n mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes.
A → B : Isothermal expansion at temperature T so that the volume is doubled from V₁ to V₂ = 2V₁ and pressure charges from P₁ to P₂
B → C : Isobaric compression at pressure P₂ to initial volume V₁.
C → A : Isochoric change leading to change of pressure from P₂ to P₁.
Total workdone in the complete cycle ABCA is :

Solution

Physics sir · 4 steps

Add up the work of the three legs one by one; the key is finding P₂V₁ from point B on the isotherm.

1. Isothermal A to B. For an isothermal expansion, W_(AB) = nRT ln V₂/V₁ = nRT ln 2. It is positive because the gas expands.
2. Isobaric B to C. At constant pressure, W_(BC) = P₂(V₁ - 2V₁) = -P₂V₁. Point B lies on the isotherm, so P₂(2V₁) = nRT, which gives P₂V₁ = nRT/2. So W_(BC) = -nRT/2, negative because the gas is compressed.
3. Isochoric C to A. The volume does not change, so W_(CA) = 0.

   W_(cycle) = nRT ln 2 - nRT/2 + 0 = nRT(ln 2 - 1/2)

4. So. The net work is nRT(ln 2 - 1/2), which is option (c). It is positive (0.693 > 0.5), as it should be for a cycle that runs clockwise on the P-V diagram.

The idea to keep. For a cycle, net work is the sum of the work in each leg. Use PV = nRT at a shared point to write unknown pressures in terms of nRT, and watch the sign of compression work.

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