- Tracy Gunn

# Understanding Boyle`s Law in Physics IDC Dive Theory

Updated: Jun 27, 2021

**Boyle`s law ***states that "If temperature remains constant both volume and density of a gas are affected in proportion to the atmospheric pressure and inversely to each other."*

**Sir Roberto Boyle**

__Sir Roberto Boyle__ was a devout seventeenth cenutry Irish scientist and his work was greatly influenced by Torricelli. __Evangelista Torricelli__ who was student of Galileo (whose work Boyle also studied) and is best known for inventing the Barometer. While Toricelli determined the pressure of the atmosphere, Boyle aimed to discover what happened to volumes of air when the pressure changed. In 1662 he formulated Boyles Law.

**Boyle`s Law**

But that is not his only contribution to diving. In 1670 he was the first person to document decompression sickness. He placed a viper (one of his many experiements with animals) in a vacuum jar and decreased the pressure by removing the air, obviously rather rapidly. "I once observed a viper furiously tortured in our exhausted receiver… that had manifestly a conspicuous bubble moving to and fro in the waterish humour of one of its eyes."(3) Upon increasing the pressure, the bubble disappeared and the snake seemed happy as it could be in a small jar. He put forward no theory as he had no idea what had happened and it would be a good two centuries until there were any further discoveries on decompression sickness but it was from this experiment that the idea began to arise .

Boyles law is perhaps the most useful and important gas law in diving. So much so that it is the first question in the first knowledge review in your open water course (although you were probably not aware of that at the time).

**Applications**

This law affects you on every dive. It affects your BCD, mask, sinuses, ears and lungs (think of these as little gas-filled containers). It affects your wetsuit, with its millions of tiny bubbles in its neoprene, and your dry suit with its one big one. It can help you calculate how fast you will consume air in varying depths (and hence how much time you have) or how much usable gas volume you have in a cylinder at a given pressure. It explains why you should not hold your breath while scuba diving, why you need to add air to your BCD upon descent and release air from it upon ascent and why you need to equalise your ears.

Understanding this law will also help you understand and avoid diving injuries such as ear squeezes, lung over expansion injuries and decompression sickness.

If a diver takes a lungful of air underwater, that air will expand in the lungs upon ascent. From 10msw to surface the pressure halves. If a dive is holding their breath, the air in their lungs will double in volume causing a ruptured lung. This is Boyles Law in action and the reason why the golden rule of diving is: Never, never hold your breath!

**Understanding Boyles`s Law**

The law states the **PRESSURE AND VOLUME OF A GAS ARE INVERSELY PROPORTIONAL TO EACH OTHER ** and **PRESSURE AND DENSTIY ARE DIRECTLY PROPORTIONAL. **We assume that there is no change in temperature as temperature changes the pressure/volume relationship which we discuss in Charle`s law.

Put simply, the greater the pressure, the less the volume and the greater the density. As you increase the pressure, the volume will decrease in proportion to the amount of pressure and the same number of molecules must occupy less space, meaning it becomes denser. The density is therefore in direct proportion to the pressure.

This explains why we beathe more air at depth. The air volume and density inside a scuba tank will not change when we descend as the tank is inflexible, but the moment it leaves that tank, the air compresses and becomes more dense, entering our flexible lungs. Physiologically, we need to be able to breathe at the same pressure as that which surrounds us.

At 2 bar, we will breathe twice the amount of molecules and our air supply willl only last half as long. At 3 bar, three times the molecules so one-third the time, and so on.

If we pump air down from the surface, the same rules will apply to fill a tank at depth in that the molecules of air compress in direct relation to the pressure

Capillary gauges are tubes with a very small diameter, smaller than the surface tension of a drop of water which means you cannot pour water in or out of the tube without force. They are very inexpensive, being a simple clear piece of tubing that is sealed at one end and open at the other. This tube acts the same way an inverted container filled with air would. Ambient pressure pushes water into the tube and at 10mt (2 bar) it is half the length of the tube (10 mt being double the pressue of the surface and half the volume), 20mt a third and so on, in accordance with Boyles Law. They are hard to read deeper than 10 mt. Some divers prefer capillary gauges for altitude diving because it takes into account variations in atmospheric pressure.

