---
title: "Pressure: From Sport to Diving"
book: "High School Physics"
subject: physics
language: en
chapter: 7
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/7-pressure-from-sport-to-diving
---

# Chapter 7 — Pressure: From Sport to Diving

A skier glides over snow that a walker sinks into; a $5\,\mathrm{N}$ push drives a needle through leather a fist cannot dent. Same [forces](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force), wildly different effects: the missing quantity is *[pressure](#def-g10-pressure-pressure)*, [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) divided by the area receiving it. This chapter defines it, follows it down a pool, up a barometer, into a squeezed syringe — and ends forty [metres](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) under the sea, where it turns vital.

## 7.1 Pressure: force spread over an area

**Definition 7.1 (Pressure).**

When a [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) of magnitude $F$ (in $\mathrm{N}$) presses perpendicularly on a surface of area $S$ (in $\mathrm{m}^{2}$), the *pressure* exerted on that surface is

$$
P = \frac{F}{S}.
$$

Its [unit](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit), the $\mathrm{N}/\mathrm{m}^{2}$, is the *pascal* ($\mathrm{Pa}$) — a tiny [unit](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit), an apple’s [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) spread over a square [metre](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit), so $\mathrm{kPa}$ and $\mathrm{MPa}$ are the working [units](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit).

**Example 7.2 (Heel versus snowshoe).**

A $60\,\mathrm{kg}$ person weighs $F = 60 \times 9.81 \approx 590\,\mathrm{N}$. On one stiletto heel of $1.0\,\mathrm{cm}^{2}$ = $1.0 \times 10^{-4}\,\mathrm{m}^{2}$: $P = 590/1.0 \times 10^{-4} \approx 5.9 \times 10^{6}\,\mathrm{Pa} = 5.9\,\mathrm{MPa}$. On two snowshoes of $0.20\,\mathrm{m}^{2}$ each: $P = 590/0.40 \approx 1.5\,\mathrm{kPa}$. The *same* [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) presses $4000$ times harder under the heel: heels pockmark wooden floors, snowshoes float on powder snow.

![The same force on two areas: the pressure, not the force, decides whether the surface yields.](https://one-course.com/images/onecourse/chapters/physics-2/g10-pressure/fig-6bcafb92682d.svg)

![The same force on two areas: the pressure, not the force, decides whether the surface yields.](https://one-course.com/images/onecourse/chapters/physics-2/g10-pressure/fig-ca1ef2327aa6.svg)

*The same [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on two areas: the [pressure](#def-g10-pressure-pressure), not the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force), decides whether the surface yields.*

**Remark 7.3 (Concentrate or spread).**

A sewing needle pushed with $5\,\mathrm{N}$ on a tip of $0.01\,\mathrm{mm}^{2}$ = $1 \times 10^{-8}\,\mathrm{m}^{2}$ exerts $5 \times 10^{8}\,\mathrm{Pa} = 500\,\mathrm{MPa}$ — enough to part leather fibres. Piercing tools concentrate [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force); load-bearing designs (skis, caterpillar tracks, foundations) spread it. Sport is applied [pressure](#def-g10-pressure-pressure) management: skates bite into ice, skis glide on snow.

## 7.2 Pressure in a liquid at rest

Why does water press harder lower down? Isolate an imaginary vertical column of water of cross-section $S$, from the surface to depth $h$: volume $Sh$, mass $\rho S h$ ($\rho$ the liquid’s density, in $\mathrm{kg}/\mathrm{m}^{3}$), [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) $\rho S h g$. The bottom of the column supports that [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) *plus* the atmosphere’s push $P_0 S$ on the top. Dividing the total [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) by $S$: the [pressure](#def-g10-pressure-pressure) at depth $h$ should be $P_0 + \rho g h$ — and it is, everywhere in the liquid.

**Theorem 7.4 (Pressure at depth).**

In a liquid of density $\rho$ at rest, open to the air, the [pressure](#def-g10-pressure-pressure) at depth $h$ below the surface is

$$
P = P_0 + \rho g h,
$$

where $P_0$ is the [atmospheric pressure](#def-g10-pressure-atmospheric) at the surface. The [pressure](#def-g10-pressure-pressure) depends only on the depth, not on the container’s shape, and at a given point the liquid presses equally hard in all directions.

