One mole of a monatomic ideal gas is taken along two cyclic processes `E to F to G to E and E to F to H to E` as shown in the PV diagram. The processes involved are purely isochoric , isobaric , isothermal or adiabatic .

Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists.

Options

A. a.`{:(P,Q,R,S),(4,3,2,1):}`
B. b.`{:(P,Q,R,S),(4,3,1,2):}`
C. c.`{:(P,Q,R,S),(3,1,2,4):}`
D. d.`{:(P,Q,R,S),(3,1,2,4):}`

Answer

A

Explanation

Process `FG` is isothermal so work done =`nRT "In"((P_(t))/(P_(f))) = 32 P_(0)V_(0) "In" ((32P_(0))/(P_(0)))= 160 P_(0)V_(0) "In"2`.
Process `GE` is isobaric
So work done = `P|DeltaV| = P_(0)|(V_(G)- V_(E))|`
=`P_(0)|(32V_(0) - V_(0))|`
=`31P_(0)V_(0)`
Process `FH` is adiabatic so `(32P_(0))V_(0)^(5//3) = (P_(0))V_(H)^(5//3) implies V_(H) = 8V_(0)`
Since process `FH` is adiabatic so `|((P_(H)V_(H) - P_(F)V_(F)))/((8-1))|= (|(P_(0)8V_(0))-32P_(0)V_(0)|)/(((5)/(3)-1))=36P_(0)V_(0)`
Process `G rarr H` is isobaric so work done =`P_(0)|(32V_(0)- 8V_(0))| = 24 P_(0)V_(0)`