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Uttarakhand PMT Uttarakhand 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: A

    Solution :

     For monoatomic gas at constant pressure \[\frac{\Delta U}{Q}=\frac{1}{3}\]or\[\Delta U=\frac{Q}{3}\] Now, applying first law of thermodynamics \[W=\Delta Q-\Delta U\] \[=Q-\frac{Q}{3}=\frac{2Q}{3}\]


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