The liquid pressure equation

Definition

Liquid pressure

The pressure at a point in a liquid caused by the weight of the liquid above that point.

Definition

Gravitational field strength

The gravitational force acting on each kilogram of mass, which is $N/kg10\ \text{N/kg}$ near the Earth's surface.

The pressure caused by a column of liquid depends on its height, the density of the liquid and the gravitational field strength.
The equation is $strength\text{pressure due to a column of liquid} = \text{height of column} \times \text{density of liquid} \times \text{gravitational field strength}$ .
In symbols, this is $p=hρgp = h\rho g$ $p = h ρ g$ .
1. $p$ is the pressure due to the liquid in pascals, $Pa\text{Pa}$ .
2. $h$ is the height of the liquid column, or the depth, in metres, $m\text{m}$ .
3. $ρ\rho$ is the density of the liquid in kilograms per cubic metre, $kg/m3\text{kg/m}^3$ .
4. $g$ is the gravitational field strength in newtons per kilogram, $N/kg\text{N/kg}$ .
Near the Earth's surface, $10\ \text{N/kg}$ .
You do not need to memorise this equation, but you must be able to choose it from a list of equations and use it.
The equation comes from the weight of the liquid column: a column of height $h$ standing on an area $A$ has a volume of $h A$ , a mass of $hAρhA\rho$ and a weight of $hAρghA\rho g$ .
Dividing this weight by the area $A$ that it rests on gives the pressure, $h\rho g$ , so the area cancels out.
This is why the pressure depends only on depth and density, not on the width or shape of the container.
The equation gives the pressure due to the liquid alone, because the total pressure on a submerged object also includes the atmospheric pressure acting on the liquid surface.

Water jets from holes at three depths in a tall container. The deeper the hole, the taller the column of water above it, so the greater the pressure forcing the water out.

Pressure differences between depths

Definition

Depth

The vertical distance of a point below the surface of a liquid.

Definition

Pressure difference

The difference between the pressures on the two sides of a surface, which produces a net force on the surface towards the side at lower pressure.

The difference in pressure between two points in the same liquid depends only on the vertical distance between them.
For a difference in depth $Δh\Delta h$ , the difference in pressure is $g\Delta p = \Delta h\,\rho\,g$ .
The pressure difference does not depend on how far below the surface the two points are, only on how far apart they are vertically.
In fresh water, the pressure increases by $Pa10\,000\ \text{Pa}$ for every extra metre of depth, because $\times 1000 \times 10 = 10\,000$ .
A diver $m10\ \text{m}$ down therefore feels an extra $Pa100\,000\ \text{Pa}$ from the water, which is about the same as the whole of atmospheric pressure.
Always convert depths to metres before substituting, because a depth in centimetres gives an answer $100$ times too large.
Large pressures are often given in kilopascals, where $Pa1\ \text{kPa} = 1000\ \text{Pa}$ .
Rearranging the equation gives $Δh=Δpρg\Delta h = \dfrac{\Delta p}{\rho g}$ for a difference in depth and $g\rho = \dfrac{\Delta p}{\Delta h\,g}$ for a density.

Example

Pressure at the bottom of a pool

A swimming pool is $m2.5\ \text{m}$ deep and is filled with water of density $kg/m31000\ \text{kg/m}^3$ .

h\rho g

\times 1000 \times 10

25\,000\ \text{Pa}

The water adds $Pa25\,000\ \text{Pa}$ , or $kPa25\ \text{kPa}$ , to the atmospheric pressure acting on the surface of the pool.

Example

Difference between two depths

A diver swims down from $m3.0\ \text{m}$ to $m12.0\ \text{m}$ below the surface of sea water of density $kg/m31030\ \text{kg/m}^3$ .
Find the difference in depth first.

\Delta h = 12.0 - 3.0 = 9.0\ \text{m}

\Delta p = \Delta h\,\rho\,g = 9.0 \times 1030 \times 10

\Delta p = 92\,700\ \text{Pa} \approx 9.3 \times 10^{4}\ \text{Pa}

The pressure on the diver increases by $Pa92\,700\ \text{Pa}$ .

Example

Finding a depth

The pressure difference between the top and the bottom of a water tank is $kPa18\ \text{kPa}$ , and the water has a density of $kg/m31000\ \text{kg/m}^3$ .
Convert the pressure difference into pascals.

