Question

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

Answer 1

(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.