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 (O
2
) and carbon dioxide (CO
)
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 (CO
) 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.
2
2
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 H
2
O at end-expiration to about -10 cm H
O at
end-inspiration. The graph depicted for answer B shows that
change from -5 cm H
2
DIF: 1 REF: pg. 4
O to -10 cm H
2
O.
4. During spontaneous inspiration alveolar pressure (P
) is about:
________________.
a. - 1 cm H
O
b. + 1 cm H
O
c. 0 cm H
O
d. 5 cm H
ANS: A
-1 cm H
2
2
2
2
O
2
A
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 H
2
O.
DIF: 1 REF: pg. 3
5. The pressure required to maintain alveolar inflation is known as
which of the following?
a. Transairway pressure (P
TA
)
b. Transthoracic pressure (P
TT
)
c. Transrespiratory pressure (P
TR
2
)
d. Transpulmonary pressure (P
L
ANS: D
The definition of transpulmonary pressure (P
L
) is the pressure
required to maintain alveolar inflation. Transairway pressure (P
)
is the pressure gradient required to produce airflow in the
conducting tubes. Transrespiratory pressure (P
TR
) 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 (P
Aw
) is 35 cm H
2
O and the alveolar pressure (P
) is 5 cm
H
O.
a. 7 cm H
2
O
b. 30 cm H
O
c. 40 cm H
O
d. 175 cm H
ANS: B
The transairway pressure (P
TA
2
2
2
2
O
) is used to calculate the pressure
required to overcome airway resistance during mechanical
ventilation. This formula is P
DIF: 2 REF: pg. 3
TA
= P
aw
- P
A.
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
A
)
TA
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 H
O.
a. 6 cm H
2
O
b. 26.7 cm H
O
c. 37.5 cm H
O
d. 41.5 cm H
ANS: B
C = V/P then P = V/C
DIF: 2 REF: pg. 4
2
2
2
2
O
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
) given the following
information about a patient receiving mechanical ventilation:
peak inspiratory pressure (PIP) is 56 cm H
2
s
O, plateau pressure
(P
plateau
) is 40 cm H
2
O, exhaled tidal volume (V
) is 650 mL, and
positive-end expiratory pressure (PEEP) is 10 cm H
2
O.
a. 14.1 mL/cm H
2
O
b. 16.3 mL/ cm H
O
c. 21.7 mL/cm H
O
d. 40.6 mL/cm H
2
T
ANS: C
The formula for calculating effective static compliance is C
2
2
O
/
(P
plateau
– 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 (V
) is 675
mL, peak inspiratory pressure (PIP) is 28 cm H
O, plateau
pressure (P
plateau
) is 8 cm H
2
2
O, and PEEP is set at 5 cm H
O.
a. 0.02 L/cm H
2
O
b. 0.03 L/cm H
O
c. 0.22 L/cm H
O
d. 0.34 L/cm H
2
2
2
O
ANS: C
The formula for calculating effective static compliance is C
/
(P
plateau
– EEP).
DIF: 2 REF: pg. 4| pg. 5
12. A patient receiving mechanical ventilation has an exhaled tidal
volume (V
) of 500 mL and a positive-end expiratory pressure
setting (PEEP) of 5 cm H
2
T
O. Patient-ventilator system checks
reveal the following data:
Time PIP (cm H
2
O) P
plateau
2
(cm H
2
T
s
s
= V
= V
O)
T
T
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 (P
) by subtracting the
plateau pressure from the peak inspiratory pressure. Analyzing
the P
TA
TA
will show any changes in the pressure needed to
overcome airway resistance. Analyzing the P
will
demonstrate any changes in compliance. The P
plateau
remained the
same for the first two checks and then actually dropped at the
1000 hour check. Analyzing the P
TA
plateau
, however, shows a slight
increase between 0600 and 0800 (from 12 cm H
2
O to 14 cm H
O)
and then a sharp increase to 23 cm H
2
O at 1000. Increases in P
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
2
TA
13. The values below pertain to a patient who is being mechanically
ventilated with a measured exhaled tidal volume (V
T
Time Peak Inspiratory
Pressure (cm
H
2
O)
) of 700 mL.
