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2.5 kg of air at 150 kPa and 12°C is contained in a gas-tight, frictionless piston-cylinder device. The air is now compressed to a final pressure of 600 kPa. During the process, heat is transferred from the air such that the temperature inside the cylinder remains constant. Calculate the work input during this process.

Answer :

Answer:

Work input =283.47 KJ

Explanation:

Given that

[tex]P_1=150\ KPa[/tex]

[tex]P_2=600\ KPa[/tex]

T=12°C=285 K

m= 2.5 kg

Given that this is the constant temperature process.

e know that work for isothermal process  

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

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

So now putting the values

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

[tex]W=2.5\times 0.287\times 285\ln \dfrac{150}{600}[/tex]

W=-283.47 KJ

Negative sign indicates that work is done on the system.

So work input =283.47 KJ

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