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JEE Main & Advanced Physics Thermodynamical Processes Question Bank Self Evaluation Test - Thermodynamics

  • question_answer
    An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and work (W) involved in each of these states \[{{Q}_{1}}=6000J;{{Q}_{2}}=-5500J;{{Q}_{3}}=-3000J\] \[{{Q}_{4}}=+\,3500J\] \[{{W}_{1}}=2500J;{{W}_{2}}=-1000J;{{W}_{3}}=-1200J\] \[{{W}_{4}}=xJ\] The ratio of the net work done by the gas to the total heat absorbed by the gas is \[\eta \]. The values of x and n respectively are

    A) 500; 7.5%

    B) 700; 10.5%

    C) 1000; 21%

    D) 1500; 15%

    Correct Answer: B

    Solution :

    [b] \[Q={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}}\] \[=6000-5500-3000+3500=+1000J\] \[W={{W}_{1}}+{{W}_{2}}+{{W}_{3}}+{{W}_{4}}\] \[=2500-1000-1200+x=+300+x\] In cyclic process, \[\Delta U=0\] Now, \[Q=\Delta U+W\] \[\text{or }1000=0+\left( 300+x \right)\] \[\therefore x=700J\] \[\eta =\frac{W}{{{Q}_{1}}+{{Q}_{4}}}\] \[=\frac{1000}{6000+3500}=10.5%\]


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