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

Prerequisite ideas we lean on: First Law of Thermodynamics, Entropy and the Second Law, the Legendre Transform, and later Maxwell Relations. This is the visual companion to the parent topic.

Step 1 — The one thing we start with: a tiny energy budget

WHAT. A system (say a gas in a cylinder) holds some hidden energy we call ==internal energy $U$ == — the sum of all the jiggling and bonding energy of its molecules. We never track its total; we track how it changes when we nudge the system a little. A tiny change is written $d U$ ("a small change in $U$ "). The little $d$ just means "an infinitesimal step," like moving one pixel.

WHY. Every potential we build is $U$ plus or minus something. If we understand how $U$ changes, we understand all of them. So we need the rule for $d U$ .

The First Law of Thermodynamics says: energy change = heat you put in, minus work the system does.

$d U = heat in δ Q - work out δ W$

$δ Q$ — a small blob of heat flowing into the gas.
$δ W$ — a small blob of work the gas does on the outside (pushing the piston).

PICTURE. The gas is a bucket. Heat pours in from the top ( $+ δ Q$ ), work spills out the side ( $- δ W$ ). Whatever is left changes the level $U$ .

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

Step 2 — Naming the two blobs: $T d S$ and $p d V$

WHAT. We now say what those two blobs are made of, for a gentle (reversible) process.

Heat blob: $δ Q = T d S$ .
- $T$ — the temperature (how hot, in kelvin).
- $d S$ — a tiny change in ==entropy $S$ ==, a number measuring how "spread out" or disordered the energy is. Adding heat spreads energy out, so heat and $d S$ travel together; $T$ is the exchange rate.
Work blob: $δ W = p d V$ .
- $p$ — the pressure the gas pushes with.
- $d V$ — a tiny change in ==volume $V$ ==. If the gas expands ( $d V > 0$ ) it pushes the piston out and does work.

WHY these tools? We use $T d S$ and not just " $δ Q$ " because $δ Q$ alone is not a property of the state — it depends on the path taken. But $T$ and $S$ are honest state properties. Rewriting heat as $T d S$ turns a path-dependent blob into state variables, which is exactly what lets us build clean potentials later. (This is the payoff of Entropy and the Second Law.)

Substituting both blobs:

$d U = hotness T disorder step d S - push p expansion step d V$

This is the mother equation. Notice its shape: (something)·(change in something) + (something)·(change in something).

PICTURE. Two labelled arrows feeding the bucket: a red $T d S$ arrow in, a blue $p d V$ arrow out, with the resulting $d U$ shown as the net change.

Step 3 — Reading slopes off the equation

WHAT. Any smooth function of two variables obeys the chain rule: a small step is (slope in $S$ )· $d S$ + (slope in $V$ )· $d V$ :

$d U = (\frac{\partial U}{\partial S})_{V} d S + (\frac{\partial U}{\partial V})_{S} d V$

$(\frac{\partial U}{\partial S})_{V}$ — how fast $U$ climbs if you nudge $S$ while holding $V$ fixed (that little subscript $V$ means "keep $V$ frozen").

WHY. We have $d U$ written two ways: once physically ( $T d S - p d V$ ), once mathematically (chain rule). If two expressions equal each other for every choice of $d S, d V$ , their matching coefficients must be equal. Match term by term:

$(\frac{\partial U}{\partial S})_{V} = T, (\frac{\partial U}{\partial V})_{S} = - p$

The slope of energy versus entropy is temperature. The slope versus volume is minus the pressure. Symbols we introduced abstractly now have physical meaning.

PICTURE. The surface $U (S, V)$ as a hill; walking east (increasing $S$ ) the steepness reads off as $T$ ; walking north (increasing $V$ ) the downhill steepness reads off as $- p$ .

Step 4 — The problem: we can't grab $S$ in the lab

WHAT. $U (S, V)$ wants us to control entropy $S$ and volume $V$ . Volume is fine (move a piston). But there is no "entropy knob" — you cannot dial disorder directly. What you can set is temperature $T$ (a thermostat) and pressure $p$ (an open beaker feels the atmosphere).

WHY a new tool. We need to swap the awkward variable $S$ for its friendly partner $T$ , and $V$ for $p$ , without losing information. The tool that does exactly this trade is the Legendre Transform. It answers the precise question: "How do I rewrite an energy so that a slope becomes the variable, and the old variable disappears?"

The recipe: subtract the product of the two partners you want to swap.

To trade $S \to T$ (partners, since $T = \partial U / \partial S$ ): subtract $T S$ .
To trade $V \to p$ (partners, since $- p = \partial U / \partial V$ ): add $p V$ .

PICTURE. A knob panel: the " $S$ " knob is crossed out (no such knob exists), the " $T$ " and " $p$ " knobs glow — these are what a real lab can turn.

Step 5 — First swap: add $p V$ to get enthalpy $H$

WHAT. Define enthalpy $H = U + p V$ . Take its small change with the product rule $d (p V) = p d V + V d p$ :

$d H = d U + p d V + V d p$

Now plug in $d U = T d S - p d V$ :

$d H = (T d S - p d V) + p d V + V d p$

The $- p d V$ from the seed and the $+ p d V$ from $d (p V)$ are equal and opposite → they cancel.
What survives: $T d S$ (untouched) and a brand-new $+ V d p$ .

