Pressure in liquids

Key idea: Pressure in a liquid increases with depth: the extra pressure is hρg. The total pressure also includes the pressure acting on the liquid's surface.

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Dive to the bottom of a deep pool and you feel the water pressing on your ears. The deeper you go, the more water there is above you, and the greater its weight pressing down. That is why dams are built thicker at the bottom and why submarines need strong hulls. A single equation gives the pressure increase with depth in a liquid at rest, provided its density and the local gravitational field strength are effectively constant.

The pressure of a liquid column

Picture a column of liquid of cross-sectional area A standing a height h above a point. The liquid’s weight presses down on the area A below it:

volume of the column = Ah;
mass of the column = ρ Ah, where ρ is the density of the liquid;
weight of the column = ρ Ahg.

Pressure is force divided by area, so the pressure due to the column is ρ Ahg/A. The area cancels:

p = hρ g

where h is the vertical depth below the surface in metres, ρ is the density of the liquid in kg m⁻³ and g is the gravitational field strength in N kg⁻¹.

Because the area cancels, the pressure depends only on the depth, the density of the liquid and g. It does not depend on how wide the container is or what shape it has. For water (ρ = 1000 kg m⁻³, g = 10 N kg⁻¹), each metre of depth adds 10 kPa.

Deeper water gives greater pressure; add surface pressure for the total. At B, equal tiny areas feel equal force magnitudes in every orientation.

Check your understanding 1 Predict the pressure probe

Before changing the model below, predict the liquid-column pressure for water at 0.40 m. At the same depth compare Straight, Wide top and Narrow neck tanks, then move the probe sideways using Probe position across the tank. Finally switch to oil at 0.60 m. Record your predictions and explanations, then compare the model readings if it is available. The model uses g = 9.81 N/kg, not the rounded 10 N/kg used in the worked examples. Optional: enable Include atmospheric pressure and explain what changes.

Water-column pressure at 0.40 m

Unit: Pa

Oil-column pressure at 0.60 m

Unit: Pa

Show answer

Water: 0.40 × 1000 × 9.81 = 3924 Pa, about 3.92 × 10³ Pa. Oil: 0.60 × 800 × 9.81 = 4708.8 Pa, about 4.71 × 10³ Pa. Record these predictions before comparing the model.

Depth, h: 0.40 m
Pressure due to the liquid, hρg: 3924 Pa
Total pressure: 1.049 × 10⁵ Pa
Manometer height difference: 0.400 m
Pressure in the liquid, F₁/A₁: 75000 Pa
Force on the large piston, F₂: 1500 N
Effort moves down, d₁: 0.0 cm
Load rises, d₂: 0.00 cm

Pressure due to the liquid against depth

Try this

0 of 4 done

Measure the pressure at the same depth in two differently shaped tanks. (not done yet)
Record the pressure at five different depths in one liquid. (not done yet)
Lift the 2000 N load with an effort of 200 N or less. (not done yet)
Find the smallest effort, to the nearest 10 N, that lifts the load. (not done yet)

Your readings

#	h / m	p / kPa	Remove
No readings yet. Set up a measurement, then record it.

Check your understanding 2 Compare the pressure readings

Compare both pressures with your predictions. Explain what happens when you change tank shape or move the probe sideways at the same depth. Explain how the atmospheric-pressure option changes the reading. If the model is unavailable, use the stated conditions and pressure equation.

Predictions, shape and sideways comparisons, oil reading and surface-pressure explanation

Show answer

Water at 0.40 m gives 0.40 × 1000 × 9.81 = 3924 Pa, displayed to three significant figures as about 3.92 × 10³ Pa. At unchanged depth, density and surface pressure, all three shapes and sideways probe positions give the same pressure. Oil at 0.60 m gives 0.60 × 800 × 9.81 = 4708.8 Pa, about 4.71 × 10³ Pa: the increased depth outweighs the lower density in this comparison.

The model’s liquid-column reading excludes atmospheric pressure. Enabling the atmospheric option adds its fixed 101000 Pa to give total pressure: 104924 Pa for the original water case, displayed at four significant figures as about 104900 Pa. It does not change the liquid-column contribution. If you cannot use the model, evaluate these predictions with p = hρg and the supplied conditions; no observed readings are required.

Between two points in the same liquid, one a vertical height Δ h below the other, the pressure difference is

Δ p = Δ h ρ g

Always use the vertical depth. A sloping path through the liquid to the same depth does not change the pressure.

Column pressure and total pressure

hρ g is the pressure due to the liquid column: the extra pressure compared with the surface. The surface of an open tank has the atmosphere pressing on it, and that pressure is passed down through the liquid too. The total pressure at depth h is therefore

p total = p atm + hρ g

Two straight lines versus depth: one shows pressure due to the water column only; the other includes atmospheric pressure as a constant offset. — Illustrative values for water using ρ = 1000 kg m⁻³, g = 10 N kg⁻¹ and atmospheric pressure = 100 kPa. Both lines have the same gradient; atmospheric pressure adds a constant 100 kPa.

