JEE Main & Advanced Sample Paper JEE Main Sample Paper-38

  • question_answer The root mean square speed of gas molecules at 25K & \[1.5\times {{10}^{5}}N{{m}^{-2}}\] is \[100.5\,m{{s}^{-1}}\]. If the temperature is raised to 100 K & pressure to \[6.0\times {{10}^{5}}N{{m}^{-2}},\] the root mean square speed becomes.

    A)  \[100.5\,m{{s}^{-1}}\]           

    B)  \[201.0\,m{{s}^{-1}}\]

    C)                             \[402\,m{{s}^{-1}}\]             

    D)  \[1608\,m{{s}^{-1}}\]

    Correct Answer: A

    Solution :

     In such a case there is no change in velocity\[u=\sqrt{(3RT/M)}=\sqrt{(3PV/M)}\] The increase in temperature = 4 times & also the increase in pressure = 4 times. Both of these reinforce each other


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