---
title: "Nitrogen in piston cylinder: work and heat"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 4
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:nitrogen-piston-work-heat"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/nitrogen-piston-work-heat
site: BCT Engineering
---

# Nitrogen in piston cylinder: work and heat

Fundamental of Thermodynamics & Heat Transfer · Chapter 4: First Law of Thermodynamics

- Asked in: 2075 Baisakh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/nitrogen-piston-work-heat
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-4

## Question

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.]

## 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.

## Answer

**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. 1 → 2: the piston rests on the stops (800 < 1500 kPa), so heating is at constant volume until P = 1500 kPa.
2. 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

```text
 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).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why does the first process occur at constant volume?
   A. The piston rests on stops until pressure reaches the 1500 kPa needed to lift it
   B. The gas is incompressible
   C. Temperature is held constant
   D. The heat is zero
   Answer: A. 800 < 1500 kPa so the piston cannot move. (Key point: Process)

2. T2 is found from:
   A. T2 = T1 V2/V1
   B. T2 = T1 P2/P1 = 750 K (constant volume)
   C. T2 = T3
   D. T2 = 527 °C
   Answer: B. Since 750 < 800 K, the piston lifts and heating continues at 1500 kPa. (Key point: State 2)

3. Total work is:
   A. P1(V3 - V1) = 39.6 kJ
   B. Zero
   C. P(V3 - V2) = 74.25 kJ, only in the constant-pressure part
   D. 1500 × 0.7425 = 1114 kJ
   Answer: C. Constant volume 1-2 does no work. (Key point: Work)

4. Heat transfer is:
   A. Q = W = 74.25 kJ
   B. Q = m c_v ΔT = 1486 kJ only
   C. Q = 1486 - 74.25 kJ
   D. Q = m c_v (T3 - T1) + W = 1560 kJ
   Answer: D. First law: Q = ΔU + W. (Key point: Heat)

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