Diving Instructor Courses on Koh Tao

Dive Physics During the IDC or Divemaster Course

Why Physics Matters

Physics is the study of how matter and energy behave whilst chemistry is simply the study of the composition, structure and properties of substances. The chemistry and physics of diving deal with how the behavior of matter and energy underwater affect us.

There are five forms of energy that we are concerned with regarding diving:

Heat (the more heat the greater the motion of molecules), Light (a form of electromagnetic radiation), Electrical, Chemical (stored within a substances molecular composition) and Mechanical (resulting from motion- kinetic energy, or the possibility of motion- potential energy).

Heat:

Water is able to conduct heat more efficiently than air because it is far denser than air. The molecules within water are closer together making it easier for heat (energy) to be transferred or conducted (more about this below). Water has a higher heat capacity compared to air.

Two hydrogen atoms bonded with one oxygen atom create one water molecule- H2O. Although a water molecule has an overall neutral charge, the position of the two hydrogen atoms on one side give it positive and negative ends. This makes water a polar molecule, which causes many of its unique properties. This polar nature also enables water to absorb more heat than other liquids.

Heat transmits from one substance to another by three means; conduction, convection and radiation. Conduction and convection affect you underwater.
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Conduction: refers to heat transmission by direct contact. Air is a good insulator – a substance that resists conduction. Water is an excellent conductor that not only holds more heat than air, but conducts heat away faster. This is because the molecules within water are denser and so heat is able to pass between molecules more easily in water than in air.

Convection: refers to heat transmission via fluids. Body heat also dissipates faster in water than it does in air because of convection. When a fluid becomes heated, it becomes less dense and tends to rise. As it rises cooler fluids flow in to replace it. This is a continuous flow that draws heat away from whatever the fluid surrounds causing heat transmission that’s faster than conduction alone. In diving, the water becomes heated by the skin within the wetsuit.

Radiation: is the heat transmission via electromagnetic waves. Heat in the form of radiation affects a diver the least underwater.
Divers are most affected by conduction underwater.
Water dissipates body heat more than 20 times faster than air or the body loses heat approximately 20 times faster in water than in air. Inversely, it can be said that air is able to conduct heat 0.05 times as fast as water; water is 20 times faster than air, so inversely it would be 1/20th or 0.05.

Light:

The behaviour of light changes as it passes through water, and so water can affect how far a diver can see, clarity, colours and apparent distances. Light is a form of electromagnetic energy.

Colour Absorption: Wavelengths relate to the amount of energy in light. Colours towards the red end of the spectrum have less energy than the colours at the blue end- infrared light is very low energy and ultraviolet is very high energy.

Even when very clear, water absorbs light passing through it, transforming it into heat. Wavelengths with less energy are absorbed more easily, and this is why red will disappear more quickly with depth. Following red, water readily absorbs orange, yellow and green.

Colour absorption affects how well a diver can see something underwater by altering contrast as well as colour because it is easier to see something that stands out against its background. For example, a red object will contrast distinctly with a green/blue at the surface, but at 30m, they may both appear to be the same colour. The red object becomes harder to see because it no longer contrasts with its background.

Refraction: This is a change in direction when light passes from a medium of one density to a medium of differing density e.g. air into water which is denser, or vice versa. The light refracts due to a change in speed with entering a different density, causing a shift in the light path.

As a result of refraction, objects underwater are magnified so they appear closer by a ratio of about 4:3 for their actual and apparent distances i.e. a fish that is actually four metres away will appear to be three metres away.

It would be fair to say then, that an objects position appears 25% closer or 33% larger.

Visual Reversal: refers to an object’s tendency to appear further away than its actual distance. The visual reversal phenomenon is more a function of human perception than of physics. In air, we see objects that are far away as hazy or blurry and darker. And so underwater, an object that is blurry would be perceived as being far away, when in reality it is not.
In essence, the lack of contrast and other familiar visual references we are used to seeing on the surface are not present underwater. As a result, we can be tricked into perceiving objects as further away than their actual distance from us.

