Chapter 01: Basic Terms and Concepts of Mechanical Ventilation
Cairo: Pilbeam’s Mechanical Ventilation: Physiological and Clinical Applications, 6th
Edition
MULTIPLE CHOICE
1. The body’s mechanism for conducting air in and out of the lungs is known as which of
the following?
a. External respiration
b. Internal respiration
c. Spontaneous ventilation
d. Mechanical ventilation
ANS: C
The conduction of air in and out of the body is known as ventilation. Since the question asks
for the body’s mechanism, this would be spontaneous ventilation. External respiration
involves the exchange of oxygen (O2) 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.
REF: pg. 2
2. Which of the following are involved in external respiration?
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.
REF: pg. 2
3. The graph that shows intrapleural pressure changes during normal spontaneous breathing
is depicted by which of the following?
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.
REF: pg. 3
4. During spontaneous inspiration alveolar pressure (PA) is about: .
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.
REF: pg. 4
5. The pressure required to maintain alveolar inflation is known as which of the following?
a. Transairway pressure (PTA)
b. Transthoracic pressure (PTT)
c. Transrespiratory pressure (PTR)
d. Transpulmonary pressure (PL)
ANS: D
, The definition of transpulmonary pressure (PL) is the pressure required to maintain alveolar
inflation. Transairway pressure (PTA) is the pressure gradient required to produce airflow in
the conducting tubes. Transrespiratory pressure (PTR) is the pressure to inflate the lungs and
airways during positive-pressure ventilation. Transthoracic pressure (PTT) represents the
pressure required to expand or contract the lungs and the chest wall at the same time.
REF: pg. 4
6. Calculate the pressure needed to overcome airway resistance during positive-pressure
ventilation when the proximal airway pressure (PAw) is 35 cm H2O and the alveolar
pressure (PA) is 5 cm H2O.
a. 7 cm H2O
b. 30 cm H2O
c. 40 cm H2O
d. 175 cm H2O
ANS: B
The transairway pressure (PTA) is used to calculate the pressure required to overcome airway
resistance during mechanical ventilation. This formula is PTA = Paw - PA.
REF: pg. 4
7. 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?
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.
REF: pg. 5
8. Calculate the pressure required to achieve a tidal volume of 400 mL for an intubated
patient with a respiratory system compliance of 15 mL/cm H2O.
a. 6 cm H2O
b. 26.7 cm H2O
c. 37.5 cm H2O
d. 41.5 cm H2O
ANS: B
C = V/P then P = V/C
REF: pg. 5
, 9. Which of the following conditions causes pulmonary compliance to increase?
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.
REF: pg. 6 | pg. 7
10. Calculate the effective static compliance (Cs) given the following information about a
patient receiving mechanical ventilation: peak inspiratory pressure (PIP) is 56 cm H2O,
plateau pressure (Pplateau) is 40 cm H2O, exhaled tidal volume (VT) is 650 mL, and positive
end expiratory pressure (PEEP) is 10 cm H2O.
a. 14.1 mL/cm H2O
b. 16.3 mL/cm H2O
c. 21.7 mL/cm H2O
d. 40.6 mL/cm H2O
ANS: C
The formula for calculating effective static compliance is Cs = VT/(Pplateau EEP).
REF: pg. 6 | pg. 7
11. Based upon the following patient information, calculate the patient’s static lung
compliance: exhaled tidal volume (VT) is 675 mL, peak inspiratory pressure (PIP) is 28 cm
H2O, plateau pressure (Pplateau) is 8 cm H2O, and PEEP is set at 5 cm H2O.
a. 0.02 L/cm H2O
b. 0.03 L/cm H2O
c. 0.22 L/cm H2O
d. 0.34 L/cm H2O
ANS: C
The formula for calculating effective static compliance is Cs = VT/(Pplateau EEP).
REF: pg. 5 | pg. 6
12. A patient receiving mechanical ventilation has an exhaled tidal volume (VT) of 500 mL and a
positive end expiratory pressure setting (PEEP) of 5 cm H2O. Patient-ventilator system checks
reveal the following data:
Time PIP (cm H2O) Pplateau (cm H2O)
0600 27 15
0800 29 15
1000 36 13
The respiratory therapist should recommend which of the following for this patient?