Plateau Pressure
(cm H
0800 35 30
1000 39 34
1100 45 39
1130 50 44
2
O)
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 P
TA
= 5 cm H
= 5 cm H
2
O, 1100 P
TA
= 6 cm H
2
O, and 1130 P
TA
2
O, 1000 P
= 6 cm H
O. This
data shows that there is no significant increase or decrease in
this patient’s airway resistance. Analysis of the patient’s plateau
pressure (P
plateau
) reveals an increase of 15 cm H
O over the three
and one half hour time period. This is directly related to a
decrease in lung compliance. Calculation of the lung compliance
(C
S
= V
T
/(P
plateau
-EEP) at each time interval reveals a steady
decrease from 20 mL/cm H
2
DIF: 3 REF: pg. 6
O to 14 mL/cm H
2
O.
2
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
(P
plateau
)
2
TA
b. A decreased elastic resistance
c. A low peak inspiratory
pressure (PIP)
d. A large transairway pressure
(P
TA
) gradient
ANS: A
ARDS is a pathological condition that is associated with a
reduction in lung compliance. The formula for static compliance
(C
S
) utilizes the measured plateau pressure (P
plateau
) in its
denominator (C
S
= V
T
/(P
- EEP). Therefore, with a consistent
exhaled tidal volume (V
T
plateau
) , an elevated P
DIF: 2 REF: pg. 5| pg. 6
plateau
will decrease C
15. The formula used for the calculation of static compliance (C
S
) is
which of the following?
a. (Peak pressure (PIP) –
EEP)/tidal volume (V
)
C
b. (Plateau pressure (P
S
T
= (PIP-EEP)/V
) –
EEP/tidal volume (V
)
C
S
plateau
T
= (P
plateau
–
EEP)/V
T
c. Tidal volume/(plateau pressure
– EEP) C
S
= V
T
/
(P
- EEP)
d. Tidal volume /(peak pressure
ANS: C
C
S
= V
T
/(P
plateau
- EEP)
DIF: 1 REF: pg. 7
16. Plateau pressure (P
plateau
plateau
(PIP) – plateau pressure
(P
plateau
)) C
/
(PIP- P
plateau
)
) is measured during which phase of the
ventilatory cycle?
a. Inspiration
S =
V
S
T
.
T
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 noflow
conditions.
The
airway
pressure
(P
) is measured at endinspiration.
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
endexpiration
is
the
baseline
pressure;
this
reading
is
either
at
zero
(atmospheric
pressure)
or
at
above
atmospheric
pressure
(PEEP).
DIF: 1 REF: pg. 6
aw
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
18. An increase in peak inspiratory pressure (PIP) without an increase
in plateau pressure (P
plateau
) is associated with which of the
following?
a. Increase in static compliance
(C
S
)
b. Decrease in static compliance
(C
S
)
c. Increase in airway resistance
d. Decrease in airway resistance
ANS: C
The PIP represents the amount of pressure needed to overcome
both elastance and airway resistance. The P
plateau
is the amount of
pressure required to overcome elastance alone. Since the P
has remained constant in this situation, the static compliance is
unchanged. The difference between the PIP and the P
is the
transairway pressure (P
) and represents the pressure required to
overcome the airway resistance. If P
TA
increases, the airway
resistance is also increasing, when the gas flow rate remains the
same.
DIF: 2 REF: pg. 5| pg. 6
TA
19. The patient-ventilator data over the past few hours demonstrates
an increased peak inspiratory pressure (PIP) with a constant
transairway pressure (P
TA
). The respiratory therapist should
conclude which of the following?
a. Static compliance (C
plateau
) has
increased.
b. Static compliance (C
S
) has
decreased.
c. Airway resistance (R
S
) has
increased.
d. Airway resistance (R
aw
aw
)has
decreased.
ANS: B
The PIP represents the amount of pressure needed to overcome
both elastance and airway resistance. The P
plateau
is the amount of
pressure required to overcome elastance alone, and is the
plateau
pressure used to calculate the static compliance. Since P
TA
has
stayed the same, it can be concluded that R
aw
has remained the
same. Therefore, the reason the PIP has increased is because of
an increase in the P
plateau
DIF: 2 REF: pg. 5
. This correlates to a decrease in C
S.