$d H = T d S + V d p natural variables (S, p)$

WHY. The cancellation is the whole point: we traded the $d V$ term for a $d p$ term, so pressure is now a natural variable. Watch the sign flip — $- p d V$ became $+ V d p$ . This is the classic mistake trap; the plus sign is forced by the cancellation, it is not optional.

PICTURE. Two stacked bars: $d U$ 's $- p d V$ block and $d (p V)$ 's $+ p d V$ block annihilate; the leftover $+ V d p$ block slides in.

Step 6 — Second swap: subtract $T S$ to get Helmholtz $F$

WHAT. Define Helmholtz free energy $F = U - T S$ . Its change uses the product rule $d (T S) = T d S + S d T$ :

$d F = d U - T d S - S d T$

Plug in the seed:

$d F = (T d S - p d V) - T d S - S d T$

The $+ T d S$ from the seed and the $- T d S$ from $- d (T S)$ cancel.
Survivors: $- p d V$ (untouched) and a new $- S d T$ .

$d F = - S d T - p d V natural variables (T, V)$

WHY. Killing $T d S$ promotes $T$ to a natural variable — perfect for a thermostat + rigid box. Note the sign pattern: $S$ rides with $T$ as $- S d T$ . (Compare Step 5 where $V$ rode with $p$ as $+ V d p$ .)

PICTURE. Mirror of Step 5, but now the $T d S$ blocks annihilate and $- S d T$ is left.

Step 7 — Both swaps at once: Gibbs $G$

WHAT. Do both tricks: add $p V$ and subtract $T S$ . Define Gibbs free energy $G = U + p V - T S$ . Then

$d G = d U + p d V + V d p - T d S - S d T$

$d G = (T d S - p d V) + p d V + V d p - T d S - S d T$

Both the $p d V$ pair and the $T d S$ pair cancel. Survivors:

$d G = - S d T + V d p natural variables (T, p)$

WHY. $G$ 's natural variables are $T$ and $p$ — the two things a chemist working in an open flask on a bench actually controls. That is why $G$ rules chemistry (see Gibbs Free Energy and Chemical Equilibrium).

PICTURE. Four blocks feeding in; both cancelling pairs vanish, leaving $- S d T + V d p$ .

Step 8 — The degenerate cases: what if a step is zero?

WHAT. The differentials must still make sense when a nudge is switched off. Check each edge:

Constant pressure ( $d p = 0$ ): $d H = T d S = δ Q$ . Enthalpy change is the heat absorbed — the reason $H$ is "heat content" for open-flask chemistry, and why $C_{p} = (\partial H / \partial T)_{p}$ (see Heat Capacities Cp and Cv).
Constant volume ( $d V = 0$ ): $d U = T d S = δ Q$ , so $C_{v} = (\partial U / \partial T)_{V}$ .
Constant temperature ( $d T = 0$ ): $d F = - p d V$ . The change in Helmholtz energy equals minus the work — that leftover is the useful work you can extract at fixed $T$ ; hence "free" energy.
Constant $T$ and $p$ ( $d T = d p = 0$ ): $d G = 0$ . At equilibrium under bench conditions $G$ stops changing — it sits at its minimum.
Fully isolated & rigid ( $d S$ set by system, $d V = 0$ , no heat): $U$ is the boss; nothing external does anything.

WHY. These aren't new physics — they're the same four boxed equations with one term deleted. A reader who meets any real experiment can look up which term dies and read the meaning straight off.

PICTURE. A 2×2 grid of the four differentials; in each cell one term is greyed out to show which constraint kills it.

The one-picture summary

Everything lives on a square. The four potentials sit at the corners; the moves between them are the Legendre trades ( $+ p V$ across, $- T S$ down). Each potential's two natural variables hang off its corner, and its differential is read directly.

Once these are visual, the Maxwell Relations fall out for free by equating mixed second derivatives of each corner.

Recall Feynman: tell the whole story in plain words

We started with one honest sentence: energy change = heat in minus work out. We renamed heat as "temperature times a change in disorder" and work as "pressure times a change in size," giving one seed equation, $d U = T d S - p d V$ . Trouble: it asks us to control disorder, which no lab can. So we play a trick — to swap an awkward variable for the friendly slope it hides, we subtract (or add) the product of the two. Add $p V$ : the volume term cancels and enthalpy $H$ appears, ruled by heat and pressure. Subtract $T S$ : the disorder term cancels and Helmholtz $F$ appears, ruled by temperature and volume. Do both: Gibbs $G$ appears, ruled by temperature and pressure — the chemist's world. Delete any single term and you read off a real experiment: constant pressure gives heat as $Δ H$ , constant temperature gives extractable work as $- Δ F$ , and at fixed $T, p$ the system slides to the bottom of $G$ . Four wallets, one seed, and a square that remembers them all.