DKW 1 Physics

Section One: Diving Physics 1-1

Section OneDiving PhysicsThe mere mention of the word “physics” is usually enough to send shivers down the spines of most people. However, as it relates to diving, the term simply refers to two basic phenomena:

1. How energy is affected underwater (such as sound, light and buoyancy).

2. How matter, primarily gases, are affected in a high pressure environment.

Both of these phenomena are important in understanding the underwater world, and how we are affected by it. In fact, many of the objectives examined in later sections of the workbook require an understanding of concepts described in this section on diving physics. As you work through the following exercises, bear in mind that most of what is asked requires nothing more from you than some common sense, the knowledge of a few constants and a few arithmetic skills.

Objective 1.1Explain why water is able to dissipate body heat faster than air, at what rate this occurs and what effect this has upon the diver.

Resources:• Encyclopedia, Chapter Four, under the heading of “Water and Heat”

Exercises:1. Water is able to conduct heat far more efficiently than air because it is:

a. less dense than air.

b. more dense than air.

c. more fluid than air.

d. less fluid than air.

2. Divers are most affected by what form of heat transmission?

a. Conduction

b. Convection

c. Radiation

d. All of the above are equally important

1-2 Diving Knowledge Workbook

3. Water is able to conduct heat about ______ times faster than air.

a. 3200

b. 775

c. 100

d. 20

Objective 1.2 Explain the behavior of light as it passes from an air/water interface and what effect this has upon the diver.

Resources:• Encyclopedia, Chapter Four, under the heading of “Water and Light”

Exercises:1. Refraction is caused by the process of:

a. light traveling at different speeds as it passes through different substances.

b. water absorbing various wave lengths of light beginning at the red end of the spectrum.

c. the changing speed of light due to sunspot activity.

d. light traveling at the same speed as sound once it enters water and encounters resistance.

2. When viewed underwater, objects normally appear ____ by a ratio of about _____ (actual to apparent distance).

a. closer/2:1

b. further away/4:3

c. closer/4:3

d. further away/2:1

3. When viewed underwater, objects tend to be magnified by a factor of about:

a. 10%

b. 33%

c. 50%

d. 74%


Objective 1.3 Define the “visual reversal” phenomenon and explain its effect upon the diver.

Resources:• Encyclopedia, Chapter Four, under the heading of “Water and Light”

Exercises:1. “Visual reversal” refers to an object’s tendency to appear:

a. as a mirror image of itself.

b. upside down, as though viewed through a magnifying lens.

c. like a photographic negative.

d. further away than its actual distance.

2. The single most important factor affecting the “visual reversal” phenomenon is:

a. depth.

b. turbidity.

c. time of day.

d. All of the above are equally important.

Objective 1.4 Explain why sound travels faster in water than in air, by approximately how much and what effect this has upon the diver.

Resources:• Encyclopedia, Chapter Four, under the heading of “Water and Sound”

Exercises:1. Light waves contain _____, while sound waves are comprised of _____.

a. heat energy/air

b. electromagnetic energy/mechanical energy

c. infrared energy/ultraviolet energy

d. kinetic energy/potential energy


2. The density and elasticity of a medium has what effect upon the transmission of sound?

a. The denser and more elastic the medium, the better sound is transmitted.

b. The denser and more elastic the medium, the poorer sound is transmitted.

c. The denser and more elastic the medium, the less sound can be transmitted.

d. Density and elasticity of a medium has no effect upon the transmission of sound.

3. Sound travels approximately _____ times faster in water than it does in air.

a. two

b. four

c. ten

d. twenty

4. Divers have difficulty determining the direction of sound underwater because:

a. there is an insufficient delay between the sound striking one ear before the other.

b. water filling the ear canal reduces the ear’s sensitivity to sound.

c. the wet suit hood makes it difficult to perceive sound as accurately as on land.

d. sound waves travel less efficiently underwater.

Objective 1.5 State Archimedes’ Principle and calculate the buoyancy required to either lift or sink an object in both fresh and seawater.

Resources:• Encyclopedia, Chapter Four, under the heading of “Buoyancy and the

“Weightless” World”

Exercises:1. According to Archimedes’ Principle, “Any object wholly or partially immersed in a fluid is

buoyed up by a force ______.”

a. equal to the weight of the object

b. equal to the weight of the fluid displaced by the object

c. equal to the weight of both the object and the weight of the fluid displaced by the object

d. slightly less than the weight of the object


2. The specific gravity of pure water is:

a. 0.0

b. 1.0

c. It varies according to where on earth it is measured.

d. It cannot be determined except in a vacuum.

3. 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.

a. 620 litres/19 cubic feet

b. 600 litres/18 cubic feet

c. 580 litres/16 cubic feet

d. The answer cannot be determined from the data provided.

4. Approximately how much water must be displaced to bring a 500 kilogram/900 pound object to the surface if the object displaces 300 litres/10 cubic feet? The object lies in 40 metres/132 feet of seawater.

a. Slightly more than 300 litres/8 cubic feet

b. Slightly more than 185 litres/4 cubic feet

c. Slightly more than 29 litres/1 cubic foot


5. An object weighing 350 kilograms/750 pounds and displacing 300 litres/10 cubic feet is lying in 15 metres/50 feet of fresh water. If a drum is to be used to lift the object to the surface, how much water must be displaced from the drum?

a. 41 litres/1.72 cubic feet


c. For any inflexible drum to be used, it must always be completely filled.



Objective1.6Define the terms “absolute,” “ambient” and “gauge” pressure and calculate pressure at any depth as expressed by these terms in both fresh and seawater.

Resources:• Encyclopedia, Chapter Four, under the heading of “Under Pressure”

Exercises:1. The absolute pressure at a depth of 90 metres/300 feet of seawater is:

a. 8 ata/118.8 psia

b. 9 ata/133.5 psia

c. 10 ata/148.2 psia


2. The ambient pressure at 30 metres/100 feet of seawater is:

a. 3 ata/44.5 psia

b. 4 ata/59.2 psia

c. 5 ata/73.9 psia


3. The gauge pressure at 23 metres/75 feet of depth in fresh water is:

a. 2.23 atm gauge/32.4 psig

b. 3.23 atm gauge/47.1 psig

c. 4.23 atm gauge/61.8 psig


4. At a depth of 34 metres/112 feet in fresh water the absolute pressure is __ _____ , the gauge pressure is _______ and the ambient pressure is __________.

Metric Responses Imperial Responses a. 4.5 ata/3.5 atm gauge/4.5 ata a. 64.54 psia/49.84 psig/64.54

psia

b. 3.4 ata/4.4 atm gauge/3.4 ata b. 48.38 psia/33.68 psig/48.38 psia

c. 4.3 ata/3.3 atm gauge/4.3 ata c. 63.08 psia/48.38 psig/63.08 psia

d. 4.3 ata/3.3 atm gauge/3.3 ata d. 63.08 psia/48.38 psig/48.38 psia


Objective 1.7 Explain the relationship between pressure and volume on a flexible gas-filled container, and calculate (in increments of whole atmospheres) the changes that will occur to that container as it is raised and lowered in the water column.

