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

Definition Three heat symbols you will keep meeting:

$δ Q$ , $δ Q_{rev}$ , $δ Q_{p}$ $d$ (as in $d H$ , $d U$ ) is an exact differential — a tiny change in a state function, a quantity that depends only on where the system is, not how it got there. The total change $\int d H = H_{final} - H_{initial}$ is path-independent. $δ Q$ (with the curly $δ$ , not $d Q$ ) is an inexact differential — a bit of heat crossing the boundary. Heat $Q$ is not a state function; how much flows depends on the path, so there is no function $Q$ to difference. The $δ$ is the warning sign. $δ Q_{rev}$ is that heat measured along a reversible path — the only case for which $δ Q_{rev} = T d S$ holds exactly. $δ Q_{p}$ (used in L2.2) is the same inexact heat $δ Q$ , but with the extra tag " $p$ " meaning measured while pressure is held constant. It is not a new object — just $δ Q$ specialised to a constant-pressure path (and taken reversible, so $δ Q_{p} = T d S$ there). Think of the subscript as naming the experimental condition, exactly like the subscript on a partial derivative.

Intuition The one picture behind every exercise (read before Level 1)

Look at the two panels below. Left panel: the Legendre transform is not algebra magic — it is a picture. The orange curve is $U (S)$ ; its slope at each point is the temperature $T$ . To swap the variable $S$ for its slope $T$ , we slide a straight line of slope $T$ down until it just touches (is tangent to) the curve, and record where it crosses the vertical axis — that intercept is $F = U - T S$ . The magenta double-arrow is the subtracted product $T S$ . That is the whole content of " $F = U - T S$ ": trade a point for a tangent line. Right panel: why "minimum of a potential = equilibrium." The violet bowl is a potential (say $F$ ) plotted against some internal coordinate the system can adjust; a ball placed anywhere rolls to the bottom. The bottom is where the slope is zero — equilibrium. Everything on this page is either (a) reading a slope off a differential, or (b) finding such a minimum.

Level 1 — Recognition

Here we only ask you to read off the right formula, name, or variable. No calculation, no trap-doors — just knowing which wallet is which.

L1.1 — Name the natural variables

For each potential, state its two natural variables (the two whose differentials appear in $d (\cdot)$ ): $U$ , $H$ , $F$ , $G$ .

Recall Solution L1.1

Read them straight off the four differentials: $d U = T d S - p d V \Rightarrow (S, V)$ $d H = T d S + V d p \Rightarrow (S, p)$ $d F = - S d T - p d V \Rightarrow (T, V)$ $d G = - S d T + V d p \Rightarrow (T, p)$ Rule of thumb: the natural variables are exactly the two things whose differentials ( $d S, d V, d p, d T$ ) sit on the right-hand side.

L1.2 — Which potential is minimized?

A reaction runs in an open beaker on a lab bench — the atmosphere fixes the pressure, the room fixes the temperature. Which potential does the system minimize at equilibrium?

Recall Solution L1.2

Constant $T$ and constant $p$ → the matching potential is Gibbs free energy $G$ , whose natural variables are $(T, p)$ . This is the chemistry case (see Gibbs Free Energy and Chemical Equilibrium). Picture the right-hand bowl in the figure above with $F$ replaced by $G$ : the system rolls to the bottom.

L1.3 — Spot the definition

Fill in each blank: $H = U + \underline{p V}$ , $F = U - \underline{T S}$ , $G = U + \underline{p V} - \underline{T S}$ .

Recall Solution L1.3

$H = U + p V, F = U - T S, G = U + p V - T S .$ Mnemonic from the parent note: add $p V$ → enthalpy; subtract $T S$ → free; both → Gibbs.

Level 2 — Application

Now plug into the differentials to extract a derivative, a heat, or a heat capacity.

L2.1 — Entropy from $F$

Given $F (T, V)$ , write $S$ as a partial derivative of $F$ .