\Delta p = 18 \times 1000 = 18\,000\ \text{Pa}

Rearrange the equation to make the depth the subject.

\Delta h = \dfrac{\Delta p}{\rho g} = \dfrac{18\,000}{1000 \times 10} = 1.8\ \text{m}

The water in the tank is $m1.8\ \text{m}$ deep.

Common Mistake

Do not substitute a depth in centimetres, because the equation needs the depth in metres.
Do not use the depth of only one point when a question asks for a pressure difference, because you must subtract the two depths first.
Do not add atmospheric pressure unless the question asks for the total pressure.
Do not use the mass of the liquid in place of its density.

Exam technique

Liquid pressure calculations

Write $h\rho g$ before you substitute, because the equation and the substitution can each earn credit even if the arithmetic slips.
Show the difference in depth on its own line, because skipping it is a common way to lose a mark.
Give the final answer with its unit, $Pa\text{Pa}$ or $kPa\text{kPa}$ , and check that its size is sensible at about $kPa10\ \text{kPa}$ for each metre of water.
If a later part of a question uses your answer, carry it forward even if you think it may be wrong, because the method marks can still be awarded.

Self review

Write the equation that links pressure due to a column of liquid, height, density and gravitational field strength.
Calculate the pressure due to the water at a depth of $m4.0\ \text{m}$ in fresh water of density $kg/m31000\ \text{kg/m}^3$ .
Explain why the pressure due to a liquid does not depend on the width of its container.
Calculate the pressure difference between depths of $m2.0\ \text{m}$ and $m7.0\ \text{m}$ in sea water of density $kg/m31030\ \text{kg/m}^3$ .
State the value of $g$ to use near the Earth's surface.

The liquid pressure equation

Definition

Liquid pressure

The pressure at a point in a liquid caused by the weight of the liquid above that point.

Definition

Gravitational field strength

The gravitational force acting on each kilogram of mass, which is $N/kg10\ \text{N/kg}$ near the Earth's surface.

The pressure caused by a column of liquid depends on its height, the density of the liquid and the gravitational field strength.
The equation is $strength\text{pressure due to a column of liquid} = \text{height of column} \times \text{density of liquid} \times \text{gravitational field strength}$ .
In symbols, this is $p=hρgp = h\rho g$ $p = h ρ g$ .
1. $p$ is the pressure due to the liquid in pascals, $Pa\text{Pa}$ .
2. $h$ is the height of the liquid column, or the depth, in metres, $m\text{m}$ .
3. $ρ\rho$ is the density of the liquid in kilograms per cubic metre, $kg/m3\text{kg/m}^3$ .
4. $g$ is the gravitational field strength in newtons per kilogram, $N/kg\text{N/kg}$ .
Near the Earth's surface, $10\ \text{N/kg}$ .
You do not need to memorise this equation, but you must be able to choose it from a list of equations and use it.
The equation comes from the weight of the liquid column: a column of height $h$ standing on an area $A$ has a volume of $h A$ , a mass of $hAρhA\rho$ and a weight of $hAρghA\rho g$ .
Dividing this weight by the area $A$ that it rests on gives the pressure, $h\rho g$ , so the area cancels out.
This is why the pressure depends only on depth and density, not on the width or shape of the container.
The equation gives the pressure due to the liquid alone, because the total pressure on a submerged object also includes the atmospheric pressure acting on the liquid surface.

Water jets from holes at three depths in a tall container. The deeper the hole, the taller the column of water above it, so the greater the pressure forcing the water out.

Pressure differences between depths

Definition

Depth

The vertical distance of a point below the surface of a liquid.

Definition

Pressure difference

The difference between the pressures on the two sides of a surface, which produces a net force on the surface towards the side at lower pressure.

The difference in pressure between two points in the same liquid depends only on the vertical distance between them.
For a difference in depth $Δh\Delta h$ , the difference in pressure is $g\Delta p = \Delta h\,\rho\,g$ .
The pressure difference does not depend on how far below the surface the two points are, only on how far apart they are vertically.
In fresh water, the pressure increases by $Pa10\,000\ \text{Pa}$ for every extra metre of depth, because $\times 1000 \times 10 = 10\,000$ .
A diver $m10\ \text{m}$ down therefore feels an extra $Pa100\,000\ \text{Pa}$ from the water, which is about the same as the whole of atmospheric pressure.
Always convert depths to metres before substituting, because a depth in centimetres gives an answer $100$ times too large.
Large pressures are often given in kilopascals, where $Pa1\ \text{kPa} = 1000\ \text{Pa}$ .
Rearranging the equation gives $Δh=Δpρg\Delta h = \dfrac{\Delta p}{\rho g}$ for a difference in depth and $g\rho = \dfrac{\Delta p}{\Delta h\,g}$ for a density.

