Flight Physiology

Question 1

Boyle’s Law (Boyle’s balloon)

Answer 1

As altitude increases, atmospheric pressure decreases and gas in an enclosed space will expand.
P = Pressure (represented as ATM
ATM = Atmospheres
V = Volume
P,V = Inverse relationship
…if pressure decreases (elevation increases), volume increases.
…if pressure increases (elevation decreases), volume decreases.
Below 5k ft, every 1500 ft elevation gain reduces ATM by 5%
P1(x)V1 = P2(x)V2

Question 2

Charles’ Law

Answer 2

The relationship between temperature and volume is proportional to constant atmospheric pressure.
As temp increases, volume of gas expands
As temp decreases, volume of gas decreases.

For every 1000’ ascended, temp will decrease 2 degrees C
For every 150 meters ascended, temp will decreased by 1 degree C

Question 3

Gay-Lussac’s Law

Answer 3

For a given mass and constant volume of gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature.

Question 4

Dalton’s law

Answer 4

The total pressure of a gas mixture is the sum of the partial pressures of all gases.
In addition, it describes how pressure is exerted by gas at various altitudes and how that pressure affects the partial pressure of the said gas.

Question 5

Henry’s Law

Answer 5

The amount of gas dissolved in a solution is directly proportional to the pressure exerted above the gas over the solution.

Question 6

Fick’s Law

Answer 6

The rate of diffusion of a gas across a permeable membrane is determined by the chemical nature of the membrane itself, the surface area of the membrane, the partial pressure gradient of the gas across the membrane, and the thickness of the membrane.

Question 7

Graham’s Law

Answer 7

The rate of diffusion of a gas is inversely proportional to the square root of its molecular weight. This means that diffusion will happen at a faster rate if the gas is thinner, and at a slower rate if the gas is heavier.

Question 8

Atmospheric Pressure

Answer 8

760 mmHg at highest point (sea level)
Measured in mmHg
OR
Torr (1/760th of a standard ATM)
Think of Torr and mmHg as the same.

Question 9

ATM (1 ATM). ***

Answer 9

=760 torr
=760 mmHg
=29.92 in Hg
=14.7 psi

Question 10

Boyle’s Law (formula)

Answer 10

P1(x) V1 = P2 (x) V2

P1 = starting atmosphere
V1 = starting volume (ml)
P2 = highest altitude
V2 = ending/highest volume (ml)

Question 11

Ruptured epigastrium example of Boyle’s Law

Answer 11

P1(x)P2 = P2(x)V2
1ATM (x) 4L = P2(x)V2
4L = 0.90 (x) V2
4.4L = V2 (epigastrium ruptured)

P1 = sea level, P2 = 3000ft (0.90 ATM)
max epigastrium vol = 4L

Question 12

ETCO2 waveform capnography

Answer 12

Gold standard for confirming ET tube placement (not colorimetric device). If waveform is lost, it indicates ET tube is lost/dislodged.

Question 13

Boyle’s Law clinical application (examples)

Answer 13

Pneumothoraces - will expand as altitude increases. (initial vs reassess lung sounds)
ETT cuffs - volume will change as altitude increases
Splits - will expand. reassess distal pulses
Flow rates (gravity) - place all gravity solutions on pressure bag (bags allow max pressure 300 mm Hg.

Question 14

Barotitis Media ***

Answer 14

aka airplane ear.
The air inside our ear expands as elevation increases (pressure drops)
DESCENT problem (the only one)
Pain can be overcome w/valsalva maneuver
In severe cases, ruptured eardrum
Application of Boyle’s Law

Question 15

Barosinusitis

Answer 15

Sinus inflammation that prevents equalization of pressure in sinus cavities
Painful
ASCENT issue
Relieved by descending
Application of Boyle’s Law

Question 16

Barobariatrauma ***

Answer 16

*Test question
Nitrogen narcosis as a result of the affinity of nitrogen for adipose tissue (fat)
Nitrogen dislodged from fat as pressure decreases (ascending)
Sx include MS change, motor impairment, visual sx, ear sx, N/V, dizziness
High incidence in morbidly obese pts (more adipose tissue = more stored nitrogen)
Tx = moving nitrogen off lipids
High flow O2 therapy is recommended to avoid this presentation (15L via NRB 10-15 min before flight, to dislodge nitrogen)
ASCENT problem
Application of Boyle’s Law

Question 17

Barodontalgia

Answer 17

Air trapped within mandible will expand upon ASCENT, causing intense pain
Descending will resolve problem
Application of Boyle’s Law

