Why does the first process occur at constant volume?

The piston rests on stops until pressure reaches the 1500 kPa needed to lift it

T2 = T1 P2/P1 = 750 K (constant volume)

P(V3 - V2) = 74.25 kJ, only in the constant-pressure part

Q = m c_v (T3 - T1) + W = 1560 kJ

2075 Baisakh · 8 marks

Nitrogen in piston cylinder: work and heat

Short answer

Heating first at constant volume (0.7425 m³) until pressure reaches 1500 kPa, then at constant pressure to 800 K. Work only during the constant-pressure step: W = 1500(0.792 - 0.7425) = 74.25 kJ. Heat transfer Q = 1486 (ΔU) + 74.25 = 1560 kJ.

Nitrogen (5 kg) is contained in a piston cylinder device shown in figure below initially at a pressure of 800 kPa and a temperature of 127°C. There is a heat transfer to the system until the temperature reaches 527°C. It takes a pressure of 1500 kPa to lift the piston. Sketch the process on P-v and T-v diagrams and determine the total work and heat transfer in the process. [Take R = 297 J/kg.K and Cv = 743 J/kg.K] [Figure: cylinder containing nitrogen with a weighted piston on stops, heat Q added.]

Given: m = 5 kg N₂, R = 0.297 kJ/kg·K, c_v = 0.743 kJ/kg·K; T₁ = 127 °C = 400 K, P₁ = 800 kPa; T₃ = 527 °C = 800 K; a pressure of 1500 kPa is needed to lift the piston.

Process:

1 → 2: the piston rests on the stops (800 < 1500 kPa), so heating is at constant volume until P = 1500 kPa.
2 → 3: the piston lifts, and heating continues at constant P = 1500 kPa until T₃ = 800 K.

Step 1: Volume at 1 (ideal gas)
V₁ = mRT₁/P₁ = 5 × 0.297 × 400/800 = 0.7425 m³

Step 2: State 2
T₂ = T₁ P₂/P₁ = 400 × 1500/800 = 750 K (below 800 K, so the piston does lift)
V₂ = V₁ = 0.7425 m³

Step 3: State 3
V₃ = mRT₃/P₃ = 5 × 0.297 × 800/1500 = 0.792 m³

 P (kPa)
 1500 |  2--------3
      |  |
      |  |   1-2: constant V
  800 |  1
      +------------------- v
       0.7425    0.792

 T (K)
  800 |           3
  750 |  2 ___---
      |  |
      |  |   1-2: constant V
  400 |  1
      +------------------- v
       0.7425    0.792

On both diagrams 1-2 is a vertical line. On P-v, 2-3 is horizontal; on T-v, 2-3 is a rising straight line (T ∝ V at constant P).

Work: only in 2 → 3:
W = P (V₃ − V₂) = 1500 × (0.792 − 0.7425) = 74.25 kJ

Heat: ΔU = m c_v (T₃ − T₁) = 5 × 0.743 × (800 − 400) = 1486 kJ
Q = ΔU + W = 1486 + 74.25 = 1560.3 kJ

Answer: W = 74.25 kJ; Q = 1560 kJ (added).

✓ Answer checked by Claude Opus 5.5 (high effort)

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