• # question_answer When the gas expands with temperature using the relation $V=K{{T}^{2/3}}$ for the temperature change of $40\text{ }K,$ the work done is A)  $20.1\,\,R$           B)  $30.2\,\,R$ C)  $26.6\,\,R$            D)  $35.6\,\,R$

$W=\int{PdV}$ $=\int{\frac{\mu \,RT}{V}}\,dV$ $=\int{\frac{\mu \,RT}{K{{T}^{\frac{2}{3}}}}}dV$ ??(i) $\because$ $V=K{{T}^{\frac{2}{3}}}$ $dV=K\frac{2}{3}{{T}^{-1/3}}dT$ ?..(ii) From Eqs. (i) and (ii) $W=\int{\frac{\mu \,RT}{K{{T}^{\frac{2}{3}}}}}\times K\frac{2}{3}{{T}^{-1/3}}dT$ $=\frac{2}{3}R\int{{{T}^{1-\frac{1}{3}-\frac{2}{3}}}}dT$ $=\frac{2}{3}R\int_{0}^{40}{{{T}^{0}}}dT=\frac{2}{3}R(T)_{0}^{40}$ $=\frac{2}{3}R(40-0)=\frac{80\,R}{3}=26.6R$