Question 18

Charles’ Law

Answer 18

At a constant pressure, there is a proportional relationship between temperature and volume of a gas.
As temp increases, volume of gas increases, and vice versa
Molecules expand and shrink w/temp.
Mostly impacts equipment
Every 1000ft of ascent = 2C temp decrease
Every 150m of ascent = 1C temp decrease

Question 19

Gay-Lussac’s Law

Answer 19

At a given mass and constant volume of gas, the pressure exerted on sides of its container is directly proportional to its absolute temperature.
Refers to exertion pressure of a gas within a container (not psi)
Smaller molecules = decreased exertion pressure

Question 20

Dalton’s Law

Answer 20

Describes pressure exerted by a gas at various altitudes (pressures)
Total pressure of gas mixture is sum of partial pressures of all the gases
P1+P2+P3+P4+P5=Ptotal (101.3kPa or 14.99 psi for atmospheric air)
Normal O2 percentage = 21%
Current atmospheric pressure (X) % oxygen in mixture
As sea level, what is partial pressure of oxygen in room air? 760 mmHg x 0.21 = 159.6
To overcome Dalton’s law, add more gas (add more O2)

Question 21

Henry’s Law (most important gas law)

Answer 21

The amount of gas dissolved in a solution is directly proportional to the pressure of the gas over the solution.
ie…The amount of oxygen dissolved in capillary blood is directly proportional to the pressure of the oxygen over the capillary blood.
Example: decompression sickness

Question 22

Solubility of O2

Answer 22

0.0244 mL/mmHg/mLH2O
(less soluble than CO2)

Question 23

Solubility of CO2

Answer 23

0.592 mL/mmHg/mLH2O
(24x more soluble than O2)

Question 24

Henry’s Law in practice

Answer 24

Increase FiO2 (100%) - changes concentration (Dalton’s Law)
Add PEEP - changes surface area (Fick’s Law)
Place gas under pressure (NC< NRB HFNC, BVM, (Henry’s Law)

Question 25

Graham’s Law

Answer 25

Gasses will defuse/migrate from a region of higher concentration (pressure) to a lower concentration until equilibrium is reached.

Question 26

Pressure Altitudes

Answer 26

Altitude 0 Atmospheres 1 (760mmHg)
Altitude 18k Atmospheres 0.5
Altitude 34k Atmospheres 0.25
Altitude 48k Atmospheres 1/8
Altitude 63k Atmospheres 1/16

Question 27

Water Pressures ***

Answer 27

Altitude 0 = 14.7psi = 1 ATM
Altitude -33 = 29.2psi = 2 ATM
Altitude -66 = 44.1psi = 3 ATM
Altitude -99 = 58.8psi = 4 ATM
Altitude -132 = 73.5psi = 5 ATM
Every additional 33ft under water = 1 additional ATM
SEA LEVEL = 1 ATM
The elevation of a body of water does not affect the pressure below the water surface

Question 28

Bedside PaO2

Answer 28

(700 torr (x) FiO2) - 50
700 (x) 50% (example) - 50
350-50 = 300 PaO2

Question 29

Flight environment stressors

Answer 29

Drugs (not dehydration)
Exhaustion
Alcohol
Tobacco
Hypoglycemia

Question 30

Effects of Hypoxia (subjective)

Answer 30

fatigue, anxiety, nausea, dizziness, headache, numbness/tingling, hot/cold flashes, blurred vision, agitation, euphoria

Question 31

Effects of Hypoxia (objective)

Answer 31

hyperventilation, confusion, poor decision making, cyanosis, lack of coordination, unconsciousness

Question 32

Stages of Hypoxia

Answer 32

Indifference (0-10k ft, SpO2 90-98%, decreased night vision over 4k ft, HR + RR increases

Compensatory (10-15k ft, SpO2 80-90%, night vision less than 50%, CNS sx ie poor judgement, irritability, decreased coordination, drowsiness

Disturbance (15-20k ft, SpO2 70-80%, impaired memory, judgement, reliability, understanding, impaired psychomotor functioning, hyperventilation/cyanosis

Critical (20-25k ft, SpO2 60-70%, inability to remain upright, lack of coordination, seizures, rapid LOC, coma/death

Question 33

Time of useful consciousness in rapid depressurization ***

Answer 33

43k ft. 9-12 sec
40k ft. 15-20 sec (likely question. answer = single digit seconds)
35k ft. 30-60 sec
30k ft. 1-2 min
28k ft. 2-3 min
25k ft. 3-5 min
22k ft. 5-10 min
18k ft. 20-30 min

Question 34

Desired O2 per altitude ***

Answer 34

(%FiO2 x P1) / P2 = FiO2 at new altitude
P1: current barometric pressure
P2: new barometric pressure at altitude