A perfect gas undergoes the following three separate and distinct process to execute a cycle
i)constant volume process during which 80 kJ of heat is supplied to the gas.
ii) constant pressure process during which 85 kJ of heat is lost to the surroundings and 20 kJ of work is done on it.
iii) adiablatic process which restores the gas back to its initial stage.
Evaluate the work done during adiabatic process and the value of internal energy at all the state points if initially its value is 95 kJ.

Explanation

This example illustrates the application of non-flow energy equation and the `1^(st)` law applied to a cyclic process.
Process 1-2:
`Q_(1-2)=80 kJ, W_(1-2) = 0`
`U_(2)-U_(1) = Q_(1-2)-W_(1-2)=80-0=80 kJ`
`U_(2)=80+95=175 kJ`

Process 2-3 :
`Q_(2-3)=-85 kJ, W_(2-3)=-20 kJ`
`U_(3)-U_(2)=Q_(2-3)-W_(2-3)=-85-(-20)=-65 kJ`
`U_(3) = -65+175 = 110 kJ`
For the complete cycle : `oint delta Q=oint delta W` . That is `Q_(1-2)+Q_(2-3)+Q_(3-1)=W_(1-2)+W_(2-3)+W_(3-1)`
`80-85+0=0+(-20)+W_(3-1)`
:. Work done during adiabatic process, `W_(3-1)=15 kJ`