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Argon in piston cylinder with stops: work

Short answer

Process: constant pressure to stops then constant volume. V₁ = 0.00164 m³, V₂ = V₃ = 0.00279 m³. Work only in 1–2: W = 6000 kPa × (0.00279 − 0.00164) m³ = 6.9 kJ.

Argon (100g) is in the piston-cylinder device shown in the figure below. The initial pressure is 6.0 MPa and temperature is 200°C. There is a heat transfer to the argon, causing the piston to rise until it hits the stops. There is an additional heat transfer until the final pressure is 8.0 MPa and temperature is 800°C. a) Draw the process on P-V and T-V diagrams b) Find the total work done in the process [Take R = 208 J/kgK]. [Figure: cylinder containing argon with a piston and upper stops.]

Given: m = 0.1 kg argon, R = 0.208 kJ/kg·K; state 1: 6 MPa, 200 °C = 473.15 K; state 3: 8 MPa, 800 °C = 1073.15 K. The piston rises at constant pressure until it hits the stops, and then the pressure rises at constant volume.

Process: 1 → 2 at constant pressure (6 MPa) up to the stops; 2 → 3 at constant volume up to 8 MPa.

a) Diagrams

 P|          3
  |          |
  | 1--------2        P-V
  +---------------- V

 T|          3
  |          |
  |          2        T-V
  |        /
  |      /
  |    1
  +---------------- V

On the P-V diagram, 1-2 is a horizontal line (constant P) and 2-3 is a vertical line (constant V). On the T-V diagram, in 1-2 T rises in proportion to V (a straight line that would pass through the origin, since V ∝ T at constant P); in 2-3 the volume is constant (vertical line) while T rises.

b) Total work

Ideal gas:
V₁ = mRT₁/P₁ = 0.1 × 0.208 × 473.15 / 6000 = 0.001640 m³
V₃ = mRT₃/P₃ = 0.1 × 0.208 × 1073.15 / 8000 = 0.002790 m³
V₂ = V₃ (piston at the stops, then constant volume) = 0.002790 m³
(T₂ = P₂V₂/mR = 6000 × 0.002790/0.0208 = 805 K = 532 °C)

Work is done only in 1 → 2 (constant P):
W = P₁ (V₂ − V₁) = 6000 × (0.002790 − 0.001640) = 6.90 kJ

Answer: Total work done by the argon ≈ 6.9 kJ.

✓ Answer checked by Claude Opus 5.5 (high effort)

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