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

These problems build from "recognise the right formula" up to "prove why reversible work is maximal." Everything here rests on the parent note Work in Expansion and its prerequisites: the First Law of Thermodynamics, the Ideal Gas Law, the Isothermal Process, and the PV Diagram.

Before you start, pin down the two workhorse formulas — everything on this page is one of these two, or a chain of them.

A conversion you will use constantly:

Level 1 — Recognition

Recall Solution

Use $w_{rev} = - n R T ln (V_{2} / V_{1})$ . Why: "Reversible" means the external pressure trails the gas pressure by only an infinitesimal $d P$ , so at every instant $P_{ext} = P_{gas} = n R T / V$ . Because $P_{ext}$ changes smoothly as $V$ changes, we cannot pull it out of the integral — we must substitute $n R T / V$ and integrate, which produces the logarithm. Why the other is forbidden: $w_{irrev} = - P_{ext} Δ V$ assumes $P_{ext}$ is a single constant number. In a reversible process $P_{ext}$ is never constant (it drops as the gas expands), so that shortcut does not describe this path.

Recall Solution

Compression means $V_{2} < V_{1}$ , so $V_{2} / V_{1} < 1$ , so $ln (V_{2} / V_{1}) < 0$ . Then $w = - n R T \times (negative number) = positive .$ Positive $w$ means work is done ON the gas by the surroundings — you had to push the piston in, spending energy on the gas. Consistent with the sign convention: energy flowing into the system is positive.

Level 2 — Application

Recall Solution

Step 1 — pick the tool. Reversible + isothermal + ideal gas ⇒ $w = - n R T ln (V_{2} / V_{1})$ . Step 2 — plug in. $w = - (3) (8.314) (400) ln (\frac{8}{2}) = - 9976.8 ln (4) .$ Step 3 — evaluate. $ln 4 = 1.3863$ , so $w = - 9976.8 \times 1.3863 = - 13831 J \approx - 13.83 kJ .$ Sense check: expansion ⇒ negative ⇒ gas did work on surroundings. ✓

Recall Solution

Step 1 — pick the tool. Constant $P_{ext}$ ⇒ $w = - P_{ext} Δ V$ . Step 2 — compute $Δ V$ . $Δ V = 8 - 2 = 6 L$ . Step 3 — multiply (in L·atm first, it's cleaner). $w = - (1.5 atm) (6 L) = - 9 L\cdotatm .$ Step 4 — convert. $w = - 9 \times 101.325 = - 911.9 J \approx - 0.91 kJ .$ Compare with L2-Q1: reversible gave $- 13.83 kJ$ , irreversible only $- 0.91 kJ$ . Reversible extracts far more.

Level 3 — Analysis

Recall Solution

Why in stages? Each stage has its own constant $P_{ext}$ , so we apply $- P_{ext} Δ V$ per stage and add. This is the path-dependent case: $P_{ext} (V)$ is a staircase. Stage 1: $w_{1} = - (3) (2 - 1) = - 3 L\cdotatm$ . Stage 2: $w_{2} = - (1) (4 - 2) = - 2 L\cdotatm$ . Total: $w = - 5 L\cdotatm = - 5 \times 101.325 = - 506.6 J \approx - 0.51 kJ$ .

Single-step against $1 atm$ : $w = - (1) (4 - 1) = - 3 L\cdotatm = - 303.98 J$ .

Insight: The two-step path extracts more work ( $- 506.6$ J) than the single step ( $- 304$ J), because part of the expansion pushed against a higher pressure ( $3 atm$ ). More sub-steps, each closer to the gas's own pressure, ⇒ more work — approaching the reversible limit. Look at the staircase figure below.

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

Recall Solution

Step 1 — volumes from the ideal gas law. Use $P V = n R T$ with $R = 0.08206 L\cdotatm mol^{- 1} K^{- 1}$ so answers come out in litres. $V_{1} = \frac{n R T}{P _{1}} = \frac{( 2 ) ( 0.08206 ) ( 300 )}{4} = 12.31 L,$ $V_{2} = \frac{n R T}{P _{2}} = \frac{( 2 ) ( 0.08206 ) ( 300 )}{1} = 49.24 L .$ Notice $V_{2} / V_{1} = P_{1} / P_{2} = 4$ (isothermal: $P V$ constant). Step 2 — reversible work. In SI, $R = 8.314$ : $w_{rev} = - (2) (8.314) (300) ln (4) = - 4988.4 \times 1.3863 = - 6916 J \approx - 6.92 kJ .$ Step 3 — irreversible work against $1 atm$ . $Δ V = 49.24 - 12.31 = 36.93 L,$ $w_{irrev} = - (1 atm) (36.93 L) = - 36.93 L\cdotatm = - 36.93 \times 101.325 = - 3742 J \approx - 3.74 kJ .$ The gap: $∣ w_{rev} ∣ - ∣ w_{irrev} ∣ = 6.92 - 3.74 = 3.18 kJ$ of work is "lost" to irreversibility. On the PV Diagram this is the area between the smooth isotherm and the flat rectangle — see the figure below. The lost work is intimately tied to Entropy and Irreversibility.