**Proof.** *Admitted at this level.* ∎

**Remark 7.5.**

The column argument only makes the law plausible; the honest derivation (any shape, all directions) is in the Year 1 volume. The water’s own term $\rho g h$ is the *gauge* [pressure](#def-g10-pressure-pressure) — what a diver’s [pressure](#def-g10-pressure-pressure) [gauge](#thm-g10-pressure-depth) displays.

**Example 7.6 (Ten metres of water).**

In fresh water ($\rho = 1000\,\mathrm{kg}/\mathrm{m}^{3}$), at $h = 10\,\mathrm{m}$: $\rho g h = 1000 \times 9.81 \times 10 \approx 98\,\mathrm{kPa}$, so $P = 101 + 98 \approx 199\,\mathrm{kPa}$. Ten [metres](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of water add almost exactly one atmosphere: the diver’s rule of thumb, *one extra atmosphere per ten [metres](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit)*.

![Fresh water at rest: pressure grows linearly with depth, one extra atmosphere every ten metres.](https://one-course.com/images/onecourse/chapters/physics-2/g10-pressure/fig-7eb3107125b7.svg)

*Fresh water at rest: [pressure](#def-g10-pressure-pressure) grows linearly with depth, one extra atmosphere every ten [metres](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit).*

## 7.3 The ocean of air

**Definition 7.7 (Atmospheric pressure).**

We live at the bottom of an ocean of air, whose [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) creates the *atmospheric pressure*, at sea level

$$
P_0 \approx 101\,\mathrm{kPa} = 1.01 \times 10^{5}\,\mathrm{Pa}.
$$

On each square centimetre this is $F = 1.01 \times 10^{5} \times 1 \times 10^{-4}
\approx 10\,\mathrm{N}$: the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) of a [one-kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) mass on every $\mathrm{cm}^{2}$ of your skin, your desk, everything.

**Example 7.8 (The invisible load).**

An A4 sheet of paper ($21.0\,\mathrm{cm}$ $\times$ $29.7\,\mathrm{cm}$, so $S \approx 0.0624\,\mathrm{m}^{2}$) receives from the air above it $F = 1.01 \times 10^{5} \times 0.0624 \approx 6.3 \times 10^{3}\,\mathrm{N}$ — the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) of a small car. Nothing tears: the air below pushes *up* just as hard. We notice the atmosphere only when one side loses it — a suction cup, a straw, an aircraft cabin.

**Remark 7.9 (Torricelli’s barometer).**

In 1643 Torricelli inverted a sealed tube of mercury over a mercury bath. The column fell to the height at which $\rho g h = P_0$: $h = 1.01 \times 10^{5}/(13600 \times 9.81) \approx 0.76\,\mathrm{m}$, leaving above it the first vacuum ever made. The height tracks the weather and shrinks with altitude: a barometer weighs the air above you (water would need $10.3\,\mathrm{m}$: [Exercise 7.9](#exo-g10-pressure-9)).

## 7.4 Gases: bombardment and Boyle’s law

A gas too presses on its container: it is a swarm of molecules in ceaseless random motion, each collision gives the wall a tiny outward push, and billions of billions of impacts per second blur into the steady [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) we measure as gas [pressure](#def-g10-pressure-pressure). Squeeze the gas into half the volume: the molecules strike twice as often, so the [pressure](#def-g10-pressure-pressure) should double. It does.

**Proposition 7.10 (Boyle’s law).**

For a fixed amount of gas held at constant temperature, [pressure](#def-g10-pressure-pressure) and volume are inversely proportional:

$$
P \, V = \text{constant},
\qquad\text{i.e.}\qquad
P_1 V_1 = P_2 V_2 .
$$

**Proof.** *Admitted at this level.* ∎

**Remark 7.11.**

An experimental fact at this level — the Year 1 volume derives it from the collision picture. The experiment (a sealed syringe of air, a [pressure](#def-g10-pressure-pressure) sensor, $20{}^{\circ}\mathrm{C}$):

| $V$ ($\mathrm{mL}$) | 60 | 50 | 40 | 30 | 20 |
| --- | --- | --- | --- | --- | --- |
| $P$ ($\mathrm{kPa}$) | 100 | 120 | 150 | 200 | 300 |
| $P \times V$ ($\mathrm{kPa}\,\mathrm{mL}$) | 6000 | 6000 | 6000 | 6000 | 6000 |