1. Tracheobronchial suctioning
2. Increase in the set tidal volume
Cairo: Pilbeam’s Mechanical Ventilation: Physiological and Clinical Applications, 6th
Edition
MULTIPLE CHOICE
1. The body’s mechanism for conducting air in and out of the lungs is known as which of
the following?
a. External respiration
b. Internal respiration
c. Spontaneous ventilation
d. Mechanical ventilation
ANS: C
The conduction of air in and out of the body is known as ventilation. Since the question asks
for the body’s mechanism, this would be spontaneous ventilation. External respiration
involves the exchange of oxygen (O2) 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.
REF: pg. 2
2. Which of the following are involved in external respiration?
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.
REF: pg. 2
3. The graph that shows intrapleural pressure changes during normal spontaneous breathing
is depicted by which of the following?
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.
REF: pg. 3
4. During spontaneous inspiration alveolar pressure (PA) is about: .
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.
REF: pg. 4
5. The pressure required to maintain alveolar inflation is known as which of the following?
a. Transairway pressure (PTA)
b. Transthoracic pressure (PTT)
c. Transrespiratory pressure (PTR)
d. Transpulmonary pressure (PL)
ANS: D
, The definition of transpulmonary pressure (PL) is the pressure required to maintain alveolar
inflation. Transairway pressure (PTA) is the pressure gradient required to produce airflow in
the conducting tubes. Transrespiratory pressure (PTR) is the pressure to inflate the lungs and
airways during positive-pressure ventilation. Transthoracic pressure (PTT) represents the
pressure required to expand or contract the lungs and the chest wall at the same time.
REF: pg. 4
6. Calculate the pressure needed to overcome airway resistance during positive-pressure
ventilation when the proximal airway pressure (PAw) is 35 cm H2O and the alveolar
pressure (PA) is 5 cm H2O.
a. 7 cm H2O
b. 30 cm H2O
c. 40 cm H2O
d. 175 cm H2O
ANS: B
The transairway pressure (PTA) is used to calculate the pressure required to overcome airway
resistance during mechanical ventilation. This formula is PTA = Paw - PA.
REF: pg. 4
7. 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?
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.
REF: pg. 5
8. Calculate the pressure required to achieve a tidal volume of 400 mL for an intubated
patient with a respiratory system compliance of 15 mL/cm H2O.
a. 6 cm H2O
b. 26.7 cm H2O
c. 37.5 cm H2O
d. 41.5 cm H2O
ANS: B
C = V/P then P = V/C
REF: pg. 5
, 9. Which of the following conditions causes pulmonary compliance to increase?
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.
REF: pg. 6 | pg. 7
10. Calculate the effective static compliance (Cs) given the following information about a
patient receiving mechanical ventilation: peak inspiratory pressure (PIP) is 56 cm H2O,
plateau pressure (Pplateau) is 40 cm H2O, exhaled tidal volume (VT) is 650 mL, and positive
end expiratory pressure (PEEP) is 10 cm H2O.
a. 14.1 mL/cm H2O
b. 16.3 mL/cm H2O
c. 21.7 mL/cm H2O
d. 40.6 mL/cm H2O
ANS: C
The formula for calculating effective static compliance is Cs = VT/(Pplateau EEP).
REF: pg. 6 | pg. 7
11. Based upon the following patient information, calculate the patient’s static lung
compliance: exhaled tidal volume (VT) is 675 mL, peak inspiratory pressure (PIP) is 28 cm
H2O, plateau pressure (Pplateau) is 8 cm H2O, and PEEP is set at 5 cm H2O.
a. 0.02 L/cm H2O
b. 0.03 L/cm H2O
c. 0.22 L/cm H2O
d. 0.34 L/cm H2O
ANS: C
The formula for calculating effective static compliance is Cs = VT/(Pplateau EEP).
REF: pg. 5 | pg. 6
12. A patient receiving mechanical ventilation has an exhaled tidal volume (VT) of 500 mL and a
positive end expiratory pressure setting (PEEP) of 5 cm H2O. Patient-ventilator system checks
reveal the following data:
Time PIP (cm H2O) Pplateau (cm H2O)
0600 27 15
0800 29 15
1000 36 13
The respiratory therapist should recommend which of the following for this patient?
1. Tracheobronchial suctioning
2. Increase in the set tidal volume
