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

  • question_answer
    A perfect gas goes from state A to another state B by absorbing \[8\times {{10}^{5}}J\] of heat and doing \[6.5\times {{10}^{5}}J\] of external work. It is now transferred between the same two states in another process in which it absorbs \[{{10}^{5}}J\] of heat. Then in the second process                                         [BHU 1997]

    A)            Work done on the gas is \[0.5\times {{10}^{5}}J\]

    B)             Work done by gas is \[0.5\times {{10}^{5}}J\]

    C)            Work done on gas is \[{{10}^{5}}J\]

    D)            Work done by gas is \[{{10}^{5}}J\]

    Correct Answer: A

    Solution :

                       In first process using \[\Delta Q=\Delta U+\Delta W\]                    Þ \[8\times {{10}^{5}}=\Delta U+6.5\times {{10}^{5}}\]Þ \[\Delta U=1.5\times 10J\]                    Since final and initial states are same in both process                    So \[\Delta U\] will be same in both process                     For second process using \[\Delta Q=\Delta U+\Delta W\]            Þ \[{{10}^{5}}=1.5\times {{10}^{5}}+\Delta W\]Þ\[\Delta W=-0.5\times {{10}^{5}}J\]


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