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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 2

  • question_answer
    An ideal gas is taken through a cyclic thermodynamic process through four steps. The amount of heat involved in these steps are \[{{Q}_{1}}=5960J,{{Q}_{2}}=-5585J,{{Q}_{3}}=-2980J\] and \[{{Q}_{4}}=3645J\] respectively. The corresponding quantities of work involved are \[{{W}_{1}}=2200J,{{W}_{2}}=-825J,\]\[{{W}_{3}}=-1100J\]  and \[{{W}_{4}}\] respectively. Find the value of \[{{W}_{4}}.\].  What is the efficiency of the cycle?

    A) 3.33%   

    B) 6.76%

    C) 5.91%             

    D) 10.83%

    Correct Answer: D

    Solution :

    [d]: As the process is cyclic, therefore, \[\Delta U=0\]According to first law of thermodynamics \[\Delta Q=\Delta U+\Delta W=\Delta W\]or\[\Delta W=\Delta Q\] or\[{{W}_{1}}+{{W}_{2}}+{{W}_{3}}+{{W}_{4}}={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}}\] \[{{W}_{4}}=({{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}})-({{W}_{1}}+{{W}_{2}}+{{W}_{3}})\] Substituting the given values, we get \[=\left( 5960-5585-2980+3645 \right)-\left( 2200-825-1100 \right)\]\[=1040-275=765\text{ }J.\] \[\text{Efficiency},\eta =\frac{\text{Net work done}}{\text{Total heat absorbed}}\] \[=\frac{{{W}_{1}}+{{W}_{2}}+{{W}_{3}}+{{W}_{4}}}{{{Q}_{1}}+{{Q}_{4}}}\] \[\eta =\frac{2200-825-1100+765}{5960+3645}=\frac{1040}{9605}=0.1083\] \[=0.1083\times 100%=10.83%\]


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