Resources: • Encyclopedia, Chapter Four, under the heading of “Boyle’s Law”

Exercises:1. A balloon containing 30 litres/1 cubic foot of air is released from 10 ata. If it does not

explode, calculate the volume of air in the balloon upon reaching the surface.



c. 300 litres/10 cubic feet


2. What would the volume of the balloon in exercise 1 be if it is taken to 50 metres / 165 feet of seawater?


b. 50 litres/1.67 cubic feet

c. 285 litres/9.5 cubic feet

d. Unchanged

3. A balloon is filled with 60 litres/2 cubic feet of air at 30 metres/100 feet of seawater. What will be the approximate volume of the balloon if it is taken to a depth of 90 metres/300 feet?

a. 24 litres/0.8 cubic feet

b. 90 litres/0.3 cubic feet

c. 20 litres/0.67 cubic feet


4. A balloon containing 300 litres/10 cubic feet of air at 7 metres /25 feet of seawater is taken to a depth of 26 metres/85 feet. What will be the exact volume of the balloon upon reaching 26 metres/85 feet?

a. 141.66 litres/4.91 cubic feet

b. 88.20 litres/2.94 cubic feet

c. 58.50 litres/1.95 cubic feet



Objective 1.8 Explain the relationship between depth and the density of the air a diver breathes, and calculate this relationship in increments of whole atmospheres.

Resources:• Encyclopedia, Chapter Four, under the heading of “Boyle’s Law”

Exercises:1. A scuba tank is filled to capacity at the surface. When this tank is used at a depth of

30 metres/99 feet in the sea, the air within the tank is four times more dense than it was at the surface.

True False

2. Because a diver’s lung volume must remain constant regardless of the depth at which he breathes, the density of the air in the diver’s lungs does not change even as he changes depth.

True False

3. The air that a diver breathes from a scuba tank at 50 metres/165 feet of seawater is ______ as dense as the air breathed from the same tank at the surface.

a. 10 times

b. 6 times

c. 5 times

d. exactly


Objective 1.9 Given a diver’s air consumption rate at one depth, calculate how that consumption rate changes when depth changes.

Resources:• Encyclopedia, Chapter Four, under the heading of “Boyle’s Law”

Exercises:1. A diver has an air consumption rate of 2 bar/25 psig per minute at the surface. If all factors, but depth

remain unchanged, what will his consumption rate be at 40 metres/132 feet of seawater?

Metric Responses Imperial Responses a. 3 bar/minute a. 50 psig/minute

b. 5 bar/minute b. 75 psig/minute

c. 7 bar/minute c. 100 psig/minute

d. 10 bar/ minute d. 125 psig/minute

2. A diver has an air consumption rate of 60 litres/2 cubic feet per minute at 10 metres/33 feet of seawater. If all factors, but depth remain unchanged, what will his consumption rate be at 30 metres/99 feet?

a. 15 litres/1/2 or 0.5 cubic foot per minute

b. 30 litres/1 cubic foot per minute

c. 90 litres/3 cubic feet per minute

d. 120 litres/4 cubic feet per minute

3. A diver has an air consumption rate of 90 litres/3 cubic feet per minute at 20 metres/66 feet of seawater. If all factors, but depth remain unchanged, what will his consumption rate be at 60 metres/200 feet?

a. 180 litres/6 cubic feet per minute

b. 210 litres/7 cubic feet per minute

c. 240 litres/8 cubic feet per minute


4. A diver has an air consumption rate of 60 litres/2 cubic feet per minute at 10 metres/33 feet of seawater. If all other factors but depth remain unchanged, what will his consumption rate be, in bar/psig per minute, at 30 metres/100 feet?

Metric Responses Imperial Responses a. 3 bar/minute a. 50 psig/minute

b. 7 bar/minute b. 100 psig/minute

c. 10 bar/minute c. 150 psig/minute

d. The answer cannot be determined from the data provided. (same for both)


Objective 1.10 Describe how the behavior of a gas within both a flexible and inflexible container is affected by changes in pressure and temperature.

Resources:• Encyclopedia, Chapter Four, under the heading of “Charles’ Law”

Exercises:1. A balloon is filled with 30 litres/1 cubic foot of air at room temperature. Describe what

would happen to that balloon if it were put into a freezer at a constant ambient pressure.

a. The volume would increase

b. The volume would decrease

c. The volume would remain unchanged, but the pressure would decrease.

d. The volume and the pressure would remain unchanged.

2. A scuba tank is filled to capacity at room temperature. Describe what would happen to that tank if it was taken on an ice dive (water at or near freezing).

a. The volume would increase

b. The volume would decrease

c. The volume would remain unchanged, but the pressure would decrease.

d. The volume and the pressure would remain unchanged.

3. An 12 litre/80 cubic foot scuba tank is filled to 200 bar/3000 psig at an ambient temperature of 27° C/80° F. If the tank is then used in water temperature of 4° C/40° F, what would be the approximate tank pressure?

a. Unchanged

b. 186 bar/2800 psig

c. 214 bar/3400 psig

d. 116 bar/2840 psig

4. An 12 litre/80 cubic foot scuba tank is filled to 200 bar/3225 psig at an ambient temperature of 26° C/78° F. What will the exact tank pressure be if the tank is used in water temperature of 7° C/44° F? (For imperial measurement calculations, assume 1 atm = 15 psi)

a. 192 bar/2885 psig

b. 187 bar/3020 psig

c. 188 bar/3035 psig



Objective 1.11 Given their percentages, calculate the partial pressures of gases in a mixture at any depth.

Resources:• Encyclopedia, Chapter Four, under the heading of “Dalton’s Law”

Exercises:1. In a mixture of air comprised of 20% oxygen and 80% nitrogen, at an ambient pressure

of 1 ata/15 psia what is the partial pressure of oxygen?

a. 0.2 ata/3 psia

b. 0.4 ata/6 psia

c. 0.9 ata/12 psia

d. 1 ata/15 psia

2. A gas mixture is comprised of 21% oxygen, 78% nitrogen and 1% carbon dioxide. At a depth of 20 metres/66 feet of seawater, what is the partial pressure of the nitrogen?

a. 1.67 ata/22.93 psia

b. 2.34 ata/34.39 psia

c. 5.69 ata/78.00 psia


3. A gas mixture is comprised of 21% oxygen, 78% nitrogen and 1% carbon dioxide. At a depth of 24 metres/78 feet of seawater, what is the partial pressure of the oxygen?

a. 3.6 ata/49.41 psia

b. 0.7 ata/10.37 psia

c. 0.5 ata/7.28 psia



Objective 1.12 Explain the effect of breathing contaminated air mixtures at depth and calculate the equivalent effect such contamination would have upon the diver at the surface.

