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 t
he 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 as
ks for the body’s mechanism, this would be spontaneous ventilation. External respiration in
volves 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 movemen
t 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 in
volves movement of oxygen from the systemic blood into the cells. Scalene and trapezius
muscles are accessory muscles of inspiration. External oblique and transverse abdominal mu
scles are accessory muscles of expiration.
REF: pg. 2
3. The graph that shows intrapleural pressure changes during normal spontaneous breathing i
s depicted by which of the following?
a.
, b.
c.
d.
ANS: B
During spontaneous breathing, the intrapleural pressure drops from 5 cm H2O at end-
about 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 ventilati
on. 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 th
e conducting tubes. Transrespiratory pressure (PTR) is the pressure to inflate the lungs and a
irways 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 alveol
ar 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 bei
ng 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 aft
er 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 pressu
re 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 patie
nt 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