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.
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The core idea
On this page
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
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.
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.
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
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
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 pCalculate the pressure difference
Reason
The vertical level difference is 15 cm = 0.15 m.Working
Δ p = (1000)(10)(0.15) = 1.5 × 10³ PaFind 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
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
Hints
Hint 1: read the liquid levels
Hint 2: subtract the liquid-column difference
View solution step by step
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 pCalculate the pressure difference
Reason
The vertical difference is 12 cm = 0.12 m.Working
Δ p = (1000)(10)(0.12) = 1.2 × 10³ PaFind 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)
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
View solution step by step
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.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
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
Identify the higher-pressure gas
1 markMethod
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_BConvert the vertical height difference
1 markReason
The liquid-column relationship requires metres.Working
Δ h = 5.0 cm = 0.050 mCalculate the signed pressure difference
1 markReason
Since A is higher pressure, p_A-p_B is positive.Working
p_A-p_B = (800)(10)(0.050) = 4.0 × 10² Pa
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark the pressure direction, height conversion and difference separately.
Challenge 5
Mercury manometer (gas pressure)
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.
- Find the gas pressure.
- 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
Hints
Hint 1: calculate the mercury pressure difference
Hint 2: hold pressure difference constant
View solution step by step
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 PaCalculate the gas pressure
Reason
The gas pressure exceeds atmosphere by Δ p.Working
p_gas = 1.00 × 10⁵ + 5.44 × 10³ = 1.05 × 10⁵ PaFind 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