Questions
n mole of a perfect gas and undergoes a cyclic process ABCA (see figure) consisting of the following processes.
: Isothermal expansion at temperature T so that the volume is doubled from to and pressure changes from to .
: Isobaric compression at pressure to initial volume
: Isochoric change leading to change of pressure from to .
Total work done in the complete cycle ABCA is :
detailed solution
Correct option is D
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=P1V1ℓn2−P1V12As per ideal gas equation, P1V1=nRT⇒WABCA=nRTln2-12Talk to our academic expert!
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