Pressure in Liquids

A liquid exerts pressure on any surface in contact with it. This pressure increases with depth because the weight of liquid above increases.

The equation (Supplement)

$p = \rho g h$

Symbol	Quantity	Unit
$p$	Pressure due to liquid column	Pa
$\rho$	Density of the liquid	kg/m$^3$
$g$	Gravitational field strength	N/kg
$h$	Depth below the surface	m

Key facts about liquid pressure

• Pressure increases with depth ($p \propto h$).
• Pressure increases with liquid density ($p \propto \rho$). Mercury exerts more pressure than water at the same depth because mercury is denser.
• At any given depth, pressure acts equally in all directions.
• Pressure does not depend on the shape of the container or the total volume of liquid, only on the depth and density.

Worked example

Calculate the pressure at a depth of 15 m in seawater of density 1030 kg/m$^3$. Use $g = 9.8$ N/kg.

$p = 1030 \times 9.8 \times 15 = 151\,410\text{ Pa} \approx 1.5 \times 10^5\text{ Pa}$

This is the pressure due to the water alone. The total pressure at that depth includes atmospheric pressure on the surface.

Transmitting pressure in liquids

Liquids are virtually incompressible. A pressure applied to one part of an enclosed liquid is transmitted equally to all parts of the liquid. This is the basis of hydraulic systems.

Common errors and how to correct them

Forgetting to add atmospheric pressure for total pressure. The equation $p = \rho g h$ gives only the pressure from the liquid. Total pressure at depth = atmospheric pressure + $\rho g h$.

Using cm instead of m for depth. Depth must be in metres for the answer to be in pascals.

How to apply this in an exam

State the equation, convert depth to metres, substitute all values, and give the answer in Pa. If asked for “total pressure,” add atmospheric pressure (approximately $1.0 \times 10^5$ Pa) to the liquid pressure.