K323 / 2027
Pressure overview

Lesson 5 of 6

Measuring atmospheric pressure

  • I can apply liquid-column pressure = height x density x gravitational field strength to new situations.Syllabus

    Syllabus K323, 5(e). Apply liquid-column pressure = height x density x gravitational field strength to new situations.

  • I can explain how a liquid-column height measures atmospheric pressure.Syllabus

    Syllabus K323, 5(f). Explain how a liquid-column height measures atmospheric pressure.

Make a guess. You are not marked.

A mercury barometer reads 760 mm. The tube is now tilted so that it leans at an angle, with its lower end still in the mercury reservoir. What happens to the vertical height of the mercury column?

  1. It rises, because the mercury has further to travel.
  2. It stays at 760 mm, measured straight up.
  3. It falls, because the tube now leans.
Show the answer

It stays at 760 mm, measured straight up.

The column balances atmospheric pressure, which has not changed. The mercury moves along the tube until its vertical height is 760 mm again.

A liquid-column barometer measures atmospheric pressure from the vertical height of a supported liquid column.

In a stationary liquid, a vertical height difference gives a pressure difference of ρgh. A pressure reference is needed to turn that difference into an atmospheric-pressure reading.

The supplied barometer diagram shows a closed-top tube connected at its lower end to a reservoir of mercury. The reservoir is open to the atmosphere. The space above the mercury column has such a small pressure that we treat it as a vacuum in this model.

Measure the column above the reservoir surface

Supplied mercury density: 13600 kg/m3; g = 10 N/kg. The pressure above the column is approximately zero.

A mercury barometer with a near-vacuum above the columnAn inverted tube is closed at its top and open beneath the surface of mercury in a reservoir. The mercury is continuous through the submerged open end. The reservoir is open to atmospheric pressure. A near-vacuum above the mercury column provides an approximately zero-pressure upper reference. The height h, 0.750 metres, is measured vertically from the reservoir surface to the top of the mercury column. It is neither the whole tube length nor the height above its lower opening. The liquid pressures at A inside the tube and B at the reservoir surface are equal. With the supplied density and gravitational field strength, atmospheric pressure is 102000 Pa.h =0.750 mNearvacuumClosed topOpen to airABA and B: atmospheric pressure

Atmospheric pressure = 13600 x 10 x 0.750 = 102000 Pa.

The labelled vertical height sets the pressure difference. A trapped gas above the column would change the upper pressure reference.

Atmospheric pressure acts on the open reservoir. The vertical height h runs from the reservoir surface to the top mercury meniscus. The space above the column provides the approximately zero-pressure reference.

Follow the two pressure references

  1. At the open reservoir surface, the pressure is atmospheric pressure.
  2. At that same horizontal level inside the connected mercury, the pressure is equal.
  3. Inside the tube, that pressure is the pressure above the column plus the mercury contribution ρgh.
  4. With the top pressure approximately zero, atmospheric pressure is approximately ρgh.
patmosphere = ptop + ρgh
If ptop is negligible, patmosphere = ρghh is the vertical column height above the reservoir surface. It is not the full tube length or the depth of the tube's lower end.

Worked example

Read a 0.750 m mercury column

Use the supplied mercury density of 13600 kg/m3 and g = 10 N/kg. The vertical height above the reservoir is 0.750 m, and the pressure at the top is negligible.

patmosphere = 13600 x 10 x 0.750 = 102000 Pa

The height supplies the pressure difference between the column top and the reservoir level. The near-vacuum at the top makes that difference approximately equal to the full atmospheric pressure.

Try it step by step

Work it out in steps: Column height in metres, then Atmospheric pressure in Pa.

  1. Problem 1. A mercury barometer shows a vertical column of 752 mm. Mercury has density 13600 kg/m3 and g = 10 N/kg. Find atmospheric pressure.
  2. Problem 2. On another day the column is 745 mm. Mercury has density 13600 kg/m3 and g = 10 N/kg. Find atmospheric pressure.
  3. Problem 3. On a third day the column is 760 mm. Mercury has density 13600 kg/m3 and g = 10 N/kg. Find atmospheric pressure.
Show the answer

Problem 1.

Column height in metres: h = 752 mm = 0.752 m

Atmospheric pressure in Pa: p = rho g h = 13600 x 10 x 0.752 = 102272 Pa, which is 102000 Pa to 3 sf

Problem 2.

Column height in metres: h = 745 mm = 0.745 m

Atmospheric pressure in Pa: p = 13600 x 10 x 0.745 = 101320 Pa, which is 101000 Pa to 3 sf

Problem 3.

Column height in metres: h = 760 mm = 0.760 m

Atmospheric pressure in Pa: p = 13600 x 10 x 0.760 = 103360 Pa, which is 103000 Pa to 3 sf

Why the liquid's density matters

For the same pressure, a less dense liquid needs a taller column. If water of density 1000 kg/m3 were used to represent the same 102000 Pa, then:

h = p / (ρg) = 102000 / (1000 x 10) = 10.2 mThe required water column is much taller than the 0.750 m mercury column because water is less dense.

For a given liquid with unchanged density, a greater atmospheric pressure supports a greater vertical height. Use the current reservoir level as the lower reference when reading h.

Put these in order

A sealed container of gas is connected to one side of a water manometer. The other side is open to the air. The whole set-up is carried up a mountain. Put the explanation of what happens to the water levels in order.

Show the answer

1. At the higher altitude, there is less air above the manometer. Less air above means a smaller weight of air pressing down on the open side.

2. The weight of air above is smaller, so atmospheric pressure is lower. The open side of the manometer now feels this lower atmospheric pressure.

3. The sealed gas pressure has not changed, so the difference between the gas pressure and atmospheric pressure is larger. A larger pressure difference needs a taller column of water to balance it.

4. The height difference between the two water levels increases.

The space above the column is part of the explanation

If appreciable gas is trapped above the column, ptop is not zero. Then atmospheric pressure is ptop + ρgh. Using ρgh alone would underestimate it.

Read the pressure unit mmHg

Millimetres of mercury, mmHg (also written mm Hg), is a conventional pressure unit based on a mercury column under fixed reference conditions. A measurement in mm alone is a length; a reading in mmHg is a pressure.

Unit conversion

Convert a pressure of 760 mmHg

Use the supplied approximation 1 mmHg = 133 Pa.

760 x 133 = 101080 Pa, or approximately 1.01 x 105 Pa.

This conversion uses the conventional pressure unit. It does not use the earlier column model's approximation g = 10 N/kg. When calculating pressure from a measured column height using ρgh, use that problem's stated density, g and vertical height.

Identify the top reference and the vertical height. The atmosphere acts on the reservoir and supports the column. The pressure above the column cannot be ignored unless the stated model makes it negligible.