Exam 2 Lecture 11 (3-6-23) The Lost Pulmonary Lecture (Alveolar Compliance, Airway Disorder, Surface Tension, Surfactant) (Andy's Cards) Flashcards
In terms of alveolar compliance, what does a steep slope tell us?
What does a horizontal slope tell us?
Steep slope = compliant
Horizontal slope = no compliance
At FRC, alveoli at the ______ of the lungs are way more compliant than the alveoli at the _____ of the lung.
At FRC, alveoli at the base of the lungs are way more compliant than the alveoli at the apex of the lung.
What is the lung volume of this graph?
1.5 Liters (Residual Volume)
What would be the average intrathoracic pressure of this graph?
Intrathoracic Pressure = +1.3 cmH2O
1.3 mmHg will be the halfway point (average ) of -2.2 cmH2O and +4.8 cmH20
At a pleural pressure of -2.2 cm H2O on the alveolar compliance graph at RV, what is the percent fullness of the alveoli?
30% full
At a pleural pressure of +4.8 cm H2O on the alveolar compliance graph at RV, why is the slope 0?
The small airways at the base of the lungs have collapsed as a result of positive intrathoracic pressure trying to force all the air out of the alveoli.
What is the reason why alveolar volume doesn’t dip below 20% fullness?
The pressure used to force the air out also forces the small airways to close. The likelihood of collapsing an airway before the alveoli are completely empty is the reason why alveoli volume does drop below 20% fullness.
This term is used to describe areas of the lung more affected by gravity in terms of blood flow and airflow.
Dependent
For an upright position, the dependent portion of the lung would be the inferior lungs.
What does PAO2 stand for?
Alveolar PO2 or Partial Pressure of Oxygen in the Alveolus
Calculate the normal Minute Alveolar Ventilation rate.
What will Alveolar PO2 be at the calculated Minute Alveolar Ventilation rate?
4.2 L/min
Respiratory Rate x Alveolar Volume
12 breaths/min x 350 mL = 4200 mL/min
At 4.2 L/min, PAO2 will be about 100 mmHg
Calculate the normal Minute Dead Space Ventilation rate.
1.8 L/min
Respiratory Rate x Dead Space Volume
12/breaths/min x 150 = 1800 mL/min
1.8 L/min
According to this graph as alveolar ventilation increases, what happens to PAO2 assuming that the metabolic rate is unchanged?
Alveolar PO2 will increase.
If we are breathing room air, PAO2 will never be above _________.
150 mmHg
Our alveolar PO2 could never be above 150mmHg. Because we can’t take in oxygen with a higher PO2 than from our normal environment.
What is the PCO2 of inspired air?
What is the PACO2 under normal alveolar ventilation after equilibrium according to this graph?
0 mmHg
Normal alveolar ventilation is 4.2 L/min. PACO2 will be about 40 mmHg.
Normally after gas exchange, our alveolar PCO2 is about _______ if we bring in twice as much fresh air, PCO2 should drop down to ________.
40 mmHg
20 mmHg
If there is a normal amount of alveolar ventilation and an abnormally high metabolic rate. What would be expected from the alveolar PCO2?
Under the same conditions, what would be expected from alveolar PO2?
PACO2 will increase.
PAO2 will decrease.
Metabolism will result in more CO2 being produced in the body, and if alveolar ventilation is constant at 4.2 L/min, PACO2 will increase and PAO2 will decrease.
If there is a normal amount of alveolar ventilation and an abnormally low metabolic rate. What would be expected from the alveolar PCO2?
Under the same conditions, what would be expected from alveolar PO2?
PACO2 will decrease.
PAO2 will increase.
With low metabolic rates, the body is producing less CO2. If alveolar ventilation is constant at 4.2 L/min, PACO2 will decrease and PAO2 will increase.
Calculate the oxygen concentration after it has been inspired and displaced by water vapor.
Partial Pressure of O2 = [O2] x Total Pressure
[O2] = Partial Pressure of O2 / Total Pressure
[O2] = 149 mmHg / 760 mmHg
[O2] = 0.1961
[O2]= 19.61%