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VMMC VMMC Medical Solved Paper-2003

  • question_answer
    If \[{{\upsilon }_{rms}}\]of          molecules of a diatomic gas is1930 m/s at room temperature \[\left( 27{}^\circ C \right)\], then the gas will be:

    A) \[{{H}_{2}}\]                      

    B)        \[{{F}_{2}}\]                      

    C) \[{{O}_{2}}\]                     

    D)        \[C{{l}_{2}}\]

    Correct Answer: A

    Solution :

    Using the relation, we have \[{{\upsilon }_{rms}}=\frac{\sqrt{3RT}}{m}\] where, R = gas constant \[=8.31\,J/mole-K\] \[\therefore \]  \[{{(1930)}^{2}}=\frac{3\times 8.31\times 300}{m}\] \[\therefore \]  \[m=2\times {{10}^{-3}}kg=2g\] Hence, the gas having 2 g molecular weight is hydrogen.


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