Thank you for reporting, we will resolve it shortly
Q.
When the gas expands with temperature using the relation $ V=KT^{2/3} $ for the temperature change of $ 40\,K, $ the work done is
J & K CETJ & K CET 2014Thermodynamics
Solution:
$W=\int P d V=\int \frac{\mu R T}{V} d V $
$=\int \frac{\mu R T}{K T^{\frac{2}{3}}} d V$ ...(i)
$\because V=K T^{\frac{2}{3}} $
$d V=K \frac{2}{3} T^{-1 / 3} d T $ ...(ii)
From Eqs. (i) and (ii)
$W=\int \frac{\mu R T}{K T^{\frac{2}{3}}} \times K \frac{2}{3} T^{-1 / 3} d T$
$=\frac{2}{3} R \int T^{1-\frac{1}{3}-\frac{2}{3}} d T $
$=\frac{2}{3} R \int_{0}^{40} T^{0} d T=\frac{2}{3} R(T)_{0}^{40}$
$=\frac{2}{3} R(40-0)=\frac{80 R}{3}=26.6 \,R$