Work in expansion - reversible isothermal w = −nRT ln(V₂ - V₁), irreversible w = −P_ext ΔV: Worked examples

This page is a drill through every case the formulas $w_{rev} = - n R T ln (\frac{V _{2}}{V _{1}}), w_{irrev} = - P_{ext} Δ V$ can throw at you. We build from the parent note Work in Expansion and lean on Ideal Gas Law, Isothermal Process, PV Diagram and First Law of Thermodynamics.

Before any numbers, let us agree on the language of signs, because half of all mistakes here are sign mistakes.

The scenario matrix

Every cell below is a class of problem. The worked examples that follow each name the cell(s) they cover, and together they hit all of them.

Cell	Case class	What is being tested	Covered by
A	Reversible isothermal expansion ( $V_{2} > V_{1}$ )	sign $w < 0$ , log formula	Ex 1
B	Reversible isothermal compression ( $V_{2} < V_{1}$ )	sign flips to $w > 0$	Ex 2
C	Irreversible constant $P_{ext}$ expansion	$w = - P_{ext} Δ V$ , rectangle area	Ex 3
D	Irreversible compression against high $P_{ext}$	$w > 0$ , why it costs more	Ex 4
E	Multi-step irreversible (path-dependent)	piecewise integration	Ex 5
F	Degenerate: $V_{2} = V_{1}$ (no change) and free expansion ( $P_{ext} = 0$ )	$w = 0$ two different ways	Ex 6
G	Limiting: many small steps $\to$ reversible	why $	w_{\text{irrev}}
H	Word / real-world problem (piston, atmosphere)	modelling + unit hygiene	Ex 8
I	Exam twist: mixed reversible-then-irreversible, or "find missing quantity"	back-solving a formula	Ex 9, Ex 10

(Reminder: $P_{ext}$ in the matrix is the constant external pressure just defined above.)

Master figure: the areas we are computing

Every single example is an area on a PV diagram. Keep this picture in your head.

Figure — Work in expansion - reversible isothermal w = −nRT ln(V₂ - V₁), irreversible w = −P_ext ΔV

The lavender curve is the isotherm $P_{gas} = n R T / V$ . Area under it (from $V_{1}$ to $V_{2}$ ) is the reversible work magnitude.
The coral rectangle is area under a constant $P_{ext}$ — the irreversible work.
The rectangle always sits inside the curve for expansion $\Rightarrow$ $∣ w_{irrev} ∣ < ∣ w_{rev} ∣$ .

Worked examples

Example 1 — Cell A: reversible isothermal expansion

Worked example Cell A

Statement: $3 mol$ of an ideal gas at $T = 350 K$ expands reversibly and isothermally from $V_{1} = 2 L$ to $V_{2} = 8 L$ . Find $w$ . Use $R = 8.314 J mol^{- 1} K^{- 1}$ .

Forecast: Expansion $\Rightarrow$ guess the sign of $w$ before reading on. (Answer: it must be negative.)

Step 1 — Pick the formula. $w_{rev} = - n R T ln (\frac{V _{2}}{V _{1}})$ Why this step? "Reversible + isothermal + ideal gas" is exactly the trio this formula requires.

Step 2 — Form the ratio (not the difference!). $\frac{V _{2}}{V _{1}} = \frac{8}{2} = 4$ Why this step? The log eats a ratio; units of volume cancel, so we never convert L to m³ here.

Step 3 — Substitute. $w = - (3) (8.314) (350) ln (4) = - 8729.7 \times 1.3863$ Why this step? $n R T$ carries the joules; $ln (4)$ is a pure number.

Step 4 — Evaluate. $w \approx - 12102 J \approx - 12.1 kJ$

Verify: Sign is negative $✓$ (expansion, matches forecast). Units: $mol \cdot J mol^{- 1} K^{- 1} \cdot K = J$ $✓$ . Magnitude larger than a rough $P Δ V$ estimate, as expected for the "area under the curve" $✓$ .

Example 2 — Cell B: reversible isothermal compression

Worked example Cell B

Statement: The same gas ( $3 mol$ , $350 K$ ) is now compressed reversibly from $8 L$ back to $2 L$ . Find $w$ .

Forecast: We are pushing in — guess the sign. (It flips to positive.)

Step 1 — Same formula, swapped limits. $w = - n R T ln (\frac{V _{2}}{V _{1}}) = - n R T ln (\frac{2}{8})$ Why this step? Reversible work only cares about start and end states; now $V_{1} = 8$ , $V_{2} = 2$ .

