Chapter 01
1. The values below pertain to a patient who is being mechanically ventilated with a measured exhaled tidal
volume (VT) of 700 mL. TimePeak Inspiratory Pressure (cm H2O)Plateau Pressure (cm
H2O)08003530100039341100453911305044Analysis of this data points to which of the following conclusions?
[Multiple Choice]
a. Airway resistance is increasing.
b. Airway resistance is decreasing.
c. Lung compliance is increasing.
d. Lung compliance is decreasing. ✔
ANS: D
To evaluate this information the transairway pressure (PTA) is calculated for the different times: 0800 PTA = 5 cm
H2O, 1000 PTA = 5 cm H2O, 1100 PTA = 6 cm H2O, and 1130 PTA = 6 cm H2O. This data shows that there is no
significant increase or decrease in this patients airway resistance. Analysis of the patients plateau pressure
(Pplateau) reveals an increase of 15 cm H2O over the three and a half hour time period. This is directly related to
a decrease in lung compliance. Calculation of the lung compliance (CS = VT/(Pplateau EEP) at each time interval
reveals a steady decrease from 20 mL/cm H2O to 14 mL/cm H2O.
2. Evaluate the combinations of compliance and resistance and select the combination that will cause the lungs to
empty slowest. [Multiple Choice]
a. CS = 0.05 L/cm H2O Raw = 2 cm H2O/(L/sec)
b. CS = 0.05 L/cm H2O Raw = 6 cm H2O/(L/sec) ✔
c. CS = 0.03 L/cm H2O Raw = 5 cm H2O/(L/sec)
d. CS = 0.03 L/cm H2O Raw = 8 cm H2O/(L/sec)
ANS: B
Use the time constant formula, TC = C × R, to determine the time constant for each choice. The combination with
the longest time constant will empty the slowest. The time constant for A is 0.1 second, B is 0.3 second, C is 0.15
second, and D is 0.24 second. To find out how many seconds for emptying, multiply the time constant by 5.
3. During spontaneous inspiration alveolar pressure (PA) is about: ________________. [Multiple Choice]
a. -1 cm H2O ✔
b. +1 cm H2O
c. 0 cm H2O
,d. 5 cm H2O
ANS: A
-1 cm H2O is the lowest alveolar pressure will become during normal spontaneous ventilation. During the
exhalation of a normal spontaneous breath the alveolar pressure will become +1 cm H2O.
4. Use this figure to compute the static compliance (CS) for an intubated patient with an exhaled tidal volume (VT)
of 500 mL. [Multiple Choice]
a. 14 mL/cm H2O
b. 20 mL/cm H2O
c. 33 mL/cm H2O
d. 50 mL/cm H2O ✔
ANS: D
Cs = Pplateau – EEP; the Pplateau in the figure is 20 cm H2O and the PEEP is 10 cm H2O.
5. The formula used for the calculation of static compliance (CS) is which of the following? [Multiple Choice]
a. (Peak inspiratory pressure (PIP) – EEP)/tidal volume (VT)
b. (Plateau pressure (Pplateau) – EEP)/tidal volume (VT)
c. Tidal volume/(plateau pressure – EEP) ✔
d. Tidal volume/(peak pressure (PIP) – plateau pressure (Pplateau))
ANS: C
CS = VT/(Pplateau – EEP)
6. Use this figure to compute the static compliance for an intubated patient with an inspiratory flow rate set at 70
L/min. [Multiple Choice]
a. 0.2 cm H2O/(L/sec)
b. 11.7 cm H2O/(L/sec) ✔
c. 16.7 cm H2O/(L/sec)
d. 20 cm H2O/(L/sec)
ANS: B
Use the graph to determine the PIP (34 cm H2O) and the Pplateau (20 cm H2O). Convert the flow into L/sec (70
L/min/60 = 1.2 L/sec). Then, Raw = (PIP – Pplateau)/flow.
,7. Which of the following are involved in external respiration? [Multiple Choice]
a. Red blood cells and body cells
b. Scalenes and trapezius muscles
c. Alveoli and pulmonary capillaries ✔
d. External oblique and transverse abdominal muscles
ANS: C
External respiration involves the exchange of oxygen and carbon dioxide (CO2) between the alveoli and the
pulmonary capillaries. Internal respiration occurs at the cellular level and involves movement of oxygen from the
systemic blood into the cells. Scalene and trapezius muscles are accessory muscles of inspiration. External oblique
and transverse abdominal muscles are accessory muscles of expiration.
8. The graph that shows intrapleural pressure changes during normal spontaneous breathing is depicted by which
of the following? [Multiple Choice]
a.
b. ✔
c.
d.