20. Calculate airway resistance (R
aw
) for a ventilator patient, in cm
H
2
O/L/sec, when the peak inspiratory pressure (PIP) is 50 cm H
O,
the plateau pressure (P
plateau
) is 15 cm H
2
O, and the set flow rate is
60 L/min.
a. 0.58 R
b. 1.2 R
c. 35 R
aw
d. 50 R
ANS: C
R
aw
= P
TA
/flow; or R
aw
= (PIP – P
DIF: 2 REF: pg. 5| pg. 6
21. Calculate airway resistance (R
plateau
aw
aw
aw
aw
)/flow
) for a ventilator patient, in cm
H
O/L/sec, with the following information: Peak inspiratory
pressure (PIP) is 20 cm H
2
2
O, plateau pressure (P
) is 15 cm
H
2
O, PEEP is 5 cm H
O, and set flow rate is 50 L/min.
a. 5 R
aw
2
b. 6 R
c. 10 R
d. 15 R
aw
ANS: B
R
aw
= (PIP – P
plateau
aw
aw
)/flow and flow is in Liters/second.
DIF: 2 REF: pg. 5| pg. 6
22. Calculate the static compliance (C
S
), in mL/cm H
2
plateau
O, when PIP is
47 cm H
2
O, plateau pressure (P
plateau
) is 27 cm H
2
O, baseline
pressure is 10 cm H
2
O, and exhaled tidal volume (V
T
) is 725 mL.
a. 43 C
S
2
b. 36 C
S
c. 20 C
S
d. 0.065 C
ANS: A DIF: 2 REF: pg. 5| pg. 6
S
23. Calculate the inspiratory time necessary to ensure 98% of the
volume is delivered to a patient with a C
s
= 40 mL/cm H
O and
the R
aw
= 1 cm H
2
O/(L/sec).
a. 0.04 sec
b. 0.16 sec
c. 1.6 sec
d. 4.0 sec
ANS: B
Time constant = C (L/cm H
2
O) x R (cm H
O/(L/sec)). 98% of the
volume will be delivered in 4 time constants. Therefore, multiply
4 times the time constant.
DIF: 2 REF: pg. 6
2
24. How many time constants are necessary for 95% of the tidal
volume (V
) to be delivered from a mechanical ventilator?
a. 1
b. 2
c. 3
d. 4
T
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; 3
time 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
2
volume is delivered to an intubated patient with a C
s
= 60 mL/cm
H
2
O and the R
aw
= 6 cm H
2
O/(L/sec).
a. 0.36 sec
b. 0.5 sec
c. 1.4 sec
d. 1.8 sec
ANS: D
Time constant (TC) = C (L/cm H
2
O) x R (cm H
2
O/(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. C
s
= 0.1 L/cm H
2
O R
= 1
cm H
O/(L/sec)
b. C
s
2
= 0.1 L/cm H
2
O R
aw
= 10
cm H
O/(L/sec)
c. C
s
2
= 0.03 L/cm H
2
O R
aw
= 1
cm H
O/(L/sec)
d. C
s
2
= 0.03 L/cm H
2
O R
aw
=
10 cm H
2
O/(L/sec)
ANS: C
Use the time constant formula, TC = C x R, to determine the time
constant 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 D
is 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 is
which of the following?
aw
1. Requires more time to fill than
a normal alveolus
2. Fills more quickly than a
normal alveolus
3. Requires more volume to fill
than a normal alveolus
4. More pressure is needed to
achieve a normal volume
a. 1 and 3 only
b. 2 and 4 only
c. 2 and 3 only
d. 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 units
that require more volume to fill than normal are high-compliance
units.
DIF: 1 REF: pg. 9
28. Calculate the static compliance (C
S
), in mL/cm H
O, when PIP is
26 cm H
2
O, plateau pressure (P
plateau
) is 19 cm H
2
O, baseline
pressure is 5 cm H
2
O and exhaled tidal volume (V
2
) is 425 mL.
a. 16
b. 20
c. 22
d. 30
ANS: D
C
S
= V
T
/(P
plateau
- EEP)
DIF: 2 REF: pg. 5
29. What type of ventilator increases transpulmonary pressure (P
T
) by
mimicking the normal mechanism for inspiration?
L
a. Positive pressure ventilation
(PPV)
b. Negative pressure ventilation
(NPV)
c. High frequency oscillatory
ventilation (HFOV)
d. 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 ventilation
is known as which of the following?
a. Plateau pressure (P
)
b. Functional residual capacity
(FRC)
c. Extrinsic positive end
plateau
expiratory pressure (extrinsic
PEEP)
d. Intrinsic positive end
expiratory pressure (intrinsic
PEEP)
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
amount of gas that normally remains in the lung after a quiet
exhalation. The plateau pressure is the pressure measured in the
lungs at no flow during an inspiratory hold maneuver.
