In a simple U-tube manometer containing one liquid, the greater gas pressure pushes the liquid surface lower on its side. Use the vertical difference between the two liquid levels in Δp = ρgh. The level difference gives the size of the pressure difference, while the lower surface identifies the higher-pressure side.
The 2027 SEC G3 Physics K323 syllabus includes manometers and pressure due to a liquid column. This guide shows how to add or subtract the measured difference when the other arm is open to the atmosphere.
- Find the lower liquid surface and identify which pressure acts above it.
- Measure the vertical gap between the two surfaces.
- Convert the height to metres before calculating pascals.
- Add or subtract the difference depending on which side has greater pressure.
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Compare pressures at one horizontal level
For the simple setup here, the same stationary liquid fills both arms and effects from narrow tubing are neglected. Two points at the same height in the connected liquid have the same pressure.
Draw a horizontal line through the lower surface. On the other side that line is below the higher surface, so the pressure there is the gas pressure above the higher surface plus the pressure of the extra liquid column.
That extra column is why the gas at the lower surface has greater pressure. In ρgh, ρ is the density of the manometer liquid and g is the gravitational field strength.
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Worked example | The gas-side liquid is lower
One arm contains gas above the water. The other arm is open to the atmosphere. The gas-side water surface is 6.0 cm below the open-side surface. Use water density 1,000 kg/m³, g = 10 N/kg and atmospheric pressure 100,000 Pa.
| Gas-side level compared with open side | Gas pressure compared with atmosphere | Operation |
|---|---|---|
| Lower | Greater | Add ρgh |
| Higher | Smaller | Subtract ρgh |
| Equal | Equal | No correction |
The height difference is 0.060 m, so Δp = 1,000 × 10 × 0.060 = 600 Pa. The gas pressure is greater than atmospheric pressure by 600 Pa. Its absolute pressure is 100,600 Pa.
The water level tells you which side is higher in pressure. The calculation ρgh gives a positive magnitude. Assign the sign only when comparing the pressures.
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Practise when the gas side is higher
With the same water and atmospheric pressure, suppose the gas-side surface is now 4.0 cm above the open side. The pressure difference is 1,000 × 10 × 0.040 = 400 Pa.
The atmosphere has the higher pressure in this case. Subtract 400 Pa from 100,000 Pa to obtain an absolute gas pressure of 99,600 Pa.
If the diagram gives heights of 12 cm and 8 cm above the same reference, subtract them to get the 4 cm gap. The distance around the bend does not enter the calculation.
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Frequently asked questions
Does the wider arm automatically have lower pressure?
In this model, compare the liquid levels and the pressures acting at the surfaces. Arm width is absent from the pressure-difference formula.
Can I use the density of the gas?
Use the density of the manometer liquid in ρgh.
What is the pressure relative to the atmosphere in the second example?
It is −400 Pa because the gas pressure is 400 Pa below atmospheric pressure.
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