Test Bank For Pilbeams Mechanical Ventilation 5th Edition Cairo

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Pilbeams Mechanical Ventilation 5th Edition Cairo

 

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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

 

  1. 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

DC = DV/DP then DP = DV/DC

 

DIF:    2                      REF:    pg. 4

 

  1. 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

 

  1. 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

 

  1. 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 Cs = VT/(Pplateau – EEP).

 

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

 

  1. 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

 

  1. 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 Pressure (cm H2O) Plateau Pressure (cm 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 (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 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

 

  1. 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 (Pplateau) 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

 

  1. 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

 

  1. 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 (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

 

  1. 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

 

  1. An increase in peak inspiratory pressure (PIP) without an increase in plateau pressure (Pplateau) is associated with which of the following?
a. Increase in static compliance (CS)
b. Decrease in static compliance (CS)
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 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 to overcome 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

 

  1. The patient-ventilator data over the past few hours demonstrates an 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. Static compliance (CS) has decreased.
c. Airway resistance (Raw) has increased.
d. 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 of pressure required to overcome elastance alone, and is the pressure used to calculate the static compliance. Since PTA has stayed the same, it can be concluded that Raw has remained the same. Therefore, the reason the PIP has increased is because of an increase in the Pplateau. This correlates to a decrease in CS.

 

DIF:    2                      REF:    pg. 5

 

  1. 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 is 60 L/min.
a. 0.58 Raw
b. 1.2 Raw
c. 35 Raw
d. 50 Raw

 

 

ANS:   C

Raw = PTA/flow; or Raw = (PIP – Pplateau)/flow

 

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

 

  1. Calculate airway resistance (Raw) for a ventilator patient, in cm H2O/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. 6 Raw
c. 10 Raw
d. 15 Raw

 

 

ANS:   B

Raw = (PIP – Pplateau)/flow and flow is in Liters/second.

 

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

 

  1. 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
b. 36 CS
c. 20 CS
d. 0.065 CS

 

 

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

 

  1. 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. 0.16 sec
c. 1.6 sec
d. 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, multiply 4 times the time constant.

 

DIF:    2                      REF:    pg. 6

 

  1. How many time constants are necessary for 95% of the tidal volume (VT) to be delivered from a mechanical ventilator?
a. 1
b. 2
c. 3
d. 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; 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

 

  1. Compute the inspiratory time necessary to ensure 100% of the volume is delivered to an intubated patient with a Cs = 60 mL/cm H2O and the Raw = 6 cm H2O/(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 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

 

  1. 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      Raw = 1 cm H2O/(L/sec)
b. Cs = 0.1 L/cm H2O      Raw = 10 cm H2O/(L/sec)
c. Cs = 0.03 L/cm H2O      Raw = 1 cm H2O/(L/sec)
d. Cs = 0.03 L/cm H2O      Raw = 10 cm H2O/(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

 

  1. The statement that describes the alveolus shown in Figure 1-1 is which of the following?

 

 

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

 

  1. Calculate the static compliance (CS), in mL/cm H2O, when PIP is 26 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. 20
c. 22
d. 30

 

 

ANS:   D

CS = VT/(Pplateau – EEP)

 

DIF:    2                      REF:    pg. 5

 

  1. What type of ventilator increases transpulmonary pressure (PL) by mimicking the normal mechanism for inspiration?
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

 

  1. Air accidently trapped in the lungs due to mechanical ventilation is known as which of the following?
a. Plateau pressure (Pplateau)
b. Functional residual capacity (FRC)
c. Extrinsic positive end 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

 

  1. The transairway pressure (PTA) shown in this figure is which of the following?

 

a. 5 cm H2O
b. 10 cm H2O
c. 20 cm H2O
d. 30 cm H2O

 

 

ANS:   B

PTA = PIP – Pplateau, where the PIP is 30 cm H2O and the Pplateau is 20 cm H2O. The PEEP is 5 cm H2O.

 

DIF:    2                      REF:    pg. 12

 

  1. Use this figure to compute the static compliance (CS) for an intubated patient with an exhaled tidal volume (VT) of 500 mL.

 

a. 14 mL/cm H2O
b. 20 mL/cm H2O
c. 33 mL/cm H2O
d. 50 mL/cm H2O

 

 

ANS:   D

Cs = Pplateau – EEP; The Pplateau in the figure is 20 cm H2O and the PEEP is 10 cm H2O.

 

DIF:    2                      REF:    pg. 12

 

  1. Evaluate the combinations of compliance and resistance and select the combination that will cause the lungs to empty slowest.
a. CS = 0.05 L/cm H2O      Raw = 2 cm H2O/(L/sec)
b. CS = 0.05 L/cm H2O      Raw = 6 cm H2O/(L/sec)
c. CS = 0.03 L/cm H2O      Raw = 5 cm H2O/(L/sec)
d. CS = 0.03 L/cm H2O      Raw = 8 cm H2O/(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

 

  1. 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 H2O/(L/sec)
b. 11.7 cm H2O/(L/sec)
c. 16.7 cm H2O/(L/sec)
d. 20 cm H2O/(L/sec)

 

 

ANS:   B

Use the graph to determine the PIP (34 cm H2O) and the Pplateau (20 cm H2O). Convert the flow into L/sec (70 L/min/60 = 1.2 L/sec). Then, Raw = (PIP – Pplateau)/flow.

