An ideal gas undergoes a cyclic process as sh | vedclass

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(B) From the graph, process $AB$ is a straight line passing through the origin, meaning $\rho \propto P$. Since $\rho = \frac{m}{V}$, we have $\frac{1

Vedclass · Accepted Answer

The efficiency of the cyclic process is $66\%$.

Vedclass · Answer

(B) From the graph, process $AB$ is a straight line passing through the origin, meaning $ho \propto P$. Since $ho = \frac{m}{V}$, we have $\frac{1}{V} \propto P$, or $PV = 	ext{constant}$. Thus, $AB$ is an isothermal process, so $\Delta U_{AB} = 0$. Given $|W_{AB}| = 70\,J$ and it is compression, $W_{AB} = -70\,J$. By $FLOT$, $Q_{AB} = \Delta U_{AB} + W_{AB} = 0 - 70 = -70\,J$.Process $BC$ is a horizontal line in the $ho-P$ graph, meaning $ho = 	ext{constant}$, so $V = 	ext{constant}$. Thus, $W_{BC} = 0$. Given $Q_{BC} = 150\,J$, then $\Delta U_{BC} = Q_{BC} - W_{BC} = 150 - 0 = 150\,J$.For a cyclic process, $\Delta U_{total} = 0$, so $\Delta U_{AB} + \Delta U_{BC} + \Delta U_{CA} = 0 \implies 0 + 150 + \Delta U_{CA} = 0 \implies \Delta U_{CA} = -150\,J$. Given $W_{CA} = 210\,J$, then $Q_{CA} = \Delta U_{CA} + W_{CA} = -150 + 210 = 60\,J$.Total work $W_{net} = W_{AB} + W_{BC} + W_{CA} = -70 + 0 + 210 = 140\,J$. Total heat absorbed $Q_{in} = Q_{BC} + Q_{CA} = 150 + 60 = 210\,J$. Efficiency $\eta = \frac{W_{net}}{Q_{in}} 	imes 100\% = \frac{140}{210} 	imes 100\% = 66.67\% \approx 66\%$. Thus, option $B$ is correct.

$List-I$	$List-II$
$(P)$ Work done in the complete cyclic process	$(1)$ $R T_0 - 4 R T_0 \ln 2$
$(Q)$ Change in the internal energy of the gas in the process $JK$	$(2)$ $0$
$(R)$ Heat given to the gas in the process $KL$	$(3)$ $3 R T_0$
$(S)$ Change in the internal energy of the gas in the process $MJ$	$(4)$ $-2 R T_0 \ln 2$
	$(5)$ $-3 R T_0 \ln 2$

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