JEE Main 24 Feb 2021 Shift 2 Solved Paper

Let pi,pf,Vi and Vf be the initial and final pressure and volume.
Given, AB is isothermal (∆T=0),
BC is isochoric (∆V=0)
and CA is adiabatic (∆Q=0)
Since, isothermal work (WAB)=p1V1‌ln‌

Vf

Vi

where, Vi and Vf are volume at A and B, respectively.
∴‌‌WAB=p1V1‌ln‌

2V1

V1

=p1V1‌ln‌2
Since, at constant volume, work done is zero.
∴‌‌WBC=0
Since, WCA is an adiabatic work done,
i.e. WCA‌‌=‌

1

1−γ

(pfVf−piVi)
⇒‌‌WCA‌‌=‌

1

1−γ

(p1V1−‌

p1

4

×2V1)
‌‌=‌

1

1−γ

(p1V1−p1V1∕2)=‌

1

1−γ

‌

p1V1

2

∴ Net work done, W‌net ‌=WAB+WBC+WCA
‌‌=p1V1‌ln‌2+0+‌

1

1−γ

‌

p1V1

2

‌‌=p1V1[ln‌2+1∕2(1−γ)]
From ideal gas law, pV=nRT
∴W‌net ‌=RT[ln‌2−1∕2(γ−1)]
(∵n=1)