 

DIF:    2                      REF:    pg. 9

 

  1. 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 3; How a Breath Is Delivered

Test Bank

 

MULTIPLE CHOICE

 

  1. The equation of motion describes the relationships between which of the following?
a. Pressure and flow during a mechanical breath
b. Pressure and volume during a spontaneous breath
c. Flow and volume during a mechanical or spontaneous breath
d. Flow, volume, and pressure during a spontaneous or mechanical breath

 

 

ANS:   D

The mathematical model that relates pressure, volume, and flow during ventilation is known as the equation of motion for the respiratory system. This means that: Muscle pressure + Ventilator pressure = (Elastance x Volume) + (Resistance x Flow)

 

DIF:    1                      REF:    pg. 30

 

  1. The equation of motion is represented by which of the following?
a. PTA = PA x Raw
b. PTR = Paw + PA
c. Pvent + Pmus = Raw + PTA
d. Pvent + Pmus = Raw x

 

 

ANS:   B

The transrespiratory pressure (PTR) is the pressure generated by either the patient contracting the respiratory muscles or by the ventilator pushing the volume into the patient. This pressure is opposed by the elastic recoil pressure (PE) and the flow resistance pressure (PR). The transairway pressure (PTA) is the pressure gradient between the airway opening and the alveolus. This produces airway movement in the conductive airways. It represents only part of the equation of motion, the pressure needed to overcome the airway resistance. The equation of motion may be represented, on one side, by Pvent + muscle pressure (Pmus). However, this is equal to the elastic recoil pressure (V/C) plus the flow resistance pressure (Raw x ) or Pvent + Pmus = V/C + (Raw x ).

 

DIF:    1                      REF:    pg. 30

 

  1. How many variables can a ventilator control at one time?
a. One
b. Two
c. Three
d. Four

 

 

ANS:   A

As the equation of motion shows, the ventilator can control four variables: pressure, volume, flow, and time. It is important to recognize that the ventilator can control only one variable at a time.

 

DIF:    1                      REF:    pg. 30

 

  1. Calculate the transrespiratory pressure given the following information: volume 0.6 L; compliance 1 L/cm H2O; airway resistance 3 cm H2O/L/sec; flow 1 L/sec.
a. 0.9 cm H2O
b. 1.8 cm H2O
c. 3.6 cm H2O
d. 4.6 cm H2O

 

 

ANS:   C

Transrespiratory pressure (PTR) = Pvent + Pmus = V/C + ( Raw x ).

 

DIF:    2                      REF:    pg. 30

 

  1. An increase in airway resistance during volume-controlled ventilation will have which of the following effects?
a. Volume increase
b. Flow decrease
c. Pressure increase
d. Rate decrease

 

 

ANS:   C

When a ventilator is volume-controlled the ventilator will maintain the volume, which will remain unchanged, along with the flow, but the pressure will vary with changes in lung characteristics. An increase in airway pressure will require more pressure to deliver the set volume. The set rate is independent of the changes in pressure.

 

DIF:    2                      REF:    pg. 32

 

  1. An increase in airway resistance during pressure-targeted ventilation will have which of the following effects?
a. Volume decrease
b. Flow increase
c. Pressure increase
d. Rate decrease

 

 

ANS:   A

During pressure-targeted (pressure-controlled) ventilation, pressure is unaffected by changes in lung characteristics. However, an increase in airway resistance will cause less volume to be delivered and will change the flow waveform. The set pressure will not be able to overcome the increased resistance, resulting in less volume delivery and a decrease in flow (V/TI).

 

DIF:    2                      REF:    pg. 32

 

  1. A patient who has a decrease in lung compliance due to acute respiratory distress syndrome during volume-limited ventilation will cause which of the following?
a. Decreased volume delivery
b. Increased peak pressure
c. Decreased flow delivery
d. Decreased peak pressure

 

 

ANS:   B

When a patient is being ventilated in a volume-limited mode the ventilator will maintain the volume, which will remain unchanged, along with the flow, but the pressure will vary with changes in lung characteristics. A decrease in lung compliance will cause the amount of pressure needed to overcome elastance to increase. This will increase the peak pressure needed to deliver the set volume. Flow and volume will remain constant.

 

DIF:    2                      REF:    pg. 30| pg. 31

 

  1. During pressure-targeted ventilation the patient’s airway resistance decreases to normal due to medication delivery. The ventilator will respond with which of the following changes?
  2. Altered flow waveform
  3. Increased pressure
  4. Increased volume
  5. Decrease volume
a. 1 and 3 only
b. 2 and 4 only
c. 1 and 4 only
d. 1, 2 and 3 only

 

 

ANS:   A

During pressure-targeted ventilation the pressure remains constant and the flow and volume will respond to changes in the patient lung and airway characteristics. An improvement in airway resistance will make it easier to put more volume into the lungs with the same pressure setting as compared to volume delivery with increased airway resistance. Since volume and flow waveform will vary with changes in airway resistance, the volume will increase and the flow waveform will change with improvements in airway resistance. In pressure-targeted ventilation the pressure does not change. A decreased volume would be the result of worsening airway resistance.

 

DIF:    2                      REF:    pg. 30| pg. 31

 

  1. High-frequency oscillators control which of the following variables?
a. Flow
b. Time
c. Volume
d. Pressure

 

 

ANS:   B

High-frequency oscillators control both inspiratory and expiratory time.

