Test Bank For Pilbeams Mechanical Ventilation 5th Edition Cairo

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Test Bank For Pilbeams Mechanical Ventilation 5th Edition Cairo

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WITH ANSWERS

Pilbeams Mechanical Ventilation 5th Edition Cairo

Chapter 1; Basic Terms and Concepts of Mechanical Ventilation

Test Bank

MULTIPLE CHOICE

The bodys 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 bodys 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

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

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

During spontaneous inspiration alveolar pressure (P_A) 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

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)
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_TA ) 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

Calculate the pressure needed to overcome airway resistance during positive pressure ventilation when the proximal airway pressure (P_Aw) is 35 cm H₂O and the alveolar pressure (P_A) 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 (P_TA) is used to calculate the pressure required to overcome airway resistance during mechanical ventilation. This formula is P_TA = P_aw P_A.

DIF: 2 REF: pg. 3

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

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

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

Calculate the effective static compliance (C_s) given the following information about a patient receiving mechanical ventilation: peak inspiratory pressure (PIP) is 56 cm H₂O, plateau pressure (P_plateau) is 40 cm H₂O, exhaled tidal volume (V_T) 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 C_s = V_T/(P_plateau EEP).

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

Based upon the following patient information calculate the patients static lung compliance: exhaled tidal volume (V_T) is 675 mL, peak inspiratory pressure (PIP) is 28 cm H₂O, plateau pressure (P_plateau) is 8 cm H₂O, and PEEP is set at 5 cm H₂O.

a.	0.02 L/cm H₂O
b.	0.03 L/cm H₂O
c.	0.22 L/cm H₂O
d.	0.34 L/cm H₂O

ANS: C

The formula for calculating effective static compliance is C_s = V_T/(P_plateau EEP).

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

A patient receiving mechanical ventilation has an exhaled tidal volume (V_T) 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)	P_plateau (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 (P_TA) by subtracting the plateau pressure from the peak inspiratory pressure. Analyzing the P_TA will show any changes in the pressure needed to overcome airway resistance. Analyzing the P_plateau will demonstrate any changes in compliance. The P_plateauremained the same for the first two checks and then actually dropped at the 1000 hour check. Analyzing the P_TA, 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 P_TA 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 Poiseuilles law. Increasing the PEEP will not address the root of this patients problem; the patients compliance is normal.

DIF: 3 REF: pg. 6

The values below pertain to a patient who is being mechanically ventilated with a measured exhaled tidal volume (V_T ) 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 (P_TA) is calculated for the different times: 0800 P_TA = 5 cm H₂O, 1000 P_TA = 5 cm H₂O, 1100 P_TA = 6 cm H₂O, and 1130 P_TA = 6 cm H₂O. This data shows that there is no significant increase or decrease in this patients airway resistance. Analysis of the patients plateau pressure (P_plateau ) 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 (C_S = V_T/(P_plateau-EEP) at each time interval reveals a steady decrease from 20 mL/cm H₂O to 14 mL/cm H₂O.

DIF: 3 REF: pg. 6

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)
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_plateau EEP). Therefore, with a consistent exhaled tidal volume (V_T) , an elevated P_plateauwill decrease C_S.

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

The formula used for the calculation of static compliance (C_S) is which of the following?

a.	(Peak pressure (PIP) EEP)/tidal volume (V_T) C_S = (PIP-EEP)/V_T
b.	(Plateau pressure (P_plateau) EEP/tidal volume (V_T) C_S = (P_plateau EEP)/V_T
c.	Tidal volume/(plateau pressure EEP) C_S = V_T/(P_plateau EEP)
d.	Tidal volume /(peak pressure (PIP) plateau pressure (P_plateau )) C_{S =}V_T /(PIP- P_plateau)

ANS: C

C_S = V_T/(P_plateau EEP)

DIF: 1 REF: pg. 7

Plateau pressure (P_plateau) is measured during which phase of the ventilatory cycle?

a.	Inspiration
b.	End-inspiration
c.	Expiration
d.	End-expiration

ANS: B

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

DIF: 1 REF: pg. 6

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 patients 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 patients lung compliance. Ascites causes fluid buildup in the peritoneal cavity and increases tissue resistance, not airway resistance.

DIF: 1 REF: pg. 5

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_plateau has remained constant in this situation, the static compliance is unchanged. The difference between the PIP and the P_plateau is the transairway pressure (P_TA) 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

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_S) has increased.
b.	Static compliance (C_S) has decreased.
c.	Airway resistance (R_aw) has increased.
d.	Airway resistance (R_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 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. This correlates to a decrease in C_S.