The single most important factor affecting the “visual reversal” phenomenon is turbidity. This will blur the outline and definition of an object due to the suspended particles causing loss of clarity.

Sound:

Sound is an energy form that also travels in waves. It is a form of mechanical energy (Mechanical energy is energy of motion or of potential for motion on a macroscopic scale) Sound results when something sets in motion a wave or pattern of waves in matter.

Sound travels best in dense media like solids and liquids. This is because denser material has more tightly packed molecules to transmit the sound wave from one to the next. The elasticity of a substance also allows it to transmit sound more effectively and it can be said that the elastic property of water is more important than its density in making it a good conductor of sound. For example, a sound wave would travel faster along a rubber/elastic band than a rope due to its elasticity.

Therefore because water is a liquid and is denser than air (which is a gas), and because water has good elasticity, sound travels faster than it does in air.

Sound travels about four times faster in water than it does in air!

The brain determines sound direction based on the slight difference in intensity and time at which a sound reaches each of your ears e.g. the sound hits the left ear first so the brain knows the sound is coming from the left direction. Your brain assumes that intensity and time are the same underwater as they are in air. However, underwater the intensity and time are the same (as far as the brain is concerned), hitting both ears the same time. Therefore, the brain often cannot determine the source of the sound and the direction from where it is coming.

This is why you can’t tell where the boat is from the sound of its engines, but you can hear it from much further away.

The brain will often perceive a sound as if it is coming from above, but depending on the sound frequency, distance from the source, intensity and other variables, you can sometimes determine a sound’s direction reasonably accurately underwater. For example a buddy’s tank banger if reasonably close to the other buddy.

Buoyancy:

“An object wholly or partially immersed in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the object” Archimedes.

Buoyancy is the upward force exerted on any object placed in a fluid (liquid or gas). The amount of buoyancy – upward force- is based on how much fluid the object displaced and the weight of fluid displaced. The denser the fluid, the more upward force (buoyancy) is for a given displacement.

Seawater/saltwater is denser and weighs more than fresh water because of the dissolved salts. This makes salt water more buoyant than fresh water;

1 litre of seawater = 1.03 kg
1 litre of freshwater = 1.00 kg

An object that is positively buoyant weighs less than the water it displaces.
An object that is negatively buoyant weighs more than the water it displaces.
A neutrally buoyant object weighs exactly the same as the water it displaces.

To determine the buoyancy of an object in water, you need to know:

1. The object’s weight out of water.
2. How much water the object displaces (the object’s volume).
3. The weight of the displaced water.

(you do not need to know the depth at which the object lies. If this information is included in a question ignore it, as it is just there to confuse)

To determine an object’s buoyancy, subtract the weight of the water the object displaces from its weight: object weight – weight of water displaced = buoyancy

A positive number means the object is negatively buoyant (the object is heavier than the water it displaces), a negative number means it’s positively buoyant (the object weighs less than the displaced water), and zero means it is neutrally buoyant.

To make a negatively buoyant object neutral, the buoyancy must increase by the amount it is negative (usually by adding air to a lifting device to increase the volume displaced). To make it positive, it must increase by more than that.

To make a positively buoyant object neutral, the buoyancy must decrease by the amount it is positive (usually by adding weight to the object). To make it negative, it must increase by more than that.

It is very important to note the type of water the question is referring to i.e. seawater or freshwater. This is because both weigh different to each other and this affects the answer.

Sample problems:

You plan to recover a 150kg outboard motor in sea water that displaces 60 litres. How much air must you put in a lifting device to make the motor neutrally buoyant?

