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One mole of an ideal gas at (P₁,V₁) is allowed to expand reversibly and isothermally from A to B, so that its pressure is reduced to one-half of the original pressure, as shown in the figure. This is followed by a constant-volume cooling from B to C until the pressure is reduced to one-fourth of the initial value. It is then restored to its initial state by a reversible adiabatic compression from C to A. The net work done by the gas is equal to

Asked in JEE Main 24th Feb 2nd Shift 2021 · Cycles with an Adiabatic Leg

Figure: Cycles with an Adiabatic Leg
Answer: (4) RT(ln 2-1/(2(γ-1)))

Step-by-step solution

A to B, isothermal, pressure halved so volume doubles: W_AB=RT ln 2.

B to C, constant volume: W_BC=0, and the pressure falls to (P₁)/4.

C to A, adiabatic compression from ((P₁)/4,2V₁) back to (P₁,V₁):

W_CA=(P_CV_C-P_AV_A)/(γ-1)=((P₁V₁)/2-P₁V₁)/(γ-1)=-(RT)/(2(γ-1)).

W=RT ln 2-(RT)/(2(γ-1)).

Why the other options are wrong

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