![The syringe data fall on the hyperbola P = 6000/V: halve the volume, double the pressure.](https://one-course.com/images/onecourse/chapters/physics-2/g10-pressure/fig-500a8f85aedb.svg)

*The syringe data fall on the hyperbola $P = 6000/V$: halve the volume, double the [pressure](#def-g10-pressure-pressure).*

**Example 7.12 (A squeezed syringe).**

Air occupies $4.0\,\mathrm{L}$ at $100\,\mathrm{kPa}$; compressed slowly (so the temperature stays constant) to $250\,\mathrm{kPa}$, it occupies $V_2 = P_1 V_1 / P_2 = 400/250 = 1.6\,\mathrm{L}$.

## 7.5 Diving: pressure put to work

**Method 7.13 (Absolute pressure at depth).**

For a diver at depth $h$ in the sea ($\rho \approx 1025\,\mathrm{kg}/\mathrm{m}^{3}$, so $\rho g h \approx 101\,\mathrm{kPa}$ per $10\,\mathrm{m}$), compute $P = P_0 + \rho g h$ — and in [Boyle’s law](#prop-g10-pressure-boyle) always use this *absolute* [pressure](#def-g10-pressure-pressure), never the [gauge](#thm-g10-pressure-depth) term alone. A diver sits under $2$ atmospheres at $10\,\mathrm{m}$, $3$ at $20\,\mathrm{m}$, $5$ at $40\,\mathrm{m}$.

![Absolute pressure in the sea: one atmosphere at the surface, plus one more for every ten metres.](https://one-course.com/images/onecourse/chapters/physics-2/g10-pressure/fig-2025d52a8678.svg)

*[Absolute pressure](#met-g10-pressure-dive) in the sea: one atmosphere at the surface, plus one more for every ten [metres](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit).*

**Remark 7.14 (Never hold your breath).**

A regulator delivers air at ambient [pressure](#def-g10-pressure-pressure): lungs fill normally at any depth. But fill $6.0\,\mathrm{L}$ at $20\,\mathrm{m}$ ($302\,\mathrm{kPa}$), hold that breath and surface: Boyle demands $6.0 \times 302/101 \approx 18\,\mathrm{L}$ — three times what a chest holds, and lungs tear well before that, even on a few [metres](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of ascent. Hence scuba’s first rule: *never hold your breath while ascending* — keep breathing and the expanding air simply flows out. A free diver is safe: her air was taken at the surface, so on the way up it only re-expands to its original volume.

**Remark 7.15 (Decompression).**

Breathing [high-pressure](#def-g10-pressure-pressure) air also dissolves extra nitrogen in the blood, like gas in a capped soda bottle; surface too fast and it fizzes into bubbles inside the body — decompression sickness. Hence the slow ascent (about $10\,\mathrm{m}$ per minute) and shallow stops, which let the gas leave quietly through the lungs. The theory waits for the Year 1 volume; the rule of conduct belongs to every diver.

## 7.6 Exercises

**Exercise 7.1 ★.**

A crate weighing $600\,\mathrm{N}$ rests on a face of $0.50\,\mathrm{m} \times 0.30\,\mathrm{m}$, then on a face of $0.30\,\mathrm{m} \times 0.20\,\mathrm{m}$. Compute both [pressures](#def-g10-pressure-pressure) on the ground. Which orientation is better on soft sand?

**Solution of Exercise 7.1.**

Large face: $S = 0.15\,\mathrm{m}^{2}$, $P = 600/0.15 = 4.0\,\mathrm{kPa}$. Small face: $S = 0.060\,\mathrm{m}^{2}$, $P = 10\,\mathrm{kPa}$. On sand, the large face: lower [pressure](#def-g10-pressure-pressure), less sinking.

**Exercise 7.2 ★.**

Express $0.25\,\mathrm{MPa}$ in $\mathrm{kPa}$ and in $\mathrm{Pa}$. Which presses harder: $3.0\,\mathrm{N}$ on $2.0\,\mathrm{cm}^{2}$, or $60\,\mathrm{N}$ on $500\,\mathrm{cm}^{2}$?