Firstly, litres cannot be subtracted from kilograms or vice versa. So they must be converted to the same constant either litres to kg, or kg to litres;

Litres to kg = x (multiply)
Kg to litres = ÷ (divide)

The question tells us that the object lies in seawater, which weighs 1.03kg per litre. To work in the same constants we need to convert the 60litres into KG;

60 (litres) x 1.03 = 61.8 kg

We now have the three pieces of information needed to work out the answer:

  1. The weight of the object = 150 kg
  2. The volume of the object (the amount of water it displaces) = 60 litres
  3. The weight of the displaced water = 61.8 kg

Remember: object weight – weight of water displaced = buoyancy

150 kg – 61.8 kg = 88.2kg (a positive number, so the object is negatively buoyant i.e. it is 88.2kg heavier than the water it displaces)

Finally, the 88.2kg needs to be converted back into liters (how air is measured) to work out how much air is needed in the lifting device to make the object weigh the same as the water it displaces to make it neutrally buoyant:

88.2 (kg) ÷ 1.03 = 85.6 litres

Approximately how much air must be added to a lifting device to bring a 600 kilogram/1200 pound object to the surface? The object lies in 30 metres/100 feet of fresh water.

The answer to this question cannot be determined from the data provided.– remember the 3 pieces of information needed to work out the answer. Unless the displacement is known, the amount of buoyant force on an object cannot calculated.

Some questions will require you to work out how much weight needs to be added to make the object a given amount negatively buoyant, or how much air needs to be added to make the object a given amount negatively buoyant.

To do this, simply work out how much air needs to be added to make the object neutrally buoyant and then add the given amount of weight or air.

You are assisting a research study and must sink into fresh water an object that weighs
50 kg and displaces 300 litres. Disregarding the minimal displacement of lead, how much lead weight do you need to affix to the object to make it 10 kg negative on the bottom?

The first thing to notice is that the question is using fresh water, so the constant 1.00kg is used, not the 1.03kg which is salt water. Therefore, the sum to convert the 300 litres into kg is

300 litres x 1.00 = 300 kg

We now have the three pieces of information needed to work out the answer:

  1. The weight of the object = 50 kg
  2. The volume of the object (the amount of water it displaces) = 300 litres
  3. The weight of the displaced water = 300 kg

Remember: object weight – weight of water displaced = buoyancy

50 kg – 300kg = -250kg (a negative number so the object is positively buoyant/floats as the object weighs less than the water it displaces).

To make the object neutrally buoyant 250kg needs to be added. Therefore to make the object
10 kg negatively buoyant an extra 10 kg needs to be added to the 250 kg =

260kg of lead weight needs to be affixed to the object to make it 10 kg negatively buoyant

An object lying in 12 metres of sea water weighs 487 kg and displaces 389 litres. How much air needs to be added to a lifting device to make it 42 kg positively buoyant?

Firstly, disregard the 12 metres, this has no relevance to the question.
Notice, the question is using salt water, so the constant 1.03 kg is used:

389 litres x 1.03 = 400.67 kg

We now have the three pieces of information needed to work out the answer:

  1. The weight of the object = 487 kg
  2. The volume of the object (the amount of water it displaces) = 389 litres\
  3. The weight of the displaced water = 400.67 kg

Remember: object weight – weight of water displaced = buoyancy

487 kg – 400.67 = 86.33 kg (a positive number so the object sinks)

To make the object neutrally buoyant, 86.33 kg needs to be added. Therefore to make the object 42 kg positively buoyant an extra 42 kg needs to be added to the 86.33 kg = 128.33 kg

The question does not finish here, not it asks for how much air needs to be added. We know that air is measured in litres, and so the final part of the question is to convert the final amount back into litres: 128.33 kg ÷ 1.03 = 124.6 litres of air needs to be added to the lifting device.

Pressure:

Pressure means a force or weight acting on a unit of area. This is expressed mathematically as:

PRESSURE = FORCE OR P = F / A
AREA

Gasses compress or expand proportionally with pressure changes. How gases affect the body is also proportionate to pressure. To predict these changes using the gas laws, the pressure at a given depth must be determined.

In terms of diving, pressure is usually expressed in atmospheres (ata) or bar.

The air at sea level is 1bar/ata of pressure. The pressure in seawater and freshwater differs due to the salt particles in saltwater, which makes it heavier than freshwater.

In seawater, every 10 metres of water exerts 1 bar/ata of pressure.

In freshwater, every 10.3 metres of water exerts 1 bar/ata of pressure. This can also be expressed as every 1 metre of freshwater exerts 0.097 bar/ata of pressure.

