One mole of an ideal monoatomic gas is subjected to changes as shown in the graph. The magnitude of the work done (by the system or on the system) is __________ J (nearest integer)
Given ; log 2=0.3
ln 10=2.3

Solution

Chemistry ma’am · 4 steps

Read each leg of the cycle; the net work is the area enclosed by the loop.

1. Name the three paths. 1 → 2 is at constant pressure (1 bar), 2 → 3 is at constant volume (40 L), and 3 → 1 is a curve. At 3, PV = 0.5 × 40 = 20 L bar, and at 1, PV = 1 × 20 = 20 L bar. Since PV is the same at both ends, T is the same, so 3 → 1 is an isothermal compression.
2. Isobaric and isochoric. Along 1 → 2 the gas does work PΔ V = 1 × (40 - 20) = 20 L bar = 2000 J, using 1 L bar = 100 J. Along 2 → 3 the volume does not change, so the work is zero.

   |w_(3→ 1)| = nRT ln V₃/V₁ = (PV)ln 40/20 = 20 × 2.3 × 0.3 = 13.8 L bar

3. Isothermal leg. Here ln 2 = 2.3 log 2 = 0.69, so the work done on the gas during 3 → 1 is 13.8 L bar = 1380 J.
4. So. Net work by the gas = 2000 - 1380 = 620 J. The cycle runs clockwise, so this work is done by the system, and its magnitude is 620.

The common slip: 2000. Counting only the isobaric work and ignoring the isothermal compression gives 2000, not 620.

The idea to keep. For a cyclic process, add the work of each leg: PΔ V for isobaric, zero for isochoric, nRT ln(V₂/V₁) for isothermal. Check whether PV is the same at both ends to spot an isotherm.

Answer from the NTA final answer key. Solution by PrepWiser.