Test Bank for Pilbeams MechanicalVentilation 7th Edition by Cairo
Contents Chapter 1; Basic Terms and Concepts of Mechanical Ventilation Test Bank
... [Show More] ........................ 2 Chapter 2; How Ventilators Work ................................ ............... 20 Chapter 3; How a Breath Is Delivered ................................ ....................... 25 Chapter 4; Establishing the Need for Mechanical Ventilation ................................ .... 37 Chapter 5; Selecting the Ventilator and the Mode ................................ ............. 48 Chapter 6; Initial Ventilator Settings ................................ ........................ 62 Chapter 7; Final Considerations in Ventilator Setup ................................ ........... 74 Chapter 8; Initial Patient Assessment ................................ ....................... 87 Chapter 9; Ventilator Graphics ................................ ............................ 101 Chapter 10: Assessment of Respiratory Function ................................ ............ 117 Chapter 11; Hemodynamic Monitoring ................................ ......... 137 Chapter 12; Methods to Improve Ventilation in Patient-Ventilator Management... 168 Chapter 13; Improving Oxygenation and Management of ARDS ................ 180 Chapter 14; Ventilator-Associated Pneumonia ................................ ... 199 Chapter 15; Sedatives, Analgesics, and ParalyticsTest Bank ................................ .... 218 Chapter 16; Extrapulmonary Effects of Mechanical VentilationTest Bank ........................ 223 Chapter 17; Effects of Positive Pressure Ventilation on the Pulmonary SystemTest Bank ........... 228 Chapter 18; Troubleshooting and Problem SolvingTest Bank ................................ .. 238 Chapter 19; Basic Concepts of Noninvasive Positive Pressure Ventilation ........ 250 Chapter 20; Discontinuation and Weaningfrom Mechanical Ventilation ......... 266 Chapter 21; Long Term Ventilation ................................ ............. 288 Chapter 22; Neonatal and Pediatric VentilationTest Bank ................................ ..... 306 Chapter 23; Special Techniques in Ventilatory Support ................................ ....... 319
1 | P a g eChapter 1; Basic Terms and Concepts of
Mechanical Ventilation
MULTIPLE CHOICE
1. The body’s mechanism for conducting air in and out of the lungsis known as
which of the following?
a.
External respiration
b.
c.
d.
Internal respiration
Spontaneous ventilation
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.
c.
d.
Scalenes and trapezius
muscles
Alveoli and pulmonary
capillaries
External oblique and
transverse abdominal muscles
ANS: C
External respiration involves the exchange of oxygen and carbondioxide (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 musclesare accessory muscles of expiration.
2 | P a g eDIF: 1
REF: pg. 3
3. The graph that shows intrapleural pressure changes duringnormal
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 atend-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.
b.
c.
d.
- 1 cm H2O
+ 1 cm H2O
0 cm H2O
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 +1cm H2O.
DIF: 1
REF: pg. 3
5. The pressure required to maintain alveolar inflation is known aswhich of the
following?
a.
Transairway pressure (PTA )
b.
c.
Transthoracic pressure (PTT)
Transrespiratory pressure (PTR)
3 | P a g ed.
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 thesame 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 cmH2O.
a.
7 cm H2O
b.
c.
d.
30 cm H2O
40 cm H2O
175 cm H2O
ANS: B
The transairway pressure (PTA ) is used to calculate the pressurerequired 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 toits original form
after being stretched or acted on by an outside force is which of the following?
a.
Elastance
b.
c.
d.
Compliance
Viscous resistance
Distending pressure
4 | P a g eANS: 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, forexample 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 complianceof 15 mL/cm H2O.
a.
6 cm H2O
b.
c.
d.
ANS: B
C = V/ P then P = V/ C
DIF: 2
REF: pg. 4
9. The condition that causes pulmonary compliance to increase iswhich of the
following?
a.
Asthma
b.
c.
d.
