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Uttarakhand PMT Uttarakhand PMT Solved Paper-2005

  • question_answer
    At\[25{}^\circ C\]the average kinetic energy of an ideal gas per molecule in S.I. unit\[[N=6.02\times {{10}^{23}}]\]is:

    A)  \[6.17\times {{10}^{-21}}J\]      

    B)  \[61.7\times {{10}^{-23}}J\]

    C)  \[6.17\times {{10}^{-20}}J\]      

    D)  \[6.17\times {{10}^{-20}}J\]

    Correct Answer: A

    Solution :

     The average kinetic energy per molecule \[=\frac{3RT}{2n}\] Here, \[T=25+273=298K\] \[R=8.314\] \[=\frac{3\times 8.314\times 298}{2\times (6.02\times {{10}^{23}})}\] \[=6.17\times {{10}^{-21}}J\]


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