 

DIF:    1                      REF:    pg. 41

 

  1. The ventilator variable that begins inspiration is which of the following?
a. Cycle
b. Limit
c. Trigger
d. Baseline

 

 

ANS:   C

The trigger mechanism ends the expiratory phase and begins the inspiratory phase. Limit is the maximum value that a variable may reach during inspiration. Cycle terminates the inspiratory phase. The baseline variable is applied during exhalation and is the pressure level from which a ventilator breath begins.

 

DIF:    1                      REF:    pg. 32

 

  1. The trigger variable in the controlled mode is which of the following?
a. Flow
b. Time
c. Pressure
d. Volume

 

 

ANS:   B

In the controlled mode the ventilator initiates all the breathing because the patient cannot. All ventilator initiated breaths are time triggered. Flow, pressure, and volume triggers are patient initiated.

 

DIF:    1                      REF:    pg. 34

 

  1. A patient who has been sedated and paralyzed by medications is being controlled by the ventilator. The set rate is 15 breaths/min. How many seconds does it take for inspiration and expiration to occur?
a. 2 seconds
b. 4 seconds
c. 6 seconds
d. 8 seconds

 

 

ANS:   B

60 sec/min divided by 15 breaths/min = 4 seconds

 

DIF:    2                      REF:    pg. 34

 

  1. The most commonly used patient-trigger variables include which of the following?
  2. Flow
  3. Time
  4. Pressure
  5. Volume
a. 1 and 3 only
b. 2 and 4 only
c. 1 and 4 only
d. 2 and 3 only

 

 

ANS:   A

The patient trigger variables are flow, pressure, and volume. Time is the ventilator trigger variable. The most common of the three patient triggers are flow and pressure. Very few ventilators use volume as a patient trigger.

 

DIF:    1                      REF:    pg. 34| pg. 35

 

  1. A patient is receiving volume-controlled ventilation. The respiratory therapist notes the pressure-time scalar on the ventilator screen, shown in the figure. The most appropriate action to take includes which of the following?
a. Increase the rate setting.
b. Increase the baseline setting.
c. Decrease the volume setting.
d. Increase the sensitivity setting.

 

 

ANS:   D

What is being shown in the figure is a trigger pressure of 5 cm H2O below the baseline setting of 5 cm H2O. This is seen during the pressure trigger dropping down to 0 cm H2O during the trigger. In this situation the machine is not sensitive enough to the patient’s effort. The patient is working too hard to trigger the ventilator breath. The respiratory therapist needs to increase the ventilator sensitivity control. Changing any of the other parameters will not decrease the work that the patient is doing to trigger inspiration.

 

DIF:    3                      REF:    pg. 35

 

  1. The inspiratory and expiratory flow sensors are reading a base flow of 5 liters per minute (L/min). The flow trigger is set to 2 L/min. The expiratory flow sensor must read what flow to trigger inspiration?
a. 1 L/min
b. 2 L/min
c. 3 L/min
d. 4 L/min

 

 

ANS:   C

Base flow minus flow trigger setting is equal to the flow needed to be sensed at the expiratory flow sensor to trigger inspiration.

 

DIF:    2                      REF:    pg. 35

 

  1. The patient trigger that requires the least amount of work of breathing for the patient is which of the following?
a. Time
b. Flow
c. Pressure
d. Volume

 

 

ANS:   B

When set properly, flow triggering has been shown to require less work of breathing than pressure triggering.

 

DIF:    1                      REF:    pg. 35

 

  1. The limit variable set on a mechanical ventilator will do which of the following?
a. End inspiration
b. Begin inspiration
c. Control the maximum value allowed
d. Control the minimum value allowed

 

 

ANS:   C

A limit variable is the maximum value a variable can attain. It limits the variable during inspiration but does not end the inspiratory phase. The cycle setting ends inspiration. The trigger variable begins inspiration, and there is no control over the minimum value.

 

DIF:    1                      REF:    pg. 36

 

  1. The control variables most often used to ventilate infants are which of the following?
a. Volume limited, time cycled ventilation
b. Pressure limited, time cycled ventilation
c. Pressure limited, pressure cycled ventilation
d. Volume limited, volume cycled ventilation

 

 

ANS:   B

Infant ventilators most often limit the pressure delivered and end inspiration using inspiratory time. Volume limited, volume cycled ventilation is volume-controlled ventilation. Pressure limited, pressure cycled ventilation is the type of breath used during intermittent positive pressure breathing (IPPB).

 

DIF:    1                      REF:    pg. 37

 

  1. The respiratory therapist enters the room of a patient being mechanically ventilated with volume ventilation. The high pressure alarm is sounding and the measured exhaled tidal volume is significantly lower than what is set. The variable that is ending inspiration is which of the following?
a. Time
b. Flow
c. Pressure
d. Volume

 

 

ANS:   C

Volume ventilation is cycled by volume. However, to protect the patient’s lungs from high pressures a maximum high pressure limit is set (usually 10 cm H2O above the average peak inspiratory pressure). Inspiration ends prematurely when the high pressure limit is reached, independent of the set volume. This is the reason why the exhaled tidal volume reading is significantly lower than the set volume. Therefore, the variable ending inspiration in this instance is pressure.

