JEE Main & Advanced Physics Thermodynamical Processes JEE PYQ-Thermodynamical Processes

  • question_answer
    Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from \[{{\tau }_{1}}\] to\[{{\tau }_{2}}\]. If \[\frac{{{C}_{P}}}{{{C}_{v}}}=\gamma \]for this gas then a good estimate for \[\frac{{{\tau }_{2}}}{{{\tau }_{1}}}\]is given by [JEE MAIN Held on 07-01-2020 Evening]

    A)  \[{{\left( \frac{1}{2} \right)}^{\frac{\gamma +1}{2}}}\]

    B)       \[{{\left( \frac{1}{2} \right)}^{\gamma }}\]

    C)  \[2\]                

    D)       None of these

    Correct Answer: D

    Solution :

    [d]
    \[\tau \propto \frac{1}{n<v>},\,<v>\,\propto \sqrt{T}\]
    \[\Rightarrow \,\,\,\tau \propto \frac{1}{n\sqrt{T}}\Rightarrow \,\frac{{{\tau }_{2}}}{{{\tau }_{1}}}=\frac{{{n}_{1}}}{{{n}_{2}}}\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\]
    \[=2\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\]
    \[{{T}_{1}}{{V}_{1}}^{\gamma -1}={{T}_{2}}{{(2{{V}_{1}})}^{\gamma -1}}\Rightarrow \,\frac{{{T}_{1}}}{{{T}_{2}}}={{2}^{\gamma -1}}\]
    \[\Rightarrow \,\,\frac{{{\tau }_{2}}}{{{\tau }_{1}}}\,=2\times {{2}^{\frac{(\gamma -1)}{2}}}=2{{\,}^{\left( \frac{\gamma +1}{2} \right)}}\]  
    Answer does not match with given options.


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