BHU PMT BHU PMT (Screening) Solved Paper-2007

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

    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 :

                     Key Idea: Average kinetic energy per molecule is equal to product of mass of 1 g molecule and square of mean square velocity. The kinetic energy of 1 g mol is                 \[E=\frac{1}{2}m{{\overline{v}}^{2}}=\frac{1}{2}M\left( \frac{3RT}{M} \right)\]                                                 \[\because \left[ \overline{v}=\sqrt{\frac{3RT}{M}} \right]\]                 \[E=\frac{3}{2}RT\] where R is gas constant. Putting the numerical values, we have                 \[E=\frac{3}{2}\times 8.31\times 273\]                 \[E=3.4\times {{10}^{3}}J\]


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