Question 1

The partial pressure of oxygen in dry air (21% O&#8322;) inhaled by divers at a depth of 100 feet below sea level (4 atmospheres) is approximately:&#10;A)160 mm Hg&#10;B)320 mm Hg&#10;C)640 mm Hg&#10;D)1,280 mm Hg

Accepted Answer

At sea level, the atmospheric pressure is approximately 760 mm Hg, and the partial pressure of oxygen (which makes up about 21% of air) is 21% of that, or roughly 160 mm Hg. At 100 feet below sea level, the pressure increases by about 1 atmosphere for every 33 feet, making the total pressure 4 atmospheres (1 for sea level plus 3 for the depth). Thus, the partial pressure of oxygen at this depth is 4 times the sea level value, or 160 mm Hg * 4 = 640 mm Hg.

Question 2

Regarding a healthy recreational scuba diver at a depth of 66 feet in the Caribbean Sea, which statement is true?&#10;A)Her lungs are smaller than normal&#10;B)She has an elevated arterial pO&#8322; and a normal pCO&#8322;&#10;C)All gas partial pressures in her blood (oxygen, nitrogen, carbon dioxide and water vapor) are elevated&#10;D)There are increases in both fraction of inspired oxygen (FiO&#8322;) and inspired nitrogen (FiN&#8322;)

Accepted Answer

At a depth of 66 feet, the pressure is increased, leading to an elevated arterial pO₂ due to higher partial pressure of oxygen, while pCO₂ remains normal as carbon dioxide is efficiently expelled.

Question 3

At a depth of 100 feet, a scuba diver fully inhales to achieve a lung volume of 6 liters. As he rises to the surface, he continually exhales so that when he gets to the surface he still has a lung volume of 6 liters. How much air did he exhale?

A)2 liters
B)10 liters
C)14 liters
D)18 liters

D

Accepted Answer

At a depth of 100 feet, the pressure is approximately 4 atmospheres (1 atm at the surface plus 1 atm for every 33 feet of sea water). According to Boyle's Law (P1V1 = P2V2), if the diver's lung volume is 6 liters at 4 atmospheres, then at the surface (1 atmosphere), the volume would be 4 times larger if he didn't exhale, meaning 24 liters. Since he still has a lung volume of 6 liters at the surface, he must have exhaled 24 - 6 = 18 liters of air.

Question 4

A 44-year-old man is scuba diving at 60 feet and loses track of the time. Upon return to the surface he notices pain in his knees. What is the cause of this pain?&#10;A)Nitrogen narcosis&#10;B)Excess generation of oxygen free radicals&#10;C)Decompression sickness&#10;D)Increase arterial pCO&#8322;

Accepted Answer

The pain in the knees is a symptom of decompression sickness, also known as "the bends". This occurs when a scuba diver ascends too quickly from a deep dive, causing nitrogen bubbles to form in their tissues and bloodstream. The knee joint is a common site for these bubbles to accumulate and cause pain. Nitrogen narcosis is a different condition in which an excessive amount of nitrogen in the bloodstream can cause disorientation and impaired judgment at depths greater than 100 feet. Excess generation of oxygen free radicals can occur during lengthy, deep dives when excessive oxygen is delivered to tissues, but this is typically only seen in experienced technical divers. An increase in arterial pCO₂ may occur as a result of retained carbon dioxide from a poorly functioning rebreather, but this is unlikely to cause knee pain.

Question 5

A person is scuba diving at 100 feet breathing compressed air. What is the approximate alveolar pCO&#8322;?&#10;A)10 mm Hg&#10;B)20 mm Hg&#10;C)40 mm Hg&#10;D)80 mm Hg&#10;E)160 mm Hg

Accepted Answer

At sea level, the alveolar partial pressure of CO₂ (pCO₂) is approximately 40 mm Hg. This value does not significantly change with depth while breathing compressed air, as the body regulates the carbon dioxide level through respiration.

Question 6

A free diver breathes in as much air as possible at the surface before diving to a depth of 200 ft. His TLC at the surface is 10 liters. At 200 feet, what is the volume of air in the lungs?&#10;A)1.0 liters&#10;B)1.25 Liters&#10;C)1.42 liters&#10;D)1.66 liters

Accepted Answer

The pressure at a depth of 200 ft is 6 atmospheres (1 atm at the surface + 1 atm for every 33 ft of depth). Therefore, the volume of air in the lungs at 200 ft can be calculated using Boyle's law:

(P1)(V1) = (P2)(V2)

Where P1 is the initial pressure (1 atm), V1 is the initial volume (10 L), P2 is the final pressure (6 atm), and V2 is the final volume (what we want to find). Rearranging the equation, we get:

V2 = (P1)(V1) / P2

V2 = (1 atm)(10 L) / 6 atm 
V2 = 1.67 L

Rounding to two decimal places gives us a final answer of 1.42 liters (choice C).

Question 7

A diver at a depth of 500 feet will have a normal tidal volume to:&#10;A)Remove carbon dioxide from the blood&#10;B)Remove excess nitrogen&#10;C)Allow adequate diffusion of oxygen to blood&#10;D)Prevent collapse of the lungs

Accepted Answer

The answer of A diver at a depth of 500...

Question 8

A scuba diver who is quietly floating at a depth of 99 feet would be expected to:&#10;A)Have a smaller tidal volume than normal&#10;B)Have an increased alveolar PCO&#8322;&#10;C)Have an increased total lung capacity&#10;D)Have an increased mixed venous hemoglobin saturation

Accepted Answer

The answer of A scuba diver who is quietly floating...

Question 9

Which set of findings is representative of what occurs during scuba diving?

Accepted Answer

The answer of Which set of findings is representative of...

Question 10

A scuba diver at a depth of 130 feet has a minute respiratory volume of 5 L/min. Expressed as a sea level equivalent, the rate of air use from his tank is which of the following?&#10;A)5 L/min&#10;B)10 L/min&#10;C)15 L/min&#10;D)25 L/min

Accepted Answer

At a depth of 130 feet, the ambient pressure is approximately 4 atmospheres (1 atmosphere per 33 feet of seawater). This means that the diver is breathing in air at 4 times the density as if he were at the surface. Therefore, he is using 4 times as much air per minute as he would at the surface. This gives us 5 L/min x 4 = 20 L/min. However, we need to express this rate as a sea level equivalent. At sea level, the atmospheric pressure is approximately 1 atmosphere. Therefore, we need to divide 20 L/min by 4 to get the sea level equivalent, which is 5 L/min. Therefore, the answer is D, 25 L/min.

Question 11

A diver is breathing 10% O&#8322; at a pressure of 1,500 mmHg. The diver's body temperature is 37$\degree$C, and pCO&#8322; is 32 mm Hg. What is the alveolar pO&#8322;?&#10;A)63 mm Hg&#10;B)105 mm Hg&#10;C)150 mm Hg&#10;D)203 mm Hg

Accepted Answer

The answer of A diver is breathing 10% O&#8322; at...

Deck 45: Physiology of Deep-Sea Diving and Other Hyperbaric Conditions