Thermodynamics

JEE Main level, single correct, medium

83 of 190

Question 83 of 190

A real gas within a closed chamber at 27∘C27^{\circ} \mathrm{C} undergoes the cyclic process as shown in figure. The gas obeys PV3=RTP V^3=R T equation for the path AA to BB. The net work done in the complete cycle is (assuming R=8 J/molKR=8 \mathrm{~J} / \mathrm{mol} \mathrm{K} ):

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Answer: (B) 205 J205 \mathrm{~J}

WAB=∫PdV\mathrm{W}_{\mathrm{AB}}=\int \mathrm{PdV} \quad (Assuming T\mathrm{T} to be constant)
=∫RTdVV3=RT∫24 V−3dV=8×300×(−12[142−122])=225 J\begin{aligned} & =\int \frac{\mathrm{RTdV}}{\mathrm{V}^3} \\ & =\mathrm{RT} \int_2^4 \mathrm{~V}^{-3} \mathrm{dV} \\ & =8 \times 300 \times\left(-\frac{1}{2}\left[\frac{1}{4^2}-\frac{1}{2^2}\right]\right) \\ & =225 \mathrm{~J}\end{aligned}
$\begin{aligned}& \mathrm{W}_{\mathrm{BC}}=\mathrm{P} \int_4^2 \mathrm{dV}=10(2-4)=-20 \mathrm{~J} \\& \mathrm{~W}_{\mathrm{CA}}=0 \\& \therefore \mathrm{W}_{\text {cycle }}=205 \mathrm{~J}
\end{aligned}$

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