Find the efficeincy of the thermodynamic cycle shown in figure for an ideal diatomic gas.

Explanation

Let n be the number of moles of the gas and the temperature be `T_(0)` in the state A.
Now, work done during the cycle
`W=(1)/(2)xx(2V_(0)-V_(0))(2P_(0)-P_(0))=(1)/(2)P_(0)V_(0)`
For the heat `(DeltaQ_(1))` given during the process
`ArarrB`, we have
`DeltaQ_(1)=DeltaW_(AB)+DeltaU_(AB)`
`DeltaW_(AB)=` area under the straight line AB
`=(1)/(2)(P_(0)+2P_(0))(2V_(0)-V_(0))=(3P_(0)V_(0))/(2)`
Applying equation of state for the gas in the state A `&` B.
`(P_(0)V_(0))/(T_(0))=((2P_(0))(2V_(0)))/(T_(B))rArrT_(B)=4T_(0)`
`thereforeU_(AB)=nC_(v)DeltaT=n((5R)/(2))(4T_(0)-T_(0))`
`=(15nRT_(2))/(2)=(15P_(0)V_(0))/(2)`
`therefore DeltaQ_(1)=(3)/(2)P_(0)V_(0)+(15)/(2)P_(0)V_(0)=9P_(0)V_(0)`
Obviously, the processes `BrarrC` and `CrarrA`
involve the abstraction of heat from the gas.
`"Efficiency"=("Work done per cycle")/("Total heat supplied per cycle")`
i.e., `eta=((1)/(2)P_(0)V_(0))/(9P_(0)V_(0))=(1)/(18)`