A thermodynamic system is taken through the cycle abcda. (a) calculate the work done by the gas during the parts ab, bc, cd and da. (b) find the total heat rejected by the gas during the process.

Explanation

(a) The work done during the part ab,
`=int_(a)^(b)pdV=(100 kPa)int_(a)^(b) dV`
`=(100 kPa)(300 cm^(3)-100 cm^(3))
`=20 J`
The work done during bc is zero as the volume does not change. The work done during cd
`=int_(c)^(d) pdV=(200 kPa)(100 cm^(3)-300 cm^(3)) ==-40 J)`.
The work done during da is zero as the volume does not change.
(b) The total work done by the system during the cycle abcda
`DeltaW=20 J-40 J=-20 J`.
The change in internal energy `DeltaU=0` as the initial state is the same as the final state. thus `DeltaQ=DeltaU+DeltaW=20 J`. so the system reject 20 J of heat during the cycle.