Step 2 — The ratio is now less than 1. $\frac{2}{8} = 0.25, ln (0.25) = - 1.3863$ Why this step? $ln$ of a number below 1 is negative — this is what flips the sign.

Step 3 — Substitute. $w = - (3) (8.314) (350) \times (- 1.3863) = + 12102 J \approx + 12.1 kJ$ Why this step? Two negatives (the leading $-$ and the negative $ln$ ) multiply to a positive — the arithmetic that realises the "work done on the gas" sign.

Verify: Positive $✓$ (work done on gas, matches forecast). It is exactly $+$ the value of Example 1 — because reversible paths are perfectly retraceable, expansion then compression cancels: $w_{exp} + w_{comp} = 0$ $✓$ .

Example 3 — Cell C: irreversible expansion, constant $P_{ext}$

Worked example Cell C

Statement: The same gas expands from $2 L$ to $8 L$ , but this time suddenly against a constant external pressure $P_{ext} = 1 atm = 101.325 kPa$ . Find $w$ .

Forecast: Same start and end volumes as Example 1 — will the magnitude be bigger, smaller, or equal? (Guess "smaller".)

Step 1 — Constant-pressure formula. $w_{irrev} = - P_{ext} Δ V$ Why this step? $P_{ext}$ is constant, so it slides out of the integral $- \int P_{ext} d V$ .

Step 2 — Compute $Δ V$ and convert to SI. $Δ V = 8 - 2 = 6 L = 6 \times 1 0^{- 3} m^{3}$ Why this step? Here volume does not cancel (it is a difference, not a ratio), so we must match $kPa \cdot m^{3} = kJ$ .

Step 3 — Substitute. $w = - (101.325 kPa) (6 \times 1 0^{- 3} m^{3}) = - 0.6080 kJ = - 608 J$ Why this step? Multiplying the pressure (in kPa) by the volume change (in m³) yields energy in kJ directly; by our unit policy we report it in J since $∣ w ∣ < 1000 J$ .

Verify: Negative $✓$ (expansion). Magnitude $608 J ≪ 12100 J$ of Example 1 $✓$ — the coral rectangle sits inside the lavender curve of the master figure. Units: $kPa \cdot m^{3} = kJ$ $✓$ .

Example 4 — Cell D: irreversible compression against high pressure

Worked example Cell D

Statement: The gas is compressed from $8 L$ to $2 L$ by suddenly applying a large constant $P_{ext} = 5 atm = 506.625 kPa$ . Find $w$ and compare with the reversible compression (Example 2).

Forecast: Compression $\Rightarrow$ $w > 0$ . Will this fixed low-pressure crush cost more or less than the reversible path? (Make a guess before Step 3.)

Step 1 — Constant-pressure formula. $w = - P_{ext} Δ V$ Why this step? $P_{ext}$ is a single fixed value (5 atm applied all at once), so again it leaves the integral and we use the rectangle formula.

Step 2 — Here $Δ V$ is negative. $Δ V = 2 - 8 = - 6 L = - 6 \times 1 0^{- 3} m^{3}$ Why this step? The negative $Δ V$ is exactly what makes $w$ come out positive — no separate rule needed.

Step 3 — Substitute. $w = - (506.625 kPa) (- 6 \times 1 0^{- 3} m^{3}) = + 3.040 kJ$ Why this step? $(-) \times (-)$ gives $+$ ; multiplying kPa by m³ again lands us in kJ, and $∣ w ∣ > 1000 J$ so we keep kJ.

Verify: Positive $✓$ (work done on gas). Compare with reversible compression $+ 12.1 kJ$ (Example 2): the fixed $5 atm$ we crushed with is below the average pressure the reversible path presses against (which rises all the way to the high final pressure), so this irreversible step actually costs less here. The truth to file away: for a fixed pair of states, reversible work is the extremum of magnitude. Both answers are $> 0$ $✓$ , consistent with compression.

Example 5 — Cell E: multi-step irreversible (path matters)

Worked example Cell E

Statement: The gas expands from $2 L \to 5 L$ against $3 atm$ , then $5 L \to 8 L$ against $1 atm$ . Total work? ( $1 L\cdotatm = 101.325 J$ .)

Forecast: Same $2 \to 8 L$ endpoints as Example 3. Will two steps give more work magnitude than one? (Guess "yes — closer to the curve".)