Absolute pressure: is the total pressure exerted by air at the surface and the pressure exerted by the water i.e. at 20 metres in seawater the absolute pressure is 3 bar/ata, that is 1 bar of surface pressure plus 2 bar of water pressure. The absolute pressure at 20 metres in freshwater is 2.94 bar/ata, that is 0.097 x 20 metres = 1.94 + 1 bar of surface pressure, or:

Seawater Absolute Pressure = Depth / 10 + 1
Freshwater Absolute Pressure = Depth / 10.3 + 1

Gauge pressure: is the pressure exerted by the water only i.e. at 20 metres the gauge pressure is 2 bar/ata in seawater. This is the pressure reading used by most gauges. At 20 metres in freshwater the gauge pressure is 1.94 bar/ata, that is 0.097 x 20 metres = 1.94 bar/ata, or:

Seawater Absolute Pressure = Depth / 10
Freshwater Absolute Pressure = Depth / 10.3

Ambient pressure: is the surrounding pressure (essentially the same as absolute pressure), i.e. at 20 metres in seawater the ambient pressure is 3 bar/ata – 1 bar of surface pressure plus 2 bar of water pressure.

Pressure And Volume:

Dive Physics Padi Idc Koh Tao Thailand

Sir Robert Boyle was a 17th century Irish scientist who wanted to find out what happens to a volume of air when the pressure changes. What he ultimately demonstrated (through a now-famous experiment involving a u-shaped glass tube) was that if the temperature remains constant, the volume of a gas is inversely proportional to the absolute pressure. This law today is called Boyle’s Law:

If the pressure increases, the volume must decrease proportionally.
If the pressure decreased, the volume must increase proportionally.

If a given gas is taken from the surface to 10 metres of sea water (msw) (2 bar/ata) its volume will be reduced to half the volume. The volume will reduce to one third at 20 msw, to one fourth at 30 msw and so on.

To work out questions involving volume, density and pressure – always bring the object back to the surface.

It is important to remember that volume decreases with increased depth and that density increases with increased depth. Inversely, volume increases as depth is decreased and that density decreases as depth decreases.

Examples:

A flexible container contains 12 litres of air at 13 metres of seawater, what will its new volume be at 53 metres?

Firstly, note that the question relates to seawater, and so every 10 metres exerts 1 bar/ata of pressure.

The container needs to be brought back up to surface, and so the volume of 12 litres will increase. It will increase by the amount equal to that of the pressure exerted at 13 metres:

13 metres of seawater =
1.3 bar/ata of water pressure + 1 bar of surface pressure = 2.3bar/ata

Because volume increases as the depth decreases, the volume needs to be multiplied.
12 litres x 2.3 bar/ata = 27.6 litres. Therefore the original volume of the container at the surface is 27.6 litres.

Now the container needs to be taken “back down” to 53 metres. Volume decreases as depth increases so the new surface volume needs to be divided by the amount equal to that of the pressure exerted at 53 metres:

53 metres of seawater =
5.3 bar of water pressure + 1 bar of surface pressure = 6.3 bar/ata

And so the new volume will be the surface volume divided by the pressure exerted at 53 metres:

27.6 litres ÷ 6.3 bar/ata = 4.4 litres

A flexible container contains 84 litres of air at 27 metres of freshwater, what would its volume be at 53 metres?

Note that the question is in freshwater and so the sum to work out the pressure is:

27 metres x 0.097 (the pressure exerted for each metre) = 2.6 bar/ata + 1 bar of surface pressure = 3.6 bar/ata (it is very important to remember to add the surface bar)

53 metres x 0.097 = 5.1 bar/ata + 1 bar/ata = 6.1 bar/ata

The container needs to be brought up from 27 metres: 84 litres x 3.6 bar/ata = 302.4 litres

Now it needs to be taken “back down” to its new depth: 302.4 litres ÷ 6.1 = 49.57 litres.

Depth and Density:

If the volume of a gas is reduced by raising the pressure, the same number of gas molecules must occupy less space, that is to say the gas becomes more compact or denser.

