Punjab Medical Punjab - MET Solved Paper-2003

  • question_answer
    The kinetic energy of one molecule of a gas at normal temperature and pressure will be \[(k=8.31\,\,\text{J}\]/mole\[k\]):

    A) \[1.7\times {{10}^{3}}\text{J}\]

    B) \[10.2\times {{10}^{3}}\text{J}\]

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

    D) \[6.8\times {{10}^{3}}\text{J}\]

    Correct Answer: C

    Solution :

    According to kinetic theory that \[KE\] of \[1\,\,g-\]mole of an ideal gas,\[E=\frac{3}{2}RT\] Hence, \[KE\] at normal temperature \[{{0}^{o}}C\]                 \[=273\,\,K\] or            \[E=\frac{3}{2}\times 8.31\times 273\]                 \[=3.4\times {{10}^{3}}\text{J}\]


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