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JAMIA MILLIA ISLAMIA Jamia Millia Islamia Solved Paper-2013

  • question_answer
        A gas under goes a change of state during which 100 J of heat is supplied to it and it does 20 J work. The system is brought back to its original state through a process during which 20 J of heat is released by the gas. The work done by the gas in the second process is

    A)  60 J                       

    B)  40 J

    C)  80 J                       

    D)  20 J

    Correct Answer: C

    Solution :

                    In a cyclic process\[\Delta U=0\] \[\Delta U=Q-W\] \[\therefore \]  \[\Delta Q=\Delta W\] \[\therefore \]  \[(100-20)+20+{{W}_{2}}\] \[\therefore \]  \[{{W}_{2}}=60\,J\]


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