Question

A gas contained within a piston-cylinder undergoes the follow change in states: Process 1: Constant volume from p1 = 1 bar V1 = 2.6 m3 to state 2 with p2 = 2.7 bar Process 2: Compression to V3 = 1.5 m3, which the pressure-volume relationship is pV = constant. Process 3: Constant pressure to state 4, where V4 = 0.5 m3. Sketch the processes on p-V graph and evaluate the work for each process in kJ.

Answer 1

Process 1:W=0

Process 2:W= -386.13 KJ

Process 3:W= -468 KJ

Step-by-step explanation:

Process 1:
`P_1=1 bar,V_1=2.6m^3`

Process 2:
`P_2=2.7bar,V_2=2.6m^3`

Process 3:
`V_3=1.5 m^3`

`V_4=0.5 m^3`

Process 1:

Work (W)=0 ,because it is constant volume process.

Process 2:

It is constant temperature process so PV=C

`P_2V_2=P_3V_3`

`P_3=(P_2V_2)/(V_3)`

`P_3=(2.7* 2.6)/(1.5)`

`P_3=4.68`bar

So work in constant temperature process

W=
`P_2V_2\ ln(V_3)/(V_2)`

W=
`270* 2.6\ ln(1.5)/(2.6)` (1 bar=100KPa)

W= -386.13 KJ

Negative sign means it is compression process.

Process 3:

It is a constant pressure.

So work W=
`P_3(V_4-V_3)`

W=468(0.5-1.5) KJ

W= -468 KJ

Negative sign means it is compression process.