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MULTIPLE CHOICE
1. What instrument is used to measure Patm ?
a. barometer c. altimeter
b. hygrometer d. dynameter
ANS: A
Feedback
A A barometer is used to measure barometric (PB) or atmospheric (Patm) pressure.
B A barometer is used to measure barometric (PB) or atmospheric (Patm) pressure.
C A barometer is used to measure barometric (PB) or atmospheric (Patm) pressure.
D A barometer is used to measure barometric (PB) or atmospheric (Patm) pressure.
PTS: 1 DIF: Recall REF: The Airways
2. What is the term for the movement of gas from the external environment to the alveoli ?
a. ventilation c. internal respiration
b. external respiration d. osmosis
ANS: A
Feedback
A The movement of gas from the external environment to the alveoli is called ventilation.
B The movement of gas from the external environment to the alveoli is called ventilation.
C The movement of gas from the external environment to the alveoli is called ventilation.
D The movement of gas from the external environment to the alveoli is called ventilation.
PTS: 1 DIF: Recall REF: Introduction OBJ: 1
3. At sea level under standard conditions, what would the PB equal in mm Hg ?
a. 760 c. 14.7
b. 1034 d. 29.9
ANS: A
Feedback
A At sea level under standard conditions, the normal barometric pressure is 760 mm Hg.
B At sea level under standard conditions, the normal barometric pressure is 760 mm Hg.
C At sea level under standard conditions, the normal barometric pressure is 760 mm Hg.
D At sea level under standard conditions, the normal barometric pressure is 760 mm Hg.
PTS: 1 DIF: Recall REF: Mechanisms of Ventilation
OBJ: 2
4. What is the general term for a pressure difference between two points in a system?
a. pressure gradient c. system pressure variation
b. diffusion d. osmotic gradient
ANS: A
Feedback
A A pressure gradient is defined as the difference in pressures occuring between two
points.
B A pressure gradient is defined as the difference in pressures occuring between two
points.
C A pressure gradient is defined as the difference in pressures occuring between two
points.
D A pressure gradient is defined as the difference in pressures occuring between two
points.
PTS: 1 DIF: Recall REF: Pressure Gradients
OBJ: 2
5. At sea level, what would the alveolar pressure at end-expiration equal?
a. 760 mm Hg c. 756 mm Hg
b. 764 mmHg d. 0 mm Hg
ANS: A
Feedback
A Because the alveolar and atmospheric pressure are identical at end-expiration, no air
movement occurs.
B Because the alveolar and atmospheric pressure are identical at end-expiration, no air
movement occurs.
C Because the alveolar and atmospheric pressure are identical at end-expiration, no air
movement occurs.
D Because the alveolar and atmospheric pressure are identical at end-expiration, no air
movement occurs.
PTS: 1 DIF: Recall REF: Pressure Gradients
OBJ: 2
6. At what point in the ventilatory cycle would the intra-alveolar pressure be higher than the atmospheric
pressure?
a. expiration c. inspiration
b. end-expiration d. pre-inspiration
ANS: A
Feedback
A For gas to leave the lungs during exhalation, the intra-alveolar pressure must be higher
than the atmospheric pressure.
B For gas to leave the lungs during exhalation, the intra-alveolar pressure must be higher
than the atmospheric pressure.
C For gas to leave the lungs during exhalation, the intra-alveolar pressure must be higher
than the atmospheric pressure.
D For gas to leave the lungs during exhalation, the intra-alveolar pressure must be higher
than the atmospheric pressure.
PTS: 1 DIF: Recall REF: Pressure Gradients
OBJ: 2
7. Which gas law states that at constant temperature, a volume of gas varies inversely proportional to its
pressure?
a. Boyle’s c. Gay-Lussac’s
b. Charles d. Henry’s
ANS: A
Feedback
A Boyle’s law states that at a constant temperature P1 x V1=P2 x V2 .
B Boyle’s law states that at a constant temperature P1 x V1=P2 x V2 .