Resources:• Encyclopedia, Chapter Four, under the heading of “Dalton’s Law”

Exercises:1. A scuba tank is accidentally filled with 1% carbon monoxide. If a diver breathes air from

this tank at a depth of 30 metres/100 feet of seawater, approximately what percentage of carbon monoxide will he be breathing?

a. 1%

b. 2%

c. 3%

d. 4%

2. Referring to the previous question, it is determined that at the surface the diver inhales 500,000 molecules of carbon monoxide with each breath. Therefore, when breathing the air at 30 metres/100 feet of seawater (assuming all other factors but depth are unchanged), he would breathe approximately how many molecules?

a. 500,000

b. 1,000,000

c. 1,500,000

d. 2,000,000

3. Again referring to question 1, breathing the contaminated air at 30 metres/100 feet of seawater would have the same effect on the diver as breathing what percentage of carbon monoxide at the surface?

a. 1%

b. 2%

c. 3%

d. 4%


4. Breathing from a contaminated air source, with 1.5% carbon monoxide, at a depth of 90 metres/300 feet of seawater, would have the same effect as breathing approximately what percentage of carbon monoxide at the surface?

a. 1.5%

b. 13.5%

c. 15%


Objective 1.13 Explain what will occur to a liquid saturated with a gas at high pressure when the pressure of the gas in contact with the liquid is quickly reduced.

Resources:• Encyclopedia, Chapter Four, under the heading of “Henry’s Law”

Exercises:1. A glass of water has been placed in a vacuum for several days. It no longer contains any

dissolved gas within it. If it is then placed in a pressure pot and pressurized to 2 ata for several days, what will the gas pressure be within the liquid?

a. 1 ata

b. 2 ata

c. 4 ata

d. The answer is impossible to determine.

2. If the pressure within the pot in the previous question is increased, the pressure of the gas within the liquid will:

a. increase.

b. decrease.

c. remain unchanged.


3. If a vacuum is created within the pressure pot in question 1, the pressure of the gas within the liquid will:

a. increase.

b. decrease.

c. remain unchanged.



Objective 1.14 Define “supersaturation” and explain what conditions are necessary for gas bubbles to form in a supersaturated liquid.

Resources:• Encyclopedia, Chapter Four, under the heading “Henry’s Law”

Exercises:1. Assume that the atmosphere contains 80% nitrogen. At sea level, the nitrogen partial

pressure within our tissues would be:

a. 1.0 ata

b. 0.8 ata

c. 0.2 ata


2. From the above question, the total gas pressure within our tissues is:

a. 1.0 ata

b. 0.8 ata

c. 0.2 ata


3. Referring to question 2, because our tissues cannot hold any more gas pressure at sea level, our body is referred to as being:

a. pressurized.

b. saturated.

c. supersaturated.

d. unsaturated.

4. When breathing at depth, the tissues of a diver begin to take on additional gas pressure. If the diver remains at depth long enough, his tissues will again equalize with the ambient pressure.

True False

5. The condition described in the above question is referred to as supersaturation.

True False

Section One: Answer Key 1-15

Correct: Confident Guess

Incorrect: Simple Mistake Lack of Knowledge





Section OneAnswer Key

Objective 1.1 Explain why water is able to dissipate body heat faster than air, at what rate this occurs and what effect this has upon the diver.

1. (b) Water is able to conduct heat far more efficiently than air because it is: more dense than air. Water conducts heat more effectively than air because the molecules are closer together. This makes it easier for the heat (energy) to be transferred or conducted. In addition, the “polar” nature of the water molecule, resulting from the hydrogen bonds, enables water to absorb more heat than other liquids.

2. (a) Divers are most affected by what form of heat transmission? Conduction. Conduction refers to the transmission of heat via direct contact. Convective and radiant transmission of heat are relatively unimportant to divers.

3. (d) Water is able to conduct heat about 20 times faster than air. This is why submerged divers become chilled very quickly in water — even in water at a temperature that might be considered comfortable in air. (See the explanation under item 1).


Objective 1.2 Explain the behavior of light as it passes from an air/water interface and what effect this has upon the diver.

1. (a) Refraction is caused by the process of: light traveling at different speeds as it passes through different substances. As light passes through transparent substances of differing densities, the speed at which it travels changes. This change of speed causes the light wave to bend or “refract.”

2. (c) When viewed underwater, objects normally appear closer by a ratio of ab out 4:3 (actual to apparent distance). Light passing through the water/air interface refracts (bends). The effect is similar to an object being viewed under a magnifying glass. If no other factors come into play, objects will appear closer by a ratio of about 4:3.

3. (b) When viewed underwater, objects tend to be magnified by a factor of about: 33%. (See above explanation).

Objective 1.3 Define the “visual reversal” phenomenon and explain its effect upon the diver.

1. (d) “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 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.

2. (b) The single most important factor affecting the “visual reversal” phenomenon is: turbidity. Studies have shown several factors affecting the phenomenon. However, the turbidity of the water appears to be one of the major factors.












Objective 1.4 Explain why sound travels faster in water than in air, by approximately how much and what effect this has upon the diver.

1. (b) Light waves contain electromagnetic energy, while sound waves are comprised of mechanical energy. Light is a form of electromagnetic energy that is visible to us. Most forms of electromagnetic energy — cosmic rays, X-rays, radio waves — are not. Sound results when an object sets in motion a series of vibrations. These vibrations, in turn, create mechanical energy (sound waves) that are transmitted and perceived by us as sound.

2. (a) The density and elasticity of a medium has what effect upon the transmission of sound? The denser and more elastic the medium, the better sound is transmitted. Sound results from vibrations. Air cannot conduct vibrations efficiently, so sounds don’t tend to travel very well in air. However, in denser media — steel or water, for example — vibrations can be transmitted quite efficiently. The molecules of these media are much closer together than in air, making it easier to transmit vibrations. Also, if sound travels so well in water, one may ask, why can’t humans talk in that environment? The problem stems from the inability to initially produce sound while underwater. Human vocal cords must operate in an air environment.

3. (b) Sound travels approximately 4 times faster in water than it does in air. In air sound travels approximately 340 metres/1,100 feet per second. In water it travels 1,475 metres/4,800 feet per second.

4. (a) Divers have difficulty determining the direction of sound underwater because: there is an insufficient delay between the sound striking one ear before the other. Because of the increased speed of sound underwater, the human brain is unable to perceive the delay from one ear to the other. The brain interprets the sound as coming from all directions equally. Therefore, the origin of sound cannot be determined well.
















Objective 1.5 State Archimedes’ Principle and calculate the buoyancy required to either lift or sink an object in both fresh and seawater.

1. (b) According to Archimedes’ Principle, “Any 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.” This is why it’s important to know the displacement of an object. By knowing its displacement, and using a constant “density” you will be introduced to later, you can easily find out the buoyant force on any object immersed in fresh or salt water.

2. (b) The specific gravity of pure water is: 1.0. Pure water is used as the basis of the specific gravity scale. Therefore, it is assigned the value of 1.0. Substances with a value of less than 1.0 are, by definition, less dense than pure water and will float. Substances with values greater than 1.0 are more dense and will sink.

3. (d) 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 cannot be determined from the data provided. Buoyancy is a function of both the weight of an object and its displacement. Unless the displacement is known, you cannot calculate the amount of buoyant force on an object.