 

DIF:    2                      REF:    pg. 39

 

  1. The variable that a ventilator uses to end inspiration is known as which of the following?
a. Cycle
b. Limit
c. Trigger
d. Baseline

 

 

ANS:   B

Cycle is the term used to call the variable that is used to end inspiration. Limit is the maximum setting for a variable. Trigger is the term used to call the variable that is used to begin inspiration. Baseline is the pressure at the end of inspiration.

 

DIF:    1                      REF:    pg. 39

 

  1. When the maximum pressure limit is reached during volume ventilation, which of the following occurs?
  2. Inspiratory time is decreased.
  3. Volume delivered is decreased.
  4. Inspiration continues until volume is delivered.
  5. Pressure is held and the breath is volume cycled.
a. 3 only
b. 4 only
c. 1 and 2 only
d. 2 and 4 only

 

 

ANS:   C

The maximum pressure limit is a safety mechanism used during volume ventilation to avoid excessive pressure in the lungs. When the pressure measured by the ventilator reaches the maximum pressure limit inspiration ends. This means that the inspiratory time will be decreased and the volume delivered will be less than the set volume. Therefore, reaching maximum pressure limit causes the delivered breath to be pressure cycled.

 

DIF:    1                      REF:    pg. 39

 

  1. The respiratory therapist is called to a patient’s room because the “alarms are ringing.” When the respiratory therapist arrives at the bedside, the high pressure limit, low exhaled tidal volume, and low exhaled minute volume alarms are active. The cause of these alarms is which of the following?
a. Disconnection from the ventilator
b. Critical leak in the ventilator circuit
c. Lung compliance has improved.
d. Airway resistance has increased.

 

 

ANS:   D

The low exhaled tidal volume and minute volume alarms are active when the high pressure limit alarm is active. This occurs because reaching the set high pressure limit setting will end inspiration immediately by pressure cycling and thereby will decrease the volume delivered to the patient. When the high pressure limit alarm is active for several breaths, the low exhaled tidal volume and then the minute ventilation alarms will become active. The high pressure alarm will sound when airway resistance is elevated (for example: asthma). A disconnect from the ventilator or a critical leak would cause the low peak inspiratory pressure alarms to ring. Improved lung compliance will lower the peak inspiratory pressure and may trigger a low pressure alarm.

 

DIF:    2                      REF:    pg. 38

 

  1. The most common method of terminating inspiration during pressure support ventilation is which of the following?
a. Flow
b. Time
c. Pressure
d. Volume

 

 

ANS:   A

During pressure support ventilation, when a breath is delivered, the flow will begin to taper down after a very short period of time. When flow drops to a certain percentage of the initial peak flow the ventilator flow cycles out of inspiration. A time cycled breath is usually a breath that is controlled by the ventilator during pressure control ventilation. Pressure cycling is used for intermittent positive pressure breathing (Bird Mark 7). Volume cycling is utilized during ventilator breaths on certain ventilators.

 

DIF:    1                      REF:    pg. 39

 

  1. What is the flow-cycle setting for the following pressure supported breath?
a. 20%
b. 30%
c. 40%
d. 50%

 

 

ANS:   D

The peak flow for this pressure supported breath is 40 L/min. The breath flow cycled at 20 L/min, which is 50% of the peak flow. Therefore, the flow-cycle setting is 50%.

 

DIF:    2                      REF:    pg. 39

 

  1. Identify the pressure-time scalar for a pressure supported breath.
a.
b.
c.
d.

 

 

ANS:   C

The pressure support breath is pressure limited. Therefore, it will have a “flat top” such as that in option C. Option A is the pressure-time scalar for a volume controlled breath that has an inspiratory hold. Option B is the pressure-time scalar for a volume controlled breath. Option D is the pressure-time scalar for a continuous positive airway pressure (CPAP) 10 cm H2O breath.

 

DIF:    1                      REF:    pg. 39

 

  1. Which maneuver will maintain air in the lungs at the end of inspiration, before the exhalation valve opens?
a. Pressure limit
b. Inspiratory hold
c. Expiratory hold
d. Expiratory resistance

 

 

ANS:   B

The inspiratory hold, inspiratory pause, or end-inspiratory pause is the maneuver that will maintain air in the lungs and extend inspiration. Pressure limit allows pressure to rise but not exceed a pressure limit setting. Expiratory hold is the maneuver that will obtain the unintended positive-end expiratory pressure (Auto-PEEP) measurement. The ventilator pauses before delivering the next machine breath. Expiratory resistance is a resistance added to exhalation to mimic pursed-lip breathing.

 

DIF:    1                      REF:    pg. 26

 

  1. The ventilator that can provide a negative pressure during the very beginning of the exhalation phase is which of the following?
a. Servoi
b. VIASYS Avea
c. Puritan Bennett 840
d. Cardiopulmonary Venturi

 

 

ANS:   D

The Cardiopulmonary Venturi applies a negative pressure to the airway only during the very beginning of the exhalation phase. This facilitates removal of air from the patient circuit and is intended to reduce the resistance to exhalation throughout the circuit at the start of exhalation.

