BHU PMT BHU PMT (Screening) Solved Paper-2010

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

    A)  \[1.3\times {{10}^{2}}J\]                             

    B)  \[2.7\times {{10}^{2}}J\]

    C)  \[0.56\times {{10}^{4}}J\]                           

    D)  \[3.4\times {{10}^{3}}J\]

    Correct Answer: D

    Solution :

                     Kinetic energy g/mol,\[E=\frac{f}{2}RT\] If nothing is said about gas then we should calculate the translational kinetic energy.                 \[E{{ & }_{trans}}=\frac{3}{2}RT=\frac{3}{2}\times 8.31\times (273+0)\]                 \[=3.4\times {{10}^{3}}J\]


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