C Boyle’s law states that at a constant temperature P1 x V1=P2 x V2 .
D Boyle’s law states that at a constant temperature P1 x V1=P2 x V2 .
PTS: 1 DIF: Recall
REF: Boyle’s Law and Its Relationship to Pressure Gradients OBJ: 2
8. At what point in the respiratory cycle is the equilibrium point reached?
I. Inspiration
II. End-inspiration
III. Expiration
IV. End-expiration
a. II and IV only c. IV only
b. II only d. 1 and III only
ANS: A
Feedback
A At end-inspiration and end-expiration, no gas movement occurs because the pressure
gradient is zero.
B At end-inspiration and end-expiration, no gas movement occurs because the pressure
gradient is zero.
C At end-inspiration and end-expiration, no gas movement occurs because the pressure
gradient is zero.
D At end-inspiration and end-expiration, no gas movement occurs because the pressure
gradient is zero.
PTS: 1 DIF: Recall
REF: The Primary Mechanism of Ventilation Applied to the Human Airways
OBJ: 3
9. What is the general term for the inward movement of tissue between the ribs during inspiration due to
increased negative intrapleural pressure generated during respiratory distress?
a. intercostal retractions c. dyspnea
b. pectus excavatum d. supraclavicular retractions
ANS: A
Feedback
A Intercostal retractions are the inward movement of tissue between ribs during
inspiration due the high negative intapleural pressure generated during respiratory
distress, especially in newborns and infants.
B Intercostal retractions are the inward movement of tissue between ribs during
inspiration due the high negative intapleural pressure generated during respiratory
distress, especially in newborns and infants.
C Intercostal retractions are the inward movement of tissue between ribs during
inspiration due the high negative intapleural pressure generated during respiratory
distress, especially in newborns and infants.
D Intercostal retractions are the inward movement of tissue between ribs during
inspiration due the high negative intapleural pressure generated during respiratory
distress, especially in newborns and infants.
PTS: 1 DIF: Recall
REF: The Primary Mechanism of Ventilation Applied to the Human Airways|Clinical Connection
2-1: Inspiratory Intercostal Retractions OBJ: 4
10. What is the general term for the force required to move gas or fluid through a tube or vessel?
a. driving pressure c. transpulmonary pressure
b. transmural pressure d. transthoracic pressure
ANS: A
Feedback
A The driving pressure is the pressure difference between two points in a tube or vessel.
B The driving pressure is the pressure difference between two points in a tube or vessel.
C The driving pressure is the pressure difference between two points in a tube or vessel.
D The driving pressure is the pressure difference between two points in a tube or vessel.
PTS: 1 DIF: Recall REF: Driving Pressure
OBJ: 5
11. Which pressure is represented by Prs = PB – Palv ?
a. transrespiratory pressure c. transthoracic pressure
b. transmural pressure d. transpulmonary pressure
ANS: A
Feedback
A Transrespiratory pressure is the difference between the atmospheric pressure and
alveolar pressure.
B Transrespiratory pressure is the difference between the atmospheric pressure and
alveolar pressure.
C Transrespiratory pressure is the difference between the atmospheric pressure and
alveolar pressure.
D Transrespiratory pressure is the difference between the atmospheric pressure and
alveolar pressure.
PTS: 1 DIF: Recall REF: Transrespiratory Pressure
OBJ: 5
12. What is the term for the pressure difference that occurs across the airway wall ?
a. Transmural pressure c. Transpulmonary pressure
b. Transrespiratory pressure d. Transthoracic pressure
ANS: A
Feedback
A The transmural pressure is derived by subtracting the pressure on the inside of the
airway from the pressure on the ouside of the airway.
B The transmural pressure is derived by subtracting the pressure on the inside of the
airway from the pressure on the ouside of the airway.
C The transmural pressure is derived by subtracting the pressure on the inside of the
airway from the pressure on the ouside of the airway.