ANS: B
During spontaneous breathing, the intrapleural pressure drops from about 5 cm H2O at end-expiration to about
10 cm H2O at end-inspiration. The graph depicted for answer B shows that change from 5 cm H2O to 10 cm
H2O.
9. The term used to describe the tendency of a structure to return to its original form after being stretched or
acted on by an outside force is which of the following? [Multiple Choice]
a. Elastance ✔
b. Compliance
c. Viscous resistance
d. Distending pressure
ANS: A
The elastance of a structure is the tendency of that structure to return to its original shape after being stretched.
The more elastance a structure has, the more difficult it is to stretch. The compliance of a structure is the ease
with which the structure distends or stretches. Compliance is the opposite of elastance. Viscous resistance is the
, opposition to movement offered by adjacent structures such as the lungs and their adjacent organs. Distending
pressure is pressure required to maintain inflation, for example, alveolar distending pressure.
10. Which of the following statements best defines elastance? [Multiple Choice]
a. Ability of a structure to stretch.
b. Ability of a structure to return to its natural shape after stretching. ✔
c. Ability of a structure to stretch and remain in that position.
d. Ability of a structure to fill and empty during static conditions.
ANS: B
The opposite of compliance, elastance is the tendency of a structure to return to its original form after being
stretched or acted on by an outside force.
11. The statement that describes the alveolus shown in Figure 1-1 is which of the following? 1. Requires more
time to fill than a normal alveolus.2. Fills more quickly than a normal alveolus.3. Requires more volume to fill than
a normal alveolus.4. More pressure is needed to achieve a normal volume. [Multiple Choice]
a. 1 and 3 only
b. 2 and 4 only ✔
c. 2 and 3 only
d. 1, 3, and 4
ANS: B
The figure shows a low-compliant unit, which has a short time constant. This means it takes less time to fill and
empty and will require more pressure to achieve a normal volume. Lung units that require more time to fill are
high-resistance units. Lung units that require more volume to fill than normal are high-compliance units.
12. Which of the following conditions causes pulmonary compliance to increase? [Multiple Choice]
a. Asthma
b. Kyphoscoliosis
c. Emphysema ✔
d. Acute respiratory distress syndrome (ARDS)
ANS: C
Emphysema causes an increase in pulmonary compliance, whereas ARDS and kyphoscoliosis cause decreases in
pulmonary compliance. Asthma attacks cause increase in airway resistance.
1. The values below pertain to a patient who is being mechanically ventilated with a measured exhaled tidal
volume (VT) of 700 mL. TimePeak Inspiratory Pressure (cm H2O)Plateau Pressure (cm
H2O)08003530100039341100453911305044Analysis of this data points to which of the following conclusions?
[Multiple Choice]
a. Airway resistance is increasing.
b. Airway resistance is decreasing.
c. Lung compliance is increasing.
d. Lung compliance is decreasing. ✔
ANS: D
To evaluate this information the transairway pressure (PTA) is calculated for the different times: 0800 PTA = 5 cm
H2O, 1000 PTA = 5 cm H2O, 1100 PTA = 6 cm H2O, and 1130 PTA = 6 cm H2O. This data shows that there is no
significant increase or decrease in this patients airway resistance. Analysis of the patients plateau pressure
(Pplateau) reveals an increase of 15 cm H2O over the three and a half hour time period. This is directly related to
a decrease in lung compliance. Calculation of the lung compliance (CS = VT/(Pplateau EEP) at each time interval
reveals a steady decrease from 20 mL/cm H2O to 14 mL/cm H2O.
2. Evaluate the combinations of compliance and resistance and select the combination that will cause the lungs to
empty slowest. [Multiple Choice]
a. CS = 0.05 L/cm H2O Raw = 2 cm H2O/(L/sec)
b. CS = 0.05 L/cm H2O Raw = 6 cm H2O/(L/sec) ✔
c. CS = 0.03 L/cm H2O Raw = 5 cm H2O/(L/sec)
d. CS = 0.03 L/cm H2O Raw = 8 cm H2O/(L/sec)
ANS: B
Use the time constant formula, TC = C × R, to determine the time constant for each choice. The combination with
the longest time constant will empty the slowest. The time constant for A is 0.1 second, B is 0.3 second, C is 0.15
second, and D is 0.24 second. To find out how many seconds for emptying, multiply the time constant by 5.
3. During spontaneous inspiration alveolar pressure (PA) is about: ________________. [Multiple Choice]
a. -1 cm H2O ✔
b. +1 cm H2O
c. 0 cm H2O
,d. 5 cm H2O
ANS: A
-1 cm H2O is the lowest alveolar pressure will become during normal spontaneous ventilation. During the
exhalation of a normal spontaneous breath the alveolar pressure will become +1 cm H2O.