Remember this fo your exams

**A -Altitude**

**B -Boyle`s Law**

**C -Capillay**

**pressure=density**

** **

Boyles law is of great importance to tech and cave divers who dive long and/or deep dives. They must use twin tanks, high capacity and/or station extra tanks along their dive. Imagine; a tank that lasts 2 hours at the surface will only last around 10 minutes at 100 meters, so gas supply is of vital importance. Even then, calculating air supply is only an estimate as it can be greatly influenced by factors such as cold, stress, exertion and depth changes. Closed circuit scuba recirculates gas exhaled and that is why there is a growing interest in this type of diving.

DO NOT CONFUSE BOYLE`S LAW WITH __ARCHIMEDES` PRINCIPLE____,__ which is about buoyancy, not pressure.

So lets start with pressure. Make sure you understand how to calculate pressure and the difference between salt water (10m/atm) and fresh water (10.3m/atm) before going any further. For calculating changes in volume and density, we always use absolute pressure so don`t foget to add in that 1atm at the surface. "If we desire to isolate a specific amount of pressure inside a process, regardless of what happens in the atmosphere, then we should use a system of absolute pressure." (5)

If you are unsure, refresh your knowledge with our blog on __Understanding Pressure__ and also test your knowledge with the inbuilt exam.

Scientists largely use the __metric system__ (International Systems of Units or SI) as it uses only one unit for each type of measurerment (meters for example covering kilo-meters, centi-meters, etc). The metric system is used by 96% of the world’s population and by all countries for scientific use. As it is the simpler system to use, I will mostly be using it here.

**PART 1 **

**To the surface and back - single depth changes**

**How to examine the questions for Boyles Law**

**STEP 1**

**Find the unit of measurement**

Look at the answers. The unit of measurement will be stated there (always multiple choice in IE final exams)

What is the question asking for?

Minutes

PSI or Bar

Litres

Find the number from inside the question that is in that unit and write it down

**STEP 2**

**Think about the question**

**More = ×**

**Less = ÷**

__TIME (minutes)__

The shallower you are, the slower you consume air (you have more time) **×**

The deeper you go, the faster you consume air. (you have less time). **÷**

**More minutes ×, less minutes ÷**

__Volume (Litres)__

If you take a balloon up from depth, its size increases (more volume) **×**

If you take a balloon down to depth, its size decreases (less volume) **÷**

**Bigger volume ×, less volume ÷**

__Air (PSI or Bar)__

The shallower you are, the less molecules of air (less dense) you breathe with each breath, the amount of air consumed in PSI/Bar will decrease **÷**

The deeper you go, the more molecules of air (more dense) you breathe with each breath, the amount of air consumed in PSI/Bar will increase **×**

**More PSI/Bar ×, Less PSI/Bar ÷**

**This is THE most important step and where the most errors occur. Think it through first. **

**STOP, BREATHE, THINK, ACT**

**STEP 3**

**Find the surrounding pressure**

READ THE QUESTION and find the depth in meters(or feet)

Determine the pressure at that depth

think Saltwater (10mt/33ft) vs Freshwater (10.3mt/34ft).

__Always +1 __(you are determining absolute pressure)

Write that pressure down

**STEP 4**

**Find the answer**

Use the unit of measurement (step 1) and multiply or divide (step 2) the surrounding pressure (step 3) = The answer

**EXAMPLES**

__Air consumption in minutes__

**Q** It takes **60 minutes** to breathe through a tank of air at the surface. If all factors remain the unchanged, how long will it take to breathe through the same tank at 20 mt?

