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

Efficiency `=("work done in cycle")/("heat absorbed")xx100`
`=("Area under P-V "diagram")/(DeltaQ_(AB)+DeltaQ_(BC))`
`therefore eta =(P_(0)V_(0))/(etaC_(v)DeltaT_(1)+nC_(P)DeltaT_(2))`
`=(P_(0)V_(0))/((3)/(2)nR(T_(B)-T_(A))+(5)/(2)nR(T_(C)-T_(B)))`
`(P_(0)V_(0))/((3)/(2)(2P_(0)V_(0)-P_(0)V_(0))+(5)/(2)(4P_(0)V_(0)-2P_(0)V_(0)))`
`(P_(0)V_(0))/((3)/(2)P_(0)V_(0)+(5)/(2)2P_(0)V_(0))=(1)/(6.5)=15.4%`