4. Metric (b) Approximately how much water must be displaced to bring a

500 kilogram object to the surface if the object displaced 300 litres? The object lies in 40 metres of seawater. Slightly more than 185 litres. The object weighs 500 kilograms. This represents a downward gravitational force. The object also displaces 300 litres of water. Each litre of displacement represents 1.03 kilograms/litres of upward buoyant force. (Remember, we are in seawater). Therefore, the total buoyant force is 309 kilograms (1.03 x 300 = 309). We in essence have an upward buoyant force of 309 kilograms counteracting the downward gravitational force of 500 kilograms. The object, in fact, sank because there was 191 kilograms more downward force than upward (500 - 309 = 191). To equalize the forces, making the object neutrally buoyant, we must somehow add 191 kilograms of buoyancy to the object. If a lift bag is attached, each litre of air added to it will provide an additional 1.03 kilograms of buoyancy. Therefore, we can acquire the additional buoyancy by displacing a little more than 185 litres of water. (191 ÷ 1.03 = 185.4) Incidentally, because water is for all practical purposes incompressible, the depth of the water can be ignored.




Imperial (b) Approximately how much water must be displaced to bring

a 900 pound object to the surface if the object displaced 10 cubic feet? The object lies in 132 feet of seawater. Slightly more than 4 cubic feet. The object weighs 900 pounds. This represents a downward gravitational force. The object also displaces 10 cubic feet of water. Each cubic foot of displacement represents 64 pounds/cubic feet of upward buoyant force. (Remember, we are in seawater). Therefore, the total buoyant force is 640 pounds (64 x 10 = 640). We in essence have an upward buoyant force of 640 pounds counteracting the downward gravitational force of 900 pounds. The object, in fact, sank because there was 260 pounds more downward force than upward (900 - 640 = 260). To equalize the forces, making the object neutrally buoyant, we must somehow add 260 pounds of buoyancy to the object. If a lift bag is attached, each cubic foot of air added to it will provide an additional 64 pounds of buoyancy. Therefore, we can acquire the additional buoyancy by displacing a little more than 4 cubic feet of water (260 ÷ 64 = 4.06). Incidentally, because water is for all practical purposes incompressible, the depth of the water can be ignored.

5. Metric (b) An object weighing 350 kilograms and displacing 300 litres is

lying in 15 metres of fresh water. If a drum is to be used to lift the object to the surface, how much water must be displaced from the drum? 50 litres. The object weighs 350 kilograms and displaces 300 litres. Each litre of displacement represents 1 kilogram/litre of upward buoyant force. (We are now in fresh water — 1 litre of fresh water weighs 1 kilogram). Therefore, the total buoyant force is 300 kilograms (1 x 300 = 300). In essence, we have an upward buoyant force of 300 kilograms counteracting the downward gravitational force of 350 kilograms. The object sank because there was 50 kilograms more downward than upward force (350 - 300 = 50). To equalize the forces, making the object neutrally buoyant, we must now add 50 kilograms of buoyancy to the object, or displace 50 litres of water. Again, because water is for all practical purposes incompressible, the depth of the water can be ignored.

Imperial (b) An object weighing 750 pounds and displacing 10 cubic feet

is lying in 50 feet of fresh water. If a drum is to be used to lift the object to the surface, how much water must be displaced from the drum? 2 cubic feet. The objects weighs 750 pounds. As in the previous problem, the object also displaces 10 cubic feet. Each




cubic foot of displacement represents 62.4 pounds/cubic feet of upward buoyant force. (We are now in fresh water — 1 cubic foot of fresh water weighs 62.4 pounds). Therefore, the total buoyant force is 624 pounds (62.4 x 10 = 624). In essence, we have an upward buoyant force of 624 pounds counteracting the downward gravitational force of 750 pounds. The object sank because there was 126 pounds more downward than upward force (750 - 624 = 126). To equalize the forces, making the object neutrally buoyant, we must now add 126 pounds of buoyancy to the object. Displacing 2 cubic feet of water from a lifting device will provide sufficient additional buoyancy (126 ÷ 62.4 = 2). Again, because water is for all practical purposes incompressible, the depth of the water can be ignored.

Objective1.6Define the terms “absolute”, “ambient” and “gauge” pressure and calculate pressure at any depth as expressed by these terms in both fresh and seawater.

1. (c) The absolute pressure at a depth of 90 metres/300 feet of seawater is: 10 ata/148.2 psia. Remember, absolute pressure is the sum of both the pressure exerted by the water column and the atmosphere. Therefore, to find out how much pressure the water (remember, its seawater) exerts, multiply .100 atm/m times 90 metres/.445 psi/ft times 300 feet. The water pressure alone is 9 atm gauge (.100 x 90 = 9)/133.5 psig (.445 x 300 = 133.5). But, as the atmosphere exerts pressure on the water, and this pressure is evenly distributed throughout the water column, we must also account for it. Therefore, we must add 1 atm/14.7 psi to the 9/133.5 to determine the total or absolute pressure of 10 ata/148.2 psia.

2. (b) The ambient pressure at 30 metres/100 feet of seawater is: 4 ata/59.2 psia. Ambient means nothing more than “surrounding” pressure. Therefore, the ambient pressure is synonymous with absolute pressure. We can then proceed in the same manner as in the previous problem. To find out how much pressure the water (remember its seawater) exerts, multiply .100 atm/m times 30 metres/.445 psi/ft times 100 feet. The water pressure alone is 3 atm gauge/44.5 psig. But, just as in the previous question, we must account for the atmospheric pressure. Therefore, we must add 1 atm/14.7 psi to the 3 atm gauge/44.5 psig to determine the ambient or absolute pressure of 4 ata/59.2 psia.






3. (a) The gauge pressure at 23 metres/75 feet of depth in fresh water is: 2.23 atm gauge/32.4 psig. Remember, gauge pressure ignores the atmospheric pressure — like your depth gauge. Therefore, this time we are only concerned with finding out how much pressure is exerted by the water column. To do that we multiply the depth (remember its fresh water) of 23 metres/75 feet times .097 atm/m/.432 psi/ft to determine that 2.23 atm gauge/32.4 psig is being exerted by the water.

4. (c) At a depth of 34 metres/112 feet in fresh water the absolute pressure is 4.3 ata/63.08 psia, the gauge pressure is 3.3 atm gauge/48.38 psig and the ambient pressure is 4.3 ata/63.08 psia. To determine the absolute pressure at 34 metres/112 feet, multiply the freshwater constant .097 atm/m/.432 psi/ft with the depth. This tells us that the water pressure alone accounts for 3.3 atm gauge/48.38 psig. Since this ignores the atmospheric pressure, this represents the gauge pressure at 34 metres/112 feet. To obtain the absolute pressure merely add 1 atm/14.7 psi. As absolute and ambient are synonymous terms, both pressures are 4.3 ata/63.08 psia.

Objective 1.7 Explain the relationship between pressure and volume on a flexible gas-filled container, and calculate (in increments of whole atmospheres) the changes that will occur to that container as it is raised and lowered in the water column.

