---
title: "Absolute and gauge pressure, mercury manometer"
subject: "Fluid Mechanics"
chapter: 2
key: "semester_3/fluid_mechanics:absolute-gauge-pressure-vapour-mercury"
source: https://www.bctengineer.com/q/semester_3/fluid_mechanics/absolute-gauge-pressure-vapour-mercury
site: BCT Engineering
---

# Absolute and gauge pressure, mercury manometer

Fluid Mechanics · Chapter 2: Pressure and Head

- Asked in: 2073 Shrawan, 2+5 marks
- Video: [Absolute Pressure vs Gauge Pressure - Fluid Mechanics - Physics Problems](https://www.youtube.com/watch?v=XCiBaFp8L_M) — The Organic Chemistry Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/fluid_mechanics/absolute-gauge-pressure-vapour-mercury
- Whole chapter: https://www.bctengineer.com/md/semester_3/fluid_mechanics/chapter-2

## Question

Define absolute and gauge pressure. Determine (i) the gauge pressure reading on the pressure gauge and (ii) the height h, of the mercury manometer. Take liquid density = 800 kg/m³, vapour pressure = 120 kPa (abs) and atmospheric pressure = 101 kPa (abs). [Figure: closed vessel with vapour over liquid; pressure gauge G at 1 m below the liquid surface level; the vessel bottom is connected to a mercury U-tube manometer open to atmosphere with reading h]

## Short answer

Absolute pressure is relative to vacuum, gauge to atmosphere. Gauge pressure at G is 26.85 kPa, and the mercury manometer height h is 201 mm.

## Answer

#### Definitions
- **Absolute pressure** is measured from absolute zero (complete vacuum): $p_{abs}=p_{gauge}+p_{atm}$.
- **Gauge pressure** is measured relative to the local atmospheric pressure: $p_{gauge}=p_{abs}-p_{atm}$. It is negative for vacuum.

#### Numerical
**Reading of the figure:** gauge $G$ and the liquid-mercury interface of the U-tube are at the same level, 1 m below the liquid surface; the mercury stands $h$ above that interface in the open limb. $\rho_l=800\ \text{kg/m}^3$, $p_v=120$ kPa (abs), $p_{atm}=101$ kPa.

**(i) Gauge reading.** Absolute pressure at G:

$$
p_G=p_v+\rho_l g(1)=120+\frac{800\times9.81\times1}{1000}=127.85\ \text{kPa abs}
$$

$$
p_{G,gauge}=127.85-101=26.85\ \text{kPa}
$$

**(ii) Mercury height $h$.** The same absolute pressure is balanced by atmosphere plus the mercury column:

$$
p_G=p_{atm}+\gamma_{Hg}h\ \Rightarrow\ h=\frac{26.85}{13.6\times9.81}=0.2012\ \text{m}
$$

**Answer: (i) gauge pressure $=26.85$ kPa; (ii) $h\approx201$ mm of mercury.**

## Check yourself

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

1. How are gauge and absolute pressure related?
   A. p_gauge = p_abs + p_atm
   B. p_gauge = p_abs - p_atm; absolute is measured from a perfect vacuum
   C. p_abs is measured from atmospheric pressure
   D. p_gauge is measured from absolute zero
   Answer: B. Gauge pressure is relative to local atmosphere, absolute pressure to zero pressure, and gauge is negative for vacuum. (Key point: Absolute and gauge pressure)

2. G is 1 m below the liquid surface under vapour at 120 kPa abs. What is the absolute pressure at G?
   A. 120 kPa
   B. 112.15 kPa
   C. 127.85 kPa
   D. 7.85 kPa
   Answer: C. Add the liquid head to the vapour pressure: 120 + 800 x 9.81 x 1 / 1000 = 127.85 kPa abs. (Key point: Pressure at gauge G)

3. What does the gauge read?
   A. 26.85 kPa
   B. 127.85 kPa
   C. 6.85 kPa
   D. -26.85 kPa
   Answer: A. Subtract atmospheric pressure from the absolute pressure: 127.85 - 101 = 26.85 kPa. (Key point: Gauge reading)

4. Which expression gives the mercury column height h?
   A. h = 127.85 / (13.6 x 9.81)
   B. h = 26.85 / 9.81
   C. h = 101 / (13.6 x 9.81)
   D. h = (gauge pressure at G) / (γHg) = 26.85 / (13.6 x 9.81)
   Answer: D. The open limb is at atmosphere, so the mercury column balances only the gauge pressure: h = 0.201 m. (Key point: Mercury height)

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