A 180 minutes

B 60 minutes

C 30 minutes

**D 20 minutes**

**STEP 1 **Answer is asking for minutes. So, we look for minutes in the question and write it down **60 minutes**

**STEP 2** Think! Do you use air faster at 20meters than at the surface? We know that we use air much faster at depth, therefore the answer will be **LESS TIME** than **60 minutes** **(÷)**

**STEP 3** The depth is 20 meters. 20÷10+1=**3ATA**

**STEP 4** 60 minutes ÷ 3 ata = **20 minutes (therefore the answer is D)**

__Volume (flexible container)__

**Q** If we have a balloon with a volume of **8 litres** at the surface, what would its volume be at 30mt?

A 32 litres

B 24 litres

**C 2 litres**

D 2.6 litres

**STEP 1** Answer is asking for litres. So, we look for litres in the question. **8 litres**

**STEP 2** Think! Does the volume increase or decrease as we descend? The air becomes denser so the volume becomes smaller, therefore the answer will be **LESS VOLUME** than **8 litres** **(÷)**

**STEP 3** The depth is 30 meters. 30÷10+1=**4ATA**

**STEP 4** 8 litres ÷ 4 ata = **2litres (therefore the answer is C)**

__Air consumption in PSI/Bar__

**Q** A diver breathes through **60 Bar of air in 60 minutes **at the surface. If all factors remain unchanged, how many BAR will the diver breathe in 40 minutes at 30 meters?

**A 160 Bar**

B 120 Bar

C 10 Bar

D 45 Bar

**STEP 1** Answer is looking for BAR. So, we look for BAR in the question and write it down

**60 bar in 60 minutes **__(here is the trick. We need to know how many bar per minute)__

60÷60=**1BAR per min**

**STEP 2** Think! Do we consume more air at depth or less? The air is denser, so we use more air with each breath. Therefore, the answer will be **MORE BAR** than **1 bar per min (X)**

**STEP 3** The depth is 30meters. 30÷10+1=** 4ATA**

**STEP 4** The answer is multiply bar per min by ATA and then again for 40 min!!!

1 bar per min X 4 ata = 4 bar per min

4 bar per min X 40 min=** 160Bar (therefore the answer is A)**

Test your knowledge __CLICK HERE__

**part 2 Multiple depth changes**

The easiest way to deal with problems of changing levels, regardless of whether it is volume (or time) or density is to go from the first level to the surface, find its surface value and then go down to the second level.

**OR**

Use a transposed formula

**V1P1=V2P2 OR** **D1/P1=D2/P2**

**V2=P1V1/P2** **D2=D1/P1×P2**

__VOLUME__

__VOLUME__

Volume becomes smaller as you descend and gets larger as you ascend. Think of how you need to equalise your ears as you go down. This is why. As you increase the pressure, the volume will decrease in proportion (inverse relationship).

__ANSWER IS____LITRES__

**FILL IN ALL KNOWN VALUES (IN BOLD) **

First take volume to surface ** **

**- multiply volume by surrounding pressure** __(multiply to go up)__

Then take volume back down to new depth

**- divide surface volume by the atmospheres of the depth you are going down to** __(divide to go down, volume gets less at depth)__

**Example 1: **A flexible container with a volume of ** 35 litres** is at

**of**

__28m__**, is taken up to**

__fresh water__**of**

__24m__**, what would the new volume be?**

__Sea Water__First work the ata at each depth (make sure you note if it is salt water or fresh water and do not forget to add the 1 surface atmosphere).

To take it to the surface, we take the volume and multiply it by the surrounding pressure

**(greater volume at surface)**35 litres × 3.7 ata = 129.5 litres at surfaceWe then divide it by the ata of the depth we go down to

**(less volume at depth)**129.5 litres ÷ 3.4 ata = 38 litres

**Answer is 38 litres **

**OR**

V=Volume (or time) and P=Pressure

V1P1=V2P2

∴V2=P1V1÷P2

V1=35 litres

P1=3.7ata (28÷10.3+1=3.7ata)

**V2 = ? litres**

P2 = 3.4 ata (24÷10+1=3.4ata)

