is the backbone of the chemical potential $μ = μ^{\circ} + R T ln (p / p^{\circ})$ .

Recall Solution 2.2

Step 1: $C_{p} - C_{V} = n R$ with $n = 1$ gives $C_{p} = C_{V} + R$ . Step 2: $C_{V} = \frac{3}{2} R$ , so $C_{p} = \frac{3}{2} R + R = \frac{5}{2} R$ . Step 3: $\frac{5}{2} \times 8.314 = 20.785 J mol^{- 1} K^{- 1}$ . WHY $C_{p} > C_{V}$ : at constant pressure the gas expands as you heat it, spending some energy pushing back the atmosphere ( $p d V$ ), so you must supply more heat per degree. See Heat capacities Cp and Cv.

L3 — Analysis

Recall Solution 3.1

Step 1 — divide $d U$ by $d V$ at fixed $T$ . Treat $S = S (T, V)$ . Holding $T$ constant and dividing $d U = T d S - p d V$ by $d V$ : $(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial S}{\partial V})_{T} - p .$ Why this step? We want $U$ 's dependence on volume at fixed $T$ , so we differentiate along the $V$ -direction with $T$ pinned. Step 2 — kill the unmeasurable $\partial S / \partial V$ with a Maxwell relation. From $d F = - S d T - p d V$ , equality of mixed partials gives $(\frac{\partial S}{\partial V})_{T} = (\frac{\partial p}{\partial T})_{V} .$ Why this one? $F$ has natural variables $(T, V)$ — exactly the two we are holding/varying — so its Maxwell relation converts entropy-vs-volume into pressure-vs-temperature, which we can measure. Step 3 — assemble. $(\frac{\partial U}{\partial V})_{T} = T (\frac{\partial p}{\partial T})_{V} - p . ✓$ Step 4 — ideal gas. $p = n R T / V \Rightarrow (\frac{\partial p}{\partial T})_{V} = n R / V = p / T$ . Then $(\frac{\partial U}{\partial V})_{T} = T \cdot \frac{p}{T} - p = 0.$ Interpretation: an ideal gas's internal energy does not depend on volume at fixed $T$ — energy lives only in temperature. (This is the "Joule expansion" fact.)

Recall Solution 3.2

Step 1 — start from the parent's result. $C_{p} - C_{V} = T (\frac{\partial p}{\partial T})_{V} (\frac{\partial V}{\partial T})_{p}$ . Step 2 — rewrite $(\partial p / \partial T)_{V}$ using the triple product rule. $(\frac{\partial p}{\partial T})_{V} = - \frac{( \partial V / \partial T ) _{p}}{( \partial V / \partial p ) _{T}} .$ Why? This "cyclic" identity lets us express the awkward constant- $V$ derivative through two constant- $p$ /constant- $T$ ones we can measure. Step 3 — substitute: $C_{p} - C_{V} = - T \frac{( \partial V / \partial T ) _{p}^{2}}{( \partial V / \partial p ) _{T}} . ✓$ Step 4 — sign. The numerator has a square (≥ 0) times $T$ (> 0). Stability demands $(\partial V / \partial p)_{T} < 0$ (squeeze harder ⇒ smaller volume). A positive number divided by a negative, with an overall minus, is ≥ 0. Hence $C_{p} \geq C_{V}$ always. Step 5 — ideal-gas check. $(\partial V / \partial T)_{p} = n R / p$ , $(\partial V / \partial p)_{T} = - n R T / p^{2}$ . Then $- T \cdot \frac{( n R / p ) ^{2}}{- n R T / p ^{2}} = - T \cdot \frac{n ^{2} R ^{2} / p ^{2}}{- n R T / p ^{2}} = - T \cdot (- \frac{n R}{T}) = n R . ✓$

L4 — Synthesis

Recall Solution 4.1

Step 1 — general formula. From $d H = T d S + V d p$ , dividing by $d p$ at fixed $T$ : $(\frac{\partial H}{\partial p})_{T} = T (\frac{\partial S}{\partial p})_{T} + V .$ Step 2 — Maxwell from $G$ . $d G = - S d T + V d p$ gives $(\frac{\partial S}{\partial p})_{T} = - (\frac{\partial V}{\partial T})_{p}$ . So $(\frac{\partial H}{\partial p})_{T} = V - T (\frac{\partial V}{\partial T})_{p} .$ Why this route: we need a form built only from measurable $p, V, T$ derivatives. Step 3 — apply the equation of state. Solve for $V$ : $V = \frac{n R T}{p} + nb$ . At fixed $p$ , $(\frac{\partial V}{\partial T})_{p} = \frac{n R}{p} .$ Step 4 — assemble. $(\frac{\partial H}{\partial p})_{T} = (\frac{n R T}{p} + nb) - T \cdot \frac{n R}{p} = nb .$ Interpretation: the ideal-gas cancellation leaves a residue $nb$ . Enthalpy of this gas rises with pressure at a constant slope $nb$ — the excluded volume never lets the two terms fully cancel. (Contrast the ideal-gas parent result, where it was zero.)

