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An ideal gas after going through a series of four thermodynamic states in order, reaches the initial state again (cyclic process). The amounts of heat (Q) and work (W) involved in these states are
$Q_{1} = 6000 J; Q_{2} = - 5500 J; Q_{3} = - 3000 J; Q_{4} = 3500 J$
$W_{1} = 2500 J; W_{2} = - 1000 J; W_{3} = - 1200 J; W_{4} = x j o u l e s$
The ratio of net work done by the gas to the total heat absorbed by the gas is $η %$ . The values of x and $η$ are nearly

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500 ; 7.5

700; 10.5

1000; 21

1500; 15

answer is B.

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Detailed Solution

$Q_{T} = W_{T}$

6000 – 5500 – 3000 + 3500 = 2500 – 1000 – 1200 + W₄

$W_{4} = 700 J$

$η = \frac{W_{n e t}}{Q_{a b s o r b e d}} \times 100 = \frac{1000}{9500} \times 100 = 10.5$

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