An ideal gas is subjected to a cyclic process | vedclass

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(B) For a cyclic process,the change in internal energy over the complete cycle is zero,i.e.,$\sum \Delta U = 0$.From the first law of thermodynamics,$

Vedclass · Accepted Answer

$700; 10.5 \%$

Vedclass · Answer

(B) For a cyclic process,the change in internal energy over the complete cycle is zero,i.e.,$\sum \Delta U = 0$.From the first law of thermodynamics,$\Delta U = Q - W$.Calculating $\Delta U$ for each state:$\Delta U_1 = Q_1 - W_1 = 6000 - 2500 = 3500 	ext{ J}$$\Delta U_2 = Q_2 - W_2 = -5500 - (-1000) = -4500 	ext{ J}$$\Delta U_3 = Q_3 - W_3 = -3000 - (-1200) = -1800 	ext{ J}$$\Delta U_4 = Q_4 - W_4 = 3500 - x$Since $\sum \Delta U = 0$:$3500 - 4500 - 1800 + 3500 - x = 0$$700 - x = 0 \implies x = 700 	ext{ J}$.Net work done $W_{	ext{net}} = W_1 + W_2 + W_3 + W_4 = 2500 - 1000 - 1200 + 700 = 1000 	ext{ J}$.Total heat absorbed $Q_{	ext{in}} = Q_1 + Q_4 = 6000 + 3500 = 9500 	ext{ J}$.Efficiency $\eta = \frac{W_{	ext{net}}}{Q_{	ext{in}}} 	imes 100 = \frac{1000}{9500} 	imes 100 \approx 10.5 \%$.

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