A manometer is a U-tube containing liquid that compares two pressures. The pressure difference equals ρ × g × h, where h is the vertical difference between the two liquid surfaces. The side with the lower liquid surface is connected to the higher pressure.
Whether manometer diagrams are required depends on your syllabus year, so check the current Cambridge page. The reasoning here is liquid pressure from the earlier lessons applied to a picture.
How does a manometer work?
Imagine a U-tube with one arm open to the air and the other joined to a gas supply. The gas pushes down on the liquid in its arm. The air pushes down on the liquid in the open arm.
If the two pressures were equal, both surfaces would be level. If the gas pressure is higher, the liquid is pushed down on the gas side and rises on the open side. The difference in heights is h, and gas pressure = atmospheric pressure + ρgh.
If the liquid is lower on the open side, the gas pressure is lower than atmospheric pressure, and gas pressure = atmospheric pressure − ρgh.
How to read any manometer diagram
- Find the two liquid surfaces and draw a horizontal line through each.
- Measure the vertical distance h between the lines, in metres.
- Decide which side is pushed down more. That side has the higher pressure.
- Calculate ρgh with the liquid’s density and the stated g.
- Add or subtract from atmospheric pressure, depending on the side.
- Give the answer in pascals, and say whether it is the gauge pressure (ρgh only) or the total.
Worked example
A water manometer is joined to a gas supply.
The water surface in the open arm is 0.25 m higher than in the gas arm.
Atmospheric pressure is 1.0 × 10⁵ Pa. Find the gas pressure. Use ρ = 1000 kg/m³ and g = 10 N/kg. (Invented example data.)
Step 1, which side is pushed down: the gas arm has the lower surface, so the gas pressure is higher than atmospheric pressure.
Step 2, pressure difference: ρgh = 1000 × 10 × 0.25 = 2500 Pa.
Step 3, gas pressure: 100 000 + 2500 = 102 500 Pa, or 1.025 × 10⁵ Pa.
Check: a 0.25 m column of water causing only 2500 Pa is small compared with atmospheric pressure, so the gas is only slightly above the atmosphere. That is sensible.
The mistake to watch for
A common slip is to add when the diagram says subtract.
Mistaken answer: in a mercury manometer the open arm is 40 mm lower than the gas arm. The student writes gas pressure = 100 000 + 5440 = 105 440 Pa.
If the open arm is lower, the air side is pushed down more, so the gas pressure is below atmospheric pressure.
The correction: ρgh = 13 600 × 10 × 0.040 = 5440 Pa, and gas pressure = 100 000 − 5440 = 94 560 Pa. Before calculating, ask which side has the lower liquid surface. That side has the higher pressure.
Check yourself
Use g = 10 N/kg and atmospheric pressure 1.0 × 10⁵ Pa.
1. The liquid in a manometer is oil of density 900 kg/m³. The levels differ by 0.10 m. Find the pressure difference.
Show answer
p = ρgh = 900 × 10 × 0.10 = 900 Pa
2. The two liquid levels in a manometer are equal. What is the gas pressure?
Show answer
There is no height difference, so the gas pressure equals atmospheric pressure, 1.0 × 10⁵ Pa.
3. A mercury manometer (13 600 kg/m³) has its open arm 0.050 m higher than the gas arm. Find the gas pressure.
Show answer
ρgh = 13 600 × 10 × 0.050 = 6800 Pa. The gas arm is lower, so the gas pressure is higher: 100 000 + 6800 = 106 800 Pa.
Where this leads next
If the units in a manometer question go wrong, revisit pressure from force and area. Then test everything in the pressure practice set.
Diagrams are where one-to-one teaching helps most, because a teacher can draw and change the picture with you. That is part of online one-to-one Physics tuition.