U-tube manometer

Key idea: Learn how a U-tube manometer measures pressure difference, how liquid levels determine the sign, and how to calculate gauge and absolute pressure.

  • SEC G3 Physics 2027
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Learning objectives

  • Define pressure as force per unit area
  • Apply pressure = force ÷ area
  • Explain pressure transmission in a hydraulic press
  • Apply density = mass ÷ volume
  • Apply liquid-column pressure = height × density × gravitational field strength
  • Explain how liquid-column height measures atmospheric pressure
  • Explain how a manometer measures pressure difference

1. Definition

A. Manometer

A manometer is an instrument used to measure a pressure difference by using the difference in height between two columns of liquid in a U-tube.

2. Key Ideas

  • In a simple U-tube manometer (same liquid in both arms):
    • the higher-pressure side pushes the liquid down on its side
    • the pressure difference is: Δ p = ρ g Δ h
  • Δ h is the vertical height difference between the two liquid levels.
  • If one side is open to the atmosphere:
    • gas-side level lower → p_gas = pₐₜₘ + ρ gΔ h
    • gas-side level higher → p_gas = pₐₜₘ - ρ gΔ h
  • Gauge pressure is p_gauge = p_gas-pₐₜₘ; it is negative when the gas pressure is below atmospheric pressure.
  • Useful densities:
    • water: ρ ≈ 1000 kg m⁻³
    • mercury: ρ ≈ 13.6 × 10³ kg m⁻³
  • Convert heights to metres (e.g. 15 cm = 0.15 m).

3. Detailed Explanations

A. How a U-tube manometer works

Reading a U-tube manometer without guessing the signTwo U-tube manometers compare a gas with the atmosphere. In the first, the gas-side liquid is lower, so gas pressure exceeds atmospheric pressure. In the second, the gas-side liquid is higher, so gas pressure is below atmospheric pressure.Gas pressure above atmosphericGas pressure below atmosphericgasopen to airΔhpgas = patm + ρgΔhgasopen to airΔhpgas = patm − ρgΔh
Scroll diagram horizontally to read all labels.
The side with the lower liquid level has the higher pressure. The magnitude of the pressure difference is ρgΔh; the level pattern determines which pressure is larger.

If both sides are at the same pressure (e.g. both open to air), the liquid levels are equal.

If one side has a higher pressure, it pushes the liquid down on that side and the levels become different.

Using hydrostatic pressure (p = ρ gh), the pressure difference between the two sides is:

Δ p = ρ g Δ h

where ρ is the density of the manometer liquid and Δ h is the vertical height difference.

Recall: hydrostatic pressure

If you need a refresher on p = ρ gh, see: Hydrostatic Pressure.

B. Quick sign check (which side has higher pressure?)

For a U-tube manometer:

  • the side with the lower liquid level has the higher pressure
  • always write the relationship as: p_high = p_low + ρ g Δ h

C. Why mercury is sometimes used

For the same pressure difference Δ p:

Δ h = (Δ p)/(ρ g)

A denser liquid (larger ρ) gives a smaller height difference, so it is useful for large pressure differences.

Safety: mercury

Mercury is toxic. Do not handle mercury outside a properly equipped laboratory.

4. Common Mistakes

  • Using the wrong Δ h (it is the vertical height difference between the two liquid levels).
  • Measuring from the bottom of the U-tube instead of measuring the difference between the two levels.
  • Forgetting to convert cm to m.
  • Using the wrong density (e.g. using water’s density when the liquid is mercury).
  • Adding pₐₜₘ when the question only asks for pressure difference.

5. Exam Tips

  • Decide what the question wants:
    • pressure difference: use Δ p = ρ gΔ h
    • gas pressure (one side open to air): use p_gas = pₐₜₘ ± ρ gΔ h
  • State which side has higher pressure by looking at which side’s liquid level is lower.
  • Keep units consistent: Pa, kg m⁻³, N kg⁻¹, m.

6. Worked Examples

Modelled example 1

Gas pressure greater than atmospheric pressure

Core

Problem

A U-tube manometer contains water (ρ = 1000 kg m⁻³). One side is open to the atmosphere (pₐₜₘ = 1.01 × 10⁵ Pa). The water level on the gas side is 15 cm lower than the open side. Find the gas pressure. Take g = 10 N kg⁻¹.

Study the worked solution
  1. Decide which side has higher pressure

    Method

    Identify the gas pressure as higher than atmospheric pressure.

    Reason

    The higher-pressure gas pushes the water lower on its side of the U-tube.

    Working

    p_gas = pₐₜₘ + Δ p
  2. Calculate the pressure difference

    Reason

    The vertical level difference is 15 cm = 0.15 m.

    Working

    Δ p = (1000)(10)(0.15) = 1.5 × 10³ Pa
  3. Find the gas pressure

    Reason

    The gas exceeds atmospheric pressure by the calculated water-column pressure.

    Working

    p_gas = 1.01 × 10⁵ + 1.5 × 10³ = 1.03 × 10⁵ Pa

Guided practice 2

Gas pressure less than atmospheric pressure

About 5 min

Problem

A water manometer is open to air on one side. The water level on the gas side is 12 cm higher than the open side. Find the gas pressure if pₐₜₘ = 1.00 × 10⁵ Pa, ρ = 1000 kg m⁻³ and g = 10 N kg⁻¹.