**Solution of Exercise 7.2.**

$0.25\,\mathrm{MPa} = 250\,\mathrm{kPa} = 2.5 \times 10^{5}\,\mathrm{Pa}$. $3.0/2.0 \times 10^{-4} = 15\,\mathrm{kPa}$ beats $60/0.050 = 1.2\,\mathrm{kPa}$: the small [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) presses harder.

**Exercise 7.3 ★.**

Compute the [gauge pressure](#thm-g10-pressure-depth) $\rho g h$ and the [absolute pressure](#met-g10-pressure-dive) at the bottom of a $3.0\,\mathrm{m}$ deep swimming pool ($\rho = 1000\,\mathrm{kg}/\mathrm{m}^{3}$, $P_0 = 101\,\mathrm{kPa}$).

**Solution of Exercise 7.3.**

$\rho g h = 1000 \times 9.81 \times 3.0 \approx 29\,\mathrm{kPa}$; absolute $P = 101 + 29 = 130\,\mathrm{kPa}$.

**Exercise 7.4 ★.**

Air occupies $4.0\,\mathrm{L}$ at $100\,\mathrm{kPa}$, temperature constant. Its volume at $250\,\mathrm{kPa}$? At $50\,\mathrm{kPa}$?

**Solution of Exercise 7.4.**

$V_2 = 100 \times 4.0 / 250 = 1.6\,\mathrm{L}$; at $50\,\mathrm{kPa}$, $V_2 = 400/50 = 8.0\,\mathrm{L}$.

**Exercise 7.5 ★.**

A $65\,\mathrm{kg}$ skater glides on one blade of contact area $12\,\mathrm{cm}^{2}$, then stands on soles totalling $360\,\mathrm{cm}^{2}$. Compute both [pressures](#def-g10-pressure-pressure) and their ratio.

**Solution of Exercise 7.5.**

$F = 65 \times 9.81 \approx 638\,\mathrm{N}$. Blade: $638/1.2 \times 10^{-3} \approx 5.3 \times 10^{5}\,\mathrm{Pa} = 530\,\mathrm{kPa}$. Soles: $638/0.036 \approx 18\,\mathrm{kPa}$. Ratio $\approx 30$: the blade bites into the ice, the soles do not.

**Exercise 7.6 ★★.**

A thumb pushes a drawing pin with $20\,\mathrm{N}$. The head has area $1.2\,\mathrm{cm}^{2}$, the tip $0.010\,\mathrm{mm}^{2}$. Compute the [pressures](#def-g10-pressure-pressure) on thumb and wall; why is only the wall pierced?

**Solution of Exercise 7.6.**

Thumb: $20/1.2 \times 10^{-4} \approx 1.7 \times 10^{5}\,\mathrm{Pa} = 170\,\mathrm{kPa}$. Wall: $20/1.0 \times 10^{-8} = 2.0 \times 10^{9}\,\mathrm{Pa}$. The same $20\,\mathrm{N}$ acts on an area $12\,000$ times smaller at the tip, so only there does the [pressure](#def-g10-pressure-pressure) exceed what the material can withstand.

**Exercise 7.7 ★★.**

A submarine hatch of area $0.50\,\mathrm{m}^{2}$ sits at $30\,\mathrm{m}$ depth in sea water ($\rho = 1025\,\mathrm{kg}/\mathrm{m}^{3}$); inside, the air is at [atmospheric pressure](#def-g10-pressure-atmospheric). Compute the net [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) holding it shut, and the mass whose [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) equals it.

**Solution of Exercise 7.7.**

$\Delta P = \rho g h = 1025 \times 9.81 \times 30 \approx
3.0 \times 10^{5}\,\mathrm{Pa}$, so $F = 3.0 \times 10^{5} \times 0.50 \approx 1.5 \times 10^{5}\,\mathrm{N}$ — the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) of $F/g \approx 1.5 \times 10^{4}\,\mathrm{kg}$, fifteen tonnes. No crew opens that outward against the sea.

**Exercise 7.8 ★★.**

A sensor $5.00\,\mathrm{m}$ deep in an unknown liquid reads a [gauge pressure](#thm-g10-pressure-depth) of $66.2\,\mathrm{kPa}$. Find the density. Is it fresh water?

