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BHU PMT BHU PMT (Screening) Solved Paper-2008

  • question_answer
    An ideal gas is taken through a cyclic thermo dynamical process through four steps. The amounts of heat involved in these steps are \[{{Q}_{1}}=5960\text{ }J,\text{ }{{Q}_{2}}=-5585\text{ }J,\text{ }{{\text{Q}}_{3}}=-2980\text{ }J,\] \[{{Q}_{4}}=3645\text{ }J;\]respectively. The corresponding works involved are\[{{W}_{1}}=22000\,J,{{W}_{2}}=-825\,J,\] \[{{W}_{3}}=-1100\,J\]and\[{{W}_{4}}\]respectively. The value of\[{{W}_{4}}\]is

    A)  1315 J                                  

    B)  275 J

    C)  765J                                     

    D)  675J

    Correct Answer: C

    Solution :

                     \[\Delta Q={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}}\] \[=5960-5585-2980+3645=1040\text{ }J\] \[\Delta W={{W}_{1}}+{{W}_{2}}+{{W}_{3}}+{{W}_{4}}\] \[=2200-825-1100+{{W}_{4}}=275+{{W}_{4}}\] For a cyclic process, \[\Delta U=0\] ie,       \[{{U}_{f}}-{{U}_{i}}=0\] From first law of thermodynamics, \[\Delta Q=\Delta U+\Delta W\] \[1040=0+275+{{W}_{4}}\] or            \[{{W}_{4}}=765\,J\]


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