Kyphoscoliosis
Emphysema
Acute respiratory distress
syndrome (ARDS)
ANS: C
Emphysema causes an increase in pulmonary compliance, whereas ARDS and
kyphoscoliosis cause decreases in pulmonarycompliance. Asthma attacks
cause increase in airway resistance.
26.7 cm H2O
37.5 cm H2O
41.5 cm H2O
5 | P a g eDIF: 1
REF: pg. 5| pg. 6
10. Calculate the effective static compliance (Cs) given the followinginformation
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.
c.
d.
16.3 mL/ cm H2O
21.7 mL/cm H2O
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 675mL, 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.
c.
d.
0.03 L/cm H2O
0.22 L/cm H2O
0.34 L/cm H2O
ANS: C
The formula for calculating effective static compliance is Cs = VT/(Pplateau – EEP).
DIF: 2
REF: pg. 4| pg. 5
12. A patient receiving mechanical ventilation has an exhaled tidalvolume (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)
6 | P a g e0600
0800
1000
27
29
36
15
15
13
The respiratory therapist should recommend which of thefollowing for this
patient?
1.
Tracheobronchial suctioning
2.
3.
4.
a.
b.
c.
d.
Increase in the set tidal
volume
Beta adrenergic bronchodilator
therapy
Increase positive end
expiratory pressure
1 and 3 only
2 and 4 only
1, 2 and 3 only
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, andmucosal
inflammation. Tracheobronchial suctioning will remove any secretion buildup
and a beta adrenergic bronchodilator will reverse bronchospasm. Increasing the
tidal volume will add to theairway 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
7 | P a g e13. The values below pertain to a patient who is being mechanically ventilated with
a measured exhaled tidal volume (VT ) of 700 mL.
Time
0800
1000
1100
1130
Peak Inspiratory
Pressure (cm
H2O)
35
39
45
50
Analysis of this data points to which of the following conclusions?
a.
Airway resistance in
increasing.
b.
c.
d.
Airway resistance is
decreasing.
Lung compliance is increasing.
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 patient’s
airway resistance. Analysis of the patient’s plateau pressure (Pplateau ) reveals an
increase of 15 cm H2O over the threeand 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 distresssyndrome (ARDS)?
a.
An elevated plateau pressure
(Pplateau)
Plateau Pressure
(cm H2O)
30
34
39
44
8 | P a g eb.
c.
A decreased elastic resistance
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 (Pplateau) in its denominator (CS = VT /(Pplateau - EEP). Therefore, with a
consistentexhaled 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) iswhich of the
following?
a.
(Peak pressure (PIP) –
EEP)/tidal volume (VT)
CS = (PIP-EEP)/VT
b.
(Plateau pressure (Pplateau) –
EEP/tidal volume (VT)
CS = (Pplateau –
EEP)/VT
c.
d.
Tidal volume/(plateau pressure
– EEP) CS = VT/
(Pplateau - EEP)
Tidal volume /(peak pressure(PIP)
– plateau pressure (Pplateau ))
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 theventilatory
cycle?
a.
Inspiration
9 | P a g eb.
c.
d.
End-inspiration
Expiration
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 (Paw) 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 airwayresistance is
which of the following?
a.
Pulmonary edema
b.
c.
d.
Bronchospasm
Fibrosis
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 tochange 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 causesfluid buildup in the peritoneal cavity and increases tissue
resistance, not airway resistance.
DIF: 1
REF: pg. 5
10 | P a g e18. An increase in peak inspiratory pressure (PIP) without an increasein plateau
pressure (Pplateau) is associated with which of the following?
a.
Increase in static compliance
(CS)
b.
c.
d.
Decrease in static compliance
(CS)
Increase in airway resistance
Decrease in airway resistance
ANS: C
The PIP represents the amount of pressure needed to overcome both elastance
and airway resistance. The Pplateau is the amount of pressure required to overcome
elastance alone. Since the Pplateau has remained constant in this situation, the static
compliance is unchanged. The difference between the PIP and the Pplateau is the
transairway pressure (PTA) and represents the pressure required toovercome the
airway resistance. If PTA increases, the airway resistance is also increasing, when
the gas flow rate remains the same.
