State functions vs path functions: Visual walkthrough

Before any symbol, three plain words:

Step 1 — Draw the map every gas lives on

WHAT. We put a gas on a flat map. The horizontal direction is its volume $V$ (how much room it fills, in litres). The vertical direction is its pressure $P$ (how hard it pushes on the walls, in pascals). One dot $=$ one complete state of the gas.

WHY this map and not another? Because for a fixed amount of ideal gas, once you fix $P$ and $V$ , the temperature is forced by $P V = n R T$ — so a single dot already tells you everything. We need a picture where "the whole state" is just one point, so we can ask cleanly: does a quantity depend on the point, or on the line joining points?

PICTURE. Two dots: the start $A$ and the end $B$ . Nothing is drawn between them yet — that is deliberate. The endpoints are the state; the line is the path, and we haven't chosen one.

See 2.5.03-Internal-Energy-and-Enthalpy for what "state" packs inside it.

Step 2 — Two roads from A to B

WHAT. Now we draw two different paths from the same dot $A$ to the same dot $B$ . Path 1 (magenta) sags low; Path 2 (violet) bulges high. Same start, same end — different routes.

WHY. This is the entire experiment. If a quantity gives the same number on both roads, it cares only about $A$ and $B$ → state function. If it gives different numbers, it cares about the road → path function. We are about to find one of each.

PICTURE. The two coloured curves share endpoints but enclose different regions. Keep your eye on the shaded areas under each curve — the next step reveals what that area is.

Recall

What makes two routes count as "the same journey" for a state function? ::: They share the same start dot and end dot; the shape in between is irrelevant.

Step 3 — What "work" means, drawn as an area

WHAT. When a gas expands by a tiny amount $d V$ while pushing against an outside pressure $P_{ext}$ , it does a small slice of work. We define $δ w = P_{ext} d V .$ Reading the symbols where they sit: $P_{ext}$ is the outside pressure resisting the push (pascals), $d V$ is the sliver of volume gained (a thin vertical strip on the map), and $δ w$ is the sliver of work — a thin rectangle of area $P_{ext} \times d V$ .

WHY the symbol $δ$ and not $d$ ? We write $δ w$ (a "crooked d") to flag: this is not yet the change of any state property. A plain $d$ is a promise that a quantity is a state function; $δ$ makes no such promise. We will earn or break that promise by Step 5. The link 4.1.02-Exact-and-Inexact-Differentials treats this $d$ -vs- $δ$ split in full.

WHY area? Adding every sliver from $V_{1}$ to $V_{2}$ is the integral $\int P_{ext} d V$ , which is literally the area under the path on the map. So work = area under the route. And two routes with different shapes enclose different areas.

PICTURE. One thin strip highlighted, width $d V$ , height $P_{ext}$ . The whole shaded region is the total work.

Step 4 — Prove work really differs: two roads, two areas

WHAT. Take the concrete numbers from the parent note. One mole of ideal gas, held at $T = 300 K$ , expands from $V_{1} = 10 L$ to $V_{2} = 20 L$ by two roads.

Road R (reversible, magenta): the outside pressure always matches the gas, $P_{ext} = n R T / V$ , so $w_{R} = \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V = n R T ln \frac{V _{2}}{V _{1}} = (1) (8.314) (300) ln 2 = 1729 J .$ Here $n = 1$ mol, $R = 8.314 J mol^{- 1} K^{- 1}$ , $T = 300 K$ , and $ln 2$ comes from doubling the volume. The area under the smooth curve is large.

Road F (free expansion, violet): the gas pushes against nothing, $P_{ext} = 0$ , so every strip has zero height: $w_{F} = \int P_{ext} d V = 0.$

WHY. Same $A$ , same $B$ , yet $w_{R} = 1729 J \neq = w_{F} = 0 J$ . Work fails the test. Work is a path function. (See 3.2.04-Reversible-vs-Irreversible-Processes for why reversible squeezes out the most work.)

PICTURE. The magenta area (big) versus the violet area (a flat line hugging the axis, zero area). The size gap is the path dependence.

Step 5 — The quantity that DOES survive: internal energy

WHAT. Now meet the internal energy $U$ — the total kinetic + potential energy stored inside the gas. For an ideal gas $U$ depends only on temperature: $U = U (T)$ . On our map, points at the same temperature all share the same $U$ .

