JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature, 300 K. The mean time between two successive collisions is \[6\times {{10}^{-8}}s.\]If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) \[4\times {{10}^{-8}}s\]                                   

    B) \[3\times {{10}^{-6}}s\]

    C) \[0.5\times {{10}^{-8}}s\]                    

    D)   \[2\times {{10}^{-7}}s\]

    Correct Answer: A

    Solution :

    Mean time between two successive collisions \[\tau \propto \frac{1}{\begin{align}   & velocity\times number\text{ }ot\text{ }particles \\  & per\,unit\,volume \\ \end{align}}\] \[\Rightarrow \]\[\tau \propto \frac{1}{\sqrt{T}}\frac{T}{P}=\frac{\sqrt{T}}{P}\therefore {{\tau }_{1}}={{\tau }_{2}}\frac{{{P}_{2}}}{{{P}_{1}}}\sqrt{\frac{{{T}_{1}}}{{{T}_{2}}}}\] \[\Rightarrow \]\[{{\tau }_{2}}=\frac{6\times {{10}^{-8}}}{2}\sqrt{\frac{500}{300}}=\sqrt{15}\times {{10}^{-8}}\simeq 4\times {{10}^{-8}}s\]


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