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An ideal gas is subjected to a cyclic process involving four thermodynamic states; the amounts of heat (Q) and work (W) involved in each of these states are $Q_{1} = 6000 J, Q_{2} = - 5500 J; Q_{3} = - 3000 J$ ; $Q_{4} = 3500 J; W_{1} = 2500 J; W_{2} = - 1000 J; W_{3} = - 1200 J; W_{4} = xJ$ . The ratio of the net work done by the gas to the total heat absorbed by the gas is $' η'.$ The values of x and $η$ respectively are

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1500; 15%

700; 10.5%

1000; 21%

500; 7.5%

answer is B.

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Detailed Solution

For cyclic process: dU=0 and dQ=dw

$\Rightarrow \sum Q = \sum W;$

$\Rightarrow Q_{1} + Q_{2} + Q_{3} + Q_{4} = W_{1} + W_{2} + W_{3} + W_{4}$

$\Rightarrow 6000 - 5500 - 3000 + 3500 = 2500 - 1000 - 1200 + x$

$\Rightarrow x = 700 J$

$And, η = \frac{W_{1} + W_{2} + W_{3} + W_{4}}{Q_{1} + Q_{4}}$

$\Rightarrow η = \frac{1000}{9500} = 0.1052$

$\Rightarrow % of η = 0.1052 \times 100 = 10.5 %$

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