DIF: 2
REF: pg. 5| pg. 6
19. The patient-ventilator data over the past few hours demonstratesan increased
peak inspiratory pressure (PIP) with a constant transairway pressure (PTA). The
respiratory therapist should conclude which of the following?
a.
Static compliance (CS) has
increased.
b.
c.
d.
Static compliance (CS) has
decreased.
Airway resistance (Raw) has
increased.
Airway resistance (Raw )has
decreased.
ANS: B
The PIP represents the amount of pressure needed to overcome both elastance
and airway resistance. The Pplateau is the amount ofpressure required to overcome
elastance alone, and is the
11 | P a g epressure used to calculate the static compliance. Since PTA has stayed the same, it
can be concluded that Raw has remained thesame. Therefore, the reason the PIP
has increased is because ofan increase in the Pplateau. This correlates to a
decrease in CS.
DIF: 2
REF: pg. 5
20. Calculate airway resistance (Raw ) for a ventilator patient, in cm H2O/L/sec, when
the peak inspiratory pressure (PIP) is 50 cm H2O, the plateau pressure (Pplateau) is
15 cm H2O, and the set flow rate is60 L/min.
a.
0.58 Raw
b.
c.
d.
1.2 Raw
35 Raw
50 Raw
ANS: C
Raw = PTA/flow; or Raw = (PIP – Pplateau)/flowDIF: 2
REF: pg. 5| pg. 6
21. Calculate airway resistance (Raw) for a ventilator patient, in cmH2O/L/sec,
with the following information: Peak inspiratory pressure (PIP) is 20 cm H2O,
plateau pressure (Pplateau) is 15 cm H2O, PEEP is 5 cm H2O, and set flow rate is
50 L/min.
a.
5 Raw
b.
c.
d.
6 Raw
10 Raw
15 Raw
ANS: B
Raw = (PIP – Pplateau)/flow and flow is in Liters/second. DIF:
REF: pg. 5| pg. 6
22. Calculate the static compliance (CS), in mL/cm H2O, when PIP is 47 cm H2O,
plateau pressure (Pplateau) is 27 cm H2O, baseline pressure is 10 cm H2O, and
exhaled tidal volume (VT) is 725 mL.
a.
43 CS
2
12 | P a g eb.
c.
d.
36 CS
20 CS
0.065 CS
ANS: A
DIF: 2
REF: pg. 5| pg. 6
23. Calculate the inspiratory time necessary to ensure 98% of the volume is
delivered to a patient with a Cs = 40 mL/cm H2O and the Raw = 1 cm
H2O/(L/sec).
a.
0.04 sec
b.
c.
d.
0.16 sec
1.6 sec
4.0 sec
ANS: B
Time constant = C (L/cm H2O) x R (cm H2O/(L/sec)). 98% of the volume will
be delivered in 4 time constants. Therefore, multiply4 times the time constant.
DIF: 2
REF: pg. 6
24. How many time constants are necessary for 95% of the tidalvolume (VT)
to be delivered from a mechanical ventilator?
a.
b.
c.
d.
1
2
3
4
ANS: C
One time constant allows 63% of the volume to be inhaled; 2 time constants
allow about 86% of the volume to be inhaled; 3time constants allow about
95% to be inhaled; 4 time constants allow about 98% to be inhaled; and 5
time constants allow 100%to be inhaled.
DIF: 1
REF: pg. 6
25. Compute the inspiratory time necessary to ensure 100% of the
13 | P a g evolume is delivered to an intubated patient with a Cs = 60 mL/cmH2O and the
Raw = 6 cm H2O/(L/sec).
a.
0.36 sec
b.
c.
d.
0.5 sec
1.4 sec
1.8 sec
ANS: D
Time constant (TC) = C (L/cm H2O) x R (cm H2O/(L/sec)). 100% of the volume
will be delivered in 5 time constants. Therefore, multiply 5 times the time
constant.