1. (c) A balloon containing 30 litres/1 cubic foot of air is released from 10 ata. If it does not explode, how much air will be in the balloon upon reaching the surface? 300 litres/10 cubic feet. A simple relationship holds true when dealing with increments of whole atmospheres. Specifically, a flexible container will expand upon ascent, and reach the surface with the original quantity of gas times the number of atmospheres from which it was released. In this case, as the container started with 30 litres/1 cubic foot at 10 ata, the gas will expand 10 times by the time it reaches the surface. This problem deals with Boyle’s Law which states, “if the temperature remains constant, the volume of a given mass of gas is inversely proportional to the absolute pressure.”

2. (b) What would the volume of the balloon in exercise 1 be if it is taken to 50 metres/165 feet of seawater? 50 litres/1.67 cubic feet. This type of problem is actually quite simple to solve if we remember one simple rule: always bring the object to the surface, then return to the depth in question. The first thing we must do is determine what pressure is exerted by 50 metres/165 feet of depth. By dividing 10 metres/33 feet (which is one atmosphere










of pressure) into 50 metres/165 feet we find that the pressure is 5 atm gauge. We must then add the 1 atm from the surface to arrive at 6 atm absolute. So, the question really asks, what will the volume be as the balloon reaches 6 ata? But first, let’s take the balloon to the surface. From problem 1, we determined that the balloon would contain 300 litres/10 cubic feet at the surface. Let’s now apply a simple relationship — the quantity in question over the number of atmospheres. In this case the quantity of 300 litres/10 cubic feet over 6 ata or (300/6) 10/6. All that is left is to reduce the fraction or divide — 50 litres/1.67 cubic feet.

3. Metric (a) A balloon is filled with 60 litres of air at 30 metres of

seawater. What will be the approximate volume of the balloon if it is taken to a depth 90 metres? 24 litres. First, let’s find out how much pressure is exerted by 90 metres of depth. By dividing 10 into 90 we determine that we are at a pressure of 9 atm gauge or 10 ata. We now have a different way of expressing the problem. “A balloon is filled with 60 litres of air at 4 ata. What will be the approximate volume of the balloon if it taken to 10 ata?” By applying our rule of taking the quantity (60 litres) to the surface from 4 ata (30 metres), we discover that it will expand to 240 litres. Now all that remains is returning the quantity (240 litres) to 10 ata, divide 240 litres by 10 to get 24 litres.

Imperial (a) A balloon is filled with 2 cubic feet of air at 100 feet of

seawater. What will be the approximate volume of the balloon if it is taken to a depth 300 feet? 0.8 cubic feet. First, as we are dealing in approximations, let’s assume that 100 feet is close enough to 99 feet to use the pressure of 4 ata. Next, let’s find out how much pressure is exerted by 300 feet of depth. By dividing 33 into 300 we determine that we are at a pressure of about 9 atm gauge or 10 ata. (Actually, 297 feet is 10 ata, but close enough). We now have a different way of expressing the problem. “A balloon is filled with 2 cubic feet of air at 4 ata. What will be the approximate volume of the balloon if it taken to 10 ata?” By applying our rule of taking the quantity (2 cubic feet) to the surface from 4 ata (100 feet), we discover that it will expand to 8 cubic feet. Now all that remains is returning the quantity (8 cubic feet) to 10 ata, or 8/10. Reduce the fraction or divide and the answer is .8 cubic feet.




4. Metric (a) A balloon containing 300 litres of air at 7 metres of seawater is

taken to a depth of 26 metres. What will be the exact volume of the balloon upon reaching 26 metres? 141.66 litres. Now it’s time to earn our money because we are no longer dealing in increments of whole atmospheres. (Actually, this is beyond the scope of the objective, but give it a try, anyway). There is an equation that will enable us to be more precise and find quantities when the depth changes are not expressed in increments of whole atmospheres. This equation is P1 x V1 = P2 x V2 (P = pressure, V = volume), and although it may appear complex, it only requires simple arithmetic.

First, we must turn the data given to us expressed as “depth” into expressions of “pressure.” From the previous objective on pres-sure, we should be able to determine that 7 metres of seawater exerts 1.7 atmospheres (absolute):

(7m x 0.100 atm/m) + 1 atm = 1.7 ata. And, that 26 metres of seawater exerts 3.600 atmospheres absolute: (26m x 0.100 atm/m) + 1 atm = 3.6 ata.

Next, we need the equation: P1 X V1 = P2 X V2. As we now have three of the four variables provided to use, the rest is merely a matter of arithmetic — after we plug in the variables:

The first pressure variable (P1) is 1.7 atmospheres absolute (7 metres of depth) P1 = 1.7

The second pressure variable (P2) is 3.6 atmospheres absolute (26 metres of depth) P2 = 3.6

The first volume variable (V1) is given as 300 litres. V1 = 300 litres

The second volume variable is what we are trying to find. V2 = X

By arranging the data according to the original equation (P1 x V1 = P2 x V2), we determine: 1.7 x 300 = 3.6 x “X,” where “X” is the unknown variable.

In order to get the unknown variable alone, we divide each side by 3.6 or 510 ÷ 3.6 = 3.6 ÷ 3.6 x X

The 3.6 ÷ 3.6 equals 1, and 1 times any quantity equals itself (1 x X = X).

Therefore, the 3.6 ÷ 3.6 is irrelevant and cancels out. So, we are left with: 510 ÷ 3.6 = X

And by dividing 510 by 3.6, we determine that: X = 141.66 litres.




Imperial (a) A balloon containing 10 cubic feet of air at 25 feet of

seawater is taken to a depth of 85 feet. What will be the exact volume of the balloon upon reaching 85 feet? 4.91 cubic feet. Now it’s time to earn our money because we are no longer dealing in increments of whole atmospheres. (Actually, this is also beyond the scope of the objective, but give it a try, anyway). There is an equation that will enable us to be more precise and find quantities when the depth changes are not expressed in increments of whole atmospheres. This equation is P1 x V1 = P2 x V2 (P = pressure, V = volume), and although it may appear complex, it only requires simple arithmetic.

First, we must turn the data given to us expressed as “depth” into expressions of “pressure.” From the previous objective on pressure, we should be able to determine that 25 feet of seawater exerts 25.82 psia (absolute):

(25 ft x .445 psi/ft) + 14.7 psi = 25.82 psia. And, that 85 feet of seawater exerts 52.52 psia:

(85 ft x .445 psi/ft) + 14.7 psi = 52.52 psia. Next, we need the equation: P1 X V1 = P2 X V2. As we now have three of the four variables provided to use, the rest

is merely a matter of arithmetic — after we plug in the variables: The first pressure variable (P1) is 25.82 psia (25 feet of depth)

P1 = 25.82 psia The second pressure variable (P2) is 52.52 psia (85 feet of depth)

P2 = 52.52 psia The first volume variable (V1) is given as 10 cubic feet.

V1 = 10 cubic ft The second volume variable is what we are trying to find.

V2 = X By arranging the data according to the original equation

(P1 x V1 = P2 x V2), we determine: 25.82 x 10 = 52.52 x “X”, where “X” is the unknown variable.

In order to get the unknown variable alone, we divide each side by 52.52 or 258.2 ÷ 52.52 = 52.52 ÷ 52.52 x X

The 52.52/52.52 equals 1, and 1 times any quantity equals itself (1 x X = X).

