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VIT Engineering VIT Engineering Solved Paper-2009

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

    From first law of thermodynamics \[Q=\Delta U+W\] or \[\Delta U=Q-W\] \[\therefore \] \[\Delta {{U}_{1}}={{Q}_{1}}-{{W}_{1}}=6000-2500=3500J\] \[\Delta {{U}_{2}}={{Q}_{2}}-{{W}_{2}}=5500+1000=-4500J\] \[\Delta {{U}_{3}}={{Q}_{3}}-{{W}_{3}}=-3000+1200=-1800J\] \[\Delta {{U}_{4}}={{Q}_{4}}-{{W}_{4}}=3500-x\] For cyclic process \[\Delta U=0\] \[\therefore \]  \[3500-4500-1800+3500-x=0\] or \[x=700J\] Efficiency,  \[\eta =\frac{output}{input}\times 100\] \[=\frac{{{W}_{1}}+{{W}_{2}}+{{W}_{3}}+{{W}_{4}}}{{{Q}_{1}}+{{Q}_{4}}}\] \[=\frac{(2500-1000-1200+700)}{6000+3500}\times 100\] \[=\frac{1000}{9500}\times 100\] \[\eta =10.5%\]


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