n mole of a perfect gas and undergoes a cyclic process ABCA (see$$ $$figure) consisting of the following processes.A→B : Isothermal expansion at temperature T so that the volume is doubled from $$V1$$ to $$V2=2V1$$ and pressure changes from $$P1$$ to $$P2$$.B→C: Isobaric compression at pressure $$P2$$ to initial volume $$V1C$$→A: Isochoric change leading to change of pressure from $$P2$$ to $$P1$$. Total work done in the complete cycle ABCA is :

Question

n mole of a perfect gas and undergoes a cyclic process ABCA (see$$ $$figure) consisting of the following processes.A→B : Isothermal expansion at temperature T so that the volume is doubled from $$V1$$  to $$V2=2V1$$  and pressure changes from $$P1$$  to $$P2$$.B→C: Isobaric compression at pressure $$P2$$ to initial volume  $$V1C$$→A: Isochoric change leading to change of pressure from $$P2$$  to $$P1$$.   Total work done in the complete cycle ABCA is :

Infinity Learn · Accepted Answer

At point A and B temperature is same.$$T1=T2$$⇒$$P1V1nR=P2V2nR$$ ⇒$$P2=P12Work$$ done in isothermal process, $$WAB=nRTlnVfVi$$ ⇒$$WAB=P1$$ $$V1$$ $$ln2V1V1$$⇒$$WAB=P1V1ln2Work$$ done in isobaric process,$$WBC=P2V1-V2$$⇒$$WBC=P2V1-2V1$$⇒$$WBC=P12-V1$$⇒$$WBC=$$−$$P1V12Work$$ done in isochoric process, $$WCA=0Hence$$, Net work done, $$WABCA=P1V1$$ℓ$$n2$$−$$P1V12As$$ per ideal gas equation, $$P1V1=nRT$$⇒$$WABCA=nRTln2-12$$

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