Recall Solution L2.1

Step 1 (WHAT): start from $d F = - S d T - p d V$ . Step 2 (WHY): hold $V$ fixed so $d V = 0$ ; then $d F = - S d T$ , and matching the coefficient of $d T$ : $S = - (\frac{\partial F}{\partial T})_{V}$ The minus sign comes directly from the $- S d T$ term — "S follows T with a minus."

L2.2 — Heat capacity $C_{p}$ from $H$

Show that the constant-pressure heat capacity is $C_{p} = (\frac{\partial H}{\partial T})_{p}$ , and compute $Δ H$ when heating $2 mol$ of an ideal monatomic gas from $300 K$ to $400 K$ at constant pressure. (Use $C_{p} = \frac{5}{2} n R$ , $R = 8.314 J mol^{- 1} K^{- 1}$ .)

Recall Solution L2.2

Step 0 (notation): $δ Q$ , $δ Q_{rev}$ and $δ Q_{p}$ are all defined in the box at the top of the page. Here $δ Q_{p}$ means "the inexact heat $δ Q$ measured along a reversible, constant-pressure path," so on this path $δ Q_{p} = δ Q_{rev} = T d S$ . Step 1: $d H = T d S + V d p$ . Now hold $p$ fixed, so $d p = 0$ , leaving $d H = T d S$ . Along a reversible path $T d S = δ Q_{rev}$ , therefore at constant $p$ : $d H = δ Q_{p} (constant p) .$ In words: the exact change in enthalpy equals the inexact heat because the constant- $p$ constraint removes the only other term ( $V d p$ ). That is why $Δ H$ is called the "heat content" for open-flask chemistry. Step 2: By definition of heat capacity, $δ Q_{p} = C_{p} d T$ . Setting $d H = δ Q_{p}$ and dividing by $d T$ at fixed $p$ : $C_{p} = (\frac{\partial H}{\partial T})_{p} .$ Step 3 (numbers): $C_{p} = \frac{5}{2} n R = \frac{5}{2} (2) (8.314) = 41.57 J K^{- 1}$ . $Δ H = C_{p} Δ T = 41.57 \times (400 - 300) = 4157 J \approx 4.16 kJ .$ See Heat Capacities Cp and Cv for why $C_{p} > C_{v}$ .

L2.3 — Volume from $G$

Given $G (T, p)$ , obtain both $S$ and $V$ as derivatives.

Recall Solution L2.3

From $d G = - S d T + V d p$ : $S = - (\frac{\partial G}{\partial T})_{p}, V = + (\frac{\partial G}{\partial p})_{T} .$ Why the signs? $- S$ multiplies $d T$ , $+ V$ multiplies $d p$ — read straight off. " $V$ follows $p$ with a plus."

Level 3 — Analysis

Here you combine two facts, or reason about why a step is legal.

L3.1 — A Maxwell relation from $G$

Derive the Maxwell relation hidden inside $G$ : $(\frac{\partial S}{\partial p})_{T} = - (\frac{\partial V}{\partial T})_{p} .$

Recall Solution L3.1

Step 1: From $d G = - S d T + V d p$ , read the first derivatives (each with its held-fixed variable named): $- S = (\frac{\partial G}{\partial T})_{p}, V = (\frac{\partial G}{\partial p})_{T} .$ Step 2 (the key idea — and its licence): $G$ is a state function of $(T, p)$ whose second partial derivatives are continuous on the region of interest. Under that continuity (the regularity condition), Clairaut's / Schwarz's theorem guarantees the two mixed second derivatives are equal — the order of differentiation does not matter: $[\frac{\partial}{\partial p} (\frac{\partial G}{\partial T})_{p}]_{T} = [\frac{\partial}{\partial T} (\frac{\partial G}{\partial p})_{T}]_{p} .$ What "order doesn't matter" looks like (see the figure below): on the surface $G (T, p)$ , reaching a neighbouring point by "step in $T$ , then step in $p$ " lands you at the same height as "step in $p$ , then step in $T$ " — the little quadrilateral closes. Equal mixed partials is exactly the statement that this loop closes. If $G$ had a kink or discontinuous curvature, the loop could fail to close and the Maxwell relation would break; continuity of the second derivatives is what forbids that.

