An ideal gas is subjected to a cyclic process | vedclass

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(B) For a cyclic process,the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics,$\Delta U = Q_{net} + W_{net} = 0$,

Vedclass · Accepted Answer

$1300; 10.5\%$

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(B) For a cyclic process,the change in internal energy $\Delta U = 0$.According to the first law of thermodynamics,$\Delta U = Q_{net} + W_{net} = 0$,which implies $Q_{net} = -W_{net}$.$Q_{net} = Q_1 + Q_2 + Q_3 + Q_4 = 6000 - 5500 - 3000 + 3500 = 1000 \, J$.$W_{net} = W_1 + W_2 + W_3 + W_4 = 2500 - 1000 - 1200 + x = 300 + x$.Since $Q_{net} = -W_{net}$,we have $1000 = -(300 + x)$,which gives $x = -1300 \, J$. Thus,$|x| = 1300 \, J$.The total heat absorbed by the gas is the sum of positive heat values: $Q_{abs} = Q_1 + Q_4 = 6000 + 3500 = 9500 \, J$.The net work done is $W_{net} = 300 + (-1300) = -1000 \, J$. The magnitude of net work done is $|W_{net}| = 1000 \, J$.$\eta = \frac{|W_{net}|}{Q_{abs}} = \frac{1000}{9500} \approx 0.10526 = 10.53\%$.

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