V2 = 35×3.7÷3.4 = **38 litres**

__DENSITY__ (= atmospheres)

__DENSITY__

Air becomes denser (more pressure) as you descend underwater. A diver will use more air due to the increased (more) pressure (He’s going deeper…more pressure) A diver will use less air due to the decreased (less) pressure as he ascends. (He’s going up… less pressure)

(density is directly proportional to pressure- more pressure, more dense)

__ANSWER IS____BAR__

**FILL IN ALL KNOWN VALUES FIRST (IN BOLD)**

First take air consumption (bar/psi) to surface

**- divide bar/psi by surrounding atmospheres to get Surface Air Consumption (SAC) **__(Divide to go up)__

Then take down to new depth to find air consumption at that depth

**- Multiply SAC by the atmospheres of the depth you are going down to **__(Multiply to go down, remember you use more air at depth)__

**Example: **A diver consumes **4bar per minute** at **10mt in seawater** with a given cylinder. Using the same cylinder, what is the diver’s consumption rate at **33mt?**

First work the ata at each depth (make sure you note if it is salt water or fresh water and do not forget to add the 1 surface atmosphere).

To take it to the surface we take the air consumption (bar/psi) and divide it by the surounding pressure

**(consume less bar/psi at surface)**4 bar ÷ 2 ata = 2 bar at surface (SAC)We then multiply it by the depth we go down to

**(consume more bar/psi at depth)**2 bar × 4.3 ata = 8.6 bar

**The answer is 8.6 Bar**

**OR**

Although I cannot find an official equation for density, logically the equation would be like this

D=Density (Air -Bar/PSI) and P=Pressure

D1÷P1=D2÷P2

∴D2=D1÷P1×P2

D1 = 4 bar

P1 = 2 ata (10÷10+1=2ata)

**D2 = ? bar**

P2 = 4.3 ata (33÷10+1=4.3ata)

D2 = 4÷2×4.3 = **8.6 bar**

* note:- If the question states Bar consumption in "X" minutes. Divide until you determine Air Consumption per Minute and then begin the equation i.e. 120 bar per hour is 120/60= 2 bar per min

Test your knowledge __CLICK HERE__ for part 2

Test 1 is single level depth changes (to the surface and back)

32 questions

Test 2 is multiple level changes

16 questions

Check out our other blogs on

__Understanding ____Pressur____e__

(1) Wikipedia contributors. (2021, February 28). *Robert Boyle*. Wikipedia. https://en.wikipedia.org/wiki/Robert_Boyle

(2) West, J. B. (2005, January 1). *Robert Boyle’s landmark book of 1660 with the first experiments on rarified air*. Journal of Applied Physiology. https://journals.physiology.org/doi/full/10.1152/japplphysiol.00759.2004

(3) *Robert Boyle, Hyperbaric Chambers - A history of hyperbaric chambers and dive medicine, Midlands Recompression & Hyperbaric facilities - The Midlands Diving Chamber*.

(2019). Midlands Diving Chamber. http://midlandsdivingchamber.co.uk/index.php?id=history&page=3#:%7E:text=Boyle%20also%20first%20described%20the,fro%20in%20the%20waterish%20humour

(4) Hempleman, H. V. (1984). *Decompression Theory*. SpringerLink. https://link.springer.com/chapter/10.

(5) *Absolute vs gauge: choosing the right pressure control*. (2021, January 27). Alicat Scientific. https://www.alicat.com/knowledge-base/absolute-vs-gauge-choosing-the-right-pressure-control/

(6) *The Encyclopedia of Recreational Diving* (3rd ed.). (2008). PADI.

(7) *Divemaster Course Instructor Guide* (1999 edition). (2005). PADI.

(8) *Why teach the metric system (SI)?* (2000, March 1). US Metric System. https://usma.org/why-teach-the-metric-system-si

(9) *Benefits*. (2017, July 8). Think Metric! https://thinkmetric.uk/benefits/