State functions vs path functions: Exercises — Chemistry deep dive

Before we start, one picture ties everything together. A state is a single dot on a pressure–volume map (fixed $P$ , $V$ , $T$ ). A process is a line connecting two dots. A state function only cares about the two dots; a path function cares about the whole line.

Figure — State functions vs path functions

Quick symbol dictionary (each earned before use below):

$U$ = internal energy, the total energy stored inside the system. A state function.
$H = U + P V$ = enthalpy. A state function.
$q$ = heat that flows into the system (positive = absorbed). A path function.
$w$ = work done by the system on surroundings in the sign convention $d U = δ q - δ w$ . A path function.
$Δ X = X_{final} - X_{initial}$ — the "change in" symbol.
$\oint$ = integral around a closed loop (start and end at the same dot).

L1 — Recognition

Problem 1.1

Classify each as state (S) or path (P): (a) Temperature $T$ , (b) Work $w$ , (c) Enthalpy $H$ , (d) Heat $q$ , (e) Entropy $S$ , (f) Volume $V$ .

Recall Solution 1.1

Ask the single test question: "Does the value depend only on the current dot, or on the route between dots?"

(a) $T$ — describes the current thermal state only → S
(b) $w$ — depends on how fast/against what pressure you push → P
(c) $H = U + P V$ , a sum of state functions → S
(d) $q$ — depends on thermal contact and route → P
(e) $S$ — measure of disorder at the current state → S (see 2.5.07-Entropy-as-State-Function)
(f) $V$ — a geometric property of the current state → S

Answer: S, P, S, P, S, S.

Problem 1.2

For a cyclic process (system returns to its starting dot), what is $\oint d U$ ? What can you say about $\oint δ q$ ?

Recall Solution 1.2

A cycle starts and ends at the same dot. Since $U$ depends only on the dot, $U_{final} = U_{initial}$ , so $\oint d U = U_{final} - U_{initial} = 0.$ But $q$ is a path function, so going around the loop generally moves net heat in or out: $\oint δ q \neq = 0 (in general) .$ This is exactly why an engine works — see 2.5.10-Cyclic-Processes-and-State-Functions.

L2 — Application

Problem 2.1

1 mol of an ideal gas expands isothermally and reversibly at $T = 300 K$ from $V_{1} = 5.0 L$ to $V_{2} = 15.0 L$ . Find $w$ , $q$ , and $Δ U$ . Use $R = 8.314 J mol^{- 1} K^{- 1}$ .

Recall Solution 2.1

WHY reversible work uses a logarithm: in a reversible expansion the external pressure matches the gas pressure at every instant, $P_{ext} = P_{gas} = n R T / V$ . Work by the gas is $w = \int_{V_{1}}^{V_{2}} P_{ext} d V = \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V = n R T ln \frac{V _{2}}{V _{1}} .$ The $1/ V$ integrates to a natural log — that's why $ln$ appears; it answers "what accumulates when the resisting pressure keeps dropping as volume grows?" $w = (1) (8.314) (300) ln \frac{15}{5} = 2494.2 ln 3 = 2740 J .$ Isothermal + ideal gas ⇒ $U$ depends only on $T$ , and $T$ is unchanged, so $Δ U = 0.$ First law $Δ U = q - w \Rightarrow q = w = 2740 J$ (absorbed). Answers: $w \approx + 2740 J$ , $q \approx + 2740 J$ , $Δ U = 0$ .

Problem 2.2

The same gas from Problem 2.1 goes between the same two dots ( $5 \to 15$ L, $300$ K) but by free expansion into vacuum ( $P_{ext} = 0$ ). Find $w$ , $q$ , $Δ U$ . Compare $Δ U$ across the two paths.

Recall Solution 2.2

With no opposing pressure, $P_{ext} = 0$ : $w = \int P_{ext} d V = 0.$ Still isothermal ideal gas, so $Δ U = 0$ , giving $q = Δ U + w = 0$ . Answers: $w = 0$ , $q = 0$ , $Δ U = 0$ . Comparison: $Δ U = 0$ for both paths — because $U$ is a state function and both trips share the same endpoints. But $q$ and $w$ differ completely ( $2740$ J vs $0$ ). That single contrast is the whole chapter in one line.

