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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

Respuesta :

Answer:

W=-940.36 KJ

Explanation:

Given that

[tex]P_1=1\ bar,V_1=4.25 {m^3}[/tex]

[tex]P_2=6.8\ bar[/tex]

Process follows pv=constant

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

[tex]W=P_1V_1\ln \dfrac{P_1}{P_2}[/tex]

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

[tex]W=P_1V_1\ln \dfrac{P_1}{P_2}[/tex]

[tex]W=140\times 4.25 \ln \dfrac{1.4}{6.8}[/tex]

W=-940.36 KJ

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

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