A mass of 2.4 kg of air at 150 kPa and 12ºC 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. Sketch the process on a P-v diagram

Pressure may be defined as the force that is significantly applied perpendicular to the surface of an object per unit area over which that force is distributed. The unit of pressure is Pascal (Pa).

According to the question,

Assuming air is an ideal gas at a constant temperature, the ideal gas equation will be P = mRT / V.

Now mRT is constant at a constant temperature, the equation for work will be 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°C

= 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.

Therefore, according to the scenario, the calculation of work input during this process is found to be 272 KJ.

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