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

Step 3: Substitute the Step-1 identities into each side: $(\frac{\partial ( - S )}{\partial p})_{T} = (\frac{\partial V}{\partial T})_{p} \Rightarrow (\frac{\partial S}{\partial p})_{T} = - (\frac{\partial V}{\partial T})_{p} .$ This trades an unmeasurable entropy slope for the ordinary thermal-expansion slope. See Maxwell Relations.

L3.2 — Check a candidate is a valid $G$

Someone proposes $G (T, p) = a T p^{2}$ (constants $a > 0$ ). Using $S = - (\partial G / \partial T)_{p}$ and $V = (\partial G / \partial p)_{T}$ , find $S$ and $V$ , then verify the Maxwell relation of L3.1 holds automatically.

Recall Solution L3.2

Compute (each derivative names what is held fixed): $S = - (\frac{\partial}{\partial T} (a T p^{2}))_{p} = - a p^{2}$ (hold $p$ fixed, so $p^{2}$ is a constant multiplier). $V = (\frac{\partial}{\partial p} (a T p^{2}))_{T} = 2 a T p$ (hold $T$ fixed). Check Maxwell: $(\frac{\partial S}{\partial p})_{T} = - 2 a p$ (hold $T$ ) and $- (\frac{\partial V}{\partial T})_{p} = - (2 a p) = - 2 a p$ (hold $p$ ). Equal ✓ — and note $G = a T p^{2}$ is a polynomial, so its second derivatives are continuous everywhere; the regularity condition of L3.1 is met. (Physical aside: $S = - a p^{2} < 0$ makes this a toy, not a real substance, but the calculus is consistent — that's the point: any smooth $G$ automatically satisfies its Maxwell relation.)

L3.3 — Why $F$ and not $U$ at constant $T$ ?

Explain in one clean paragraph why a system held at constant temperature and volume settles at the minimum of $F$ , not of $U$ .

Recall Solution L3.3

The second law says the entropy of the universe (system + surroundings) can only increase. When the system exchanges heat $δ Q$ with a bath at temperature $T$ , the bath's entropy changes by $- δ Q / T$ . Combine with the system's own entropy and, at fixed $T, V$ , the total-entropy condition $d S_{univ} \geq 0$ algebraically becomes $d F \leq 0$ . So the system spontaneously slides down in $F$ (roll to the bottom of the violet bowl in the opening figure) and stops at the minimum. $U$ alone ignores the entropy the system dumps into the bath, so its minimum is not the equilibrium condition here. (Full statement: Entropy and the Second Law.)

Level 4 — Synthesis

Build a multi-step result linking several potentials.

L4.1 — Gibbs–Helmholtz equation

Starting from $G = H - T S$ and $S = - (\partial G / \partial T)_{p}$ , prove $H = G - T (\frac{\partial G}{\partial T})_{p} and hence (\frac{\partial ( G / T )}{\partial T})_{p} = - \frac{H}{T ^{2}} .$

Recall Solution L4.1

Step 1: Substitute $S = - (\partial G / \partial T)_{p}$ into $G = H - T S$ : $G = H - T (- \frac{\partial G}{\partial T})_{p} = H + T (\frac{\partial G}{\partial T})_{p} .$ Rearrange: $H = G - T (\frac{\partial G}{\partial T})_{p} .$ ✓ Step 2 (why the $G / T$ form): differentiate $G / T$ with respect to $T$ holding $p$ fixed, using the quotient rule (note $T$ is the variable, so $1/ T$ is not constant): $(\frac{\partial ( G / T )}{\partial T})_{p} = \frac{1}{T} (\frac{\partial G}{\partial T})_{p} - \frac{G}{T ^{2}} = \frac{1}{T ^{2}} [T (\frac{\partial G}{\partial T})_{p} - G] .$ Step 3: the bracket is exactly $- H$ from Step 1, so $(\frac{\partial ( G / T )}{\partial T})_{p} = - \frac{H}{T ^{2}} .$ This is the workhorse that lets chemists get reaction enthalpies from the temperature-dependence of equilibrium.