Example

Pressure at the bottom of a pool

A swimming pool is $m2.5\ \text{m}$ deep and is filled with water of density $kg/m31000\ \text{kg/m}^3$ .

h\rho g

\times 1000 \times 10

25\,000\ \text{Pa}

The water adds $Pa25\,000\ \text{Pa}$ , or $kPa25\ \text{kPa}$ , to the atmospheric pressure acting on the surface of the pool.

Example

Difference between two depths

A diver swims down from $m3.0\ \text{m}$ to $m12.0\ \text{m}$ below the surface of sea water of density $kg/m31030\ \text{kg/m}^3$ .
Find the difference in depth first.

\Delta h = 12.0 - 3.0 = 9.0\ \text{m}

\Delta p = \Delta h\,\rho\,g = 9.0 \times 1030 \times 10

\Delta p = 92\,700\ \text{Pa} \approx 9.3 \times 10^{4}\ \text{Pa}

The pressure on the diver increases by $Pa92\,700\ \text{Pa}$ .

Example

Finding a depth

The pressure difference between the top and the bottom of a water tank is $kPa18\ \text{kPa}$ , and the water has a density of $kg/m31000\ \text{kg/m}^3$ .
Convert the pressure difference into pascals.

\Delta p = 18 \times 1000 = 18\,000\ \text{Pa}

Rearrange the equation to make the depth the subject.

\Delta h = \dfrac{\Delta p}{\rho g} = \dfrac{18\,000}{1000 \times 10} = 1.8\ \text{m}

The water in the tank is $m1.8\ \text{m}$ deep.

Common Mistake

Do not substitute a depth in centimetres, because the equation needs the depth in metres.
Do not use the depth of only one point when a question asks for a pressure difference, because you must subtract the two depths first.
Do not add atmospheric pressure unless the question asks for the total pressure.
Do not use the mass of the liquid in place of its density.

Exam technique

Liquid pressure calculations

Write $h\rho g$ before you substitute, because the equation and the substitution can each earn credit even if the arithmetic slips.
Show the difference in depth on its own line, because skipping it is a common way to lose a mark.
Give the final answer with its unit, $Pa\text{Pa}$ or $kPa\text{kPa}$ , and check that its size is sensible at about $kPa10\ \text{kPa}$ for each metre of water.
If a later part of a question uses your answer, carry it forward even if you think it may be wrong, because the method marks can still be awarded.

Self review

Write the equation that links pressure due to a column of liquid, height, density and gravitational field strength.
Calculate the pressure due to the water at a depth of $m4.0\ \text{m}$ in fresh water of density $kg/m31000\ \text{kg/m}^3$ .
Explain why the pressure due to a liquid does not depend on the width of its container.
Calculate the pressure difference between depths of $m2.0\ \text{m}$ and $m7.0\ \text{m}$ in sea water of density $kg/m31030\ \text{kg/m}^3$ .
State the value of $g$ to use near the Earth's surface.

1.3.10 Pressure differences with depth

The liquid pressure equation

Pressure differences between depths

Pressure at the bottom of a pool

Difference between two depths

Finding a depth

Liquid pressure calculations

Review 1.3.10 Pressure differences with depth by teaching Genie

1.3.10 Pressure differences with depth

The liquid pressure equation

Pressure differences between depths

Pressure at the bottom of a pool

Difference between two depths

Finding a depth

Liquid pressure calculations

Review 1.3.10 Pressure differences with depth by teaching Genie

1.3.10 Pressure differences with depth

The liquid pressure equation

Pressure differences between depths

Pressure at the bottom of a pool

Difference between two depths

Finding a depth

Liquid pressure calculations

Review 1.3.10 Pressure differences with depth by teaching Genie

1.3.10 Pressure differences with depth Revision Guide

1.3.10 Pressure differences with depth

The liquid pressure equation

Pressure differences between depths

Pressure at the bottom of a pool

Difference between two depths

Finding a depth

Liquid pressure calculations

Review 1.3.10 Pressure differences with depth by teaching Genie

1.3.10 Pressure differences with depth Revision Guide