A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oF. The air undergoes a process to a state where the pressure is 1.0 bar, during which the pressure–volume relationship is pV = constant. Assume ideal gas behavior for the air. Determine the work and heat transfer, in kJ.

The work and heat transfer for this process is = 270.588 kJ

Explanation:

Take properties of air from an ideal gas table.  R = 0.287 kJ/kg-k

The Pressure-Volume relation is PV = C

T = C for isothermal process

Calculating for the work done in isothermal process

W = P₁V₁

   = mRT₁      [∵pV = mRT]

   = (5) (0.287) (272.039)

   = 270.588 kJ

Since the process is isothermal, Internal energy change is zero

ΔU =

From 1st law of thermodynamics

Q = ΔU  + W

   = 0 + 270.588

   = 270.588 kJ

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