**Solution of Exercise 7.8.**

$\rho = \dfrac{P}{g h} = \dfrac{66.2 \times 10^{3}}{9.81 \times 5.00}
\approx 1350\,\mathrm{kg}/\mathrm{m}^{3}$. Not fresh water ($1000\,\mathrm{kg}/\mathrm{m}^{3}$): a markedly denser liquid, e.g. a concentrated brine.

**Exercise 7.9 ★★.**

How tall would Torricelli’s barometer be with water ($\rho = 1000\,\mathrm{kg}/\mathrm{m}^{3}$) instead of mercury ($\rho = 13\,600\,\mathrm{kg}/\mathrm{m}^{3}$)? Recover both heights from $\rho g h = P_0$; why did mercury win?

**Solution of Exercise 7.9.**

$h = P_0/(\rho g)$: water $1.01 \times 10^{5}/9810 \approx 10.3\,\mathrm{m}$; mercury $1.01 \times 10^{5}/(13600 \times 9.81) \approx 0.76\,\mathrm{m}$. A $76\,\mathrm{cm}$ tube fits on a desk; a $10\,\mathrm{m}$ water barometer needs a stairwell — mercury’s density won.

**Exercise 7.10 ★★.**

Using the syringe data of [Proposition 7.10](#prop-g10-pressure-boyle) ($PV = 6000\,\mathrm{kPa}\,\mathrm{mL}$), predict the volume at $240\,\mathrm{kPa}$ and the [pressure](#def-g10-pressure-pressure) at $75\,\mathrm{mL}$. What curve is $P$ against $V$? And $P$ against $1/V$?

**Solution of Exercise 7.10.**

$V = 6000/240 = 25\,\mathrm{mL}$; $P = 6000/75 = 80\,\mathrm{kPa}$. $P$ against $V$ is a hyperbola; $P$ against $1/V$ is a straight line through the origin, of slope $6000\,\mathrm{kPa}\,\mathrm{mL}$.

**Exercise 7.11 ★★.**

A diver at $20\,\mathrm{m}$ releases a bubble of volume $0.50\,\mathrm{cm}^{3}$. What is its volume just below the surface? (Constant temperature; sea water adds $101\,\mathrm{kPa}$ per $10\,\mathrm{m}$.)

**Solution of Exercise 7.11.**

At $20\,\mathrm{m}$: $P = 101 + 2 \times 101 = 303\,\mathrm{kPa}$, three atmospheres. At the surface, $V = 0.50 \times 303/101 = 1.5\,\mathrm{cm}^{3}$: the bubble triples.

**Exercise 7.12 ★★★.**

A suction cup of area $12\,\mathrm{cm}^{2}$ is pressed flat against a ceiling.

1. Perfect vacuum inside: what [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) does the atmosphere exert on the cup, and what hanging mass can it hold?
2. A real cup keeps $20\,\mathrm{kPa}$ of air inside. What mass now?

**Solution of Exercise 7.12.**

*1.* $F = 1.01 \times 10^{5} \times 1.2 \times 10^{-3} \approx 121\,\mathrm{N}$, holding $m = 121/9.81 \approx 12\,\mathrm{kg}$.

*2.* $\Delta P = 101 - 20 = 81\,\mathrm{kPa}$: $F = 8.1 \times 10^{4} \times 1.2 \times 10^{-3} \approx 97\,\mathrm{N}$, about $9.9\,\mathrm{kg}$.

**Exercise 7.13 ★★★.**

A scuba tank holds $12\,\mathrm{L}$ of air at $200\,\mathrm{bar}$ ($1\,\mathrm{bar}$ = $100\,\mathrm{kPa}$ $\approx 1$ atmosphere); $50\,\mathrm{bar}$ must stay as reserve. A diver breathes $15\,\mathrm{L}$ per minute at ambient [pressure](#def-g10-pressure-pressure). How many litres of [surface-pressure](#def-g10-pressure-pressure) air are usable? How long does the tank last at $30\,\mathrm{m}$ ([absolute pressure](#met-g10-pressure-dive) $\approx 4\,\mathrm{bar}$)? And at $10\,\mathrm{m}$?