The density of the gas is directly proportional to the absolute pressure e.g. at 2 bar/ata (10 msw) a given air volume is twice as dense as at the surface, at 3 bar/ata (20 msw) it is 3 times as dense and so on.

This explains why the deeper a diver is, the faster the air supply will be used, because each breath contains more molecules:
If you take a full breath at 10 msw (2 bar/ata) you are inhaling twice the number of air molecules as a full breath at the surface. Each breath takes twice as much air from your tank as it would at the surface and your air supply will only last half as long.
At 20 metres (3 bar/ata) the air supply will only last one-third as long as at the surface; at
30 metres (4 bar/ata) only one-fourth, and so on.

Examples:

A diver’s air consumption rate is 2 bar per minute at the surface. What will the diver’s consumption rate be at 30 metres in seawater?

Firstly, note the question is in seawater, and so the sum to calculate the pressure at 30 metres is:
3 bar/ata of water pressure + 1 bar of surface pressure = 4 bar/ata.

The air molecules are going to be more dense with increased depth, and so
2 bar per minute x 4 bar = 8 bar per minute

A diver breathes 50 bar in 20 minutes in 21 metres of seawater, how much air would the diver breath in 40 minutes at 14 metres of seawater?

First, calculate the pressure at depth in seawater:

21 metres: 2.1 bar/ata of water pressure + 1 bar surface pressure = 3.1 bar/ata
14 metres: 1.4 bar/ata of water pressure + 1 bar of surface pressure = 2.4 bar/ata

The air consumption rate per minute needs to be calculated and so:
50 bar ÷ 20 minutes = 2.5 bar per minute

Now, take this breathing/air consumption rate “back up” to the surface. Remember, air density decreases as the depth decreases and so: 2.5 bar per minute ÷ 3.1 bar/ata = 0.8 bar.
This means that the dive will consume 0.8 bar of air per minute at the surface. This rate must now be taken “back down” to the new depth of 14 metres. Remember that the density increases as the depth increases:

0.8 bar per minute at the surface x 2.4 bar/ata = 1.92 bar per minute at 14 metres.

The final part is asking how much air will be consumed at this new depth in 40 minutes:

1.92 bar per minute x 40 minutes = 76.8 bar of air will be consumed in 40 minutes at 14 metres.

Temperature:

Learn Dive Physics In Koh TaoA significant advance in the study of gases came in the early 1800’s in France. Hot air balloons were extremely popular at that time and scientists were eager to improve the performance of their balloons. A prominent French scientist, Jacques Charles, made detailed measurements on how the volume of a gas was affected by the temperature of the gas. Given the interest in hot air balloons at that time, it is easy to understand why there was an interest in the temperature-volume relationship for a gas.
Just as Robert Boyle made efforts to keep all properties of the gas constant except for the pressure and volume, so Jacques Charles took care to keep all properties of the gas constant except for temperature and volume. The equipment used by Jacques Charles was similar to that employed by Robert Boyle. A quantity of gas was trapped in a J-shaped glass tube that was sealed at one end. This tube was immersed in a water bath; by changing the temperature of the water Charles was able to change the temperature of the gas. The pressure was held constant by adjusting the height of mercury so that the two columns of mercury had equal height, and thus the pressure was always equal to the atmospheric pressure.

He found that the change in either volume or pressure of a given gas volume is directly proportional to the change in absolute temperature. He noted that if you keep the pressure of a gas within a container constant, the volume increases or decreases as the temperature increases or decreases. He similarly found that if you keep the volume constant, the pressure increases or decreases when the temperature increases or decreases..

Pressure, Volume and Temperature – Flexible and Inflexible Container

As a general guideline, a scuba cylinder pressure will change 0.6 bar for every 1° C change in temperature.

If the container is flexible, the volume of the gas will increase or decrease as the temperature increases or decreases and therefore the pressure of the gas increases or decreases.

