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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 7

  • question_answer
    Consider an ideal gas confined in an isolated closed chamber. As the gas undergoes an adiabatic expansion, the average time of collision between molecules increases as \[{{V}^{q}}\], where V is the volume of the gas. The value of q is \[\left( \gamma =\frac{{{C}_{P}}}{{{C}_{V}}} \right)\]

    A) \[\frac{\gamma +1}{2}\]           

    B) \[\frac{\gamma -1}{2}\]

    C) \[\frac{3\gamma +5}{6}\]         

    D) \[\frac{3\gamma -5}{6}\]

    Correct Answer: A

    Solution :

    [a] : Average time of collision between molecules, \[\tau =\frac{Mean\,free\,path\,(\lambda )}{Mean\,speed\,(\overline{v})}=\frac{1}{\left( \sqrt{2}\pi {{d}^{2}}\frac{N}{V} \right)\left( \sqrt{\frac{8{{k}_{B}}T}{m\pi }} \right)}\]  \[\therefore \,\,\,\,\tau \propto \frac{V}{\sqrt{T}}\,\,\,\,or\,\,\,\,\,T\propto \frac{{{V}^{2}}}{{{\tau }^{2}}}\]   For adiabatic expansion, \[T{{V}^{\lambda -1}}=\] constant or\[\frac{{{V}^{2}}}{{{\tau }^{2}}}{{V}^{\gamma -1}}=\]constant or \[\tau \propto {{V}^{\frac{(\gamma +1)}{2}}}\] Comparing it with \[\tau \propto {{V}^{q}}\], we get \[q=\frac{\gamma +1}{2}\]


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