A real gas within a closed chamber at 27° C undergoes the cyclic process as shown in figure. The gas obeys P V³=R T equation for the path A to B. The net work done in the complete cycle is (assuming R=8 J / mol K):

Solution

Physics sir · 4 steps

Add the work leg by leg: the curve AB gives the big positive part, and BC takes a little back.

1. Pressure on AB. The question intends the gas to stay at 27°C = 300 K along AB. Then PV³ = RT = 8 × 300 = 2400, so P = 2400/V³.

   W_(AB) = ∫₂⁴ 2400/V³ dV = 1200(1/2² - 1/4²) = 1200 × 3/16 = 225 J

2. Path B to C. Pressure stays at 10 N m⁻² while the volume falls from 4 to 2 m³. So W_(BC) = P Δ V = 10 × (2 - 4) = -20 J.
3. Path C to A. The volume is fixed at 2 m³, so W_(CA) = 0.
4. So. Wₙₑₜ = 225 - 20 + 0 = 205 J. This is positive, as it should be for a cycle that runs clockwise on the P-V diagram. The answer is (b).

The idea to keep. For a cycle, find the work on each leg separately: integrate on the curve, use PΔ V when pressure is constant, and take zero when volume is constant. A clockwise loop gives positive net work.

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