**Solution of Exercise 7.13.**

Boyle: $12 \times (200 - 50) = 1800\,\mathrm{L}$ of [surface-pressure](#def-g10-pressure-pressure) air is usable. At $30\,\mathrm{m}$ each breath is drawn at $4\,\mathrm{bar}$, so $15\,\mathrm{L}/\mathrm{min}$ at ambient [pressure](#def-g10-pressure-pressure) costs $15 \times 4 = 60\,\mathrm{L}/\mathrm{min}$ of surface air: $1800/60 =
30\,\mathrm{min}$. At $10\,\mathrm{m}$ ($2\,\mathrm{bar}$): $30\,\mathrm{L}/\mathrm{min}$, so $60\,\mathrm{min}$. Depth is paid for in air.

**Exercise 7.14 ★★★.**

A free diver’s lungs hold $6.0\,\mathrm{L}$ at the surface and cannot shrink below the residual volume $1.5\,\mathrm{L}$. At what sea depth ($\rho = 1025\,\mathrm{kg}/\mathrm{m}^{3}$) is that limit reached? (Once declared the limit of free diving; records now pass $100\,\mathrm{m}$, blood shifting into the chest taking up the missing volume.)

**Solution of Exercise 7.14.**

$P = P_0 \dfrac{V_0}{V} = 101 \times \dfrac{6.0}{1.5} =
404\,\mathrm{kPa}$. With $\rho g \approx 10.1\,\mathrm{kPa}/\mathrm{m}$, $h = (404 - 101)/10.1 \approx 30\,\mathrm{m}$.

**Exercise 7.15 ★★★.**

In a hydraulic lift, a liquid transmits [pressure](#def-g10-pressure-pressure) unchanged from a small piston ($s = 2.0\,\mathrm{cm}^{2}$) to a large one ($S = 400\,\mathrm{cm}^{2}$) carrying a $1200\,\mathrm{kg}$ car.

1. What [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on the small piston holds the car up? What [pressure](#def-g10-pressure-pressure) does the liquid carry?
2. To raise the car $2.0\,\mathrm{cm}$ , how far must the small piston travel? Compare the work done on each piston.

**Solution of Exercise 7.15.**

*1.* [Weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) $F = 1200 \times 9.81 \approx 1.18 \times 10^{4}\,\mathrm{N}$; [pressure](#def-g10-pressure-pressure) $P = F/S = 1.18 \times 10^{4}/0.040 \approx 2.9 \times 10^{5}\,\mathrm{Pa}$; small piston: $f = P s = 2.9 \times 10^{5} \times 2.0 \times 10^{-4} \approx 59\,\mathrm{N}$ — the [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) is divided by $S/s = 200$.

*2.* Liquid volume is conserved: $400 \times 2.0 = 800\,\mathrm{cm}^{3}$ must come from the small cylinder, which travels $800/2.0 = 400\,\mathrm{cm} = 4.0\,\mathrm{m}$. Work: $59 \times 4.0 \approx 235\,\mathrm{J}$ on one side, $1.18 \times 10^{4} \times 0.020 \approx 235\,\mathrm{J}$ on the other — the lift trades distance for [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force), never work.

## 7.7 Problem: One breath down, one breath up

**Problem 7.1.**

Weekend problem — one breath down, one breath up: the physics of a free dive and a scuba ascent, from mask squeeze to the rule that the last ten metres are the most dangerous

Lena free-dives: one surface breath, down to $30\,\mathrm{m}$ and back. Marco scuba-dives beside her, breathing from a tank. Same sea ($\rho = 1025\,\mathrm{kg}/\mathrm{m}^{3}$, $g = 9.81\,\mathrm{N}/\mathrm{kg}$, $P_0 = 101\,\mathrm{kPa}$), same $6.0\,\mathrm{L}$ lungs, opposite dangers — all governed by $P = P_0 + \rho g h$ and $PV$ = constant (temperature constant throughout).