Therefore, the 52.52 ÷ 52.52 is irrelevant and cancels out. So, we are left with: 258.2 ÷ 52.52 = X

And by dividing 258.2 by 52.52, we determine that: X = 4.91 cubic feet




Objective 1.8 Explain the relationship between depth and the density of the air a diver breathes, and calculate this relationship in increments of whole atmospheres.

1. A scuba tank is filled to capacity at the surface. When this tank is used at a depth of 30 metres/99 feet in the sea, the air within the tank is four times more dense than it was at the surface. False. Remember, a scuba tank is an inflexible container — it does not decrease in size as depth (pressure) increases. Therefore, the air within the tank is unaffected by the increasing water pressure outside the tank.

2. Because a diver’s lung volume must remain constant regardless of the depth at which he breathes, the density of the air in the diver’s lungs does not change even as he changes depth. False. When we breathe, we must breathe a “full” breath regardless of the pressure surrounding us. And, to be able to breathe, the external pressure must be equal to the air we are breathing. You can demonstrate this in a swimming pool using a short length of hose. Begin by breathing from one end of a one metre/three-foot length of hose. Next, submerge just below the surface and breath. As when using a snorkel, you will encounter relatively little breathing resistance. Now, while keeping one end of the hose above the water, submerge to about one metre/three feet of depth. You will quickly find that breathing becomes extremely difficult, if not impossible. Yet, the effect is caused by less than a 0.12 atm/2 psi pressure differential between the surface air pressure and the water pressure surrounding your chest.

3. (b) The air that a diver breathes from a scuba tank at 50 metres/165 feet of seawater is 6 times as dense as the air breathed from the same tank at the surface. This relates to the above question. In order to breathe underwater, the air must be at the same pressure as the surrounding water pressure. In this instance, the diver is in 50 metres/165 feet. The depth exerts 5 atm gauge pressure (50 ÷ 10 = 5)/(165 ÷ 33 = 5). Add 1 additional atm for the surface pressure and we get 6 ata. At 6 ata the pressure is six times greater than at the surface. Therefore, the air the diver breathes must be 6 times more dense in order to fully inflate his lungs at that depth.








Objective 1.9 Given a diver’s air consumption rate at one depth, calculate how that consumption rate changes when depth changes.

1. (d) A diver has an air consumption rate of 2 bar per minute/ 25 psig per minute at the surface. If all other factors but depth remain unchanged, what will his consumption rate be at 40 metres/132 feet of seawater? 10 bar per minute/ 125 psig per minute. At 40 metres/132 feet the pressure is 5 ata. Therefore, the diver must breathe air five times more dense than at the surface. If his rate at the surface (1 ata) is 2 bar per minute/25 psig per minute, then five times that rate is 10 bar per minute/125 psig per minute, assuming all other factors are unchanged.

2. (d) A diver has an air consumption rate of 60 litres/2 cubic feet per minute at 10 metres/33 feet of seawater. If all other factors but depth remain unchanged, what will his consumption rate be at 30 metres/99 feet? 120 litres/4 cubic feet per minute. As in previous problems involving pressure and volume, the key here is to first determine the rate at the surface. In this case, a diver who consumes 60 litres/2 cubic feet of air per minute at 10 metres/33 feet (2 ata), will consume half that amount — 30 litres/1 cubic foot — at the surface. (Remember, air consumption is a function of density; it decreases as pressure is reduced). Therefore, if all factors are equal, at 30 metres/99 feet (4 ata) he will consume 4 times the amount of air. Four times the surface amount (30 litres/1 cubic foot) is 120 litres/4 cubic feet.

3. (b) A diver has an air consumption rate of 90 litres/3 cubic feet per minute at 20 metres/66 feet of seawater. If all other factors but depth remain unchanged, what will his consumption rate be at 60 metres/200 feet? 210 litres/7 cubic feet per minute. Similar to the above question, we must first determine the surface consumption rate. In this case, the diver will consume one third the amount of air at the surface as at 20 litres/66 feet (3 ata), or 30 litres/1 cubic foot per minute. Next, we must determine what pressure 60 metres/200 feet of water exerts in terms of whole atmospheres. By dividing 10 metres/33 feet into 60 metres/ 200 feet, we determine that 60 metres/200 feet is 6 atm gauge, 7 atm absolute. As we must always work in absolute terms, we now know that his air consumption rate will increase to slightly more than seven fold at 60 metres/200 feet. Note: while this theoretical example assumes a dive to 60 metres/200 feet, the maximum depth for actual recreational dives is never to exceed 40 metres/130 feet.








4. (d) A diver has an air consumption rate of 60 litres/2 cubic feet per minute at 10 metres/33 feet of seawater. If all other factors but depth remain unchanged, what will his consumption rate be, in psig/bar per minute, at 30 metres/100 feet? The answer cannot be determined from the data provided. Always pay close attention to the way data is described. As the saying goes, “You can’t add apples and oranges.” If the question asks for the response to be in litres/cubic feet, and provides no way of converting it to another increment (bar/minute)/(psig/minute), then the answer must also be expressed in litres/cubic feet.

Objective 1.10 Describe how the behavior of a gas within both a flexible and inflexible container is affected by changes in pressure and temperature.

1. (b) A balloon is filled with 30 litres/1 cubic foot of air at room temperature. Describe what would happen to that balloon if it were put into a freezer at a constant ambient pressure. The volume would decrease. When placed in a freezer, the energy contained in the gas molecules within the room temperature tank will begin to dissipate. As the temperature decreases, the motion of the molecules decreases. As the motion decreases, the force of impact of their collisions with each other and the sides of the balloon decreases. Less force of impact means less pressure inside the balloon, and it would therefore shrink, or decrease in volume. This is a description of Charles’ Law which states, “the amount of change in volume of gas is directly proportional to the change in the absolute temperature at a constant pressure.”

2. (c) A scuba tank is filled to capacity at room temperature. Describe what would happen to that tank if it was taken on an ice dive (water at or near freezing). The volume would remain unchanged, but the pressure would decrease. This is similar to the above question, with one important difference — the container is inflexible. While the pressure would decrease for the same reasons described previously, the walls of the container will remain rigid. Therefore, only the pressure will decrease, not the volume of the container.

3. (b) An 12 litre/80 cubic foot scuba tank is filled to 200 bar/3000 psig at an ambient temperature of 27°C/80°F. If the tank is then used in water temperature of 4°C/40°F, what would be the approximate tank pressure? 186 bar/2800 psig. There is a way to roughly estimate the effect of temperature on the pressure within










a scuba tank. When using the Fahrenheit scale, the rule of thumb is that for every change in temperature of one degree, a corresponding change of 5 psig will occur. On the Centigrade scale, the rule of thumb is that for every change in temperature of one degree, a corresponding change of 0.6 bar will occur. The imperial question is expressed in Fahrenheit, and a 40° drop occurs, resulting in about a 200 psig reduction in pressure (40 x 5 = 200). Therefore, 3000 psig - 200 psig = 2800 psig. The metric question is expressed in Celsius, and a 23° drop occurs, resulting in about a 14 bar reduction in pressure (23 x 0.6 = 14). Therefore, 200 bar - 14 bar = 186 bar.

