• # question_answer A gas expands with temperature according to the relation $V=k{{T}^{2/3}}$. Calculate work done when the temperature changes by 60 K? A) 10 R     B) 30 R C) 40 R     D) 20 R

 [c]$dW=\,pdV=\,\frac{RT}{V}dV$  . (i) As,  $V=\,k{{T}^{2/3}},\,dV=\,k\frac{2}{3}\,\,{{T}^{-2/3}}\,\,dT$ $\frac{dV}{V}=\,\frac{k\frac{2}{3}\,\,{{T}^{-2/3}}dT}{k{{T}^{2/3}}}\,=\,\frac{2}{3}\,\frac{dT}{T}$ From Eq. (i), $W=\int_{{{T}_{1}}}^{{{T}_{2}}}{RT}\,\frac{dV}{V}=\int_{{{T}_{1}}}^{{{T}_{2}}}{RT\,\frac{2}{3}\,\,\frac{dT}{T}}$ $W\,=\,\frac{2}{3}\,R\,\,({{T}_{2}}-\,{{T}_{1}})$ $=\frac{2}{3}\,R\times \,60\,=\,40R$