 

DIF:    1                      REF:    pg. 40| pg. 41

 

  1. The pressure-time scalar shown in the figure could be caused by which of the following?

 

a. Inspiratory hold
b. Clogged expiratory filter
c. Excessive secretions in the airway
d. Negative end-expiratory pressure

 

 

ANS:   B

What is being shown in the figure is a peak pressure of 20 cm H2O and then a very slow drop in the pressure over the course of approximately 1 second. Normally, as soon as the peak inspiratory pressure is reached the pressure drops rapidly and remains at baseline until the next breath is given. A clogged expiratory filter will increase the resistance in the filter. This will cause there to be difficulty exhaling through this filter, showing up as a slow drop in pressure during the exhalation period. An inspiratory hold would cause there to be a plateau in the pressure-time scalar. Excessive secretions in the airway would elevate the peak inspiratory pressure to a point where the maximum safety pressure may be reached. The presence of negative end-expiratory pressure (NEEP) would pull the pressure down rapidly from the peak and drop it slightly below the baseline during that time.

 

DIF:    2                      REF:    pg. 41

 

  1. A ventilator is set to deliver a 600 mL tidal volume. The flow rate is set at 40 L/min and the frequency is set at 10 breaths/min. If the flow rate is doubled and the patient is not assisting, which of the following will occur?
a. The frequency will decrease.
b. The tidal volume will increase.
c. The expiratory time will increase.
d. The inspiratory time will increase.

 

 

ANS:   C

If you manipulate the formula: Flow = Volume change/Time to Volume/Flow = Time, and use the numbers from this example, it can be shown that by increasing the flow rate the inspiratory time will decrease. Since the frequency remains constant, then the inspiratory time will decrease, thereby increasing expiratory time.

 

DIF:    2                      REF:    pg. 38

 

  1. The variable that controls exhalation is known as which of the following?
a. Limit
b. Trigger
c. Pressure
d. Baseline

 

 

ANS:   D

The baseline variable is the parameter that generally is controlled during exhalation. Although either volume or flow could serve as a baseline variable, pressure is the most practical choice and is used by all modern ventilators.

 

DIF:    1                      REF:    pg. 40

 

  1. In pressure targeted ventilation the trigger variable for a patient who is sedated is which of the following?
a. Time
b. Flow
c. Volume
d. Pressure

 

 

ANS:   A

When the patient is not triggering a breath, as with sedation, the set mechanical ventilator rate will be time-triggered based on the backup set rate.

 

DIF:    2                      REF:    pg. 34

Chapter 13; Improving Oxygenation and Management of ARDS

Test Bank

 

MULTIPLE CHOICE

 

  1. During mechanical ventilation of a patient with COPD, the PaO2 = 58 mm Hg and the FIO2 = 0.5. If the desired PaO2 is 65 mm Hg, the FIO2 needs to be changed to which of the following?
a. 0.44
b. 0.56
c. 0.65
d. 0.74

 

 

ANS:   B

 

DIF:    2                      REF:    pg. 260

 

  1. During mechanical ventilation of a patient with CHF, the PaO2 = 38 mm Hg and the FIO2 = 0.6. If the desired PaO2 is 60 mm Hg, the FIO2 needs to be changed to which of the following?
a. 0.65
b. 0.75
c. 0.85
d. 0.95

 

 

ANS:   D

 

DIF:    2                      REF:    pg. 260

 

  1. Thirty minutes after intubation and initiation of mechanical ventilation, a patient’s PaO2 = 55 mm Hg and the FIO2 = 0.5. To what should the FIO2 be set to obtain a PaO2 of 80 mm Hg?
a. 0.65
b. 0.75
c. 0.85
d. 0.95

 

 

ANS:   B

 

DIF:    2                      REF:    pg. 260

 

  1. Calculate the pulmonary shunt fraction using the following data: Pb = 757 mm Hg; hemoglobin (Hb) = 11g/dL; FIO2 = 0.5; PaO2 = 86 mm Hg; PaCO2 = 40 mm Hg; SaO2 = 91%; = 40 mm Hg; = 71%; respiratory exchange quotient (R) = 0.8.
a. 19%
b. 23%
c. 26%
d. 32%

 

 

ANS:   A

 

DIF:    2                      REF:    pg. 260

 

  1. Calculate the pulmonary shunt fraction for a patient with the following data: Pb = 760 mm Hg; Hb = 10 g/dL; respiratory quotient = 0.8; FIO2 = 0.6; PaO2 = 100 mm Hg; SaO2 = 93%; PaCO2 = 45 mm Hg;  = 36 mm Hg;  = 70%.
a. 13 %
b. 26%
c. 30 %
d. 41%

 

 

ANS:   B

 

DIF:    2                      REF:    pg. 266

 

  1. Calculate the pulmonary shunt fraction for a patient with the following data: CAO2 = 17 vols%; CaO2 = 16.5 vols %; = 11 vols%.
a. 8%
b. 15%
c. 30%
d. 37%

 

 

ANS:   A

 

 

DIF:    2                      REF:    pg. 260

 

  1. PEEP therapy is indicated for patients with which of the following?
a. PaO2 of 95 mm Hg while receiving an FIO2 of 0.3
b. PaO2 of 100 mm Hg while receiving an FIO2 of 0.8
c. Returned VT of 600 mL with a Pplateau of 12 cm H2O
d. Returned VT of 800 mL with a Pplateau of 15 cm H2O

 

 

ANS:   B

The PaO2/FIO2 for answer A is 317, which shows no ALI or ARDS. The PaO2/FIO2 for answer B is 125; this is an indication for PEEP therapy (PaO2/FIO2 <200 for ARDS). The compliance for answers C and D is 50 mL/cm H2O and 53 mL/cm H2O, respectively. These compliances are normal for intubated patients.