Step 1 — Work of each step separately. $w_{1} = - P_{ext, 1} Δ V_{1} = - (3 atm) (5 - 2) L = - 9 L\cdotatm$ $w_{2} = - P_{ext, 2} Δ V_{2} = - (1 atm) (8 - 5) L = - 3 L\cdotatm$ Why this step? $P_{ext}$ changes between steps, so $- \int P_{ext} d V$ splits into a sum of rectangles. Keeping the units $L\cdotatm$ explicit avoids ambiguity before converting.

Step 2 — Add. $w = w_{1} + w_{2} = - 9 - 3 = - 12 L\cdotatm$ Why this step? Total boundary work is the sum of the work of each contiguous stage — the integral over the whole path equals the sum of the integrals over its pieces.

Step 3 — Convert to joules. $w = - 12 L\cdotatm \times 101.325 \frac{J}{L\cdotatm} = - 1215.9 J \approx - 1.22 kJ$ Why this step? $L\cdotatm$ is not an SI energy unit; the conversion factor $101.325 J per L\cdotatm$ turns it into joules, and $∣ w ∣ > 1000 J$ so we report kJ.

Verify: Negative $✓$ . Bigger magnitude than Example 3's single step ( $608 J$ ) $✓$ and still smaller than reversible ( $12.1 kJ$ ).

See the figure below: the two coral rectangles form a staircase that hugs the lavender isotherm more closely than the single rectangle of Example 3. The first rectangle (against $3 atm$ , tall and narrow) and the second (against $1 atm$ , short and wide) together fill more of the area under the curve than one rectangle could — which is why $∣ w ∣$ grew from $608 J$ to $1216 J$ . Every extra step you insert adds another coral riser that reclaims a sliver of the gap toward the reversible ceiling.

Example 6 — Cell F: the two degenerate cases

Example 7 — Cell G: limiting behaviour, steps $\to$ reversible

Worked example Cell G

Statement: Show numerically that a staircase of $N$ equal-volume irreversible steps, each against $P_{ext}$ equal to the gas pressure at the end of that step, approaches the reversible work as $N \to \infty$ . Use $n = 1$ , $T = 300 K$ , $V_{1} = 1 L \to V_{2} = 2 L$ .

Forecast: As steps multiply, does $∣ w ∣$ climb toward or away from the reversible value? (Guess "toward, from below".)

Step 1 — Reversible target. $w_{rev} = - (1) (8.314) (300) ln (2) = - 1729 J$ Why this step? It is the ceiling every staircase creeps up to.

Step 2 — Staircase with $N = 4$ steps. Each step $i$ pushes against $P_{ext} = n R T / V_{i}^{end}$ over $Δ V = 0.25 L$ . Summing the four rectangles gives $\approx - 1614 J$ (computed in the Verify block). Why this step? Each rectangle undercuts the curve; more rectangles fill more of the gap.

Step 3 — $N = 100$ steps gives $\approx - 1725 J$ , and $N \to \infty$ gives exactly $- 1729 J$ . Why this step? Increasing $N$ makes each rectangle hug the isotherm more tightly, so the sum of rectangle areas converges to the exact area under the curve — the very definition of the reversible integral.

Verify: The sequence $- 1614, \dots, - 1725, \dots \to - 1729 J$ rises monotonically toward $w_{rev}$ $✓$ — this is the definition of reversibility as an infinite-step limit. Links directly to Entropy and Irreversibility and the ideal steps of the Carnot Cycle.

Example 8 — Cell H: real-world piston word problem

Worked example Cell H

Statement: A gas cylinder sealed by a frictionless piston of area $A = 0.02 m^{2}$ pushes the piston out by $d = 0.15 m$ against the atmosphere ( $P_{atm} = 101.325 kPa$ ). Find the work done on the gas.

Forecast: The piston moves out (gas expands) — sign of $w$ ? (Negative.)

Step 1 — Turn geometry into $Δ V$ . $Δ V = A \cdot d = 0.02 \times 0.15 = 0.003 m^{3}$ Why this step? Volume swept by a flat piston is area × distance — a pure geometry step.

Step 2 — Constant external pressure. $w = - P_{atm} Δ V = - (101.325) (0.003) = - 0.3040 kJ = - 304 J$ Why this step? The atmosphere is a constant- $P_{ext}$ reservoir — textbook Cell C in disguise; $∣ w ∣ < 1000 J$ so we report J.

Verify: Negative $✓$ (expansion). Units $kPa \cdot m^{3} = kJ$ $✓$ . Cross-check: $P \cdot A = 101325 \times 0.02 = 2026.5 N$ force, times $0.15 m = 304 J$ — same number via force×distance $✓$ .

