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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2024 · 9 April · Evening Shift · Physics · Thermodynamics

A real gas within a closed chamber at 27^ C undergoes the cyclic...

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} ): A real gas within a closed chamber at 27^ C undergoes the cyclic process as shown in figure. The gas obeys P V^3=R T

Options

  1. A20 J20 \mathrm{~J}
  2. B205 J205 \mathrm{~J}
  3. C−20 J-20 \mathrm{~J}
  4. D225 J225 \mathrm{~J}

Correct answer

B. 205 J205 \mathrm{~J}

Step-by-step solution

WAB=∫PdV\mathrm{W}_{\mathrm{AB}}=\int \mathrm{PdV} \quad (Assuming T\mathrm{T} to be constant) =∫RTdVV3=RT∫24V−3dV=8×300×(−12[142−122])=225J\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}WBC=P∫42dV=10(2−4)=−20J&WCA=0&∴Wcycle =205J\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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