A mass of 24 kg of air at 150 kPa and 12 degree C is contain

A mass of 2.4 kg of air at 150 kPa and 12 degree 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.

Solution

Given, mass of air = 2.4 kg

initial pressure P₁= 150 kPa

initial temperature T₁ =12 C = (12+273) K = 285 K

final pressure P₂ =600 kPa

The process given is isothermal since the temperature of system remains the same.

Work input in an isothermal process is given by

W_1-2 =P₁V₁ln(P₁/P₂) =mRT₁ln(P₁/P₂) [R=0.286 kJ/kg K]

=2.4*286*285*ln(150/600)

= -271192.448 Joules = -271.192 KJ

(-ve sign shows that work is done on the system)