 

DIF:    2                      REF:    pg. 275

 

  1. Patients with which of the following clinical disorders may benefit from PEEP?
a. COPD
b. Asthma
c. ARDS
d. Cystic fibrosis

 

 

ANS:   C

ARDS is a clinical disorder that benefits from the use of PEEP. The other three choices are obstructive lung diseases and will not benefit from the use of PEEP therapy.

 

DIF:    1                      REF:    pg. 275

 

  1. How long after PEEP is increased should all ventilatory and available hemodynamic parameters be measured and calculated?
a. 5 minutes
b. 15 minutes
c. 25 minutes
d. 40 minutes

 

 

ANS:   B

Approximately 15 minutes after an increase in PEEP, all ventilatory and available hemodynamic parameters are measured and calculated.

 

DIF:    1                      REF:    pg. 267

 

  1. Assessing the outcome of PEEP at levels set above 15 to 20 cm H2O is best done using which of the following?
a. Static compliance measurements
b. Pressure-volume loop graphics
c. Pulmonary artery occlusion pressure
d. Central venous pressure measurements

 

 

ANS:   C

At pressures above 15 to 20 cm H2O, the compliance measurement is not a good indicator of cardiovascular function, and monitoring of the pulmonary artery pressure may be indicated.

 

DIF:    1                      REF:    pg. 270

 

  1. An absolute contraindication to PEEP is which of the following?
a. Emphysema
b. Bronchopleural fistula
c. Untreated tension pneumothorax
d. Elevated intracranial pressures

 

 

ANS:   C

An absolute contraindication to PEEP is a tension pneumothorax. PEEP must be used with care in patients with bronchopleural fistulas and elevated intracranial pressures. PEEP also must be used with care in patients with emphysema, because it may increase hyperinflation, which can lead to compression of adjacent capillaries.

 

DIF:    1                      REF:    pg. 270

 

  1. The level of applied PEEP should be set at what point on the pressure-volume curve?
a. At the upper inflection point of the inflation curve.
b. Above the lower inflection point of the deflation curve.
c. At the peak inspiration point of the inflation curve.
d. Above the upper inflection point of the deflation curve

 

 

ANS:   D

The level of applied PEEP should be set 3 to 4 cm H2O above the upper inflection point of the deflation limb of the P-V curve to help maintain an open lung.

 

DIF:    1                      REF:    pg. 281

 

  1. What is the optimal PEEP level given the following information?

 

Time 0600 0615 0625 0635 0640 0650
PEEP 0 5 8 10 12 15
PaO2 (FIO2 = 0.6) 50 62 75 81 86 103
Cs (mL/cm H2O) 16 21 28 38 34 30
BP 140/90 130/90 120/90 120/90 118/80 100/70

 

 

a. 8 cm H2O
b. 10 cm H2O
c. 12 cm H2O
d. 15 cm H2O

 

 

ANS:   B

At a PEEP of 10 cm H2O, the Cs is at its highest level, and the PaO2 and BP are at acceptable levels. At 12 cm H2O, the Cs is reduced, and it declines further at a PEEP of 15 cm H2O.

 

DIF:    2                      REF:    pg. 268

 

  1. What is the optimal PEEP level given the following information?

 

Time 0700 0710 0720 0735 0740 0755
PEEP

 

0 5 8 10 12 15
P(a-et)CO2 11 10 6 13 15 15

 

26 28 34 30 27 25

 

BP 138/90 128/93 120/86 100/70 100/70 105/75

 

 

 

a. 8 cm H2O
b. 10 cm H2O
c. 12 cm H2O
d. 15 cm H2O

 

 

ANS:   A

At a PEEP of 8 cm H2O, the P(a-et)CO2 has been minimized, meaning that a maximum number of alveoli have been recruited without overdistention. At a PEEP of 10 cm H2O, the P(a-et)CO2 increases. This means that too much PEEP has be added. At the same time, the  drops from 34 mm Hg at a PEEP of 8 cm H2O to 30 mm Hg at a PEEP of 10 cm H2O. This correlates with a decrease in cardiac output and the drop in blood pressure.

 

DIF:    3                      REF:    pg. 269

 

  1. The most appropriate PEEP level (optimum PEEP) for the patient whose information is in the table below is which of the following?

 

Time

 

0140 1050 0155  0200 0205 0210
PEEP (cm H2O) 5 8 10 12 15 18

 

CO (L/min) 3.5 3.5 3.6 3.6 3.2 3.2

 

 (mm Hg) 28 39 43 45 42 27

 

BP 140/90 133/90 124/90 120/80 115/78 100/70

 

 

 

a. 10 cm H2O
b. 12 cm H2O
c. 15 cm H2O
d. 18 cm H2O

 

 

ANS:   B

At a PEEP of 12 cm H2O, the cardiac output is at its highest level (3.6 L/min), and the mixed venous PO2 is at its highest level. This correlates with an enhancement in cardiac performance. However, at a PEEP of 15 cm H2O, both these indices, as well as the blood pressure, are reduced, meaning that the alveoli are overdistended and the point of optimum PEEP is no longer present.

