Helium gas goes through a cycle ABCDA (consisting of two isochoric and isobaric lines) as shown in figure Efficiency of this cycle is nearly: (Assume the gas to be close to ideal gas)

Explanation

Evidently, work done by the gas during the cyclic process is
`W = (2P_(0) - P_(0)) (2V_(0) - V_(0))` unit `= P_(0) V_(0)` unit For the process A to B which is iscohoric, the heat supplied will be given by
`Delta Q_(1) = Delta U + Delta W = nC_(V) Delta T + 0`
`= n ((3 R)/(2)) [(V_(0) (2 P_(0) - P_(0)))/(nR)] ( :. nR Delta T = V Delta P)`
`= (3)/(2) P_(0) V_(0)` unit
And for the process B to C which is isobaric, the heat abosrbed will be given by
`Delta Q_(2) = n C_(P) Delta T`
`= n ((5 R)/(2)) [(2 P_(0) (2V_(0) - V_(0)))/(nR)] ( :. nR Delta T = P (Delta V))`
`= 5P_(0) V_(0)` unit
For the processes C to D to A, the heat given will be obviously negative, which implies that heat is abstracted from the system. Therefore, efficiency `(eta)` of the cycle
`= ("Work done per cycle")/("Total heat given per cycle") %`
`= (P_(0) V_(0))/(((3 P_(0) V_(0))/(2)) + 5 P_(0) V_(0)) = (2)/(3) xx 100 = (200)/(13) % = 15.38%`