D The transmural pressure is derived by subtracting the pressure on the inside of the
airway from the pressure on the ouside of the airway.
PTS: 1 DIF: Recall REF: Transmural Pressure
OBJ: 5
13. What is the term for the difference between the alveolar pressure and the pleural pressure?
a. transpulmonary pressure c. transrespiratory pressure
b. transmural pressure d. transthoracic pressure
ANS: A
Feedback
A The transpulmonary pressure is the difference between the alveolar pressure and the
pleural pressure.
B The transpulmonary pressure is the difference between the alveolar pressure and the
pleural pressure.
C The transpulmonary pressure is the difference between the alveolar pressure and the
pleural pressure.
D The transpulmonary pressure is the difference between the alveolar pressure and the
pleural pressure.
PTS: 1 DIF: Recall REF: Transpulmonary Pressure
OBJ: 5
14. What is the term for the difference between the alveolar pressure and the body surface pressure?
a. transthoracic pressure c. transrespiratory pressure
b. transmural pressure d. transpulmonary pressure
ANS: A
Feedback
A Transthoracic pressure is the difference between the alveolar pressure and the body
surface pressure.
B Transthoracic pressure is the difference between the alveolar pressure and the body
surface pressure.
C Transthoracic pressure is the difference between the alveolar pressure and the body
surface pressure.
D Transthoracic pressure is the difference between the alveolar pressure and the body
surface pressure.
PTS: 1 DIF: Recall REF: Transthoracic Pressure
OBJ: 5
15. In a flail chest, which pressure gradients are responsible for the inward movement on inspiration of the
section of unattached ribs?
I. Transpulmonary
II. Transmural
III.Transthoracic
IV.Transrespiratory
a. I and III only c. I and II only
b. II and IV only d. II and III only
ANS: A
Feedback
A When a flail chest occurs, the section of unattached ribs moves inward on inspiration
due to the transpulmonary and transthoracic pressure gradients.
B When a flail chest occurs, the section of unattached ribs moves inward on inspiration
due to the transpulmonary and transthoracic pressure gradients.
C When a flail chest occurs, the section of unattached ribs moves inward on inspiration
due to the transpulmonary and transthoracic pressure gradients.
D When a flail chest occurs, the section of unattached ribs moves inward on inspiration
due to the transpulmonary and transthoracic pressure gradients.
PTS: 1 DIF: Recall
REF: Lung Compliance|Clinical Connection 2-2: The Harmful Effects of Pressure Gradients When
the Thorax is Unstable OBJ: 6
16. Which clinical measurement is used to evaluate the elastic forces of the lungs?
a. lung compliance c. elastance
b. airway resistance d. surface tension
ANS: A
Feedback
A The elastic forces of the lungs can be evaluated by measuring lung compliance.
B The elastic forces of the lungs can be evaluated by measuring lung compliance.
C The elastic forces of the lungs can be evaluated by measuring lung compliance.
D The elastic forces of the lungs can be evaluated by measuring lung compliance.
PTS: 1 DIF: Recall REF: Elastic Properties of the Lung and Chest Wall
OBJ: 7
17. What of the following is used to calculate lung compliance?
a. V/ P c. P1V1=P2V2
b. P/ V d. P=(2ST) / r
ANS: A
Feedback
A Lung compliance is defined as the change in lung volume per unit of pressure change.
B Lung compliance is defined as the change in lung volume per unit of pressure change.
C Lung compliance is defined as the change in lung volume per unit of pressure change.
D Lung compliance is defined as the change in lung volume per unit of pressure change.
PTS: 1 DIF: Recall REF: Lung Compliance
OBJ: 8
18. What would the lung compliance equal if a pressure change of 4 cm H20 resulted in a volume change
of 600 mL?
a. 0.15L/cm H20 c. 1.5 L/cm H20
b. 0.066 L/cm H20 d. 0.24 L/cm H20
ANS: A
Feedback
A A volume change of 0.6 L from pressure change of 4 cm H20 would result in a lung
compliance of 0.15 L/cm H20 (0.6L/4 cm H20).