 

DIF:    3                      REF:    pg. 270

 

  1. The level of PEEP that is most appropriate for a patient with the information shown below is which of the following?

 

Time 0725 0730 0742 0748 0755 0800

 

PEEP (cm H2O) 5 8 10 12 15 18
Cs (mL/cm H2O) 27 28 34 29 29 25
P(a-et)CO2 (mm Hg) 8.8 6.2 4.8 4.9 6.5 10
PaO2/FIO2 159 190 237 237 221  204

 

 

 

a. 8 cm H2O
b. 10 cm H2O
c. 12 cm H2O
d. 15 cm H2O

 

 

ANS:   B

At a PEEP of 10 cm H2O, the static compliance is at its highest level (34 mL/cm H2O), and the arterial to end-tidal CO2 tension gradient is at its lowest (4.8 mm Hg). This means that at a PEEP of 10 cm H2O, this patient’s FRC has been restored and the effectiveness of ventilation has been optimized. With PEEP higher than 10 cm H2O, these values show that the alveoli are being overstretched. The PaO2/FIO2 is also at its highest level of 237

 

DIF:    3                      REF:    pg. 268

 

  1. A patient is being ventilated with a PEEP of 10 cm H2O and an FIO2 of 0.4. The arterial blood gas results show that the patient remains hypoxemic, and the respiratory therapist increases the PEEP to 18 cm H2O, maintaining the FIO2 at 0.4. The patient’s static compliance changes from 28 mL/cm H2O to 22 mL/cm H2O just after this change. The respiratory therapist should do which of the following?
a. Decrease PEEP to 10 cm H2O and increase the FIO2 to 0.6.
b. Decrease PEEP to 15 cm H2O and measure static compliance.
c. Keep PEEP at 18 cm H2O and increase the FIO2 to 0.6.
d. Increase PEEP to 20 cm H2O and measure static compliance.

 

 

ANS:   B

The increase in PEEP from 10 to 18 cm H2O caused overdistention of the gas exchange units. This is evidenced by the decrease in compliance from 28 to 22 mL/cm H2O. PEEP should be increased in increments of 3 to 5 cm H2O at a time. Therefore, the respiratory therapist should step the PEEP back to 13 to 15 cm H2O and check the static compliance.

 

DIF:    3                      REF:    pg. 268

 

  1. A patient with ARDS has the slow pressure-volume loop shown below. Based on this loop, at what level should PEEP be set?

 

a. 10 cm H2O
b. 12 cm H2O
c. 18 cm H2O
d. 22 cm H2O

 

 

ANS:   D

The rapid change in the slope of the deflation curve identifies the upper inflection point on the deflation portion of the curve as approximately 20 cm H2O. PEEP should be set 2 to 3 cm H2O above the upper inflection point, which is 22 cm H2O.

 

 

DIF:    3                      REF:    pg. 281

 

  1. What is the Pflex on the following pressure-volume loop?

 

a. 10 cm H2O
b. 15 cm H2O
c. 19 cm H2O
d. 23 cm H2O

 

 

ANS:   C

The Pflex, or lower inflection point on the inspiratory limb, is the point on the curve where the slop of the inspiratory line changes significantly (see the following figure).

 

DIF:    2                      REF:    pg. 281

 

  1. Regardless of the procedure used to establish an appropriate PEEP level, ventilating pressures should not be allowed to exceed which of the following?
a. Upper inflation point on the inspiratory limb (UIPi)
b. 2 to 3 cm H2O above the UIPi
c. 30 cm H2O
d. Lower inflection point on the inspiratory limb

 

 

ANS:   A

Regardless of the procedure used to establish an appropriate PEEP level, ventilating pressures should not be allowed to exceed the UIP on the UIPi, because injury to lungs can occur if the lungs become overstretched. The appearance of the UIP on the graphic display may be influenced by the type of recruitment procedure used. For example, in one study, when the VT was set low (5 to 6 mL/kg), the UIP was 26 cm H2O.

 

DIF:    1                      REF:    pg. 281

 

  1. Despite the risk, it is still important to use PEEP, because it can prevent alveolar collapse during exhalation and reopening, even when a low VT is used. It now is theorized that it is important to use the pressure-volume loop to set PEEP _____________________.
a. at the upper inflection point detected during inflation of the lung
b. above the upper inflection point detected during deflation of the lung
c. at the lower inflection point detected during inflation of the lung
d. at the peak inspiration point detected during inflation of the lung

 

 

ANS:   B

Regardless of the procedure used to establish an appropriate PEEP level, ventilating pressures should not be allowed to exceed the UIP on the UIPi, because injury to lungs can occur if the lungs become overstretched. The appearance of the UIP on the graphic display may be influenced by the type of recruitment procedure used.

 

DIF:    1                      REF:    pg. 281

 

  1. During a patient case study, increasing increments of PEEP showed no significant effects until 15 cm H2O was used, at which time the PaO2 improved markedly. This represents the point at which _______________.
a. cardiac output decreased
b. airway resistance decreased
c. hemoglobin saturation improved
d. alveolar recruitment probably occurred

 

 

ANS:   D

Recruitment maneuvers can produce varying results among patients. In patients who respond to an RM, PaO2 increases, PaCO2 decreases, and the change in pressure (DP) required to cause an acceptable VT decreases.

 

DIF:    2                      REF:    pg. 272

 

  1. In which ventilator mode should a patient receiving a sustained inflation technique be placed?
a. VC-CMV
b. APRV
c. PC-IMV
d. CPAP/spontaneous

 

 

ANS:   D

To perform the sustained inflation technique, the ventilator needs to be set in the CPAP/spontaneous mode, because no mechanical breaths should be given during the procedure. The patient also requires sedation and short-term paralysis during this procedure.

