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Air: PV^3 expansion, cooling, heating; total work

Short answer

Process 1-2 (polytropic, n=3): P₂=2500(0.25/0.65)³=142.24 kPa, W₁₂=(P₂V₂-P₁V₁)/(1-3)=266.27 kJ. Process 2-3 (constant P): W₂₃=142.24(0.25-0.65)= -56.90 kJ. Process 3-1 (constant V): W=0. Total net work = 209.4 kJ.

4 kg of air contained in a piston cylinder arrangement at an initial pressure of 2500 kPa and initial volume of 0.25 m³ is allowed to expand according to pressure volume relation of PV³ = constant until its volume equals to 0.65 m³. The air is then cooled at constant pressure until the piston comes to its initial position. Then heat is supplied to the air as the piston is locked with a screw until the pressure rises to its initial pressure. Determine the total work transfer.

Given: m = 4 kg air; state 1: P₁ = 2500 kPa, V₁ = 0.25 m³; process 1-2: PV³ = C to V₂ = 0.65 m³; 2-3: constant pressure back to V₃ = V₁ = 0.25 m³; 3-1: constant volume (piston locked) until P = P₁.

 P|1
  | \
  |  \   PV^3 = C
  |   \
  |3----\2
  |     (const P)
  +-------------- V
   0.25   0.65

Step 1: Pressure at state 2 P₂ = P₁ (V₁/V₂)³ = 2500 × (0.25/0.65)³ = 2500 × 0.05690 = 142.24 kPa

Step 2: Work 1-2 (polytropic, n = 3) W₁₂ = (P₂V₂ − P₁V₁)/(1 − n) = (142.24 × 0.65 − 2500 × 0.25)/(1 − 3)
= (92.46 − 625)/(−2) = 266.27 kJ

Step 3: Work 2-3 (constant pressure, P₃ = P₂) W₂₃ = P₂ (V₃ − V₂) = 142.24 × (0.25 − 0.65) = −56.90 kJ

Step 4: Work 3-1 (constant volume) W₃₁ = 0

Step 5: Total work W_net = 266.27 − 56.90 + 0 = 209.4 kJ

Answer: Total work transfer ≈ 209.4 kJ (net work done by the air). The mass (4 kg) is not needed for the work calculation.

✓ Answer checked by Claude Opus 5.5 (high effort)

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