Question

Air is compressed slowly in a piston-cylinder assembly from an initial state where P1 = 1.4 bar, V1= 4.25 m^3, to a final state where P2= 6.8 bar. During the process, the relation between pressure and volume follows pv= constant. For the air as the closed system, determine the work, in kJ

Answer 1

W=-940.36 KJ

Step-by-step explanation:

Given that

`P_1=1\ bar,V_1=4.25 {m^3}`

`P_2=6.8\ bar`

Process follows pv=constant

So this is the isothermal process and work in isothermal process given as

`W=P_1V_1\ln (P_1)/(P_2)`

Now by putting the values (1.4 bar =140 KPa)

`W=P_1V_1\ln (P_1)/(P_2)`

`W=140* 4.25 \ln (1.4)/(6.8)`

W=-940.36 KJ

Negative sign indicates that this is a compression process and work will given to the system.