K323 / 2027
Pressure overview

Topic 3 of 6

Pressure in a liquid

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.

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.

Optional check 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/m^3, g = 10 N/kg and surface pressure 100000 Pa. What is the total pressure at P?
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/m^3, g = 10 N/kg and surface pressure 100000 Pa. What is the total pressure at P?