B A volume change of 0.6 L from pressure change of 4 cm H20 would result in a lung
compliance of 0.15 L/cm H20 (0.6L/4 cm H20).
C A volume change of 0.6 L from pressure change of 4 cm H20 would result in a lung
compliance of 0.15 L/cm H20 (0.6L/4 cm H20).
D A volume change of 0.6 L from pressure change of 4 cm H20 would result in a lung
compliance of 0.15 L/cm H20 (0.6L/4 cm H20).
PTS: 1 DIF: Application REF: Lung Compliance
OBJ: 9
19. How does air trapping and hyperinflation of the lungs affect lung compliance?
a. lung compliance is reduced
b. lung compliance is increased
c. lung compliance is normal
d. lung compliance is unaffected by hyperinflation
ANS: A
Feedback
A When air trapping and hyperinflation of the lungs occur, lung compliance decreases.
B When air trapping and hyperinflation of the lungs occur, lung compliance decreases.
C When air trapping and hyperinflation of the lungs occur, lung compliance decreases.
D When air trapping and hyperinflation of the lungs occur, lung compliance decreases.
PTS: 1 DIF: Recall
REF: Lung Compliance|Clinical Connection 2-3: Pulmonary Disorders that Force the Patient to
Breathe at the Top Flat Portion of the Volume Pressure Curve OBJ: 10
20. How do obstructive lung diseases that cause air trapping affect lung compliance?
a. Lung compliance is reduced
b. Lung compliance is increased
c. Lung compliance remains normal
d. Lung compliance is unaffected by air trapping
ANS: A
Feedback
A Lung compliance is decreased in the presence of obstructive lung diseases that cause air
trapping and hyperinflation.
B Lung compliance is decreased in the presence of obstructive lung diseases that cause air
trapping and hyperinflation.
C Lung compliance is decreased in the presence of obstructive lung diseases that cause air
trapping and hyperinflation.
D Lung compliance is decreased in the presence of obstructive lung diseases that cause air
trapping and hyperinflation.
PTS: 1 DIF: Recall
REF: Lung Compliance|Clinical Connection 2-3: Pulmonary Disorders that Force the Patient to
Breathe at the Top Flat Portion of the Volume Pressure Curve OBJ: 10
21. What effect do restrictive lung diseases have on lung compliance?
a. Lung compliance decreases
b. Lung compliance increases
c. Lung compliance remains normal
d. Restrictive lung diseases do not affect lung compliance.
ANS: A
Feedback
A Restrictive lung diseases shift the volume-pressure curve to the right so lung compliance
is reduced.
B Restrictive lung diseases shift the volume-pressure curve to the right so lung compliance
is reduced.
C Restrictive lung diseases shift the volume-pressure curve to the right so lung compliance
is reduced.
D Restrictive lung diseases shift the volume-pressure curve to the right so lung compliance
is reduced.
PTS: 1 DIF: Recall
REF: Lung Compliance|Clinical Connection 2-4: Pulmonary Disorders that Shift the Pressure
Volume Curve to the Right OBJ: 11
22. Which of the following would shift the volume-pressure curve to the right?
I. Acute asthma episode
II. Pneumothorax
III. Pleural effusion
IV. Pulmonary edema
a. II, II, and IV only c. I. III and IV only
b. I, II, and IV only d. Ii and IV only
ANS: A
Feedback
A Restrictive lung conditions, including pneumothorax, pleural effusion, and pulmonary
edema shift the volume pressure curve to the right.
B Restrictive lung conditions, including pneumothorax, pleural effusion, and pulmonary
edema shift the volume pressure curve to the right.
C Restrictive lung conditions, including pneumothorax, pleural effusion, and pulmonary
edema shift the volume pressure curve to the right.
D Restrictive lung conditions, including pneumothorax, pleural effusion, and pulmonary
edema shift the volume pressure curve to the right.