DIF: 2
REF: pg. 6
26. Evaluate the combinations of compliance and resistance and select the
combination that will cause the lungs to fill fastest.
a.
Cs = 0.1 L/cm H2O
b.
c.
d.
cm H2O/(L/sec)
Cs = 0.1 L/cm H2O
cm H2O/(L/sec)
Cs = 0.03 L/cm H2O
cm H2O/(L/sec)
Cs = 0.03 L/cm H2O
10 cm H2O/(L/sec)
Raw = 1
Raw =
Raw = 1
Raw = 10
ANS: C
Use the time constant formula, TC = C x R, to determine the timeconstant for
each choice. The time constant for answer A is 0.1 sec. The time constant for
answer B is 1 sec. The time constant for answer C is 0.03 seconds, and the time
constant for answer Dis 0.3 sec. The product of multiplying the time constant
by 5 is the inspiratory time needed to deliver 100% of the volume.
DIF: 3
REF: pg. 6
27. The statement that describes the alveolus shown in Figure 1-1 iswhich of the
following?
14 | P a g e1.
Requires more time to fill than
a normal alveolus
2.
3.
4.
a.
b.
c.
d.
Fills more quickly than a
normal alveolus
Requires more volume to fill
than a normal alveolus
More pressure is needed to
achieve a normal volume
1 and 3 only
2 and 4 only
2 and 3 only
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 unitsthat require more volume to fill than normal are
high-complianceunits.
DIF: 1
REF: pg. 9
28. Calculate the static compliance (CS), in mL/cm H2O, when PIP is26 cm H2O,
plateau pressure (Pplateau) is 19 cm H2O, baseline pressure is 5 cm H2O and
exhaled tidal volume (VT) is 425 mL.
a.
16
b.
c.
d.
ANS: D
CS = VT/(Pplateau - EEP)
DIF: 2
REF: pg. 5
29. What type of ventilator increases transpulmonary pressure (PL) bymimicking
the normal mechanism for inspiration?
20
22
30
15 | P a g ea.
Positive pressure ventilation
(PPV)
b.
c.
d.
Negative pressure ventilation
(NPV)
High frequency oscillatory
ventilation (HFOV)
High frequency positive
pressure ventilation (HFPPV)
ANS: D
Negative pressure ventilation (NPV) attempts to mimic the function of the
respiratory muscles to allow breathing through normal physiological
mechanisms. Positive pressure ventilation (PPV) pushes air into the lungs by
increasing the alveolar pressure. High frequency oscillatory ventilation (HFOV)
delivers very small volumes at very high rates in a “to-and-fro” motion by
pushing the gas in and pulling it out during exhalation. High frequency positive
pressure ventilation (HFPPV) pushes in small volumes at high respiratory rates.
DIF: 1
REF: pg. 5| pg. 6
30. Air accidently trapped in the lungs due to mechanical ventilationis known as
which of the following?
a.
Plateau pressure (Pplateau)
b.
c.
d.
Functional residual capacity
(FRC)
Extrinsic positive end expiratory
pressure (extrinsic
PEEP)
Intrinsic positive end
expiratory pressure (intrinsicPEEP)
ANS: D
The definition of intrinsic PEEP is air that is accidentally trapped in the lung.
Another name for this is auto-PEEP. Extrinsic PEEP is the positive baseline
pressure that is set by the operator.
Functional residual capacity (FRC) is the sum of a patient’s residual
volume and expiratory reserve volume, and is the
16 | P a g eamount of gas that normally remains in the lung after a quiet exhalation. The plateau pressure is the pressure measured in thelungs at no flow during an inspiratory hold maneuver.
DIF: 1 REF: pg. 7| pg. 8 31. The transairway pressure (PTA) shown in this figure is which of thefollowing?
a. b. c. d.
5 cm H2O 10 cm H2O 20 cm H2O 30 cm H2O
ANS: B [Show Less]