Q 11-11-127JEE MainJEE Main 2021 (26 Feb, Shift 2)Medium
The volume $V$ of a given mass of monoatomic gas changes with temperature $T$ according to the relation $V = KT^{\frac{2}{3}}$. The work done when temperature changes by $90$ K will be $xR$. The value of $x$ is [$R$ universal gas constant]
Numerical value type. Enter your answer.
Answer: 60
Take one mole of gas, $PV = RT$.
From $V = KT^{2/3}$: $\dfrac{dV}{V} = \dfrac{2}{3}\dfrac{dT}{T}$.
$$W = \int P\,dV = \int \frac{RT}{V}dV = \int \frac{2}{3}R\,dT = \frac{2}{3}R\times90 = 60R$$
So $x = 60$.
Solution by Sreeraj P, M.Sc Physics