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Nitrogen piston with stops: work and heat

Short answer

Heating at constant volume to 400 kPa, then constant-pressure expansion to 0.15 m³. Work W = 39.6 kJ, heat transfer Q = 164.2 kJ.

Nitrogen (0.10 kg) in the piston-cylinder device shown in Figure P.3 has an initial pressure of 0.2 MPa and temperature of 70°C. Energy is added until the piston just reaches the upper stops where the total volume is 0.015 m³. It takes a pressure of 0.4 MPa to move the piston. Calculate the total work done and heat transfer. [R = 297 J/kg.K, Cv = 743 J/kg.K] [Figure P.3: cylinder containing N2 with a piston and upper stops.]

Given: m = 0.10 kg N₂, R = 0.297, c_v = 0.743 kJ/kg·K; P₁ = 0.2 MPa = 200 kPa, T₁ = 70 °C = 343.15 K; upper stops at V_total = 0.015 m³ (as printed); 0.4 MPa (400 kPa) is needed to move the piston.

Check of data: V₁ = mRT₁/P₁ = 0.1 × 0.297 × 343.15/200 = 0.0510 m³. This is larger than the printed 0.015 m³, so the data are inconsistent: the stop volume is evidently 0.15 m³ (or the mass is 0.010 kg). The solution below uses m = 0.10 kg and V_stops = 0.15 m³.

Process:

  1. 1 → 2: the piston rests on the lower stops, so heating is at constant volume V = 0.0510 m³ until P = 400 kPa:
    T₂ = T₁ × 400/200 = 686.3 K
  2. 2 → 3: the piston moves at constant P = 400 kPa until it reaches the upper stops, V₃ = 0.15 m³:
    T₃ = P₃V₃/(mR) = 400 × 0.15/(0.1 × 0.297) = 2020 K
 P (kPa)
 400 |  2------------3
     |  |
     |  |  1-2: constant V
 200 |  1
     +------------------- v
      0.051         0.15

Work (only in 2 → 3):
W = P (V₃ − V₂) = 400 × (0.15 − 0.0510) = 39.6 kJ

Heat:
ΔU = m c_v (T₃ − T₁) = 0.1 × 0.743 × (2020 − 343.15) = 124.6 kJ
Q = ΔU + W = 124.6 + 39.6 = 164.2 kJ

Answer: W ≈ 39.6 kJ; Q ≈ 164 kJ. (If instead m = 0.010 kg with V = 0.015 m³ is meant, every quantity is one tenth: W = 3.96 kJ, Q = 16.4 kJ.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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