In a certain thermodynamical process, the pressure of a gas depends on its volume as $\mathrm{kV}^{3}$. The work done when the temperature changes from $100^{\circ} \mathrm{C}$ to $300^{\circ} \mathrm{C}$ will be_________ $\mathrm{nR}$, where $\mathrm{n}$ denotes number of moles of a gas.
$\mathrm{P}=\mathrm{kV}^{3}$
$\mathrm{T}_{\mathrm{i}}=100^{\circ} \mathrm{C} \& \mathrm{~T}_{\mathrm{f}}=300^{\circ} \mathrm{C}$
$\Delta \mathrm{T}=300-100$
$\Delta \mathrm{T}=200^{\circ} \mathrm{C}$
$P=k V^{3}$
now $\mathrm{PV}=\mathrm{nRT}$
$\therefore \mathrm{kV}^{4}=\mathrm{nRT}$
now $4 \mathrm{kV} 3 \mathrm{dV}=\mathrm{nRdT}$
$\therefore \mathrm{PdV}=\mathrm{nRdT} / 4$
$\therefore$ Work $=\int \mathrm{PdV}=\int \frac{\mathrm{nRdT}}{4}=\frac{\mathrm{nR}}{4} \Delta \mathrm{T}$
$=\frac{200}{4} \times \mathrm{nR}=50 \mathrm{nR}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.