JEE MAIN - Physics (2024 - 9th April Evening Shift - No. 7)
A real gas within a closed chamber at $$27^{\circ} \mathrm{C}$$ undergoes the cyclic process as shown in figure. The gas obeys $$P V^3=R T$$ equation for the path $$A$$ to $$B$$. The net work done in the complete cycle is (assuming $$R=8 \mathrm{~J} / \mathrm{mol} \mathrm{K}$$):
_9th_April_Evening_Shift_en_7_1.png)
Explanation
For $A$ to $B$ :
Given $\mathrm{PV}^3=\mathrm{RT} \Rightarrow \mathrm{P}=\frac{\mathrm{RT}}{\mathbf{V}^3}$
Work done $\mathrm{W}_{\mathrm{AB}}=\int \mathrm{PdV}$
$\begin{aligned} & =\int \frac{R T}{V^3} d V=R T\left[\frac{V^{-2}}{-2}\right]_{V_1=2}^{V_B=4} \\\\ & =-\frac{R T}{2}\left[\frac{1}{V^2}\right]=-\frac{R T}{2}\left[\frac{1}{16}-\frac{1}{4}\right] \\\\ & =-\frac{R T}{2}\left(\frac{-3}{16}\right)=\frac{3 R T}{32}\end{aligned}$
$\begin{aligned} & =\frac{3}{32} \times 8 \times 300=225 \mathrm{~J} \\\\ & \mathrm{~W}_{\mathrm{B} . \mathrm{C}}(\text { Isobaric process ) } \\\\ & \mathrm{W}_{\mathrm{BC}}=\mathrm{P} \Delta \mathrm{V}=10(2-4)=-20 \mathrm{~J} \\\\ & \mathrm{~W}_{\mathrm{B} . \mathrm{C}}(\text { Isochoric process })=0 \\\\ & \mathrm{~W}_{\text {nct }}=\mathrm{W_{AB}}+\mathrm{W}_{\mathrm{BC}}+\mathrm{W}_{C A}=225-20+0=205 \mathrm{~J}\end{aligned}$
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