K323 / 2027
Pressure overview

Lesson 3 of 6

Pressure in a liquid

  • 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.

Make a guess. You are not marked.

A wide tank and a narrow tube are joined at the bottom and filled with water to the same level. Point A is 0.50 m below the surface in the tank. Point B is 0.50 m below the surface in the tube. Which point has the greater pressure?

  1. Neither, because pressure depends on depth, not width.
  2. B, because the narrow tube squeezes the water.
  3. A, because the tank holds much more water.
Show the answer

Neither, because pressure depends on depth, not width.

Pressure in a liquid depends on vertical depth, density and g, not on the width of the container. Same depth, same liquid, same pressure.

In a stationary liquid, pressure increases with vertical depth. The increase depends on the liquid's density and gravitational field strength.

Pressure is force per unit area, measured in Pa. Density tells us the mass per unit volume. Use density in kg/m3 and depth in m in the calculation below.

Pressure is a scalar quantity. The force that a liquid exerts on a surface acts perpendicular to that surface, so liquid can push on a container's side walls as well as its bottom.

Pressure increase below the surface = ρgh
Δp = ρghΔp means the pressure difference between the surface and the point below it. ρ = liquid density in kg/m3; g = gravitational field strength in N/kg; h = vertical depth in m. The result is in Pa.

This relationship assumes a stationary liquid of uniform density in a uniform gravitational field. The height h is measured vertically from the liquid surface, not along a sloping wall or tube.

Why width is not in the relationship

Consider a vertical liquid column of area A and height h. Its volume is Ah, its mass is ρAh, and its weight is ρAhg. Dividing that weight by area A gives the pressure contribution ρgh. A wider column has more weight, but that weight is spread over a larger area.

At the same horizontal level in the same connected stationary liquid, the pressure is equal. Different widths or shapes of the connected regions do not change that equality.

Same vertical depth, same pressure

Connected, stationary water: density 1000 kg/m3; g = 10 N/kg. Both open surfaces have pressure 101000 Pa.

Equal-depth points in differently sized connected regionsA wide left chamber and a narrower right chamber contain the same connected, stationary water. Both open surfaces lie at the same horizontal level and have pressure 101000 Pa. Points A and B lie on a second horizontal level, at vertical depth h of 0.80 metres below those surfaces. The bracket runs vertically from the surface level to the points, not from the container rim. The water contribution at either point is 8000 Pa, so total pressure at A and B is 109000 Pa.hSurface pressure: 101000 PaABVertical depth h = 0.80 m

Water contribution = 1000 x 10 x 0.80 = 8000 Pa.

Total at A and B = 101000 + 8000 = 109000 Pa.

The connected water regions have different widths. Points at the same 0.80 m vertical depth below the open surface have the same pressure. The marked height is vertical, not a distance along the container.

Extra pressure and total pressure are different

The liquid contribution ρgh is an increase below the surface. There may already be pressure acting on that surface. For an open container, the surrounding atmosphere supplies that surface pressure.

Worked example

Water 0.80 m below an open surface

Use the supplied values: water density 1000 kg/m3, g = 10 N/kg and atmospheric pressure at the surface of 101000 Pa.

  1. Find the liquid contribution: Δp = 1000 x 10 x 0.80 = 8000 Pa.
  2. Add the surface pressure: p = 101000 + 8000 = 109000 Pa.

8000 Pa is the difference between the surface and the point. 109000 Pa is the total pressure there. Any point at the same depth in this connected water has the same total pressure.

Try it step by step

Work it out in steps: Depth in metres, then Pressure increase from the liquid in Pa, then Total pressure in Pa.

  1. Problem 1. A point is 60 cm below the open surface of oil of density 800 kg/m3. Atmospheric pressure at the surface is 100000 Pa and g = 10 N/kg. Find the total pressure at the point.
  2. Problem 2. A diver is 2.5 m below the surface of sea water of density 1030 kg/m3. Atmospheric pressure at the surface is 101000 Pa and g = 10 N/kg. Find the total pressure on the diver.
  3. Problem 3. A point is 45 cm below the open surface of water of density 1000 kg/m3. Atmospheric pressure at the surface is 100000 Pa and g = 10 N/kg. Find the total pressure at the point.
Show the answer

Problem 1.

