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  • PILBEAM'S MECHANICAL VENTILATION: PHYSIOLOGICAL AND CLINICAL APPLICATIONS 8TH EDITION BY JAMES M. CAIRO PHD
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Test Bank for Pilbeams Mechanical Ventilation

8th Edition by Cairo

Chapters 1 - 23 (Complete Download)

,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 H2 O 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 H2 O
c. 0 cm H2O
d. 5 cm H2O


ANS: A
-1 cm H2 O 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 H2 O.

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 (P TT) 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 H2 O


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 H2 O
c. 37.5 cm H2 O
d. 41.5 cm H2 O


ANS: B
C = V/ P 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 (C s) 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 (V T) is 650 mL, and
positive-end expiratory pressure (PEEP) is 10 cm H2 O.
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 C s = 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


ANS: C
The formula for calculating effective static compliance is C s = VT/
(Pplateau – EEP).

DIF: 2 REF: pg. 4| pg. 5

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
3. Beta adrenergic bronchodilator
therapy
4. Increase positive end
expiratory pressure

a. 1 and 3 only
b. 2 and 4 only
c. 1, 2 and 3 only
d. 2, 3 and 4 only


ANS: A
Calculate the transairway pressure (PTA) by subtracting the
plateau pressure from the peak inspiratory pressure. Analyzing
the PTA will show any changes in the pressure needed to
overcome airway resistance. Analyzing the Pplateau will
demonstrate any changes in compliance. The Pplateau remained the
same for the first two checks and then actually dropped at the
1000 hour check. Analyzing the PTA, however, shows a slight
increase between 0600 and 0800 (from 12 cm H2O to 14 cm H2O)
and then a sharp increase to 23 cm H2O at 1000. Increases in PTA
signify increases in airway resistance. Airway resistance may be
caused by secretion buildup, bronchospasm, mucosal edema, and
mucosal inflammation. Tracheobronchial suctioning will remove
any secretion buildup and a beta adrenergic bronchodilator will
reverse bronchospasm. Increasing the tidal volume will add to the
airway resistance according to Poiseuille’s law. Increasing the
PEEP will not address the root of this patient’s problem; the
patient’s compliance is normal.

DIF: 3 REF: pg. 6

,13. The values below pertain to a patient who is being mechanically
ventilated with a measured exhaled tidal volume (VT ) of 700 mL.

Time Peak Inspiratory Plateau Pressure
Pressure (cm (cm H2O)
H2O)
0800 35 30
1000 39 34
1100 45 39
1130 50 44


Analysis of this data points to which of the following conclusions?
a. Airway resistance in
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 (P TA) 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 patient’s airway resistance. Analysis of the patient’s plateau
pressure (Pplateau ) reveals an increase of 15 cm H2O over the three
and one 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.

DIF: 3 REF: pg. 6

14. The respiratory therapist should expect which of the following
findings while ventilating a patient with acute respiratory distress
syndrome (ARDS)?
a. An elevated plateau pressure
(Pplateau)

, b. A decreased elastic resistance
c. A low peak inspiratory
pressure (PIP)
d. A large transairway pressure
(PTA) gradient


ANS: A
ARDS is a pathological condition that is associated with a
reduction in lung compliance. The formula for static compliance
(CS) utilizes the measured plateau pressure (P plateau) in its
denominator (CS = VT /(Pplateau - EEP). Therefore, with a consistent
exhaled tidal volume (VT) , an elevated Pplateau will decrease CS.

DIF: 2 REF: pg. 5| pg. 6

15. The formula used for the calculation of static compliance (CS) is
which of the following?
a. (Peak pressure (PIP) –
EEP)/tidal volume (VT)
<equation> CS = (PIP-EEP)/VT
b. (Plateau pressure (Pplateau) –
EEP/tidal volume (VT)
<equation> CS = (Pplateau –
EEP)/VT
c. Tidal volume/(plateau pressure
– EEP) <equation> CS = VT/
(Pplateau - EEP)
d. Tidal volume /(peak pressure
(PIP) – plateau pressure
(Pplateau )) <equation> CS = VT /
(PIP- Pplateau)


ANS: C
CS = VT/(Pplateau - EEP)

DIF: 1 REF: pg. 7

16. Plateau pressure (Pplateau) is measured during which phase of the
ventilatory cycle?
a. Inspiration

, b. End-inspiration
c. Expiration
d. End-expiration


ANS: B
The calculation of compliance requires the measurement of the
plateau pressure. This pressure measurement is made during no-
flow conditions. The airway pressure (P aw) is measured at end-
inspiration. The inspiratory pressure is taken when the pressure
reaches its maximum during a delivered mechanical breath. The
pressure that occurs during expiration is a dynamic measurement
and drops during expiration. The pressure reading at end-
expiration is the baseline pressure; this reading is either at zero
(atmospheric pressure) or at above atmospheric pressure (PEEP).

DIF: 1 REF: pg. 6

17. The condition that is associated with an increase in airway
resistance is which of the following?
a. Pulmonary edema
b. Bronchospasm
c. Fibrosis
d. Ascites


ANS: B
Airway resistance is determined by the gas viscosity, gas density,
tubing length, airway diameter, and the flow rate of the gas
through the tubing. The two factors that are most often subject to
change are the airway diameter and the flow rate of the gas. The
flow rate of the gas during mechanical ventilation is controlled.
Pulmonary edema is fluid accumulating in the alveoli and will
cause a drop in the patient’s lung compliance. Bronchospasm
causes a narrowing of the airways and will, therefore, increase
the airway resistance. Fibrosis causes an inability of the lungs to
stretch, decreasing the patient’s lung compliance. Ascites causes
fluid buildup in the peritoneal cavity and increases tissue
resistance, not airway resistance.

DIF: 1 REF: pg. 5

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