L4.2 — Relate $C_{p}$ and $C_{v}$ via a Maxwell relation

First motivate the identity $C_{p} - C_{v} = T (\frac{\partial p}{\partial T})_{V} (\frac{\partial V}{\partial T})_{p}$ , then evaluate it for one mole of an ideal gas ( $p V = R T$ ) and confirm $C_{p} - C_{v} = R$ .

Recall Solution L4.2

Where the identity comes from (the "why"). Both heat capacities measure "heat per degree," but under different constraints, so they must differ by whatever extra energy the constant-pressure path spends. Step A: By definition $C_{v} = T (\frac{\partial S}{\partial T})_{V}$ and $C_{p} = T (\frac{\partial S}{\partial T})_{p}$ (from $δ Q_{rev} = T d S$ , divided by $d T$ under each constraint). Step B: Regard entropy as $S (T, V)$ and expand its total differential, then read the constant- $p$ temperature slope with the chain rule: $(\frac{\partial S}{\partial T})_{p} = (\frac{\partial S}{\partial T})_{V} + (\frac{\partial S}{\partial V})_{T} (\frac{\partial V}{\partial T})_{p} .$ Step C: Multiply by $T$ and subtract $C_{v}$ : $C_{p} - C_{v} = T (\frac{\partial S}{\partial V})_{T} (\frac{\partial V}{\partial T})_{p} .$ Step D (bring in Maxwell): the Maxwell relation from $F$ (parent note) is $(\frac{\partial S}{\partial V})_{T} = (\frac{\partial p}{\partial T})_{V}$ . Substitute to reach the advertised identity: $C_{p} - C_{v} = T (\frac{\partial p}{\partial T})_{V} (\frac{\partial V}{\partial T})_{p} .$ This turns an entropy quantity (hard) into two easy gas-law slopes. Now evaluate for the ideal gas $p V = R T$ (one mole): Step 1: Hold $V$ fixed and write $p = R T / V \Rightarrow (\frac{\partial p}{\partial T})_{V} = \frac{R}{V}$ . Step 2: Hold $p$ fixed and write $V = R T / p \Rightarrow (\frac{\partial V}{\partial T})_{p} = \frac{R}{p}$ . Step 3: Multiply and prepend $T$ : $C_{p} - C_{v} = T \cdot \frac{R}{V} \cdot \frac{R}{p} = \frac{T R ^{2}}{p V} = \frac{T R ^{2}}{R T} = R,$ since $p V = R T$ cancels the denominator. So $C_{p} - C_{v} = R = 8.314 J mol^{- 1} K^{- 1}$ . ✓ (Mayer's relation.)

Level 5 — Mastery

Full-power problems: reconstruct a potential, or reason about a limiting/degenerate case.

L5.1 — Reconstruct $F$ for an ideal gas

For one mole of ideal monatomic gas the entropy is (Sackur–Tetrode, schematically) $S (T, V) = C_{v} ln T + R ln V + s_{0},$ with $C_{v} = \frac{3}{2} R$ and constant $s_{0}$ . Using $F = U - T S$ with $U = C_{v} T + u_{0}$ , and then $p = - (\partial F / \partial V)_{T}$ , recover the ideal-gas law $p = R T / V$ .

Recall Solution L5.1

Step 1: Assemble $F$ : $F = U - T S = (C_{v} T + u_{0}) - T (C_{v} ln T + R ln V + s_{0}) .$ Step 2: Take $- (\partial F / \partial V)_{T}$ , i.e. differentiate with respect to $V$ holding $T$ fixed. Only the $- T R ln V$ term depends on $V$ : $(\frac{\partial F}{\partial V})_{T} = - T R \cdot \frac{1}{V} = - \frac{R T}{V} .$ Step 3: Therefore $p = - (\frac{\partial F}{\partial V})_{T} = \frac{R T}{V} .$ The ideal-gas law falls out of the potential. ✓ (Notice $U$ dropped out of the $V$ -derivative because $U$ here has no $V$ -dependence — internal energy of an ideal gas depends only on $T$ .)

