In a certain thermodynamical process, the pressure of a gas depends on its volume as $kV ^{3}$. The work done when the temperature changes from $100^{\circ} C$ to $300^{\circ} C$ will be .......... $nR$, where $n$ denotes number of moles of a gas.
JEE MAIN 2021, Diffcult
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$P=k V^{3}$
$T_{i}=100^{\circ} C $
$T _{ f }=300^{\circ} C$
$\Delta T =300-100$
$\Delta T =200^{\circ} C$
$P = kV ^{3}$
now $PV = nRT$
$\therefore kV ^{4}= nRT$
now $4 kV ^{3} dV = nRdT$
$\therefore PdV = nRdT / 4$
$\therefore$ Work $=\int PdV =\int \frac{ nRdT }{4}=\frac{ nR }{4} \Delta T$
$=\frac{200}{4} \times nR =50 nR$
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