A Carnot engine uses an ideal monoatomic gas | vedclass

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(C) The efficiency of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$.Here,the source temperature $T_1 = 411 + 273 = 684 \, K$ a

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$500$

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(C) The efficiency of a Carnot engine is given by the formula: $\eta = 1 - \frac{T_2}{T_1}$.Here,the source temperature $T_1 = 411 + 273 = 684 \, K$ and the sink temperature $T_2 = 69 + 273 = 342 \, K$.Substituting these values into the efficiency formula: $\eta = 1 - \frac{342}{684} = 1 - 0.5 = 0.5$.The work done per cycle is equal to the area enclosed by the $PV$ diagram,which is given by $W = \eta 	imes Q$.Given that the heat extracted $Q = 1000 \, J$,we have $W = 0.5 	imes 1000 = 500 \, J$.

$A$ Carnot engine uses an ideal monoatomic gas first and then an ideal diatomic gas. If the source and sink temperatures are $411^{\circ}C$ and $69^{\circ}C$ respectively,and the engine extracts $1000 \, J$ of heat in each cycle,then the area enclosed by the $PV$ diagram is ........ $J$.

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