VMMC VMMC Medical Solved Paper-2013

  • question_answer
    The kinetic energy of 1 g molecule of a gas at normal temperature and pressure is\[(R=8.31\,J/mol-K)\]

    A) \[3.4\times {{10}^{3}}J\]

    B)  \[2.97\times {{10}^{3}}J\]

    C)  \[1.2\times {{10}^{2}}J\]

    D)  \[0.66\times {{10}^{4}}J\]

    Correct Answer: A

    Solution :

    \[E=\frac{l}{2}m{{v}^{-2}}\] \[=\frac{1}{2}M\left( \frac{3RT}{M} \right)\] \[E=\frac{3}{2}RT\] \[E=\frac{3}{2}\times 8.31\times 273\] \[E=3.4\times {{10}^{3}}J\]


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