4. Metric (b) A 12 litre scuba tank is filled to 200 bar at an ambient

temperature of 26°C. What will the tank pressure be if the tank is used in water temperature of 7°C? 187 bar. Now it’s time for another challenge. In order to be as precise as the question asks we need to use a special equation. (As in the previous objective on pressure/volume relationships, this is actually beyond the scope of this objective, but let’s give it a try). First, let’s look at the equation:

P1 x V1/T1 = P2 x V2/T2 Where P = pressure, V = volume and T = temperature As we are dealing with a scuba tank, its volume will remain

unchanged. Therefore, we can ignore the volume variable altogether, restructuring the equation as follows:

P1/T1 = P2/T2 Next, let’s determine what we know from the problem: The

pressure the tank is filled to is 200 bar (P1), and it asks us to determine what the pressure will become (P2). But, these are not yet proper variables to plug into the equation. To predict the behavior of gases we must always work in absolute terms. So, we must add 1 atm (or bar) to the 200 bar for 201bar. This then becomes the quantity to insert into the equation.

The temperature of the tank is 26°C, and it will be used in 7°C water. But, as stated before, we must work in absolute terms. With temperature, this means working in terms of absolute zero. Readings in degrees Celsius must be converted into degrees Kelvin. To convert a Celsius reading into a Kelvin reading, just add 273.° Therefore, 26°C, becomes 299°K (T1); and 7°C, becomes 280°K (T2).

Now we can plug the values into the formula: 201 ÷ 299 = X ÷ 280, where “X” is the unknown.

This can be rearranged to become: (280 x 201) ÷ 299 = X




And by doing the simple arithmetic, we determine that: 188 bar = X. But this still is not the correct answer since

188 bar is absolute pressure. We need to subtract 1 bar from 188 bar to arrive at the tank’s real pressure. Therefore, 188 - 1 = 187 bar.

Imperial (b) An 80 cubic foot scuba tank is filled to 3225 psig at an

ambient temperature of 78°F. What will the tank pressure be if the tank is used in water temperature of 44°F? (Assume 1 atm = 15 psi) 3020 psig. Now it’s time for another challenge. In order to be as precise as the question asks we need to use a special equation. (As in the previous objective on pressure/volume relationships, this is actually beyond the scope of this objective, but let’s give it a try). First, let’s look at the equation:

P1 x V1/T1 = P2 x V2/T2 Where P = pressure, V = volume and T = temperature. As we are dealing with a scuba tank, its volume will remain

unchanged. Therefore, we can ignore the volume variable altogether, restructuring the equation as follows:

P1/T1 = P2/T2 Next, let’s determine what we know from the problem: The

pressure the tank is filled to is 3225 psig (P1), and it asks us to determine what the pressure will become (P2). But, these are not yet proper variables to plug into the equation. To predict the behavior of gases we must always work in absolute terms. So, we must add 15 psi to the 3225 psi for 3240 psia. This then becomes the quantity to insert into the equation.

The temperature of the tank is 78°F, and it will be used in 44°F water. But, as stated before, we must work in absolute terms. With temperature, this means working in terms of absolute zero. Readings in degrees Fahrenheit must be converted into degrees Rankin. To convert a Fahrenheit reading into a Rankin reading, just add 460.° Therefore, 78°F, becomes 538°R (T1); and 44°F, becomes 504°R (T2).

Now we can plug the values into the formula: 3240 ÷ 538 = X ÷ 504, where “X” is the unknown.

This can be rearranged to become: 504 x 3240 ÷ 538 = X

And by doing the simple arithmetic, we determine that: 3035 psia = X. But this still is not the correct answer since

3035 psia is absolute pressure. We need to subtract 15 psi from 3035 psi to arrive at the tank’s real pressure. Therefore, 3035 - 15 = 3020 psig.




Objective 1.11Given their percentages, calculate the partial pressures of gases in a mixture at any depth.

1. (a) In a mixture of air comprised of 20% oxygen and 80% nitrogen, at an ambient pressure of 1 ata/15 psia, what is the partial pressure of oxygen? 0.2 ata/3 psia. The partial pressure of a gas is dependent upon two variables: the absolute pressure, and the percentage of the gas in question. In this case, we are told to assume the absolute pressure is 1 ata/15 psia, and the gas — oxygen —comprises 20% of the total. So, by simply multiplying the absolute pressure (1 ata)/(15 psia) times the percentage of oxygen (20%), we find that the partial pressure is 0.2 ata (1 x 0.2 = 0.2)/3 psia (15 x .20 = 3). The English scientist John Dalton summarized his experience with the behavior of gases within mixtures via Dalton’s Law, which states, “the total pressure exerted by a mixture of gases is equal to the sum of the pressure of each of the different gases making up the mixture — each gas acts as if it alone were present and occupied the total volume.”

2. (b) A gas mixture is comprised of 21% oxygen, 78% nitrogen and 1% carbon dioxide. At a depth of 20 metres/66 feet of seawater, what is the partial pressure of the nitrogen? 2.34 ata/34.39 psia. Similar to the above question, we must first determine the absolute pressure. At 20 metres/66 feet this is 3 ata (1 x 3 = 3)/44.1 psia (14.7 x 3 = 44.1). Next, determine the percentage of the gas in question. In this case, nitrogen comprises 78% of the total. So, 3 x .78 = 2.34 ata/44.1 x .78 = 34.39 psia.

3. (b) A gas mixture is comprised of 21% oxygen, 78% nitrogen and 1% carbon dioxide. At a depth of 24 metres/78 feet of seawater, what is the partial pressure of the oxygen? 0.7 ata/10.37 psia. This problem is identical to the two previous, except we must determine the absolute pressure using a different method. As the depth given — 24 metres/78 feet of seawater — is not an increment of a whole atmosphere, we must derive the pressure using methods we explored in Objective 1.6. Therefore, to find the absolute pressure, we multiply 24 metres/78 feet times the seawater constant of .100 atm/m/.445 psi/ft. Then, add the atmospheric pressure of 1 atm/14.7 psi to determine the pressure as 3.4 ata/49.41 psia: imperial calculation — (78 x .445) + 14.7 = 49.41; metric calculation — (24 x .100) + 1 = 3.4. Next, we use the percentage of oxygen in the mixture, which is given as 21%, to find that the partial pressure is 0.7 ata/10.37 psia: imperial calculation — .21 x 49.41 = 10.37; metric calculation — .21 x 3.4 = 0.7.








Objective 1.12Explain the effect of breathing contaminated air mixtures at depth and calculate the equivalent effect such contamination would have upon the diver at the surface.

1. (a) A scuba tank is accidentally filled with 1% carbon monoxide. If a diver breathes air from this tank at a depth of 30 metres/100 feet of seawater, approximately what percentage of carbon monoxide will he be breathing? 1%. Once a tank is filled, the percentages of gases within it cannot change (unless more air is added). Therefore, if only 1% of a gas is put into a tank, no more or less than 1% can be in the tank regardless of the pressure to which the tank is subjected. It may be noted however, that the partial pressure of each gas breathed at depth does increase.