DIF: 1 REF: pg. 7| pg. 8
31. The transairway pressure (P
TA
) shown in this figure is which of the
following?
a. 5 cm H
2
O
b. 10 cm H
O
c. 20 cm H
O
d. 30 cm H
2
ANS: B
P
TA
= PIP - P
plateau
, where the PIP is 30 cm H
2
2
O
2
O and the P
is 20
cm H
2
O. The PEEP is 5 cm H
DIF: 2 REF: pg. 12
2
O.
32. Use this figure to compute the static compliance (C
) for an
intubated patient with an exhaled tidal volume (V
a. 14 mL/cm H
2
O
b. 20 mL/cm H
O
c. 33 mL/cm H
O
d. 50 mL/cm H
2
ANS: D
C
s
= P
plateau
– EEP; The P
plateau
2
2
O
in the figure is 20 cm H
T
O and the
PEEP is 10 cm H
2
O.
DIF: 2 REF: pg. 12
33. Evaluate the combinations of compliance and resistance and
select the combination that will cause the lungs to empty
slowest.
S
plateau
) of 500 mL.
2
a. C
S
= 0.05 L/cm H
2
O R
aw
= 2
cm H
2
O/(L/sec)
b. C
S
= 0.05 L/cm H
2
O R
= 6
cm H
2
O/(L/sec)
c. C
S
= 0.03 L/cm H
2
O R
aw
aw
= 5
cm H
O/(L/sec)
d. C
S
2
= 0.03 L/cm H
2
O R
aw
= 8
cm H
2
O/(L/sec)
ANS: B
Use the time constant formula, TC = C x R, to determine the time
constant for each choice. The combination with the longest time
constant will empty the slowest. The time constant for A is 0.1
sec, B is 0.3 sec, C is 0.15 sec, and D is 0.24 sec. To find out how
many seconds for emptying, multiply the time constant by 5.
DIF: 3 REF: pg. 7
34. Use this figure to compute the static compliance for an intubated
patient with an inspiratory flow rate set at 70 L/min.
a. 0.2 cm H
O/(L/sec)
b. 11.7 cm H
2
2
O/(L/sec)
c. 16.7 cm H
O/(L/sec)
d. 20 cm H
ANS: B
Use the graph to determine the PIP (34 cm H
2
2
2
O/(L/sec)
O) and the P
plateau
(20 cm H
O). Convert the flow into L/sec (70 L/min/60 = 1.2
L/sec). Then, R
aw
2
= (PIP – P
plateau
DIF: 2 REF: pg. 9
)/flow.
35. The ventilator that functions most physiologically uses which of
the following?
a. Open loop
b. Double circuit
c. Positive pressure
d. Negative pressure
ANS: D
Air is caused to flow into the lungs with a negative pressure
ventilator because the ventilator generates a negative pressure
at the body surface that is transmitted to the pleural space and
then to the alveoli. The transpulmonary pressure becomes
greater because the pleural pressure drops. This closely
resembles how a normal spontaneous breath occurs.
DIF: 2 REF: pg. 5| pg. 6
Chapter 2; How Ventilators Work
Test Bank
MULTIPLE CHOICE
1. The respiratory therapist enters modes and parameters into the ventilator
with which of the following?
a. Control logic
b. Input power
c. User interface
d. Drive mechanism
ANS: C
The user interface or control panel contains certain knobs, dials, or keypads
where the ventilator operator sets or enters certain information to establish
how the pressure and pattern of gas flow is delivered by the machine. Inside
the ventilator is the control logic or control system which interprets the
operator settings and produces and regulates the desired output. The input
power is the ventilator’s power source that provides the energy to enable the
ventilator to perform the work of ventilating the patient. The drive
mechanism is a mechanical device that produces gas flow to the patient.
DIF: 1 REF: pg. 18
2. Which of the following ventilators is pneumatically powered?
a. LTV 1000
b. Bio-Med MVP-10
c. Lifecare PLV-102
d. Intermed Bear 33
ANS: B
The Bio-Med MVP-10 is a fluidic ventilator and uses only gases for its
operation. The LTV 1000, Lifecare PLV-102, and Intermed Bear 33 are
electrically controlled and powered ventilators.
DIF: 1 REF: pg. 18
3. A patient being transferred from a hospital to a skilled nursing facility
requires mechanical ventilation with a fractional inspired oxygen (F
I
) of
0.21. The skilled nursing facility has no piped in gases. Which of the following
ventilators will be able to function in the skilled nursing facility without any
extra equipment?
a. Servo
b. LTV 1000
c. Bird Mark 7
d. Bio-Med MVP-10
i
O
2