Choose the sign before calculating

How does the gas pressure compare with atmospheric pressure?
Unit: Pa

Hints

Hint 1: read the liquid levels
The gas-side liquid is higher, so the gas pressure is the lower pressure.
Hint 2: subtract the liquid-column difference
Use p_gas = pₐₜₘ-ρ gΔ h with Δ h = 0.12 m.
View solution step by step
  1. Choose the pressure relationship

    Method

    Subtract the manometer pressure difference from atmospheric pressure.

    Reason

    The gas-side liquid level is higher, so the gas pressure is lower than the open-side pressure.

    Working

    p_gas = pₐₜₘ-Δ p
  2. Calculate the pressure difference

    Reason

    The vertical difference is 12 cm = 0.12 m.

    Working

    Δ p = (1000)(10)(0.12) = 1.2 × 10³ Pa
  3. Find the gas pressure

    Reason

    The gas is below atmosphere by the calculated pressure difference.

    Working

    p_gas = 1.00 × 10⁵-1.2 × 10³ = 9.88 × 10⁴ Pa

Common misconception 3

Reading a diagram (cm of water)

Find and correct the mistake

Learner response

A water manometer is connected to a gas supply. The gas-side liquid level is lower, and the vertical difference between the levels is 8 cm. A student chooses 8 cm of water less than atmospheric pressure because the gas-side level is lower.

Locate the first error and select the correct description.

Correct the diagram reading

Which description is correct?

View solution step by step
  1. Locate the direction error

    Method

    Associate the lower liquid level with the higher pressure.

    Reason

    The higher-pressure gas pushes the water down on its side and up on the open side.

    Working

    The gas pressure is greater than atmospheric pressure.
  2. Use the measured vertical difference

    Reason

    The stated separation is 8 cm, so the pressure difference is 8 cm of water.

    Working

    The gas pressure is 8 cm of water more than atmospheric pressure.

Examiner practice 4

Measure pressure difference between two gases

3 marks

Examination question

A U-tube manometer containing oil (ρ = 800 kg m⁻³) connects gas containers A and B. The liquid level on the A side is 5.0 cm lower than the B side. Calculate p_A-p_B and identify which gas has the higher pressure. Take g = 10 N kg⁻¹. [3 marks]

Show the direction, conversion and pressure difference

View solution step by step
  1. Identify the higher-pressure gas

    1 mark

    Method

    Select gas A as the higher-pressure side.

    Reason

    The liquid level is lower on the side exerting the greater pressure.

    Working

    p_A > p_B
  2. Convert the vertical height difference

    1 mark

    Reason

    The liquid-column relationship requires metres.

    Working

    Δ h = 5.0 cm = 0.050 m
  3. Calculate the signed pressure difference

    1 mark

    Reason

    Since A is higher pressure, p_A-p_B is positive.

    Working

    p_A-p_B = (800)(10)(0.050) = 4.0 × 10² Pa

Challenge 5

Mercury manometer (gas pressure)

Minimal support

Instrument-liquid transfer

A U-tube manometer contains mercury (ρ = 13.6 × 10³ kg m⁻³). One side is open to atmosphere at 1.00 × 10⁵ Pa. The mercury level on the gas side is 4.0 cm lower than the open side. Take g = 10 N kg⁻¹.

  1. Find the gas pressure.
  2. Find the height difference a water manometer (ρ = 1000 kg m⁻³) would show for the same pressure difference.

Calculate and compare the two manometer liquids

Unit: Pa
Unit: m

Hints

Hint 1: calculate the mercury pressure difference
Convert 4.0 cm to metres and use the mercury density in Δ p = ρ gΔ h.
Hint 2: hold pressure difference constant
For water, rearrange to Δ h = Δ p/(ρ g) using the same Δ p.
View solution step by step
  1. Find the mercury-column pressure difference

    Method

    Use the vertical mercury separation of 0.040 m.

    Reason

    The gas-side level is lower, so this difference is added to atmospheric pressure.

    Working

    Δ p = (13.6 × 10³)(10)(0.040) = 5.44 × 10³ Pa
  2. Calculate the gas pressure

    Reason

    The gas pressure exceeds atmosphere by Δ p.

    Working

    p_gas = 1.00 × 10⁵ + 5.44 × 10³ = 1.05 × 10⁵ Pa
  3. Find the equivalent water height

    Reason

    Water is less dense, so it needs a greater height difference for the same pressure difference.

    Working

    Δ h_water = (5.44 × 10³)/(1000)(10) = 0.544 m

7. Mind Stretchers

Mind stretcher 1: Changing the manometer liquidExtension

For the same pressure difference, a water manometer gives a height difference Δ h. If the manometer liquid is replaced with a liquid of twice the density, what happens to Δ h? Explain.

Show Answer

Δ h halves.

For a fixed Δ p: Δ p = ρ g Δ h ⇒ Δ h = (Δ p)/(ρ g)

If ρ doubles, Δ h becomes half.

Mind stretcher 2: Why must Δh be vertical?Extension

In a manometer, why do we use the vertical height difference (not the length along a tilted tube)?

Show Answer

Hydrostatic pressure depends on vertical depth: p = ρ gh

The pressure difference comes from the difference in vertical height, so only the vertical component matters.

8. Practice and next step

Complete the Pressure structured questions, then return to the Pressure hub for the full topic checklist or continue to Thermal Physics.

Continue with the next resource in this course.

Course and syllabus information
Course
SEC G3 Physics
Edition
SEC G3 Physics 2027