An ideal gas undergoes a cyclic process as shown. Part of the process ab is isothermal expansion and work done by the gas in this expansion is 700 J. in adiabatic expansion be it again does 400 J work. If it rejects 100 J heat when it returns to a from c via b' then find efficiency of the cycle.

Explanation

ab is an isothermal process, therefore `DeltaU_(ab)=0` ltBrgt Now from first law of thermodynamics
`Q_(ab)=W_(ab)=700J` . . . . (i)
Again bc is an adiabatic process, therefore `Q_(bc)=0`
as given, `W_(bc)=400J`
therefore
`DeltaU_(bc)=-W_(bc)=-400J`
Now for the complete cycle, `DeltaU=0`
`impliesDeltaU_(ab)+DeltaU_(bc)+DeltaUc_(ca)=0`
.
From first law of thermodynamics
`Q_(ab)+Q_(bc)+Q_(ca)=W_(ab)+W_(bc)+W_(ca)`
Substituting values from above, `700+0-100=700+400+DeltaW_(ca)`
`impliesDeltaW_(ca)=-500J`
This negative sign indicates that the work is done on the gas.
Now net work done on the gas
`DeltaW_(net)=W_(ab)+W_(bc)+W_(ca)=[700+400+500]J=600J`
total heat supplied to the gas during the cycle
`phi_(sup)=700J`
`eta=(DeltaW_("net"))/(Q_("sup"))xx100=(600)/(700)xx100=85.71%`.