Depth in metres: h = 60 cm = 0.60 m

Pressure increase from the liquid in Pa: change in p = rho g h = 800 x 10 x 0.60 = 4800 Pa

Total pressure in Pa: p = 100000 + 4800 = 104800 Pa, which is 105000 Pa to 3 sf

Problem 2.

Depth in metres: h = 2.5 m, already in metres

Pressure increase from the liquid in Pa: change in p = 1030 x 10 x 2.5 = 25750 Pa

Total pressure in Pa: p = 101000 + 25750 = 126750 Pa, which is 127000 Pa to 3 sf

Problem 3.

Depth in metres: h = 45 cm = 0.45 m

Pressure increase from the liquid in Pa: change in p = 1000 x 10 x 0.45 = 4500 Pa

Total pressure in Pa: p = 100000 + 4500 = 104500 Pa, which is 105000 Pa to 3 sf

Pressure at depth = surface pressure + ρghAt the surface, h = 0: the liquid contribution is zero, but the surface pressure need not be zero.

For the same liquid and g, doubling depth doubles the liquid contribution. It does not double the total pressure when the surface pressure stays unchanged. At a fixed depth, a denser liquid gives a larger pressure increase.

Choose and measure the vertical height

Use a vertical scale and identify the two levels being compared. Read both levels from the same reference and subtract. The distance along an inclined tube is longer than the vertical separation and would give too large a pressure difference if substituted for h.

Identify the requested pressure before calculating. ρgh gives a pressure difference. Add the stated surface pressure when the question asks for total pressure; do not silently assume an open surface has zero pressure.

Check your understanding

A water-filled tube runs 1.0 m along a slope from its open surface to point P. P is only 0.50 m vertically below that surface. Use water density 1000 kg/m3, g = 10 N/kg and surface pressure 100000 Pa. What is the total pressure at P?

  1. 5000 Pa
  2. 110000 Pa
  3. 105000 Pa
Show the answer

105000 Pa

Use the vertical depth: the increase is 1000 x 10 x 0.50 = 5000 Pa. Add the supplied surface pressure: 100000 + 5000 = 105000 Pa.

Practise exam questions

Write your answer first, then tick the mark points it makes.

Exam-style question

A diver swims 12 m below the surface of the sea. The density of sea water is 1030 kg/m3, g = 10 N/kg and atmospheric pressure is 1.0 x 105 Pa. Calculate the total pressure on the diver. [4]

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Exam-style question

A student measures the pressure due to a liquid at different depths. Her graph of pressure against depth is a straight line through the origin, passing through (0.50 m, 4000 Pa). Taking g = 10 N/kg, use the graph to find the density of the liquid. [3]

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Exam-style question

Data-based question: Water pressure from a roof tank [9]

A water tank on the roof of a block of flats supplies the taps below it. A plumber turns off the taps and uses a gauge to measure the water pressure at a tap on three floors. The gauge reads the pressure above atmospheric pressure. The density of water is 1000 kg/m3 and g = 10 N/kg.

Table 1. Gauge pressure at taps on different floors
floorheight of tap above ground / mgauge pressure / kPa
26.0255
515.0165
824.075
  1. (a) Use two readings to show that the pressure falls by 10 kPa for each metre of height, and show that this agrees with p = ρgh. [2]
  2. (b) Calculate the height of the water surface in the tank above the ground. [2]
  3. (c) Predict the gauge pressure at a tap on floor 10, 30.0 m above the ground. [1]
  4. (d) State one assumption you made in (b) or (c). [1]
  5. (e) A washing machine needs a gauge pressure of at least 50 kPa. A resident on floor 10 says it will work if the tank is raised by 3.0 m. Evaluate her suggestion. [3]

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