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JEE Main & Advanced Physics Thermodynamical Processes Question Bank First Law of Thermodynamics

  • question_answer
    110 J of heat is added to a gaseous system, whose internal energy change is 40 J,  then the amount of external work done is    [CBSE PMT 1993; DPMT 1996, 03; AFMC 1999; JIPMER 2000; MH CET 2000; Pb. PMT 2003]

    A)            150 J                                        

    B)            70 J

    C)            110 J                                        

    D)            40 J

    Correct Answer: B

    Solution :

               \[\Delta Q=\Delta U+\Delta W\]Þ\[\Delta W=\Delta Q-\Delta U=100-40=70J\]


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