---
title: "Three pressure gauges on two compartments"
subject: "Fundamental of Thermodynamics & Heat Transfer"
chapter: 1
key: "semester_2/fundamental_of_thermodynamics_and_heat_transfer:three-pressure-gauges-two-compartments"
source: https://www.bctengineer.com/q/semester_2/fundamental_of_thermodynamics_and_heat_transfer/three-pressure-gauges-two-compartments
site: BCT Engineering
---

# Three pressure gauges on two compartments

Fundamental of Thermodynamics & Heat Transfer · Chapter 1: Introduction

- Asked in: 2073 Magh, 6 marks
- Video: [THERMODYNAMICS || PRESSURE NUMERICAL (TU-2068, Magh) || CHAPTER 2](https://www.youtube.com/watch?v=hMhZPbhIX3k) — Institution of Engineering
- 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/three-pressure-gauges-two-compartments
- Whole chapter: https://www.bctengineer.com/md/semester_2/fundamental_of_thermodynamics_and_heat_transfer/chapter-1

## Question

Three pressure gauges are connected to a container consisting of two compartments as shown in figure below. If the local barometer reads 750 mm of Hg and pressure gauges A and B reads 300 kPa and 200 kPa respectively. Determine the absolute pressure in each compartment and reading of pressure gauge C. [Take ρHg = 13600 kg/m³ and g = 9.81 m/s²]
[Figure: a container with two compartments 1 and 2; gauge A is connected to compartment 1, gauge B is connected between compartments 1 and 2, gauge C is connected to compartment 2.]

## Short answer

Atmospheric pressure = 13600 × 9.81 × 0.75 = 100.06 kPa. Absolute pressure in compartment 1 = 100.06 + 300 = 400.06 kPa. Compartment 2 = 400.06 - 200 = 200.06 kPa absolute. Gauge C reads 200.06 - 100.06 = 100 kPa.

## Answer

**Given:** barometer = 750 mm Hg, P_A = 300 kPa (gauge, compartment 1), P_B = 200 kPa (difference between compartments 1 and 2), ρ_Hg = 13600 kg/m³.

**Assumption:** Gauge A reads compartment 1 above atmosphere, gauge B reads the pressure difference P₁ − P₂, and gauge C reads compartment 2 above atmosphere.

**Step 1: Atmospheric pressure**
P_atm = ρ g h = 13600 × 9.81 × 0.750 = 100 062 Pa = 100.06 kPa

**Step 2: Compartment 1**
P₁ = P_atm + P_A = 100.06 + 300 = **400.06 kPa (abs)**

**Step 3: Compartment 2**
P₂ = P₁ − P_B = 400.06 − 200 = **200.06 kPa (abs)**

**Step 4: Gauge C**
P_C = P₂ − P_atm = 200.06 − 100.06 = **100 kPa (gauge)**

**Answer: P₁ = 400.06 kPa, P₂ = 200.06 kPa (absolute); gauge C reads 100 kPa.**

## Check yourself

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

1. With a 750 mm Hg barometer, rho = 13600 and g = 9.81, P_atm is about
   A. 100.06 kPa
   B. 750 kPa
   C. 13.6 kPa
   D. 10.0 kPa
   Answer: A. P_atm = rho g h = 13600 x 9.81 x 0.75 = 100 062 Pa. (Key point: Atmospheric pressure from barometer)

2. How is the absolute pressure of compartment 1 found from gauge A (300 kPa)?
   A. P1 = P_A - P_atm
   B. P1 = P_A + P_atm
   C. P1 = P_A only
   D. P1 = P_atm - P_A
   Answer: B. Gauge pressure is relative to atmosphere, so P_abs = P_atm + P_gauge = 400.06 kPa. (Key point: Gauge to absolute)

3. Gauge B reads 200 kPa as the difference between compartments. How is P2 found?
   A. P2 = P1 + 200
   B. P2 = P_atm + 200
   C. P2 = P1 - 200 = 200.06 kPa
   D. P2 = 200 kPa exactly
   Answer: C. B reads P1 - P2, so P2 = 400.06 - 200 = 200.06 kPa abs. (Key point: Differential gauge)

4. What does gauge C read?
   A. 200.06 kPa
   B. 300 kPa
   C. 400.06 kPa
   D. 100 kPa
   Answer: D. P_C = P2 - P_atm = 200.06 - 100.06 = 100 kPa (gauge), equal to A - B. (Key point: Gauge C reading)

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