J & K CET Engineering J and K - CET Engineering Solved Paper-2014

  • 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)  \[35.6\,\,R\]

    Correct Answer: C

    Solution :

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


You need to login to perform this action.
You will be redirected in 3 sec spinner