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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2006

  • question_answer
    A monoatomic gas supplied the heat Q very slowly keeping the pressure constant. The work done by the gas will be :

    A)  \[\frac{2}{3}Q\]                              

    B)         \[\frac{3}{5}Q\]                              

    C)         \[\frac{2}{5}Q\]                                              

    D)         \[\frac{1}{5}Q\]

    Correct Answer: C

    Solution :

    For monoacomic gas at constant pressure                 \[\frac{\Delta U}{Q}=\frac{3}{5}\] or            \[\Delta U=\frac{3Q}{5}\] Now, applying first law at thermodynamics                 \[W=\Delta Q-\Delta U\]                     \[=Q-\frac{3Q}{5}\]                                 \[=\frac{2Q}{5}\]


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