15.13P-15.14P - Liquid pressure calculations

Pressure below a liquid surface

Imagine two small pressure sensors in the same still liquid. The deeper sensor has a taller column of liquid above it. That column contains more liquid, so its weight produces a greater force on each unit area: pressure increases with depth.

Changing the liquid also matters. At the same depth, a denser liquid has more mass, and therefore more weight, in the same volume. It produces greater pressure than a less dense liquid at that depth. This is why both depth and liquid density appear in the pressure relationship.

This equation is supplied on the Pearson 2026 equation sheet and marked Higher tier. It gives the pressure caused by the liquid column. If the surface is open to the atmosphere, atmospheric pressure also acts, but it is not included in the value of P calculated here.

Calculating pressure and pressure differences

Worked example: pressure at one depth

A diver is 2.4 m below the surface of fresh water. Take $\rho = 1000 \mathrm{kg}/m^{3}$ and $g = 9.8 \mathrm{N}/\mathrm{kg}$. Calculate the pressure due to the water column.