L3 — Analysis

Problem 3.1

Take 1 mol ideal gas around a cycle on the $P$ – $V$ diagram: $A \to B$ isothermal reversible expansion ( $5 \to 15$ L at $300$ K, giving $w_{A B} = + 2740$ J), then $B \to A$ isothermal reversible compression back to $5$ L. Show $\oint d U = 0$ while $\oint δ w \neq = 0$ per step but the net closes.

Recall Solution 3.1

Look at the loop in the figure.

Leg $A \to B$ (expand): $w_{A B} = n R T ln (15/5) = + 2740 J$ . Leg $B \to A$ (compress, same isotherm): $w_{B A} = n R T ln (5/15) = - 2740 J$ . Net work around cycle: $\oint δ w = 2740 + (- 2740) = 0$ here, because we retraced the same isotherm. Internal energy: $T$ returns to $300$ K, so $U$ returns to its start: $\oint d U = 0$ always. Key point: For this special retraced path net work is $0$ , but if the return leg used a different path (e.g. irreversible), $\oint δ w \neq = 0$ even though $\oint d U$ stays $0$ . That gap is what an engine harvests.

Problem 3.2

A gas is taken $A \to B$ by two different paths. Path 1: $q_{1} = + 500$ J, $w_{1} = + 200$ J. Path 2: $q_{2} = + 350$ J. Find $w_{2}$ . What principle forces your answer?

Recall Solution 3.2

$Δ U$ is a state function, so it is the same for both paths (same endpoints): $Δ U = q_{1} - w_{1} = 500 - 200 = 300 J .$ For path 2, $Δ U = q_{2} - w_{2} \Rightarrow w_{2} = q_{2} - Δ U = 350 - 300 = 50 J .$ Answer: $w_{2} = 50 J$ . The principle: even though $q$ and $w$ are path functions and differ between paths, their combination $q - w$ must reconstruct the same state-function change $Δ U$ .

L4 — Synthesis

Problem 4.1

Heat 100 g of water from $2 5^{\circ} C$ to $7 5^{\circ} C$ at constant pressure ( $c = 4.18 J g^{- 1} K^{- 1}$ ). Path A heats directly. Path B heats to $7 5^{\circ}$ , cools to $4 0^{\circ}$ , reheats to $7 5^{\circ}$ — all at the same constant pressure. Compute $q$ for each path and explain the outcome using $q_{P} = Δ H$ .

Recall Solution 4.1

WHY $q_{P} = Δ H$ : at constant pressure, first law gives $Δ U = q_{P} - P Δ V$ , and since $H = U + P V$ , at constant $P$ we get $Δ H = Δ U + P Δ V = q_{P}$ . So constant-pressure heat equals a state-function change — see 2.5.03-Internal-Energy-and-Enthalpy.

Path A: $q_{A} = m c Δ T = (100) (4.18) (75 - 25) = (100) (4.18) (50) = 20900 J .$ Path B (three legs, all constant $P$ ):

$25 \to 75$ : $q_{1} = (100) (4.18) (50) = 20900 J$
$75 \to 40$ : $q_{2} = (100) (4.18) (- 35) = - 14630 J$
$40 \to 75$ : $q_{3} = (100) (4.18) (35) = 14630 J$
$q_{B} = 20900 - 14630 + 14630 = 20900 J .$

Answer: $q_{A} = q_{B} = 20900 J$ . Explanation: Both paths are entirely at constant pressure, so along each, $q = Δ H$ . Since $H$ is a state function and both paths share the same endpoints ( $2 5^{\circ} \to 7 5^{\circ}$ ), the net heat matches. This is a conditional coincidence: heat behaves state-like only because the constraint (constant $P$ ) welds it to $Δ H$ .

Problem 4.2

Show algebraically that internal energy passes the exactness test while heat need not. For $d U = C_{V} d T + (\frac{\partial U}{\partial V})_{T} d V$ , the exactness condition (Euler reciprocity) is $\partial M / \partial V = \partial N / \partial T$ . State what each partial derivative means in words and why $δ q$ fails to have a fixed function behind it.