For example, if a balloon (flexible container) is filled with 30 litres of air at room temperature, then put in a freezer at a constant ambient pressure, then the volume would decrease. This is because as the temperature decreases, the motion of the molecules decreases. As this motion decreases, the force of impact with the side of the balloon decreases. Less force of impact means less pressure inside the balloon, and it would therefore shrink, or decrease in volume.

If the container is inflexible the volume of the gas will remain the same, while the pressure will increase or decrease as the temperature increases or decreases. An example of an inflexible container is a scuba tank. If a scuba tank is filled to capacity at room temperature then used for an ice dive, then the volume would remain unchanged but the pressure would decrease.

Examples:

A scuba tank is filled to 200 bar at a temperature of 27 °C. If the tank is then used in water temperature of 4°C, what would be the approximate tank pressure?

Firstly, work out the difference in temperature: from 27 °C to 4°C = 23°C drop in temperature.

Multiply this figure by 0.6 (the change in pressure (per bar) for every 1°C change in temperature) : 0.6 x 23°C = 13.8 bar difference in pressure.

This number must now be subtracted from this original tank pressure:
200 bar – 13.8 bar = 186.2 bar

And so it can be said that the new approximate tank pressure at this new temperature is 186 bar.

A scuba tank is filled to 200 bar at 20°C. What would happen to the pressure in the tank if it were put into a freezer at 0°C?

First, the difference in this instance is 20°C.

Multiply the difference in pressure by 0.6 : 0.6 bar x 20°C = 12 bar difference in pressure.

Finally, subtract the 12 bar from the original tank pressure of 200 bar:
200 bar – 12 bar = 188 bar

Partial Pressure:

Dalton’s Law: The total pressure exerted by a mixture of gasses is equal to the sum of the pressures of each of the different gases making up the mixture – each gas acting as if it alone were present and occupied the total volume.

Within a gas mixture each component gas in the mixture continues to demonstrate its own individual behavior in terms of its pressure; the individual pressure exerted by a component gas is referred to as a partial pressure (sometimes abbreviated “pp” or as a “P” before a gas, such as PO2 for “partial pressure of oxygen), as it is part of the total pressure exerted by the gas mixture.

As depth in water increases and the pressure increases, the percentage of the gas in the mixture remains the same. It is the partial pressure that increases proportionally to the depth.

To calculate the partial pressure of a gas component you need to first convert the percentage that it makes up in the gas mixture to bar. For example oxygen makes up 21% of air – convert this to bar by dividing by 100 – 21/100 = 0.21 bar. Then multiply that by the absolute pressure, for example the partial pressure of oxygen at sea level will be 0.21 x1 = 0.21 bar/ata.

At 20 metres of sea water the partial pressure of oxygen would be:
0.21 bar x 3 bar (the pressure at 20 metres of sea water) = 0.63 bar.

As Dalton’s Law states, the total pressure exerted by the gas mixture is equal to the sum of the pressures of each of the gases making up the mixture. The partial pressure exerted by a particular component of the gas mixture is proportional to the number of molecules of that particular gas in the mixture (This is important with regard to nitrox).

For example, at 20 metres in sea water:
0.63 bar/ata of oxygen + 2.37 bar/ata of nitrogen = 3 bar/ata which is the total pressure at 20 metres of sea water.

Example:

A diver is using Enriched Air Nitrox that is 34% oxygen and 66% nitrogen. What will the partial pressure of oxygen and nitrogen be at a depth of 16 metres of sea water? (round to two decimal places)

Firstly, we need to know the pressure are 16 metres of sea water = 2.6 bar/ata

And so the PO2 would be 0.34 x 2.6 = 0.88 ata and the PN2 would be 0.66 x 2.6 = 1.72 ata.

Note that the partial pressures of the oxygen and nitrogen equal the sum of the total pressure exerted by the gas mixture, that is 0.88 + 1.72 = 2.6 ata/bar the same as the pressure at 16 metres of sea water.

How does the physiological effect of breathing a given percentage of gas at depth compare to breathing the same percentage of the gas at the surface?

Dalton’s Law becomes significant when considering impurities in breathing air. By applying Dalton’s Law, you can compare the effects of a single gas in a mix breathed at a particular depth with the effects of a greater percentage of the same gas at the surface.

