Q3(a)(ii). Reversible work along two different paths

Reading the figure: both processes take the gas from (P₁=3.00 atm, V₁=500 cm³) to (P₂=1.00 atm, V₂=2000 cm³). In (a) the pressure is first lowered at constant volume V₁ (step 1), then the gas expands at constant pressure P₂ (step 2). In (b) the gas first expands at constant pressure P₁ (step 1), then the pressure is lowered at constant volume V₂ (step 2).

1. Reversible work: w=-∫ P dV (sign: w is work done on the gas). At constant volume dV=0, so w=0. At constant pressure w=-P(V_f-Vᵢ).
2. Δ V=2000-500=1500cm³=1.500dm³; 1 L atm =101.325 J.
3. Path (a): step 1 (constant V₁): w₁=0; step 2 (constant P₂=1.00 atm): w₂=-1.00×1.500=-1.50 L atm 3pt] w₍ₐ₎=-1.50 L atm=-152 J
4. Path (b): step 1 (constant P₁=3.00 atm): w₁=-3.00×1.500=-4.50 L atm; step 2 (constant V₂): w₂=0 3pt] w_((b))=-4.50 L atm=-456 J
5. Work is the area under the path on the P–V diagram. The expansion in (b) happens at the higher pressure, so the gas does three times as much work.
6. The initial and final states are the same, yet w differs: work is a path function, not a state function. (Δ U is the same for both paths, so q also differs: q_((b))-q₍ₐ₎=304 J.)

Application. the same area rule gives the net work of a heat engine as the area enclosed by its cycle on a P–V diagram; path (b) followed by path (a) reversed is such a cycle, with net work 304 J per cycle.