A mass of 2.4 kg of air at 150 kpa and 12 c is contained in a gas-tight, frictionless pistoncylinder 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.
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Given A mass of 2.4 kg of air at 150 kpa and 12 degree c is contained in a gas-tight, friction less 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.
- Assuming air is an ideal gas at constant temperature, the ideal gas equation will be
- P = mRT / V
- Now mRT is constant at constant temperature, the equation for work will be
- V2
- W = ʃ path P dv
- =v1 ʃv1 to v2 mRT / V dv
- = ʃv1 to v2 dV/V
- = mRT (log V ) v1 to v2
- = mRT (log V2 / V1)
- We need to find V2/V1 for constant temperature.
- So V2 / V1 = mRT / P2 / mRT / P1
- = P1 / P2
- Substituting P1/P2 in V2/V1 using gas constant for air = 0.2870,
- So m = 2.4 kg
- T = 12 degree Celsius
- = 285.15 K
- So W = mRT log(P1/P2)
- = 2.4kg x 0.2870 kJ / kg K . 285.15 K log (150 kpa / 600 kpa)
- = -272 kJ
The negative sign implies that it is a work input of 272 KJ.
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