Figure data: Column pressure vs total pressure (open tank)
Depth below surface (m)	Column pressure (ρgh) in water	Total pressure (p_atm + ρgh)
0	0	100
6	60	160

Figure data: Column pressure vs total pressure (open tank)
Depth below surface (m)	Column pressure (ρgh) in water	Total pressure (p_atm + ρgh)
0	0	100
6	60	160

“Pressure due to the water” or “pressure due to the liquid column” means hρ g only. “Total pressure” or “pressure at that depth” when atmospheric pressure is given means p atm + hρ g.

Problem

A diver is 6.0 m below the surface of fresh water. Calculate:

the pressure due to the water column;
the total pressure if atmospheric pressure is 1.0 × 10⁵ Pa.

Take ρ water = 1000 kg m⁻³ and g = 10 N kg⁻¹.

Show full solution

Identify the pressure requested first
Method
Calculate the extra pressure produced by the water column.
Reason
The first part excludes atmospheric pressure, so use p column = hρ g.
Working
Answer
p column = (6.0)(1000)(10) = 6.0 × 10⁴ Pa
Add the surface pressure
Reason
Total pressure at the diver includes both atmospheric pressure at the surface and the water-column increase.
Working
Answer
p total = 1.0 × 10⁵ + 6.0 × 10⁴ = 1.6 × 10⁵ Pa

Quick check

The total pressure must exceed both atmospheric pressure and the water-column contribution.

Problem

Two points in the same liquid are separated vertically by 0.25 m. Find the pressure difference between them.

Take ρ = 1000 kg m⁻³ and g = 10 N kg⁻¹.

Explain the vertical separation and pressure difference

Vertical height selection, difference equation and surface-pressure cancellation

Show which distance belongs in the equation and why the shared surface pressure cancels. The optional check below assesses only the pressure difference.

Pressure difference

Unit: Pa

Hints

Hint 1

choose the difference relationship

Which vertical separation determines the pressure difference?

Hint 2

substitute the vertical separation

Use Δ h = 0.25 m directly because it is already a vertical SI distance.

Show solution step by step

Select the vertical depth difference
Method
Use the 0.25 m vertical separation as Δ h.
Reason
Hydrostatic pressure differences depend on vertical height, not on the path or container shape between the points.
Working
Answer
Δ h = 0.25 m
Calculate the pressure difference
Reason
Both points are in the same liquid and share the stated g.
Working
Answer
Δ p = (0.25)(1000)(10) = 2.5 × 10³ Pa

Common mistake

Use the 0.25 m vertical separation, not a sloping distance. The lower point is 2500 Pa higher in pressure. Both totals contain the same surface pressure, so it cancels on subtraction: do not add atmospheric pressure to this difference.

Examination question

In water, the pressure due to the liquid column is 3.0 × 10⁴ Pa. Calculate the depth. Take ρ water = 1000 kg m⁻³ and g = 10 N kg⁻¹. [3 marks]

Write your rearrangement and answer

Show the equation, rearrangement and depth with unit

Depth below the water surface

Unit: m

Show solution step by step

Choose the liquid-column relationship
1 mark
Method
Use p = hρ g without adding atmospheric pressure.
Reason
The given pressure is explicitly due to the liquid column.
Working
Answer
p = hρ g
Rearrange for depth
1 mark
Reason
Depth is the unknown, so divide the pressure by ρ g.
Working
Answer
h = p/(ρ g)
Calculate the depth
1 mark
Reason
The supplied quantities are in compatible SI units.
Working
Answer
h = (3.0 × 10⁴)/((1000)(10)) = 3.0 m

Self-mark with the mark scheme

Tick each mark point that your answer includes.

Self-mark the relationship, rearrangement and final value separately.

Mark points

1 mark: Uses p = hρg for the stated liquid-column pressure.
1 mark: Rearranges to h = p/(ρg).
1 mark: Obtains 3.0 m.Your answer includes the correct unit.

Common mistake

Multiplying instead of dividing gives 3.0 × 10⁸ m and loses the last two marks.

Same depth, same pressure

For the same liquid, gravitational field strength, surface pressure and vertical depth, container shape does not change the pressure at A, B or C.

In a liquid at rest, all points at the same depth have the same pressure, whatever the shape of the container. A wide tank and a narrow tube filled with the same liquid to the same depth have the same pressure at the bottom. The wide tank holds more liquid in total, but in a straight-sided tank that extra weight is spread over a larger base, so each square metre carries the same weight. Sloping walls can exert an upward or downward force on the liquid, depending on their orientation. The force on the base therefore need not equal the total liquid weight; the local pressure rule remains unchanged.

“Same depth, same pressure” needs the same liquid and the same surface pressure. Two different liquids at the same depth give different pressures, because their densities differ.