L5.2 — The degenerate case $T \to 0$

Using the third law ( $S \to 0$ as $T \to 0$ ) and $F = U - T S$ , argue what happens to $F$ and to $(\frac{\partial F}{\partial T})_{V}$ as $T \to 0$ .

Recall Solution L5.2

Step 1: As $T \to 0$ , the product $T S \to 0$ (both $T \to 0$ and $S \to 0$ ), so $F = U - T S \to U$ . The free energy merges with the internal energy at absolute zero — there is no entropy "tax" left to pay. Step 2: Recall $(\frac{\partial F}{\partial T})_{V} = - S$ . Since $S \to 0$ , the slope of $F$ versus $T$ flattens to zero: $F (T)$ approaches $U$ horizontally.

Read the figure below. The horizontal axis is temperature $T$ , the vertical axis is energy. The orange curve is the internal energy $U (T)$ (rising steadily with $T$ ); the magenta curve is the free energy $F (T) = U - T S$ . The vertical navy double-arrow marks the gap between them at one temperature — that gap is exactly the entropy tax $T S$ , which grows as you heat up. At the far left, $T = 0$ , the magenta $F$ curve touches the orange $U$ curve at the violet dot, and it arrives there tangentially (flat slope, because $- S \to 0$ ). So the two curves "kiss" at absolute zero and peel apart as $T$ increases — that is the whole content of Step 1 and Step 2 drawn as a picture.

L5.3 — Sign of $Δ G$ and spontaneity

A reaction at $298 K$ has $Δ H = - 50.0 kJ$ and $Δ S = + 100 J K^{- 1}$ . Compute $Δ G = Δ H - T Δ S$ and state whether it is spontaneous. Then find the temperature at which $Δ G = 0$ (equilibrium crossover), and describe the case $T \to \infty$ .

Recall Solution L5.3

Step 1: Convert units: $Δ H = - 50000 J$ , $T Δ S = 298 \times 100 = 29800 J$ . $Δ G = - 50000 - 29800 = - 79800 J = - 79.8 kJ .$ Since $Δ G < 0$ , the reaction is spontaneous at $298 K$ . Step 2 (crossover): set $Δ G = 0 = Δ H - T Δ S$ : $T = \frac{Δ H}{Δ S} = \frac{- 50000}{100} = - 500 K .$ A negative temperature is unphysical, meaning there is no real crossover — with $Δ H < 0$ and $Δ S > 0$ , both terms drive $Δ G$ negative, so the reaction is spontaneous at every positive temperature. Step 3 (limit): as $T \to \infty$ , the $- T Δ S$ term dominates and $Δ G \to - \infty$ : still spontaneous, ever more strongly. This is the "both favourable" quadrant of the Ellingham-type sign analysis.

Recall One-line self-check before you leave

Cover the page. Can you (a) name the natural variables of each potential, (b) get $S$ and $V$ from $G$ , (c) derive one Maxwell relation, and (d) compute a $Δ G$ with correct units? If yes to all four, you own this topic.

Which potential minimizes at constant

T, V

Helmholtz free energy

F

For one mole ideal gas,

C_{p} - C_{v} = ?

R = 8.314 J mol^{- 1} K^{- 1}

Gibbs–Helmholtz form of

\partial (G / T) / \partial T

at fixed

p

- H / T^{2}

T \to 0

F

tends to

U

(since

T S \to 0

), with slope

- S \to 0

What does

δ

(as in

δ Q

) mean versus

d

δ

= inexact differential (path-dependent, not a state function);

d

= exact differential of a state function

What condition licenses equal mixed partials in a Maxwell relation?

continuity of the second partial derivatives (Clairaut/Schwarz theorem)