---
title: "Closed vessel at critical state cooled to 1 MPa"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 3
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:critical-state-vessel-cooled-1mpa"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/critical-state-vessel-cooled-1mpa
site: BCT Engineering
---

# Closed vessel at critical state cooled to 1 MPa

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

- Asked in: 2069 Bhadra, 6 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/critical-state-vessel-cooled-1mpa
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-3

## Question

A one liter closed vessel contains water at its critical conditions. This vessel is cooled until its pressure drops to 1 MPa. Calculate the mass of water in the vessel, the final dryness fraction and final temperature. Also show the process on P-v and T-v diagrams.

## Short answer

From critical point vc=0.003106 m³/kg, mass m=V/vc=0.322 kg. Final state at 1 MPa has same specific volume; wet at Tsat=179.9°C, dryness fraction x=0.0102.

## Answer

**Given:** V = 1 L = 0.001 m³ rigid vessel (constant volume), initially at the critical point of water, cooled to P₂ = 1 MPa.

**Critical state of water (steam tables):** P_c = 22.06 MPa, T_c = 373.9 °C, v_c = 0.003106 m³/kg.

#### Mass of water
m = V/v_c = 0.001/0.003106 = **0.322 kg**

#### Final state at 1 MPa
v₂ = v_c = 0.003106 m³/kg. At 1 MPa: T_sat = 179.9 °C, v_f = 0.001127, v_g = 0.19435 m³/kg. Since v_f < v₂ < v_g, it is a wet mixture.

**Final temperature** = T_sat(1 MPa) = **179.9 °C**

**Final dryness fraction**  
x₂ = (v₂ − v_f)/(v_g − v_f) = (0.003106 − 0.001127)/(0.19435 − 0.001127) = **0.0102**

(So the vessel holds about 0.0033 kg of vapour and 0.319 kg of liquid.)

```text
 P|
  |  CP  (22.06 MPa)
  |  |
  |  |   v = vc = const
  |  |
  |  2   (1 MPa)
  +----------- v

 T|
  |  CP  (373.9 C)
  |  |
  |  |   v = vc = const
  |  2   (179.9 C)
  +----------- v
```
On both diagrams the process is a vertical (constant-volume) line that starts at the critical point, goes down into the dome and ends at 2 (1 MPa).

**Answer: m = 0.322 kg; x₂ = 0.0102; T₂ = 179.9 °C.**

## Check yourself

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

1. m =
   A. V x vc
   B. V/vc = 0.322 kg
   C. 1 kg
   D. vc/V
   Answer: B. 0.001/0.003106. (Key point: Mass)

2. Cooling is:
   A. Constant pressure
   B. Isothermal
   C. Constant volume at vc
   D. Adiabatic
   Answer: C. Rigid vessel. (Key point: Process)

3. T2 =
   A. 374 C
   B. 100 C
   C. 0
   D. Tsat(1 MPa) = 179.9 C
   Answer: D. Wet mixture. (Key point: Final T)

4. x2 =
   A. (v - vf)/(vg - vf) = 0.0102
   B. 0.5
   C. 1
   D. 0.322
   Answer: A. Mostly liquid. (Key point: Dryness)

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