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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2005

  • question_answer
    In a thermodynamics process pressure of a fixed mass of a gas is changed in such a manner that the gas molecule give out 30 joule of heat and 10 joule of work is done on the gas. If the initial internal energy of the gas was 40 joule, then the final internal energy will be:

    A)  -20 J                                     

    B)  20 J

    C)  80 J                       

    D)         3 J

    Correct Answer: B

    Solution :

                Here: \[dQ=-30\,J\] (negative sign appears as the heat is going out from the system) \[dW=-10\,J,\,U,=40\,J\] Using the relation                 \[dQ=dU+dW\] \[-30=({{U}_{f}}-40)+(-10)\]   \[{{U}_{f}}=20\,J\]


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