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

Answer:

Process 1:W=0

Process 2:W= -386.13 KJ  

Process 3:W= -468 KJ

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

$P_2V_2=P_3V_3$

$P_3=\dfrac{P_2V_2}{V_3}$

$P_3=\dfrac{2.7\times 2.6}{1.5}$

$P_3=4.68$bar

So work in constant  temperature process

W=$P_2V_2\ ln\dfrac{V_3}{V_2}$

W=$270\times 2.6\ ln\dfrac{1.5}{2.6}$    (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.