The body responds to a gas breathed based on the partial pressure, not the percentage of the gas in the mix. And so, as the pressure increases, the physiological effects increase.

This is important because the increased partial pressure of nitrogen can lead to nitrogen narcosis. Nitrogen narcosis begins to take effect when the nitrogen reaches a partial pressure of around 3.2 bar/ata. Also, the increased partial pressure of oxygen can cause oxygen toxicity. When the partial pressure of oxygen reaches 1.4 bar/ata, oxygen can become toxic ( though this is more of a concern when diving enriched air nitrox than with air within recreational limits – discussed more in physiology).

Also, if contaminants such as carbon monoxide are present in the air, then levels that may be harmful at the surface may become toxic under elevated partial pressures at depth.

Example:

A scuba tank is accidentally filled with 1% carbon monoxide. If a diver breathes air from this tank at a depth of 30 metres of sea water, approximately what percentage of carbon monoxide will he be breathing?

Once a tank is filled, the percentages of gases within it cannot change. The answer will remain the same and be 1% of carbon monoxide. It is the partial pressures of each gas that may increase not the percentage.

An air mixture has 0.5% carbon monoxide. Breathing it at 40 metres of sea water would be the equivalent of breathing what percent at the surface?

This question is asking for the equivalent. And so simply multiply:
0.5% by 5 bar (the absolute pressure at 40 metres of sea water) = 2.5%

Therefore breathing 0.5% carbon monoxide at 40 metres of sea water would be the equivalent of breathing 2.5% carbon monoxide at the surface.

Raising the Pressure of a Gas in Contact with a Liquid

When a gas is in solution (dissolved in) with a liquid the gas molecules still exert pressure from inside the liquid, this is called gas tension.

Remember that gas always goes from a higher pressure to a lower pressure until equilibrated.

If a bucket of liquid with no gas dissolved in it the gas tension would be zero. When the liquid comes into contact with a gas, the gas molecules rush to penetrate the liquid, flowing from high pressure to low pressure. The gas dissolves into the liquid until the gas tension of the gas in the liquid is the same as the partial pressure of the gas in contact with the solution.

If there is an increase in the partial pressure of the gas in contact with the solution, for example from 0.79 bar to 1.58 bar (the same as taking the nitrogen in air from sea level to 10 msw), the gas molecules will again rush to penetrate the solution until the gas tension reaches 1.58 bar, the same as the partial pressure of the gas in contact with the solution.

Inversely, if there is a reduction in the partial pressure of the gas in contact with the solution, for example from 1.58 bar to 0.79 bar (the same as taking the nitrogen in air from 10msw to sea level), then the gas molecules will leave the solution until the gas tension reaches 0.79 bar, which is the same as the partial pressure of the gas in contact with the solution.

Supersaturation:

When you reduce the pressure of a gas in contact with the liquid, the gas dissolved in the liquid will have a greater gas tension than the partial pressure of the same gas in contact with the liquid.
The liquid will be supersaturated – it will contain more gas than it can keep in solution at that pressure.

Providing that the partial pressure of the gas in contact with the solution is not reduced too quickly, then the gas particles will simply start to leave the solution until the gas tension is equilibrated with the surrounding pressure.

When divers ascend, we follow the 18 metres per minute rule, which allows the nitrogen in our systems to be eliminated from our bodies at a safe rate.

Excessive Supersaturation:

When the partial pressure of a gas in contact with solution is reduced too quickly, the gas may come out of solution faster than it can diffuse through the liquids surface. As a result, bubbles will form. An example of this is when a can of soda has been shaken up. Once the can is opened, the pressure is suddenly reduced and the gas comes out of solution too quickly and froths / bubbles over.

This is the basic cause of DCS. If a diver ascends too fast, the difference in pressure of nitrogen dissolved in the body and the pressure of surrounding pressure (pressure gradient, discussed in decompression theory) is too great. The nitrogen comes out of solution faster than the body can eliminate it in breathing air, and bubbles form in the tissues causing DCS.

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