DIF: 2 REF: pg. 5

Calculate airway resistance (R_aw ) for a ventilator patient, in cm H₂O/L/sec, when the peak inspiratory pressure (PIP) is 50 cm H₂O, the plateau pressure (P_plateau) is 15 cm H₂O, and the set flow rate is 60 L/min.

a.	0.58 R_aw
b.	1.2 R_aw
c.	35 R_aw
d.	50 R_aw

ANS: C

R_aw = P_TA/flow; or R_aw = (PIP P_plateau)/flow

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

Calculate airway resistance (R_aw) for a ventilator patient, in cm H₂O/L/sec, with the following information: Peak inspiratory pressure (PIP) is 20 cm H₂O, plateau pressure (P_plateau) is 15 cm H₂O, PEEP is 5 cm H₂O, and set flow rate is 50 L/min.

a.	5 R_aw
b.	6 R_aw
c.	10 R_aw
d.	15 R_aw

ANS: B

R_aw = (PIP P_plateau)/flow and flow is in Liters/second.

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

Calculate the static compliance (C_S), in mL/cm H₂O, when PIP is 47 cm H₂O, plateau pressure (P_plateau) is 27 cm H₂O, baseline pressure is 10 cm H₂O, and exhaled tidal volume (V_T) is 725 mL.

a.	43 C_S
b.	36 C_S
c.	20 C_S
d.	0.065 C_S

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

Calculate the inspiratory time necessary to ensure 98% of the volume is delivered to a patient with a C_s= 40 mL/cm H₂O and the R_aw = 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

How many time constants are necessary for 95% of the tidal volume (V_T) 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

Compute the inspiratory time necessary to ensure 100% of the volume is delivered to an intubated patient with a C_s= 60 mL/cm H₂O and the R_aw = 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

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

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

Calculate the static compliance (C_S), in mL/cm H₂O, when PIP is 26 cm H₂O, plateau pressure (P_plateau) is 19 cm H₂O, baseline pressure is 5 cm H₂O and exhaled tidal volume (V_T) 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

What type of ventilator increases transpulmonary pressure (P_L) 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

Air accidently trapped in the lungs due to mechanical ventilation is known as which of the following?

a.	Plateau pressure (P_plateau)
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 patients 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

The transairway pressure (P_TA) 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

P_TA = PIP P_plateau, where the PIP is 30 cm H₂O and the P_plateau is 20 cm H₂O. The PEEP is 5 cm H₂O.

DIF: 2 REF: pg. 12

Use this figure to compute the static compliance (C_S) for an intubated patient with an exhaled tidal volume (V_T) 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

C_s = P_plateau EEP; The P_plateau in the figure is 20 cm H₂O and the PEEP is 10 cm H₂O.

DIF: 2 REF: pg. 12

Evaluate the combinations of compliance and resistance and select the combination that will cause the lungs to empty slowest.

a.	C_S = 0.05 L/cm H₂O R_aw = 2 cm H₂O/(L/sec)
b.	C_S = 0.05 L/cm H₂O R_aw = 6 cm H₂O/(L/sec)
c.	C_S = 0.03 L/cm H₂O R_aw = 5 cm H₂O/(L/sec)
d.	C_S = 0.03 L/cm H₂O R_aw = 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

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 P_plateau (20 cm H₂O). Convert the flow into L/sec (70 L/min/60 = 1.2 L/sec). Then, R_aw = (PIP P_plateau)/flow.

DIF: 2 REF: pg. 9

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

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

The equation of motion is represented by which of the following?

a.	P_TA = P_A x R_aw
b.	P_TR = P_aw + P_A
c.	P_vent+ P_mus = R_aw + P_TA
d.	P_vent+ P_mus = R_aw x

ANS: B

The transrespiratory pressure (P_TR) 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 (P_E) and the flow resistance pressure (P_R). 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 P_vent+ muscle pressure (P_mus)_.However, this is equal to the elastic recoil pressure (V/C) plus the flow resistance pressure (R_aw x ) or P_vent+ P_mus = V/C + (R_aw x ).

DIF: 1 REF: pg. 30

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

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 (P_TR) = P_vent+ P_mus = V/C + ( R_aw x ).

DIF: 2 REF: pg. 30

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

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/T_I).

DIF: 2 REF: pg. 32

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

During pressure-targeted ventilation the patients airway resistance decreases to normal due to medication delivery. The ventilator will respond with which of the following changes?
Altered flow waveform
Increased pressure
Increased volume
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

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

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

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

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

The most commonly used patient-trigger variables include which of the following?
Flow
Time
Pressure
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

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 patients 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

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

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

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

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

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 patients 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