11.Thermodynamics
hard

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.

A

$25$

B

$40$

C

$50$

D

$60$

(JEE MAIN-2021)

Solution

$P=k V^{3}$

$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 $=\int PdV =\int \frac{ nRdT }{4}=\frac{ nR }{4} \Delta T$

$=\frac{200}{4} \times nR =50 nR$

Standard 11
Physics

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