**Part I — The [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) of water.**

1. Compute $\rho g$ for sea water in $\mathrm{kPa}$ per [metre](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) , and check the rule of thumb: one atmosphere per $10\,\mathrm{m}$ .
2. Compute the [absolute pressure](#met-g10-pressure-dive) at $10\,\mathrm{m}$ , $20\,\mathrm{m}$ , $30\,\mathrm{m}$ and $40\,\mathrm{m}$ .
3. Express each as a multiple of the surface [pressure](#def-g10-pressure-pressure) .
4. Lena’s mask covers $150\,\mathrm{cm}^{2}$ . At $30\,\mathrm{m}$ , if the air inside it were still at surface [pressure](#def-g10-pressure-pressure) , what net [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) would the water exert? (Divers exhale into the mask through the nose to prevent this *mask squeeze* .)
5. An eardrum has area about $0.60\,\mathrm{cm}^{2}$ . Compute the net [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) on it at $3.0\,\mathrm{m}$ if the middle ear stays at surface [pressure](#def-g10-pressure-pressure) ; why do ears hurt in a mere pool, and what must “equalizing” achieve?

**Part II — One breath down.** Lena leaves the surface with $6.0\,\mathrm{L}$ of air in her lungs.

6. Using [Boyle’s law](#prop-g10-pressure-boyle) with [absolute pressures](#met-g10-pressure-dive) , compute her lung volume at $10\,\mathrm{m}$ , $20\,\mathrm{m}$ and $30\,\mathrm{m}$ .
7. At what depth is her lung volume halved?
8. Is lung volume proportional to depth? Describe the curve of $V$ against [absolute pressure](#met-g10-pressure-dive) $P$ .
9. Her residual volume is $1.5\,\mathrm{L}$ : lungs cannot shrink further. Show that this limit is reached near $30\,\mathrm{m}$ .
10. Free-diving records nonetheless exceed $100\,\mathrm{m}$ : what fills the missing volume? (What else can flow into the chest?)

**Part III — One breath up.** At $30\,\mathrm{m}$, Marco’s regulator fills his $6.0\,\mathrm{L}$ lungs.

11. At what [pressure](#def-g10-pressure-pressure) does the regulator deliver that air? How many times denser is it than surface air?
12. Marco panics, holds his breath, and rises to $10\,\mathrm{m}$ : what volume does his trapped air demand there?
13. What volume at the surface — how many times his capacity?
14. State the scuba diver’s first rule, and why breathing normally removes the danger.
15. Lena also holds her breath from $30\,\mathrm{m}$ to the surface, yet her lungs are perfectly safe. Explain the asymmetry.

**Part IV — Bubbles, and the last ten [metres](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit).**

16. Marco releases a $1.0\,\mathrm{cm}^{3}$ bubble at $40\,\mathrm{m}$ . Compute its volume at $30\,\mathrm{m}$ , $20\,\mathrm{m}$ , $10\,\mathrm{m}$ and just below the surface.
17. For each $10\,\mathrm{m}$ stage of the rise, compute the factor by which the bubble grows. Where is the growth fastest?
18. Divers ascend at most $10\,\mathrm{m}$ per minute: how long from $40\,\mathrm{m}$ , and in which minute does any trapped or dissolved gas expand the most?
19. At depth Marco’s blood dissolves extra nitrogen, like gas in a capped soda bottle. What does a too-fast ascent do, and how do the slow ascent and a $5\,\mathrm{m}$ safety stop prevent it?
20. Punchline: in one sentence each, give the free diver’s verdict, the scuba diver’s verdict, and the quantified reason the last ten [metres](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) of any ascent deserve the most respect.

**Solution of Problem 7.1.**

**1.** $\rho g = 1025 \times 9.81 \approx 10.1\,\mathrm{kPa}$ per [metre](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit), so $10\,\mathrm{m}$ of sea water add $\approx 101\,\mathrm{kPa}$: one atmosphere per ten [metres](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit), almost exactly.

**2.** $P = 101 + 10.1\,h$: $202\,\mathrm{kPa}$ at $10\,\mathrm{m}$, $302\,\mathrm{kPa}$ at $20\,\mathrm{m}$, $403\,\mathrm{kPa}$ at $30\,\mathrm{m}$, $503\,\mathrm{kPa}$ at $40\,\mathrm{m}$.