2. (d) Referring to the previous question, it is determined that at the surface the diver inhales 500,000 molecules of carbon monoxide with each breath. Therefore, when breathing the air at 30 metres/100 feet of seawater (assuming all other factors but depth are unchanged), he would breathe approximately how many molecules? 2,000,000. This question relates back to what we explored in Objective 1.8. Remember that in order to breathe underwater (under increased pressure), the air that is breathed must be at the same pressure as the surrounding water. This makes the air more dense. As a result, the diver will inhale more molecules. In this problem the diver is breathing air at 30 metres/100 feet, which is (approximately 99 feet) or 4 ata. Therefore, the air he is breathing must be four times more dense than at the surface. He must then inhale four times the number of molecules. So, if he inhales 500,000 in each breath at the surface, he must inhale 2,000,000 at a depth of 30 metres/100 feet. Note that while significantly more molecules reach the diver’s lungs at depth, the actual percentage of the gas in the tank is still unchanged.

3. (d) Again referring to question 1, breathing the contaminated air at 30 metres/100 feet of seawater would have the same effect on the diver as breathing what percentage of carbon monoxide at the surface? 4%. In order to understand this question more clearly, ask yourself, “what percentage of gas would the diver have to breathe at the surface in order to get the 2,000,000 molecules of carbon monoxide?” 2,000,000 molecules is four times the amount of gas he was breathing at the surface, and the mixture contains only 1% carbon monoxide. Therefore, he would require 4 times the amount of carbon monoxide at the surface — 4% — in order to get the








same effect as breathing only 1% at 30 metres/100 feet. That’s why this phenomenon is referred to as “surface equivalency.” By the way, a gas mixture of 4% carbon monoxide at the surface would be extremely poisonous.

4. (c) Breathing from a contaminated air source with 1.5% carbon monoxide at a depth of 90 metres/300 feet of seawater would have the same effect as breathing approximately what percentage of carbon monoxide at the surface? 15%. This question is just a slightly more complicated version of the previous. First, we saw from questions involving previous objectives that 90 metres/ 300 feet is close enough to be considered 10 atm. Then, if we are breathing a gas of a certain percentage, it will have 10 times the effect on us at 10 ata. In this case, the percentage of contamination is 1.5%. So, it would have the same effect at 90 metres/300 feet as breathing 15% at the surface. (1.5 x 10 = 15).

Objective 1.13Explain what will occur to a liquid saturated with a gas at high pressure when the pressure of the gas in contact with the liquid is quickly reduced.

1. (b) A glass of water has been placed in a vacuum for several days. It no longer contains any dissolved gas within it. If it is then placed in a pressure pot and pressurized to 2 ata for several days, what will be the gas pressure within the liquid? 2 ata. There is a tendency for a state of equilibrium to exist between the pressure within a liquid (gas tension), and the pressure of the gas in contact with that liquid. This equilibrium will be maintained until the pressure in contact with the liquid changes. This phenomenon was first explained by chemist William Henry. Through experimentation Henry concluded — as Henry’s Law states — that, “the amount of gas that will dissolve into a liquid is almost directly proportional to the partial pressure of that gas.”

2. (a) If the pressure within the pot in the previous question is increased, the pressure of the gas within the liquid will: increase. Henry’s Law is a direct relationship. Therefore, if one factor increases, so will the other. Think of it in terms of everyday experience. When you uncap a bottle of soda, the pressure of the carbon dioxide decreases because the pressure of the gas in contact with the soda decreased. Eventually, the gas in the soda will equal the pressure of the surrounding air. In other words, the soda will “go flat.” If the phenomenon was an inverse relationship (one variable increases if one decreases), the soda would actually have to take on more gas as it sat open. This, of course, violates common sense, and our own practical experience.








3. (b) If a vacuum is created within the pressure pot in question 1, the pressure of the gas within the liquid will: decrease. This question is simply the reverse of the previous. A vacuum would represent zero pressure in contact with the liquid. Therefore, the tendency would be for any gas contained in the liquid to come out. So, the pressure would decrease.

Objective 1.14Define “supersaturation” and explain what conditions are necessary for gas bubbles to form in a supersaturated liquid.

1. (b) Assume that the atmosphere contains 80% nitrogen. At sea level, the nitrogen pressure within our tissues would be: 0.8 ata. Our tissues are primarily liquid. Therefore, if the gas in contact with our tissues is comprised of 80% nitrogen, then our tissues will contain a gas tension of 80% nitrogen. 80% of 1 ata is .8 ata (.80 x 1 = .8).

2. (a) From the above question, the total gas pressure within our tissues is: 1.0 ata. As the absolute pressure in contact with our tissues is 1.0 ata, then the absolute gas tension within our tissues must be 1.0 ata.

3. (b) Referring to question 2, because our tissues cannot hold any more gas pressure at sea level, our body is referred to as being: saturated. Saturation refers to the fact than there is no net exchange of gases between the tissues (liquid) and the gas in contact with them. This is the state of equilibrium we referred to in Objective 1.13.

4. When breathing at depth, the tissues of a diver begin to take on additional gas pressure. If the diver remains at depth long enough, his tissues will again equalize with the ambient pressure. True. This is merely a logical extension of what we have been exploring. When the diver breathes air underwater, he is breathing gas at a higher pressure than at the surface. Therefore, the pressure of the gas in contact with his tissues increases. As the gas pressure increases so must the gas tension within the tissues. Conversely, when the gas in contact with the tissues decreases, the gas tension within the tissues also decreases. This phenomenon is called decompression.












5. The condition described in the above question is referred to as supersaturation. False. We discussed previously that at the surface our tissues, or any liquid, can only hold an amount of gas equal to 1 atm. Regardless of how long we remain at 1 atm, our tissues can never achieve a gas tension of more than that amount. However, there is one way more gas can be absorbed — increase the pressure in contact with the tissues. At a greater ambient pressure the tissues will again equalize with the gas in contact with them. Therefore, as the pressure of the gas in contact with the tissue increases, so does the pressure within the tissues. But after time, if the ambient pressure is reduced, the reverse phenomenon will occur. More pressure or gas tension will exist within the tissue than the ambient pressure. In order words, a “pressure gradient” will exist in favor of the tissue. In this case the tissue is referred to as supersaturated, as it contains a higher gas tension that the ambient pressure. (Just as the gas tension within a can of soda is higher than the ambient pressure when the top is initially removed). For the diver this is an extremely important phenomenon to understand.




Section AnalysisFrom the answer key, identify any items marked "correct-guess" or "incorrect-lack of knowledge." These items represent important points of information or concepts you still might not fully understand. Check below any objectives that contained items with a "correct-guess" or "incorrect-lack of knowledge" response. Completing this section is an important step in determining your understanding of physics as it relates to recreational diving.

Confident Guess Total

Correct Responses

Simple Mistake Lack of Knowledge Total

Incorrect Responses

Objectives To Be Reviewed:

1.1 1.2 1.3 1.4 1.5 1.6 1.7

1.8 1.9 1.10 1.11 1.12 1.13 1.14

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DKW 1 Physics