WHY it's a state function. Its differential is exact. Writing $U$ as a function of $T$ and $V$ , $d U = (\frac{\partial U}{\partial T})_{V} d T + (\frac{\partial U}{\partial V})_{T} d V,$ each piece reads: $\partial U / \partial T$ is how $U$ changes when you nudge temperature (holding $V$ ), times the temperature nudge $d T$ ; likewise for volume. The test that this is exact is Euler's reciprocity — the two "cross-slopes" must match: $\frac{\partial}{\partial V} (\frac{\partial U}{\partial T})_{V} = \frac{\partial}{\partial T} (\frac{\partial U}{\partial V})_{T} .$ When cross-slopes agree, the little arrows of $d U$ knit together into one consistent height map — an elevation. That is exactly what "state function" means.

PICTURE. Instead of area under a curve, $U$ is a height at each dot, like colour-shading the map by altitude. Move A→B by any road and you climb the same net height.

More on why $S$ passes: 2.5.07-Entropy-as-State-Function.

Step 6 — Close the loop: the knockout test

WHAT. Send the gas around a cycle: A → B along magenta, then B → A along violet, back to the start. Add up each quantity around the whole loop, written $\oint$ ("integral around a closed loop").

For internal energy, since it is a height and we return to the same dot: $\oint d U = 0.$

For work, the loop encloses a nonzero area, so the two half-areas do not cancel: $\oint δ w = (area under magenta) - (area under violet) = enclosed area \neq = 0.$ The same holds for heat: $\oint δ q \neq = 0$ . (Explored further in 2.5.10-Cyclic-Processes-and-State-Functions.)

WHY it's the decisive picture. A state function must return to its starting value after a round trip — like walking a loop trail and ending at your original elevation. Work and heat do not: a steam engine returns to the same state each cycle ( $\oint d U = 0$ ) yet pumps out real work every loop ( $\oint δ w \neq = 0$ ). The enclosed area of the cycle is the net work per cycle.

PICTURE. The closed magenta-then-violet loop, its enclosed area shaded — that shaded lens is $\oint δ w$ , provably not zero, while the altitude returns home so $\oint d U = 0$ .

Step 7 — Edge cases: when a path function pretends to be state-like

WHAT. Pin down special roads where $q$ or $w$ becomes predictable — but is still, underneath, a path quantity.

Road	Condition	What collapses
Constant volume	$d V = 0$	strips have zero width, $w = 0$ , so $q_{V} = Δ U$
Constant pressure	$P$ fixed	$q_{P} = Δ H$ , a genuine state function
Adiabatic	$δ q = 0$	$Δ U = - w$ , yet $w$ still depends on the road

WHY it doesn't break our result. Fixing a condition (say constant $P$ ) chooses one road, so of course the answer becomes a single number — you removed the freedom to differ. Change the constraint (reversible vs irreversible) and $q$ , $w$ split apart again. The pretence is real only inside its cage.

PICTURE. A vertical line (constant $V$ , zero-width strips), a horizontal line (constant $P$ , a clean rectangle of area), and a steep adiabatic curve — three tamed roads side by side.

The one-picture summary

Everything compresses into a single frame: two roads between $A$ and $B$ forming a loop. Area under a road = work (path). Altitude at a dot = internal energy (state). Round the loop, altitude comes home ( $\oint d U = 0$ ) but the enclosed area survives ( $\oint δ w \neq = 0$ ).

Recall Feynman retelling — say it to a 12-year-old

Picture a hilly park drawn as a map. Each spot has an altitude — that's the gas's internal energy $U$ . If you hike from spot A to spot B and back to A, your altitude change is exactly zero, no matter which trails you took. That's a state thing: it only cares about where you are.

Now count your footsteps instead. The number of steps depends completely on which trail you chose — the long scenic loop costs way more steps than the shortcut, even though both end where they started. Steps are a path thing.

In a gas, internal energy is altitude and work (and heat) are footsteps. On the pressure–volume map, work is the area under the trail, so a fatter loop means more work — even for a round trip that returns to the exact same gas. The First Law is the park's honesty rule: however wildly your heat-footsteps and work-footsteps wander, they must wander together so that your altitude always returns home. That's why we write a straight $d$ for $U$ (an altitude exists) and a crooked $δ$ for $q$ and $w$ (only footsteps, no altitude).