Recall Solution 4.2

Write $d U = M d T + N d V$ with $M = C_{V} = (\partial U / \partial T)_{V}$ and $N = (\partial U / \partial V)_{T}$ . Euler reciprocity demands the "cross slopes" agree: $(\frac{\partial M}{\partial V})_{T} = \frac{\partial}{\partial V} (\frac{\partial U}{\partial T}) = ? \frac{\partial}{\partial T} (\frac{\partial U}{\partial V}) = (\frac{\partial N}{\partial T})_{V} .$ Because $U (T, V)$ is a genuine function of state, its mixed second derivatives are equal regardless of order (a property of any smooth function), so the two sides match and $d U$ is exact. See 4.1.02-Exact-and-Inexact-Differentials. Why $δ q$ fails: there is no function $q (T, V)$ of the state whose differential is $δ q$ . If there were, $\oint δ q$ would be $0$ around every loop — but engines exist, so $\oint δ q \neq = 0$ . With no parent function, the cross-derivative test cannot even be set up; that's precisely what "inexact" means.

L5 — Mastery

Problem 5.1

A gas undergoes $Δ U = + 150 J$ between fixed endpoints. Give three distinct $(q, w)$ pairs consistent with this, each realized by a named process, and explain why infinitely many pairs exist while $Δ U$ is pinned.

Recall Solution 5.1

First law: $q - w = Δ U = 150 J$ , i.e. $q = w + 150$ . Any $w$ you choose fixes a valid $q$ :

Constant volume: $w = 0 \Rightarrow q = 150 J$ .
Heat + expansion: $w = 100 J \Rightarrow q = 250 J$ .
Adiabatic compression: $q = 0 \Rightarrow w = - 150 J$ (surroundings do $150$ J on the gas).

Why infinitely many: $Δ U$ constrains only the difference $q - w$ . That is one equation in two unknowns — a whole line of solutions. The process (which is the path) selects one point on that line. State functions pin the change; path details choose the split.

Problem 5.2

A Carnot-style engine runs a cycle absorbing $q_{h} = + 800 J$ from a hot reservoir and rejecting $q_{c} = - 500 J$ to a cold reservoir, returning to its starting state. Find $\oint d U$ , the net work done by the engine $\oint δ w$ , and verify the path/state accounting.

Recall Solution 5.2

Cycle closes, so the state function $U$ returns to start: $\oint d U = 0.$ First law integrated around the loop: $\oint d U = \oint δ q - \oint δ w$ , hence $\oint δ w = \oint δ q = q_{h} + q_{c} = 800 + (- 500) = 300 J .$ Answers: $\oint d U = 0$ , net heat $\oint δ q = 300 J$ , net work by engine $= 300 J$ . Accounting: The state function $U$ shows zero net change — nothing "accumulated" inside. Yet the path functions $q$ and $w$ are each large and nonzero around the loop, and they balance ( $\oint δ q = \oint δ w$ ) exactly because $\oint d U = 0$ . This is the engine principle: harvest the mismatch between $q_{h}$ and $∣ q_{c} ∣$ as usable work while the working substance goes nowhere in state space.

Problem 5.3

An ideal gas is taken $A \to B$ by an irreversible path with $w_{irr} = 1800 J$ and by a reversible path with $w_{rev} = 2740 J$ (from Problem 2.1). Both share the same endpoints, so $Δ U = 0$ . Find $q$ for each path and state which path transfers more heat and why the reversible one is the maximum.

Recall Solution 5.3

Same endpoints, isothermal ideal gas ⇒ $Δ U = 0$ ⇒ $q = w$ for each path.

Reversible: $q_{rev} = w_{rev} = 2740 J$ .
Irreversible: $q_{irr} = w_{irr} = 1800 J$ . More heat via the reversible path. Why maximum: reversible expansion pushes against the largest possible opposing pressure at every instant ( $P_{ext} = P_{gas}$ ), so it extracts the most work — and since $q = w$ here, it also draws the most heat. Any irreversible path (smaller $P_{ext}$ ) does less work and needs less heat, yet lands on the same state $B$ with the same $Δ U = 0$ . State function unchanged; path functions maximised at reversibility (link: 3.2.04-Reversible-vs-Irreversible-Processes).

Recall Self-test recap

One-line category test ::: Does the value depend on the route between dots (path) or only on the dots (state)? Why $\oint d U = 0$ ::: $U$ is a state function; a closed cycle returns to the same state. Why $\oint δ q \neq = 0$ in general ::: heat has no parent state function; net heat around a loop drives engines. When does $q$ act state-like ::: only under a fixed constraint — $q_{V} = Δ U$ (const $V$ ) or $q_{P} = Δ H$ (const $P$ ). Why reversible work is maximal ::: it opposes the largest external pressure at every instant.