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JEE Main & Advanced Physics Kinetic Theory of Gases Question Bank Self Evaluation Test - Kinetic Theory

  • question_answer
    N molecules each of mass m of a gas A and 2N molecules each of mass 2m of gas B are contained in the same vessel which is maintained at temperature T. The mean square velocity of molecules of B type is \[{{v}^{2}}\] and the mean square rectangular component of the velocity of A type is denoted by \[{{\omega }^{2}}.\] Then \[6.21\text{ }\times \text{ }{{10}^{-21}}\text{ }J\]is

    A) 2

    B) 1     

    C)  1/3

    D) 2/3

    Correct Answer: D

    Solution :

    [d] Mean kinetic energy of the two types of molecules should be equal. The mean square velocity of A type molecules =\[{{\omega }^{2}}+{{\omega }^{2}}=3{{\omega }^{2}}\]  Therefore, \[\frac{1}{2}m\left( 3{{\omega }^{2}} \right)=\frac{1}{2}\left( 2m \right){{v}^{2}}\] This gives \[{{\omega }^{2}}/{{v}^{2}}=2/3\]


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