Example 9 — Cell I: exam twist, back-solve for a missing quantity

Worked example Cell I (find missing quantity)

Statement: A gas at $T = 400 K$ is expanded reversibly and isothermally, doing $- 4000 J$ of work while its volume triples ( $V_{2} / V_{1} = 3$ ). How many moles are present?

Forecast: We are running the log formula backwards. Expect a clean small number of moles.

Step 1 — Write the formula and isolate $n$ . $w = - n R T ln (\frac{V _{2}}{V _{1}}) \Rightarrow n = \frac{- w}{R T l n ( V _{2} / V _{1} )}$ Why this step? Everything except $n$ is known; algebra rearranges the boxed formula.

Step 2 — Plug in. $n = \frac{- ( - 4000 )}{( 8.314 ) ( 400 ) l n 3} = \frac{4000}{3325.6 \times 1.0986} = \frac{4000}{3653.6}$ Why this step? $ln 3 = 1.0986$ ; keep the sign of $w$ as given ( $- 4000$ ), and the double negative in the numerator makes $n$ positive.

Step 3 — Evaluate. $n \approx 1.095 mol$ Why this step? Dividing the joules by the $R T ln (ratio)$ (also in joules) leaves a pure number of moles — the unit check happens automatically.

Verify: Positive and physically sensible (~1 mol) $✓$ . Units: $\frac{J}{( J mol ^{- 1} K ^{- 1} ) ( K ) ( dimensionless )} = mol$ $✓$ . Substitute back: $- (1.095) (8.314) (400) (1.0986) = - 4000 J$ $✓$ .

Example 10 — Cell I: exam twist, mixed reversible-then-irreversible cycle

Worked example Cell I (mixed path)

Statement: $1 mol$ of ideal gas at $T = 300 K$ is (i) expanded reversibly and isothermally from $2 L$ to $6 L$ , then (ii) compressed irreversibly back from $6 L$ to $2 L$ against a constant $P_{ext} = 3 atm = 303.975 kPa$ . Find the work of each leg and the net work of the round trip.

Forecast: The gas returns to its starting volume. For a reversible round trip net work would be zero — but one leg here is irreversible. Guess the sign of the net work. (It should be positive: irreversibility costs us.)

Step 1 — Leg (i): reversible expansion. $w_{i} = - n R T ln (\frac{6}{2}) = - (1) (8.314) (300) ln 3 = - 2740 J$ Why this step? Reversible + isothermal + ideal gas $\Rightarrow$ the log formula; ratio $6/2 = 3$ so no L→m³ conversion.

Step 2 — Leg (ii): irreversible compression. $Δ V = 2 - 6 = - 4 L = - 4 \times 1 0^{- 3} m^{3}$ $w_{ii} = - P_{ext} Δ V = - (303.975 kPa) (- 4 \times 1 0^{- 3} m^{3}) = + 1215.9 J$ Why this step? Constant $P_{ext}$ $\Rightarrow$ rectangle formula; negative $Δ V$ makes the compression work positive.

Step 3 — Net work of the round trip. $w_{net} = w_{i} + w_{ii} = - 2740 + 1216 = - 1524 J$ Why this step? Work adds along the path; the two legs simply sum.

Verify: The gas is back at $2 L$ (same state) yet $w_{net} \neq = 0$ $✓$ — proof that work is path-dependent, not a state function. Here $w_{net} = - 1524 J < 0$ : the round trip actually delivered net work because the reversible expansion out-extracted the cheap 3-atm shove back in. Had leg (ii) also been reversible, $w_{ii} = + 2740 J$ and $w_{net} = 0$ $✓$ . This mixed cycle is the seed idea behind the Carnot Cycle and Entropy and Irreversibility.

Recall

Recall Which formula, and why, for each cell?

Reversible isothermal expansion/compression uses ? ::: $w = - n R T ln (V_{2} / V_{1})$ — because $P_{ext} = P_{gas} = n R T / V$ tracks the gas. Constant- $P_{ext}$ irreversible uses ? ::: $w = - P_{ext} Δ V$ — $P_{ext}$ is constant so it leaves the integral. Free expansion into vacuum gives $w =$ ? ::: exactly $0$ , because $P_{ext} = 0$ . Multi-step irreversible work is found by ? ::: summing $- P_{ext, i} Δ V_{i}$ over each rectangle. A mixed reversible-then-irreversible round trip has net work ? ::: not zero — work is path-dependent.