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(C) The ideal gas equation is $PV = \mu RT$,which implies $P = (\mu R/V)T$. In the $P-T$ graph,lines passing through the origin represent isochoric pr

Vedclass · Accepted Answer

$40$

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(C) The ideal gas equation is $PV = \mu RT$,which implies $P = (\mu R/V)T$. In the $P-T$ graph,lines passing through the origin represent isochoric processes $(V = 	ext{constant})$.Processes $AB$ and $CD$ are straight lines passing through the origin,so volume remains constant during these processes. Thus,$W_{AB} = 0$ and $W_{CD} = 0$.Processes $BC$ and $DA$ are isobaric processes $(P = 	ext{constant})$.Work done in an isobaric process is $W = P\Delta V = \mu R\Delta T$.For process $BC$ (at pressure $P_2$): $W_{BC} = \mu R(T_C - T_B) = 6 	imes R 	imes (2200 - 800) = 6R 	imes 1400 \ J$.For process $DA$ (at pressure $P_1$): $W_{DA} = \mu R(T_A - T_D) = 6 	imes R 	imes (600 - 1200) = -6R 	imes 600 \ J$.Total work done per cycle $W = W_{AB} + W_{BC} + W_{CD} + W_{DA} = 0 + 6R(1400) + 0 - 6R(600) = 6R(800) \ J$.Using $R \approx 8.314 \ J/(mol \cdot K)$,$W = 6 	imes 8.314 	imes 800 \approx 39907 \ J \approx 40 \ kJ$.

Six moles of an ideal gas perform a cycle shown in the figure. If the temperatures are $T_A = 600 \ K, T_B = 800 \ K, T_C = 2200 \ K$,and $T_D = 1200 \ K$,the work done per cycle is ..... $kJ$.

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