Level 4 — Synthesis

Recall Solution

Step 1 — internal energy. For an ideal gas, $U$ depends only on $T$ . The process is isothermal ( $T$ constant), so $Δ U = 0.$ Step 2 — apply the First Law of Thermodynamics. $Δ U = q + w$ , so $0 = q + w ⟹ q = - w = - (- 6.92) = + 6.92 kJ .$ Meaning: The gas absorbs $6.92 kJ$ of heat from the surroundings and converts all of it into work. This is only possible because $Δ U = 0$ (isothermal, ideal). Positive $q$ = heat flowing into the system.

Recall Solution

Step 1 — the key fact. Expansion into vacuum means there is nothing to push against: $P_{ext} = 0$ . Step 2 — work. $w = - \int P_{ext} d V = - \int 0 d V = 0.$ No work is done, no matter how large the volume change. Step 3 — internal energy. Isothermal ideal gas ⇒ $Δ U = 0$ . Step 4 — First Law. $0 = q + 0 ⟹ q = 0$ . Insight: Same endpoints as L3-Q2, yet $w$ ranges from $- 6.92 kJ$ (reversible) to $0$ (free expansion). This is the whole point: work is path-dependent, not a state function. The endpoints ( $T$ , $V_{1}$ , $V_{2}$ ) are identical; only the path differs.

Level 5 — Mastery

Recall Solution

Setup. Both processes start at the same state $(P_{1}, V_{1})$ and reach $(P_{2}, V_{2})$ . Magnitudes: $∣ w_{rev} ∣ = n R T ln \frac{V _{2}}{V _{1}} = \int_{V_{1}}^{V_{2}} \frac{n R T}{V} d V = \int_{V_{1}}^{V_{2}} P_{gas} (V) d V,$ $∣ w_{irrev} ∣ = P_{ext} (V_{2} - V_{1}) = \int_{V_{1}}^{V_{2}} P_{ext} d V .$ Key comparison. For an irreversible expansion the gas must actually reach $(P_{2}, V_{2})$ , so $P_{ext} = P_{2} = \frac{n R T}{V _{2}}$ (the final gas pressure). Over the whole range $V \in [V_{1}, V_{2}]$ : $P_{gas} (V) = \frac{n R T}{V} \geq \frac{n R T}{V _{2}} = P_{ext} because V \leq V_{2} .$ The reversible integrand is never below the irreversible one at any $V$ , and strictly above for $V < V_{2}$ . Integrating a larger-or-equal function over the same interval: $\int_{V_{1}}^{V_{2}} P_{gas} d V > \int_{V_{1}}^{V_{2}} P_{ext} d V ⟹ ∣ w_{rev} ∣ > ∣ w_{irrev} ∣.$ Geometrically (figure below): the reversible work is the area under the falling curve $n R T / V$ ; the irreversible work is the area of the rectangle of height $P_{2}$ . The rectangle sits entirely under the curve, so its area is smaller. The curve is the ceiling — reversible work is maximal. This is why reversible engines set the efficiency ceiling in the Carnot Cycle. ∎

Recall Solution

Step 1 — write the formula and solve for the unknown ratio. $w = - n R T ln \frac{V _{2}}{V _{1}} ⟹ ln \frac{V _{2}}{V _{1}} = \frac{- w}{n R T} .$ Step 2 — plug numbers. $ln \frac{V _{2}}{V _{1}} = \frac{- ( - 2000 )}{( 0.5 ) ( 8.314 ) ( 350 )} = \frac{2000}{1454.95} = 1.3746.$ Step 3 — exponentiate. $\frac{V _{2}}{V _{1}} = e^{1.3746} = 3.954 ⟹ V_{2} = 10 \times 3.954 = 39.54 L .$ Sense check: Expansion, so $V_{2} > V_{1}$ ✓, and $w < 0$ ✓. The exponential undoes the natural log — that is exactly why $ln$ appears in the work law, so that solving for a volume ratio means applying $e^{(\cdot)}$ .

Recall Self-test summary (reveal after attempting)

Reversible isothermal formula ::: $w = - n R T ln (V_{2} / V_{1})$ Irreversible constant-pressure formula ::: $w = - P_{ext} (V_{2} - V_{1})$ Why reversible work is the maximum magnitude ::: gas pushes against pressure equal to its own at every instant, the highest possible resistance Work of a free expansion (into vacuum) ::: $w = 0$ because $P_{ext} = 0$ For isothermal ideal gas, $Δ U$ equals ::: $0$ , so $q = - w$ $1 L\cdotatm$ in joules ::: $101.325 J$