PTS: 1 DIF: Recall
REF: Lung Compliance|Clinical Connection 2-4: Pulmonary Disorders that Shift the Pressure
Volume Curve to the Right OBJ: 11
23. What is the reciprocal of compliance?
a. elastance c. surface tension
b. resistance d. viscosity
ANS: A
Feedback
A The reciprocal of compliance is elastance.
B The reciprocal of compliance is elastance.
C The reciprocal of compliance is elastance.
D The reciprocal of compliance is elastance.
PTS: 1 DIF: Recall REF: Hooke’s Law
OBJ: 12
24. Which physical law explains elastance?
a. Hooke’s law c. Gay-Lussac’s law
b. Boyle’s law d. Charles’ law
ANS: A
Feedback
A Hooke’s law explains elastance.
B Hooke’s law explains elastance.
C Hooke’s law explains elastance.
D Hooke’s law explains elastance.
PTS: 1 DIF: Recall REF: Hooke’s Law
OBJ: 12
25. When a positive pressure breath is delivered from a mechanical ventilator, how would intra-alveolar
and intrapleural pressures be affected during inspiration?
a. Both would increase
b. The intra-alveolar pressure would rise while the intrapleural pressure remains
subatmospheric
c. Both would decrease
d. Both would remain constant at their resting levels
ANS: A
Feedback
A The intra-alveolar and intrapleural pressures would increase during a positive pressure
breath from a mechanical ventilator.
B The intra-alveolar and intrapleural pressures would increase during a positive pressure
breath from a mechanical ventilator.
C The intra-alveolar and intrapleural pressures would increase during a positive pressure
breath from a mechanical ventilator.
D The intra-alveolar and intrapleural pressures would increase during a positive pressure
breath from a mechanical ventilator.
PTS: 1 DIF: Recall
REF: Hooke’s Law|Clinical Connection 2-5: Positive Pressure Ventilation
OBJ: 13
26. When a tension pneumothorax occurs during positive pressure ventilation, how will the cardiac output
and blood pressure affected?
a. Both will decrease
b. Both will increase
c. The cardiac output will increase but the BP will decrease
d. The BP will increase but the BP will decrease
ANS: A
Feedback
A When a tesnion pneumothorax occurs, the cardiac output and blood pressure decrease
due to compression of major vessels from accumulated gas in the pleural cavity.
B When a tesnion pneumothorax occurs, the cardiac output and blood pressure decrease
due to compression of major vessels from accumulated gas in the pleural cavity.
C When a tesnion pneumothorax occurs, the cardiac output and blood pressure decrease
due to compression of major vessels from accumulated gas in the pleural cavity.
D When a tesnion pneumothorax occurs, the cardiac output and blood pressure decrease
due to compression of major vessels from accumulated gas in the pleural cavity.
PTS: 1 DIF: Recall
REF: Hooke’s Law|Clinical Connection 2-6: Hazards of Positive Pressure Ventilation
OBJ: 14
27. Which law best explains the basic operation of the negative pressure ventilator?
a. Boyle’s c. Charles’
b. Dalton’s d. Hooke’s
ANS: A
Feedback
A The basic pressure and volume relationships described by Boyle’s law are implemented
by negative pressure ventilators.
B The basic pressure and volume relationships described by Boyle’s law are implemented
by negative pressure ventilators.
C The basic pressure and volume relationships described by Boyle’s law are implemented
by negative pressure ventilators.
D The basic pressure and volume relationships described by Boyle’s law are implemented
by negative pressure ventilators.
PTS: 1 DIF: Recall
REF: Hooke’s Law|Clinical Connection 2-7: Negative Pressure Ventilation
OBJ: 15
28. Which of the following are periods of no gas flow during negative pressure ventilation?
I. Inspiration
II. End inspiration
III. Expiration
IV. End expiration
a. II and IV only c. IV only
b. II only d. I and III only
