Exam (elaborations)
Pilbeams Mechanical Ventilation TEST BANK , 8th Edition by Cairo, Chapters 1 - 23
Pilbeams Mechanical Ventilation TEST BANK , 8th Edition by Cairo, Chapters 1 - 23
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Test Bank for Pilbeams Mechanical Ventilation8th Edition by Cairo
Chapters 1 - 23 (Complete Download)
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Chapter 1; Basic Terms and Concepts of Mechanical Ventilation
Test Bank
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.
DIF: 1 REF: pg. 3
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.
, DIF: 1 REF: pg. 3
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.
DIF: 1 REF: pg. 4
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.
DIF: 1 REF: pg. 3
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.
DIF: 1 REF: pg. 3
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.
DIF: 2 REF: pg. 3
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.
DIF: 1 REF: pg. 4
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= then P = V/ C
DIF: 2 REF: pg. 4
9. The condition that causes pulmonary compliance to increase is
which of the following?
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.
, DIF: 1 REF: pg. 5| pg. 6
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).
DIF: 2 REF: pg. 4| pg. 5
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
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