Learner response

A point is 0.40 m below the surface of water (ρ = 1000 kg m⁻³) and another point is 0.40 m below the surface of oil (ρ = 800 kg m⁻³). Take g = 10 N kg⁻¹.

A student writes: The points are at the same depth, so the two liquid-column pressures must be equal.

Locate the first error and calculate both pressures.

Diagnose and explain the same-depth claim

Missing condition, both pressure calculations and valid same-depth rule

Explain the missing condition and show both calculations. The optional checks below assess only the choice and numerical pressures.

What has the student ignored?

Surface pressure must be added to each column.Density must also be the same.Container width must also be the same.

Water-column pressure

Unit: Pa

Oil-column pressure

Unit: Pa

Show solution step by step

Locate the missing variable
Method
Compare density as well as depth.
Reason
The relationship p = hρ g contains both h and ρ; equal depth alone is not enough for different liquids.
Working
Answer
Water is denser than oil: 1000 > 800 kg m⁻³.
Calculate the water pressure
Reason
Use the water density at the common depth.
Working
Answer
p water = (0.40)(1000)(10) = 4.0 × 10³ Pa
Calculate and compare the oil pressure
Reason
The lower oil density produces a smaller pressure for the same g and h.
Working
Answer
p oil = (0.40)(800)(10) = 3.2 × 10³ Pa

Common mistake

“Same depth means same pressure” applies to the same connected liquid under the same conditions, not automatically to different densities.

Problem

In an unknown liquid, pressure increases by 5.0 × 10³ Pa when the depth increases by 0.50 m. Find the density of the liquid. Take g = 10 N kg⁻¹.

Infer the liquid property from pressure data

Density inference, labelled paper pressure–depth lines and interpretation of gradients

Infer the density with units. Then use the actual open-tank plot earlier to sketch liquid-column and total pressure against depth on paper: label the axes and mark the values at 0 m, 3 m and 6 m. Explain why both gradients give the same density despite different intercepts. The optional check assesses only the inferred density value.

Density of the liquid

Unit: kg m⁻³

Hints

Hint 1

use differences rather than total pressure

The data give a pressure increase across a depth increase, so use Δ p = Δ h ρ g.

Hint 2

make density the subject

Divide Δ p by gΔ h.

Show solution step by step

Rearrange the difference relationship
Method
Make density the subject.
Reason
The pressure change, vertical separation and gravitational field strength are measured, while the liquid is unknown.
Working
Answer
ρ = (Δ p)/(gΔ h)
Infer the density
Reason
The pressure increase is caused by the additional 0.50 m liquid column.
Working
Answer
ρ = (5.0 × 10³)/((10)(0.50)) = 1000 kg m⁻³

Quick check

The inferred 1000 kg/m³ matches water’s typical density but does not identify the liquid uniquely. Self-review your paper graph against the actual plot: depth in metres horizontally and pressure in kPa vertically; column pressures 0, 30 and 60 kPa at 0, 3 and 6 m, and total pressures 100, 130 and 160 kPa. Draw parallel straight lines, with gradients 10 kPa/m = 10000 Pa/m. Dividing by the plot’s g = 10 N/kg gives density 1000 kg/m³ for both. The total-pressure intercept is the 100 kPa surface pressure; it affects the intercept, not the gradient. Equivalent sensible scales are valid.

Pressure acts in all directions

At any point in a liquid at rest, the pressure is the same in every direction. The liquid pushes down on the bottom of a tank, sideways on its walls and upwards on anything submerged in it, always at right angles to the surface it touches. Pressure itself is a scalar; it is the force it produces on a surface that has a direction.

You can see this in a can with holes punched down one side and filled with water. Water spurts sideways out of every hole, so the water pushes sideways as well as down. It comes out fastest from the lowest hole, where the pressure is greatest.

Common mistakes

Using a slanted distance instead of the vertical depth below the surface.
Forgetting to convert centimetres to metres, for example 35 cm = 0.35 m.
Using a density in g cm⁻³ with g and h in SI units. Use kg m⁻³.
Adding atmospheric pressure when the question asks for the pressure due to the liquid only, or leaving it out when the total pressure is asked for.
Thinking a wider container gives a greater pressure at the same depth.

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Pressure in liquids

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Guided steps

Reading settings

The pressure of a liquid column

Check your understanding 1 Predict the pressure probe

Try this

Your readings

Check your understanding 2 Compare the pressure readings

Column pressure and total pressure

Pressure on a diver

Identify the pressure requested first

Add the surface pressure

Pressure difference between two depths

Select the vertical depth difference

Calculate the pressure difference

Find depth from pressure

Choose the liquid-column relationship

Rearrange for depth

Calculate the depth

Same depth, same pressure

Different liquids, same depth

Locate the missing variable

Calculate the water pressure

Calculate and compare the oil pressure

Identify an unknown liquid

Rearrange the difference relationship

Infer the density

Pressure acts in all directions

Common mistakes