$n$ mole of 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 $V_{1}$ to $V_{2}=2 V_{1}$ and Pressure changes from $P_{1}$ to $P_{2}$. $B \rightarrow C$ : Isobaric compression at pressure $P_{2}$ to initial volume $V_{1}$. $C \rightarrow A$ : Isobaric change leading to change of pressure from $P_{2}$ to $P_{1}$. Total work done in the complete cycle ABCA is:

(B) $nRT\left(\ln 2-\frac{1}{2}\right)$

(A) $nRT\left(\ln 2+\frac{1}{2}\right)$

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Heat & Thermodynamics

n

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

A \to B

: Isothermal expansion at temperature T so that the volume is doubled from

V_{1}

V_{2} = 2 V_{1}

and Pressure changes from

P_{1}

P_{2}

B \to C

: Isobaric compression at pressure

P_{2}

to initial volume

V_{1}

C \to A

: Isobaric change leading to change of pressure from

P_{2}

P_{1}

.
Total work done in the complete cycle ABCA is:

Answer: (B)

For the cyclic process ABCA:

A→B (Isothermal expansion): Work done

W_{A B} = n R T ln \frac{V _{2}}{V _{1}} = n R T ln 2

B→C (Isobaric compression): Since

V_{2} = 2 V_{1}

and from isothermal process

P_{1} V_{1} = P_{2} V_{2}

, we get

P_{2} = \frac{P _{1}}{2}

. Work done

W_{B C} = P_{2} (V_{1} - V_{2}) = P_{2} (V_{1} - 2 V_{1}) = - P_{2} V_{1} = - \frac{n R T}{2}

**C→A (Isochoric process at constant volume

V_{1}

):** Work done

W_{C A} = 0

Total work =

W_{A B} + W_{B C} + W_{C A} = n R T ln 2 - \frac{n R T}{2} + 0 = n R T (ln 2 - \frac{1}{2})

Answer: (B)

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option