Recall Solution 4.2

Step 1 — coexistence condition. On the boundary the molar Gibbs energies are equal, $g_{1} (T, p) = g_{2} (T, p)$ . Move a tiny step along the boundary: both change but stay equal, so $d g_{1} = d g_{2}$ . Step 2 — write each $d g$ . Using $d g = - s d T + v d p$ (lower case = per mole) for each phase: $- s_{1} d T + v_{1} d p = - s_{2} d T + v_{2} d p .$ Step 3 — collect. $(v_{1} - v_{2}) d p = (s_{1} - s_{2}) d T \Rightarrow \frac{d p}{d T} = \frac{s _{1} - s _{2}}{v _{1} - v _{2}} = \frac{Δ s}{Δ v} .$ Step 4 — latent heat. A reversible phase change at fixed $T$ absorbs heat $L = T Δ s$ , so $Δ s = L / T$ and $\frac{d p}{d T} = \frac{L}{T Δ v} .$ This is Clausius–Clapeyron: the slope of the coexistence line is set purely by latent heat and the volume jump.

L5 — Mastery

Recall Solution 5.1

(a) $Δ U$ . For an ideal gas $U$ depends only on $T$ (Problem 3.1). The process is isothermal, so $Δ U = 0.$ (b) $Δ S$ . Use $d F = - S d T - p d V$ 's sibling: at fixed $T$ , from $d U = T d S - p d V$ and $d U = 0$ , $T d S = p d V \Rightarrow Δ S = \int_{V_{0}}^{V} \frac{p}{T} d V = \int_{V_{0}}^{V} \frac{n R}{V} d V = n R ln \frac{V}{V _{0}} .$ Numerically: $n R ln (30/10) = 8.314 \times ln 3 = 8.314 \times 1.0986 = 9.134 J K^{- 1}$ . (c) $Δ F$ . At fixed $T$ , $d F = - p d V$ , so $Δ F = - \int_{V_{0}}^{V} p d V = - \int_{V_{0}}^{V} \frac{n R T}{V} d V = - n R T ln \frac{V}{V _{0}} .$ Numerically: $- 8.314 \times 400 \times 1.0986 = - 3653.5 J$ . Cross-check via definition: $F = U - T S$ , and at fixed $T$ , $Δ F = Δ U - T Δ S = 0 - 400 \times 9.134 = - 3653.5 J$ . ✓ (Same number, two routes.) (d) Maximum work. For a reversible isothermal expansion the work done by the gas is $W = \int_{V_{0}}^{V} p d V = n R T ln \frac{V}{V _{0}} = 3653.5 J .$ And indeed $- Δ F = + 3653.5 J = W_{m a x}$ . This is the parent statement " $- Δ F$ at constant $T$ = maximum work." ✓

Figure — Thermodynamic potentials — U (internal), H (enthalpy), F (Helmholtz), G (Gibbs)

Reading the figure: the red curve is the isotherm $p = n R T / V$ ; the shaded area under it (from $V_{0}$ to $V$ ) is exactly $\int p d V = W_{m a x} = - Δ F$ .

Recall Solution 5.2

Step 1 — energy. Free expansion: $W = 0$ (vacuum, nothing to push) and $Q = 0$ (insulated). First law $Δ U = Q - W = 0$ . Step 2 — temperature. $U = \frac{3}{2} R T$ depends only on $T$ , and $Δ U = 0 \Rightarrow Δ T = 0$ . Final $T = 400 K$ , unchanged. Step 3 — entropy. $S$ is a state function, so its change depends only on the endpoints, not the path. The endpoints ( $T = 400 K$ , $V : 10 \to 30 L$ ) are identical to Problem 5.1. Therefore $Δ S = n R ln \frac{V}{V _{0}} = 9.134 J K^{- 1},$ exactly as before. Step 4 — the lesson. Work and heat are path-dependent; entropy and internal energy are state functions. The reversible isotherm and the irreversible free expansion connect the same two states, so $Δ U$ and $Δ S$ match, but only the reversible path actually extracts the $3653.5 J$ of work. The free expansion "wastes" that available work — that lost opportunity is measured by $- Δ F$ .

Recall Final self-audit (cover and recall)

Answer these before leaving the page. Which potential is minimized at fixed $T, p$ ? ::: $G$ $(\partial U / \partial V)_{T}$ for an ideal gas equals? ::: $0$ Is $C_{p} - C_{V}$ ever negative? ::: No — it is $- T (\partial V / \partial T)_{p}^{2} / (\partial V / \partial p)_{T} \geq 0$ $- Δ F$ at constant $T$ measures? ::: the maximum work obtainable Clausius–Clapeyron slope in terms of latent heat? ::: $d p / d T = L / (T Δ v)$ Why do free and reversible expansions share the same $Δ S$ ? ::: $S$ is a state function; same endpoints

Potential	Definition	Natural vars	Differential
$U$	—	$S, V$	$d U = T d S - p d V$
$H$	$U + p V$	$S, p$	$d H = T d S + V d p$
$F$	$U - T S$	$T, V$	$d F = - S d T - p d V$
$G$	$H - T S$	$T, p$	$d G = - S d T + V d p$