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

Answer:

W=-940.36 KJ

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

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

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

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

$W=140\times 4.25 \ln \dfrac{1.4}{6.8}$

W=-940.36 KJ

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