Two moles of Helium gas undergo a reversible cyclic process as shown in figure. Assuming gas to be ideal, what is the net work involved in the cyclic process ?

Options

A. 200 Rln2
B. 100 Rln2
C. 300 Rln2
D. 400 Rln2

Answer

A

Explanation

At constant pressure
`DeltaW =PDeltaV`
`=Pxx(V_(f)-V_(i))`
For an ideal gas PV = nRT
or `PV_(F)=nRT_(f) andPV_(i)=nrT_(i)`
From (i)
`therefore DeltaW=nR(T_(f)-T_(i))`
For constant temperature , PV =constatne
`P_(f)V_(f)=P_(i)V_(i)"or" ((V_(f)/(V_(i)))=((P_(i)/(P_(f)))f`
`therefore DeltaW=nRTln (P_(i))/(P_(f))`
so work dine for path AB,BC,CD and DA respectively will be
`DeltaW_(AB)=nR(T_(f)-T_(i))=(2xxR(400-300)=200R)`
`DeltaW_(CD)=nRTLn(P_(i))/(P_(F))=2xxRxx400ln2=800Rln2`
`DeltaW_(CD)=nR(T_(f)-T_(i))=2xxR[300-400]=-200R`
`DeltaW_(DA)=nRTln((P_(i))/(P_(f))=2xxRxx300ln((1)/(2))=-600Rln2`
Hence the work done in the complete cycle
`DeltaW=W_(AB)+W_(BC)+W_(CD)+W_(DA)`
=200R+800Rln2-200R-600Rln2=200Rln2