State functions vs path functions: Worked examples — Chemistry deep dive

This page belongs to the parent topic. Here we do not learn new theory — we stress-test the one big idea until no case can surprise you:

Before any symbol appears in a formula, here is the plain-word key you must carry:

The scenario matrix

Every case this topic can throw is one of these cells. The examples that follow are tagged with the cell they cover, so together they fill the whole grid.

#	Cell (case class)	What is tricky about it	Covered by
C1	Two routes, same endpoints — reversible vs irreversible	Same $Δ U$ , different $q$ and $w$	Ex 1
C2	Sign of $w$ — expansion ( $w > 0$ ) vs compression ( $w < 0$ )	The direction flips the sign	Ex 2
C3	Zero input — free expansion, $P_{ext} = 0$	$w = 0$ even though volume changes	Ex 1, Ex 5
C4	Adiabatic — $q = 0$	All energy goes to work; $Δ U = - w$	Ex 3
C5	Closed cycle — return to start	$\oint d U = 0$ but $\oint q \neq = 0$	Ex 4
C6	Constant- $V$ vs constant- $P$ — $q$ becomes state-like	$q_{V} = Δ U$ , $q_{P} = Δ H$	Ex 6
C7	Degenerate / limiting — $V_{2} \to V_{1}$ , or $ln 1 = 0$	Everything must go to zero smoothly	Ex 7
C8	Real-world word problem — a steam-engine-style cycle	Translate words into the matrix	Ex 8
C9	Exam twist — "given $Δ U$ , find $q$ and $w$ " (trap)	Impossible without the path	Ex 9

The one picture that ties them together

Figure — State functions vs path functions

Look at the map: the height axis (blue) is a state function $U$ — points $A$ and $B$ sit at fixed heights no matter which trail you take. The wiggly trails (orange and green) have different lengths — that length is the path function. Keep this in mind for every example.

Worked example Example 1 — Cell C1 & C3: two routes, same endpoints

Statement: 1 mol ideal gas, held at $T = 300 K$ , expands from $V_{1} = 10 L$ to $V_{2} = 20 L$ . Compute $Δ U$ , $q$ , $w$ for (A) reversible and (B) free expansion.

Forecast: Guess now — will $Δ U$ be the same for both? Will $q$ ?

$Δ U$ for both paths. Why this step? For an ideal gas $U$ depends only on $T$ . Both paths keep $T = 300 K$ , so $Δ T = 0 \Rightarrow Δ U = 0$ . State function → route-blind. $Δ U_{A} = Δ U_{B} = 0$
Reversible work. Why this step? Reversible means $P_{ext} = P_{gas} = n R T / V$ at every instant, so we integrate $w = \int P_{ext} d V$ : $w_{A} = n R T ln \frac{V _{2}}{V _{1}} = (1) (8.314) (300) ln 2 = + 1729 J$
Reversible heat. First Law $q = Δ U + w = 0 + 1729 = + 1729 J$ .
Free-expansion work. Why this step? "Free" means $P_{ext} = 0$ , so $w_{B} = - \int 0 d V = 0$ . Cell C3: zero input despite volume change. $w_{B} = 0, q_{B} = Δ U + w_{B} = 0$

Verify: $Δ U$ identical (0 = 0) ✔ state function. But $w_{A} = 1729 J \neq = w_{B} = 0$ and $q_{A} = 1729 J \neq = q_{B} = 0$ ✔ path functions. Same endpoints, different journey-costs — exactly the matrix's C1/C3.

The shaded area under each curve in the figure is the work. Reversible (orange) has area; free (green) hugs the axis with zero area.

Worked example Example 4 — Cell C5: a closed cycle

Statement: A gas goes $A \to B \to C \to A$ . Segment heats absorbed: $q_{A B} = + 800 J$ , $q_{B C} = - 300 J$ , $q_{C A} = - 200 J$ ; works done: $w_{A B} = + 500 J$ , $w_{B C} = 0$ , $w_{C A} = - 200 J$ . Show $\oint d U = 0$ but $\oint q \neq = 0$ .

Forecast: For a loop back to the start, which sum must vanish?

Loop heat. $\oint q = 800 - 300 - 200 = + 300 J$ . Why this step? Path functions don't cancel around a loop — this is the engine's net heat intake.
Loop work. $\oint w = 500 + 0 - 200 = + 300 J$ — net work delivered.
Loop internal energy. $\oint d U = \oint q - \oint w = 300 - 300 = 0$ . Why this step? $U$ is a state function; returning to $A$ must return to the same $U$ . See cyclic processes.

Verify: $\oint d U = 0$ ✔ (state), $\oint q = 300 \neq = 0$ ✔ (path). The gas took in $300 J$ of net heat and gave out $300 J$ of net work — a one-cycle heat engine.

Recall Quick self-test

Two paths give the same $Δ U$ but different $q$ . Which kind of function is $q$ ? ::: A path function. In a complete cycle, what is $\oint d U$ ? ::: Zero, because $U$ is a state function. Adiabatic means which path quantity is zero? ::: $q = 0$ . Why can't you find $q$ and $w$ from $Δ U$ alone? ::: Because $Δ U$ fixes only the endpoints; $q$ and $w$ need the path. Under constant pressure, $q_{P}$ equals which state function change? ::: $Δ H$ (enthalpy).