4. Use this figure to compute the static compliance (CS) for an intubated patient with an exhaled tidal volume (VT)
of 500 mL. [Multiple Choice]
a. 14 mL/cm H2O
b. 20 mL/cm H2O
c. 33 mL/cm H2O
d. 50 mL/cm H2O ✔
ANS: D
Cs = Pplateau – EEP; the Pplateau in the figure is 20 cm H2O and the PEEP is 10 cm H2O.
5. The formula used for the calculation of static compliance (CS) is which of the following? [Multiple Choice]
a. (Peak inspiratory pressure (PIP) – EEP)/tidal volume (VT)
b. (Plateau pressure (Pplateau) – EEP)/tidal volume (VT)
c. Tidal volume/(plateau pressure – EEP) ✔
d. Tidal volume/(peak pressure (PIP) – plateau pressure (Pplateau))
ANS: C
CS = VT/(Pplateau – EEP)
6. Use this figure to compute the static compliance for an intubated patient with an inspiratory flow rate set at 70
L/min. [Multiple Choice]
a. 0.2 cm H2O/(L/sec)
b. 11.7 cm H2O/(L/sec) ✔
c. 16.7 cm H2O/(L/sec)
d. 20 cm H2O/(L/sec)
ANS: B
Use the graph to determine the PIP (34 cm H2O) and the Pplateau (20 cm H2O). Convert the flow into L/sec (70
L/min/60 = 1.2 L/sec). Then, Raw = (PIP – Pplateau)/flow.
,7. Which of the following are involved in external respiration? [Multiple Choice]
a. Red blood cells and body cells
b. Scalenes and trapezius muscles
c. Alveoli and pulmonary capillaries ✔
d. External oblique and transverse abdominal muscles
ANS: C
External respiration involves the exchange of oxygen and carbon dioxide (CO2) between the alveoli and the
pulmonary capillaries. Internal respiration occurs at the cellular level and involves movement of oxygen from the
systemic blood into the cells. Scalene and trapezius muscles are accessory muscles of inspiration. External oblique
and transverse abdominal muscles are accessory muscles of expiration.
8. The graph that shows intrapleural pressure changes during normal spontaneous breathing is depicted by which
of the following? [Multiple Choice]
a.
b. ✔
c.
d.
ANS: B
During spontaneous breathing, the intrapleural pressure drops from about 5 cm H2O at end-expiration to about
10 cm H2O at end-inspiration. The graph depicted for answer B shows that change from 5 cm H2O to 10 cm
H2O.
9. The term used to describe the tendency of a structure to return to its original form after being stretched or
acted on by an outside force is which of the following? [Multiple Choice]
a. Elastance ✔
b. Compliance
c. Viscous resistance
d. Distending pressure
ANS: A
The elastance of a structure is the tendency of that structure to return to its original shape after being stretched.
The more elastance a structure has, the more difficult it is to stretch. The compliance of a structure is the ease
with which the structure distends or stretches. Compliance is the opposite of elastance. Viscous resistance is the
, opposition to movement offered by adjacent structures such as the lungs and their adjacent organs. Distending
pressure is pressure required to maintain inflation, for example, alveolar distending pressure.
10. Which of the following statements best defines elastance? [Multiple Choice]
a. Ability of a structure to stretch.
b. Ability of a structure to return to its natural shape after stretching. ✔
c. Ability of a structure to stretch and remain in that position.
d. Ability of a structure to fill and empty during static conditions.
ANS: B
The opposite of compliance, elastance is the tendency of a structure to return to its original form after being
stretched or acted on by an outside force.
11. The statement that describes the alveolus shown in Figure 1-1 is which of the following? 1. Requires more
time to fill than a normal alveolus.2. Fills more quickly than a normal alveolus.3. Requires more volume to fill than
a normal alveolus.4. More pressure is needed to achieve a normal volume. [Multiple Choice]
a. 1 and 3 only
b. 2 and 4 only ✔
c. 2 and 3 only
d. 1, 3, and 4
ANS: B
The figure shows a low-compliant unit, which has a short time constant. This means it takes less time to fill and
empty and will require more pressure to achieve a normal volume. Lung units that require more time to fill are
high-resistance units. Lung units that require more volume to fill than normal are high-compliance units.
12. Which of the following conditions causes pulmonary compliance to increase? [Multiple Choice]
a. Asthma
b. Kyphoscoliosis
c. Emphysema ✔
d. Acute respiratory distress syndrome (ARDS)
ANS: C
Emphysema causes an increase in pulmonary compliance, whereas ARDS and kyphoscoliosis cause decreases in
pulmonary compliance. Asthma attacks cause increase in airway resistance.
