Pressure in Fluids (Higher)

Dive to the bottom of a swimming pool and you feel the water pressing on your ears; the deeper you go, the harder it presses. A dam is built thick and heavy at its base but can be slimmer at the top. Both facts come from the same principle: the pressure in a liquid increases with depth. This lesson, part of Topic P1 (Matter) of OCR Gateway Science A, explains why a liquid exerts pressure in all directions and why that pressure grows with depth and with the density of the liquid, introduces the Higher-tier equation $p = h\rho g$p=hρg with worked calculations, and applies the idea to dam walls and submarines.

Higher tier: The equation $p = h\rho g$p=hρg and its calculations are Higher tier. Foundation candidates should still understand qualitatively that pressure in a liquid increases with depth and with density.

By the end of this lesson you should be able to explain why pressure in a liquid acts in all directions and increases with depth and density, use the equation $p = h\rho g$p=hρg (Higher tier), and explain applications such as dam walls and submarines.

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Why a Liquid Exerts Pressure

A fluid is anything that can flow — a liquid or a gas. Like a gas, a liquid is made of particles in motion that collide with any surface in contact with the liquid, exerting a force on it. Spread over the area of the surface, this gives a pressure. So a liquid pushes on the walls and bottom of its container, and on any object placed in it.

A key feature of fluid pressure is that it acts in all directions. At any point in a liquid, the pressure pushes equally in every direction — up, down and sideways. This is why a hole in the side of a container lets water squirt out sideways, and why a submerged object is squeezed from all sides at once.

Exam Tip: Pressure in a liquid acts in all directions at a given point, not just downward. This is why water spurts sideways out of a hole in the side of a tank.

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Why Pressure Increases with Depth

The pressure at a point in a liquid is caused by the weight of the liquid above that point pressing down (along with, near the surface, the atmosphere above). The deeper you go, the more liquid there is above you, so the greater its total weight and the higher the pressure.

This is why:

• the pressure at the bottom of a deep tank is greater than near the surface;
• water squirts furthest from a hole near the bottom of a tall container, because the pressure there is greatest;
• divers feel increasing pressure on their ears as they descend.

The pressure also depends on the density of the liquid. A denser liquid has more mass (and so more weight) in each unit of volume, so at a given depth a denser liquid produces a greater pressure. You would feel more pressure at $1 \text{ m}$1 m deep in mercury than at $1 \text{ m}$1 m deep in water, because mercury is far denser.

Exam Tip: Pressure in a liquid increases with depth (more liquid weight above) and with density (more weight per unit volume). Both make the column of liquid above heavier, raising the pressure.

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Higher Tier: The Pressure Equation $p = h\rho g$p=hρg

Higher tier only: The pressure due to a column of liquid is calculated from:

$p = h\rho g$p=hρg

where $p$p is the pressure (in $\text{Pa}$Pa, pascals), $h$h is the height (depth) of the column of liquid (in $\text{m}$m), $\rho$ρ is the density of the liquid (in $\text{kg/m}^3$kg/m3), and $g$g is the gravitational field strength (on Earth, $g = 9.8 \text{ N/kg}$g=9.8 N/kg on the OCR data sheet).

The equation captures everything in the qualitative picture: pressure rises in proportion to the depth $h$h and to the density $\rho$ρ. It rearranges to:

$p = h\rho g \qquad h = \dfrac{p}{\rho g} \qquad \rho = \dfrac{p}{hg}$p=hρgh=ρgp​ρ=hgp​

Worked Example 1 — Pressure at the bottom of a pool

Calculate the pressure due to the water at the bottom of a swimming pool $2 \text{ m}$2 m deep. (Density of water $= 1000 \text{ kg/m}^3$=1000 kg/m3; $g = 9.8 \text{ N/kg}$g=9.8 N/kg.)

Step 1 — write the equation: $p = h\rho g$p=hρg.

Step 2 — substitute: $p = 2 \times 1000 \times 9.8$p=2×1000×9.8.

Step 3 — calculate: $p = 19\,600 \text{ Pa}$p=19600 Pa.

Answer: $19\,600 \text{ Pa}$19600 Pa (about $19.6 \text{ kPa}$19.6 kPa) due to the water — this is in addition to atmospheric pressure pushing down on the surface.

Worked Example 2 — Pressure deeper down

Calculate the pressure due to seawater at a depth of $30 \text{ m}$30 m. (Density of seawater $= 1030 \text{ kg/m}^3$=1030 kg/m3; $g = 9.8 \text{ N/kg}$g=9.8 N/kg.)

Step 1 — write the equation: $p = h\rho g$p=hρg.

Step 2 — substitute: $p = 30 \times 1030 \times 9.8$p=30×1030×9.8.

Step 3 — calculate: $p = 302\,820 \text{ Pa}$p=302820 Pa.

Answer: about $3.03 \times 10^{5} \text{ Pa}$3.03×105 Pa (roughly $303 \text{ kPa}$303 kPa) — about three times atmospheric pressure, which is why deep diving requires careful pressure management.

Worked Example 3 — A denser liquid

Calculate the pressure due to a $0.76 \text{ m}$0.76 m column of mercury. (Density of mercury $= 13\,600 \text{ kg/m}^3$=13600 kg/m3; $g = 9.8 \text{ N/kg}$g=9.8 N/kg.)

Step 1 — write the equation: $p = h\rho g$p=hρg.

Step 2 — substitute: $p = 0.76 \times 13\,600 \times 9.8$p=0.76×13600×9.8.

Step 3 — calculate: $p = 101\,293 \text{ Pa} \approx 1.01 \times 10^{5} \text{ Pa}$p=101293 Pa≈1.01×105 Pa.

Answer: about $1.01 \times 10^{5} \text{ Pa}$1.01×105 Pa — this is close to atmospheric pressure, which is exactly why a mercury barometer reads atmospheric pressure as a column about $0.76 \text{ m}$0.76 m (760 mm) tall.

Worked Example 4 — Rearranging for depth

At what depth in fresh water is the pressure due to the water equal to $49\,000 \text{ Pa}$49000 Pa? (Density of water $= 1000 \text{ kg/m}^3$=1000 kg/m3; $g = 9.8 \text{ N/kg}$g=9.8 N/kg.)

Step 1 — rearrange for $h$h: $h = \dfrac{p}{\rho g}$h=ρgp​.

Step 2 — substitute: $h = \dfrac{49\,000}{1000 \times 9.8} = \dfrac{49\,000}{9800}$h=1000×9.849000​=980049000​.

Step 3 — calculate: $h = 5 \text{ m}$h=5 m.

Answer: a depth of $5 \text{ m}$5 m.

Worked Example 5 — Comparing two liquids at the same depth

At a depth of $4 \text{ m}$4 m, calculate the pressure due to (a) fresh water (density $1000 \text{ kg/m}^3$1000 kg/m3) and (b) a salt solution of density $1200 \text{ kg/m}^3$1200 kg/m3. Comment on the difference. ($g = 9.8 \text{ N/kg}$g=9.8 N/kg.)

Step 1 — fresh water: $p = h\rho g = 4 \times 1000 \times 9.8 = 39\,200 \text{ Pa}$p=hρg=4×1000×9.8=39200 Pa.