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QJEEM24Ph8ea9: JEE Main 2024 Physics - Class 11 JEEM24Ph8ea9 Solution: A real gas within a closed chamber at 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} ):

JEE Main 2024 (Online) 9th April Evening Shift Physics - Heat and Thermodynamics Question 47 English

Multiple Choice Options

  • A −20-20 J
  • B 205J
  • C 225J
  • D 20J

Step-by-Step Verified Solution

For AA to BB :

Given PV3=RT⇒P=RTV3\mathrm{PV}^3=\mathrm{RT} \Rightarrow \mathrm{P}=\frac{\mathrm{RT}}{\mathbf{V}^3}

Work done WAB=∫PdV\mathrm{W}_{\mathrm{AB}}=\int \mathrm{PdV}

amp;=∫RTV3dV=RT[V−2−2]V1=2VB=4amp;=−RT2[1V2]=−RT2[116−14]amp;=−RT2(−316)=3RT32\begin{aligned} & =\int \frac{R T}{V^3} d V=R T\left[\frac{V^{-2}}{-2}\right]_{V_1=2}^{V_B=4} \\\\ & =-\frac{R T}{2}\left[\frac{1}{V^2}\right]=-\frac{R T}{2}\left[\frac{1}{16}-\frac{1}{4}\right] \\\\ & =-\frac{R T}{2}\left(\frac{-3}{16}\right)=\frac{3 R T}{32}\end{aligned}

amp;=332×8×300=225 Jamp; WB.C( Isobaric process ) amp;WBC=PΔV=10(2−4)=−20 Jamp; WB.C( Isochoric process )=0amp; Wnct =WAB+WBC+WCA=225−20+0=205 J\begin{aligned} & =\frac{3}{32} \times 8 \times 300=225 \mathrm{~J} \\\\ & \mathrm{~W}_{\mathrm{B} . \mathrm{C}}(\text { Isobaric process ) } \\\\ & \mathrm{W}_{\mathrm{BC}}=\mathrm{P} \Delta \mathrm{V}=10(2-4)=-20 \mathrm{~J} \\\\ & \mathrm{~W}_{\mathrm{B} . \mathrm{C}}(\text { Isochoric process })=0 \\\\ & \mathrm{~W}_{\text {nct }}=\mathrm{W_{AB}}+\mathrm{W}_{\mathrm{BC}}+\mathrm{W}_{C A}=225-20+0=205 \mathrm{~J}\end{aligned}

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