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 changes 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 work done in the complete cycle ABCA is: (a) nRTln 2 (b) 0 (c) nRT(ln 2 - 1/2) (d) nRT(ln 2 + 1/2)

Answer: (c). A → B isothermal: WAB = nRTln (V₂)/(V₁) = nRTln 2 B → C isobaric: WBC = P₂(V₁ - 2V₁) = -P₂V₁. Since P₂(2V₁) = nRT, P₂V₁ = nRT/2 C → A isochoric: WCA = 0 Total W = nRTln 2 - nRT/2 = nRT(ln 2 - 1/2). So the answer is (c).

n mole of a perfect gas undergoes a cyclic process ABCA (see figure)… | JEE Main 2021 (24 Feb, Shift 1)

Q55JEE Main 2021 · 24 FEB · SHIFT 1Difficult

P-V Diagrams & Cyclic Processes

n mole of a perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes.

A $\rightarrow$ B: Isothermal expansion at temperature T so that the volume is doubled from V1 to V2 = 2V1 and pressure changes from P1 to P2

B $\rightarrow$ C: Isobaric compression at pressure P2 to initial volume V1.

C $\rightarrow$ A: Isochoric change leading to change of pressure from P2 to P1.

Total work done in the complete cycle ABCA is:

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Answer: (c)

$A \to B$ isothermal: $W_{AB} = nRT\ln\frac{V_2}{V_1} = nRT\ln 2$
$B \to C$ isobaric: $W_{BC} = P_2(V_1 - 2V_1) = -P_2V_1$ . Since $P_2(2V_1) = nRT$ , $P_2V_1 = \frac{nRT}{2}$
$C \to A$ isochoric: $W_{CA} = 0$
Total $W = nRT\ln 2 - \frac{nRT}{2} = nRT\left(\ln 2 - \frac{1}{2}\right)$ . So the answer is (c).

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