---
title: "Piston cylinder: 2 kg water 500 kPa, 20% quality"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 3
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:steam-piston-500kpa-quality-20"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/steam-piston-500kpa-quality-20
site: BCT Engineering
---

# Piston cylinder: 2 kg water 500 kPa, 20% quality

Fundamental of Thermodynamics & Heat Transfer · Chapter 3: Properties of Common Substances

- Asked in: 2073 Bhadra, 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/steam-piston-500kpa-quality-20
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-3

## Question

A piston cylinder device shown in figure below contains 2 kg of water initially at a pressure of 500 KPa with a quality of 20%. The water is heated until it becomes a saturated vapor. The volume of the system when the piston is at the upper stops is 0.4 m³. Sketch the process on P-v and T-v diagrams and determine:
a) the final pressure, and
b) the total work transfer
[Figure: cylinder with H2O under a piston with upper stops, heat Q added.]

## Short answer

The piston rises at constant pressure 500 kPa until volume reaches 0.4 m³, then heating continues at constant volume to saturated vapor. Final pressure is 970 kPa. Work done is 500 kPa × (0.4 − 0.1517) m³ = 124.2 kJ.

## Answer

**Given:** m = 2 kg; P₁ = 500 kPa, x₁ = 0.2; heated until saturated vapour; V at the upper stops = 0.4 m³.

**Steam table values at 500 kPa:** T_sat = 151.8 °C, v_f = 0.001093, v_g = 0.3748 m³/kg.

**State 1:** v₁ = 0.001093 + 0.2 (0.3748 − 0.001093) = 0.07583 m³/kg  
V₁ = m v₁ = 2 × 0.07583 = 0.1517 m³

**Process path:**
1. 1 → 2: the piston is free and floats, so heating is at constant P = 500 kPa until the piston reaches the stops at V₂ = 0.4 m³ (v₂ = 0.2 m³/kg). Check: v_g (500 kPa) = 0.3748 > 0.2, so at this point the steam is still wet (x₂ = 0.53).
2. 2 → 3: with the piston on the stops, heating continues at constant volume, v₃ = 0.2 m³/kg, until the steam is saturated vapour (v_g = 0.2).

```text
 P|
  |            3   (on sat. vapour line, ≈ 970 kPa)
  |            |
  |            |   2-3: const v = 0.2
  | 1----------2   1-2: const P = 500 kPa
  +---------------- v
  0.0758      0.2

 T|
  |            3   (≈ 178.6 C, sat. vapour)
  |            |
  | 1----------2   (151.8 C, wet region)
  +---------------- v
```

#### a) Final pressure
At state 3, v_g = 0.2 m³/kg. From the saturated tables: v_g = 0.2149 at 900 kPa and 0.1943 at 1000 kPa. Interpolating:  
P₃ = 900 + 100 × (0.2149 − 0.2)/(0.2149 − 0.1943) ≈ **970 kPa** (T₃ ≈ 178.6 °C)

#### b) Total work
Work is done only in 1 → 2 (constant pressure):  
W = P₁ (V₂ − V₁) = 500 × (0.4 − 0.1517) = **124.2 kJ**

**Answer: P₃ ≈ 970 kPa; W ≈ 124 kJ.**

## Check yourself

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

1. How is v1 found?
   A. v1 = vg at 500 kPa
   B. v1 = vf + x1 vfg at 500 kPa = 0.07583 m3/kg
   C. v1 = vf only
   D. v1 = x1 vg only
   Answer: B. For x = 0.2, v1 = 0.001093 + 0.2 x (0.3748 - 0.001093). (Key point: Initial specific volume)

2. What happens from 1 to 2?
   A. Constant volume heating
   B. Isothermal compression
   C. Constant pressure at 500 kPa until the piston reaches the stops at 0.4 m3
   D. Pressure rises while the piston is fixed
   Answer: C. The free piston floats at constant pressure until it hits the upper stops. (Key point: First stage)

3. What fixes the final pressure at state 3?
   A. P3 = 500 kPa
   B. P3 is the pressure where vf = 0.2
   C. P3 follows from T3 = 151.8 C
   D. v3 = 0.2 m3/kg equals vg, so P is found from the saturated table (about 970 kPa)
   Answer: D. With the piston on the stops, v is constant at 0.2 and the state is saturated vapour. (Key point: Final state)

4. How is the total work found?
   A. W = P1 (V2 - V1) = 124.2 kJ, only in the constant-pressure stage
   B. W = P3 (V3 - V1)
   C. W is zero
   D. W = integral of V dP over both stages
   Answer: A. Constant-volume heating does no work. (Key point: Work)

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