• # 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) $\frac{1}{2}$

Correct Answer: A , B , C , D

Solution :

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

Solution :

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

Solution :

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

Solution :

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

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