**3.** $2.0$, $3.0$, $4.0$ and $5.0$ times $P_0$.

**4.** $\Delta P = 403 - 101 = 302\,\mathrm{kPa}$ over $0.015\,\mathrm{m}^{2}$: $F = 3.02 \times 10^{5} \times 0.015 \approx
4.5 \times 10^{3}\,\mathrm{N}$ — the [weight](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-weight) of nearly half a tonne on the face.

**5.** $\Delta P = 10.1 \times 3.0 \approx 30\,\mathrm{kPa}$; $F = 3.0 \times 10^{4} \times 6.0 \times 10^{-5} \approx 1.8\,\mathrm{N}$ — a finger pressed on the eardrum, at pool depth already. Equalizing pushes air into the middle ear until the inside [pressure](#def-g10-pressure-pressure) matches the water’s.

**6.** $V = 6.0 \, P_0/P$: $6.0 \times 101/202 = 3.0\,\mathrm{L}$; $606/302 = 2.0\,\mathrm{L}$; $606/403 = 1.5\,\mathrm{L}$.

**7.** Halved when $P = 2P_0$, i.e. at $10\,\mathrm{m}$ — in the very first stretch of the dive.

**8.** No: $V = 606/P$ is a hyperbola in the [absolute pressure](#met-g10-pressure-dive) (and $P$, not $h$, is what [Boyle’s law](#prop-g10-pressure-boyle) sees); each extra $10\,\mathrm{m}$ removes less volume than the previous one.

**9.** $V = 1.5\,\mathrm{L}$ requires $P = 101 \times 6.0/1.5 = 404\,\mathrm{kPa}$, i.e. $h = 303/10.1 \approx 30\,\mathrm{m}$ ([Exercise 7.14](#exo-g10-pressure-14)).

**10.** Blood: plasma shifts into the vessels of the chest, incompressible, and occupies the volume the air no longer fills.

**11.** At ambient [pressure](#def-g10-pressure-pressure), $403\,\mathrm{kPa}$. By Boyle the same air at $101\,\mathrm{kPa}$ would fill $4$ times the volume: it is $4.0$ times denser than surface air.

**12.** $V = 6.0 \times 403/202 \approx 12\,\mathrm{L}$ — double, after only $20\,\mathrm{m}$ of rise.

**13.** $V = 6.0 \times 403/101 \approx 24\,\mathrm{L}$: four times his lung capacity.

**14.** *Never hold your breath while ascending.* With the airway open, the expanding air flows out through the regulator, and lung volume never exceeds $6.0\,\mathrm{L}$.

**15.** Lena’s $1.5\,\mathrm{L}$ at $30\,\mathrm{m}$ *is* her surface $6.0\,\mathrm{L}$ compressed: on the way up it re-expands to exactly $6.0\,\mathrm{L}$, never beyond. Marco’s $6.0\,\mathrm{L}$ were *taken* at $403\,\mathrm{kPa}$: they have $24\,\mathrm{L}$ of surface air in them.

**16.** $V = 1.0 \times 503/P$: $1.25\,\mathrm{cm}^{3}$ at $30\,\mathrm{m}$, $1.67\,\mathrm{cm}^{3}$ at $20\,\mathrm{m}$, $2.49\,\mathrm{cm}^{3}$ at $10\,\mathrm{m}$, $4.98\,\mathrm{cm}^{3}$ at the surface.

**17.** Stage factors: $503/403 = 1.25$, then $1.33$, $1.49$, and $202/101 = 2.0$ for the last ten [metres](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) — growth accelerates as the surface nears.

**18.** $40/10 = 4.0\,\mathrm{min}$. The last minute: from $10\,\mathrm{m}$ to the surface the [absolute pressure](#met-g10-pressure-dive) halves, the biggest relative drop of the whole ascent.

**19.** A fast ascent lets the dissolved nitrogen fizz into bubbles inside blood and joints (decompression sickness); ascending slowly and pausing near $5\,\mathrm{m}$ keeps the gas dissolved long enough to leave quietly through the lungs.

**20.** Free diver: her air only returns to its original volume — Boyle protects her. Scuba diver: air taken at depth multiplies on the way up — breathe, never hold. And the last ten [metres](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) double every trapped volume ($\times 2.0$, against $\times 1.25$ down at $40\,\mathrm{m}$): the closer the surface, the slower you should approach it.