 

DIF:    1                      REF:    pg. 285

 

  1. To perform a slow-flow (quasi-static) technique for determining the appropriate PEEP level, the most appropriate ventilator flow setting is which of the following?
a. 2 L/min
b. 6 L/min
c. 10 L/min
d. 14 L/min

 

 

ANS:   A

Although flow rates up to 9 L/min can be used, the higher flow rates cause a slight shift to the right of the resulting P-V curve. The single breath slow-flow should be 2 L/min.

 

DIF:    1                      REF:    pg. 280

 

  1. The highest pressure attained during the slow-flow (quasi-static) technique should be ______ cm H2O.
a. 25
b. 35
c. 45
d. 55

 

 

ANS:   C

The slow-flow for static P-V measurement uses a single breath delivered at 2 L/min until the pressure reaches 45 cm H2O.

 

DIF:    1                      REF:    pg. 280

 

  1. The point on a static pressure-volume curve (SPV) where the alveoli begin to open is referred to as which of the following?
a. Lower inflection point on the inflation limb
b. Upper inflection point on the inflation limb
c. Upper inflection point on the deflation limb
d. Lower inflection point on the deflation limb

 

 

ANS:   A

At the lower inflection point on the inflation limb, the slope of the line changes significantly. It originally was believed that this point represents the opening of most of the collapsed alveoli. However, it is the point at which the alveoli begin to open. Even at the upper inflection point on the inflation limb, the alveoli are still being recruited in some parts of the lungs.

 

DIF:    1                      REF:    pg. 281

 

  1. The “sigmoid” shape of the static pressure-volume lung recruitment maneuver indicates which of the following?
a. All parts of the lungs open with the same pressure.
b. Independent portions of the lungs open with the same pressure.
c. Dependent portions of the lungs open with different pressures.
d. Lung units open at different times with different pressures

 

 

ANS:   D

The shape of the pressure-volume curve suggests that different areas of the lung open at different pressures during the recruitment maneuver. As pressure is exerted every few seconds, different areas of the lungs are recruited. This occurs between the lower inflection point on the inspiratory limb up to the upper inflection point on the inspiratory limb.

 

DIF:    1                      REF:    pg. 283

 

  1. “Loose atelectasis,” or compression atelectasis, is most often associated with _____________.
a. ALI
b. ARDS
c. anesthesia
d. pulmonary fibrosis

 

 

ANS:   C

Compression atelectasis, the result of gravitational pressure from lung and heart tissue, often occurs with anesthesia.

 

DIF:    1                      REF:    pg. 282

 

  1. A recruitment maneuver (RM) is being performed on a patient receiving mechanical ventilation with PCV. During the maneuver the mode remains in PCV, rate = 10/min, I:E = 1:2, and PIP = 35 cm H2O. The following information is documented during the RM:

 

Time 0800 0805 0810 0815 0820 0825 0830 0835 0840

 

PEEP 10 15 20 25 30 25 20 15 10

 

Static

compliance

24 29 34 25 22 27 31 36 32

 

BP 126/68 124/66 118/58 110/55 106/58 112/60 118/62 115/60 117/62
HR 108 108 105 100 100 102 105 106 105

 

 

 

After the RM, the lungs are reinflated. At what level should the PEEP be set for this patient?

a. 12 cm H2O
b. 17 cm H2O
c. 22 cm H2O
d. 27 cm H2O

 

 

ANS:   A

This is the PCV with increased PEEP recruitment technique. The point of decreased compliance on deflation represents the UIPd of the lungs. Once the lungs have been reinflated to allow reopening of lung units, PEEP is decreased until a pressure 2 cm H2O above the UIPd is obtained. For this patient, that would be 12 cm H2O.

 

DIF:    3                      REF:    pg. 267| pg. 270

 

  1. The patient with which of the following assessment findings meets the criteria for beginning weaning from PEEP?
a. PaO2 = 85 mm Hg; FIO2 = 0.6; Cs = 20 mL/cm H2O; PEEP = 12 cm H2O
b. PaO2 = 100 mm Hg; FIO2 = 0.9; Cs = 22 mL/cm H2O; PEEP = 10 cm H2O
c. PaO2 = 150 mm Hg; FIO2 = 0.7; Cs = 25 mL/cm H2O; PEEP = 12 cm H2O
d. PaO2 = 95 mm Hg; FIO2 = 0.3; Cs = 30 mL/cm H2O; PEEP = 15 cm H2O

 

 

ANS:   D

The criteria for initiation of weaning from PEEP include (1) an acceptable PaO2 (90 mm Hg) on an FIO2 0.4; (2) hemodynamic stability; (3) absence of sepsis; (4) improved CL (e.g., CS >25 mL/cm H2O); and (5) PaO2/FIO2 ratio >250 to 300. Choice A does not meet the minimum PaO2 or the minimum FIO2 and has a low Cs. Choice B has the appropriate PaO2, but the FIO2 is high and the Cs is low. Choice C has a high PaO2, but the FIO2 is 0.7, the calculated PaO2/FIO2 ratio is 214, and the compliance is just at the acceptable level. Choice D has all acceptable criteria, including a calculated PaO2/FIO2 ratio of 317.

 

DIF:    2                      REF:    pg. 275

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