Ch. 6 Test, Initial Vent Settings

Question 1

Calculate the tubing compliance (CT) when the measured volume is 100 mL and the static pressure is 65 cm H2O.

a. 0.0015 mL/cm H2O
b. 0.65 cm H2O/mL
c. 1.5 mL/cm H2O
d. 6.5 mL/cm H2O

Answer 1

ANS: C
Calculate tubing compliance (CT) by dividing measured volume by measured static pressure.

DIF: 2 REF: pgs . 87, 88

Question 2

Calculate the tubing compliance (CT) when the measured volume is 150 mL and the static pressure is 53 cm H2O.

a. 0.003 mL/cm H2O
b. 0.35 cm H2O/mL
c. 2.8 mL/cm H2O
d. 7.95cm H2O/mL

Answer 2

ANS: C
Calculate tubing compliance (CT) by dividing measured volume by measured static pressure.

DIF: 2 REF: pgs. 88, 89

Question 3

When initially setting up a ventilator the plateau pressure (PPlateau) is measured at 47 cm H2O with a set volume of 100 mL. After applying the ventilator to the patient, the average peak pressure reached during volume delivery is 28 cm H2O. How much volume is lost in the ventilator tubing?

a. 13 mL
b. 60 mL
c. 147 mL
d. 168 mL

Answer 3

ANS: B
Calculate tubing compliance (CT) by dividing measured volume by measured static pressure. The amount of volume lost to the circuit equals the pressure reached during a tidal volume (VT) delivery multiplied by the CT factor.

DIF: 2 REF: pg. 89

Question 4

When initially setting up a ventilator, the plateau pressure (PPlateau) is measured at 68 cm H2O with a set volume of 200 mL. After applying the ventilator to the patient, the average peak pressure reached during volume delivery is 22 cm H2O. How much volume is lost in the ventilator tubing?

a. 0.064 mL
b. 7.5 mL
c. 64.7 mL
d. 95.7 mL

Answer 4

ANS: C
Calculate tubing compliance (CT) by dividing measured volume by measured static pressure. The amount of volume lost to the circuit equals the pressure reached during a tidal volume (VT) delivery multiplied by the CT factor.

DIF: 2 REF: pgs. 88-90

Question 5

Calculate the volume lost if the tubing compression factor is 2.5 mL/cm H2O and the pressure change during ventilation is 32 cm H2O.

a. 2.5 mL
b. 12.8 mL
c. 64 mL
d. 80 mL

Answer 5

ANS: D
The amount of volume lost to the circuit equals the pressure reached during a tidal volume (VT) delivery multiplied by the tubing compliance (CT) factor.

DIF: 2 REF: pgs. 89, 90

Question 6

Calculate the average tidal volume for a patient who has a minute ventilation of 10 L/min with a respiratory rate (RR) of 12 bpm.

a. 120 mL
b. 833 mL
c. 1000 mL
d. 1200 mL

Answer 6

ANS: B
Minute ventilation equals respiratory rate multiplied by tidal volume (VT). Therefore, tidal volume equals minute ventilation divided by respiratory rate.

DIF: 2 REF: pg. 87

Question 7

Calculate the inspiratory time (TI) when a ventilator is set at a tidal volume (VT) of 800 mL and a constant flow rate of 40 L/min.

a. 0.02 second
b. 0.5 second
c. 1.2 seconds
d. 3.2 seconds

Answer 7

ANS:	C
Inspiratory Time (TI) = Tidal Volume (VT)/Minute Ventilation (VE) (convert L/min to L/sec first)

DIF: 2 REF: pgs. 89. 90

Question 8

Calculate the inspiratory time (TI) when a ventilator is set at a tidal volume (VT) of 500 mL and a constant flow rate of 30 L/min.

a. 0.6 second
b. 1 second
c. 1.5 seconds
d. 1.7 seconds

Answer 8

ANS:	B
Inspiratory Time (TI) = Tidal Volume (VT)/Minute Ventilation (VE) (convert L/min to L/sec first)

DIF: 2 REF: pgs. 89, 90

Question 9

Calculate the inspiratory to expiratory (I:E) ratio for a ventilator that is set to deliver 850 mL at a frequency of 15 bpm with a flow rate of 45 L/min.

a. 1:1.1
b. 1:2.5
c. 1:3.5
d. 1:4

Answer 9

ANS: B
Minute ventilation equals respiratory rate multiplied by tidal volume.

DIF: 2 REF: pg. 91

Question 10

Calculate the inspiratory to expiratory (I:E) ratio when the inspiratory time is 0.5 seconds and the respiratory rate is 30 bpm.

a. 1:3
b. 1:4
c. 4:1
d. 3:1

Answer 10

ANS: A
Total Cycle Time (TCT) = 60 sec/f; TCT – Inspiratory Time (TI) = Expiratory Time (TE); TI:TE = 1:X

DIF: 2 REF: pg. 91

Question 11

Calculate the expiratory time (TE) when the ventilator frequency is set to 25 bpm and the inspiratory time (TI) is 0.75 second.

a. 0.75 second
b. 1.16 seconds
c. 1.65 seconds
d. 2.4 seconds

Answer 11

ANS: C
Total Cycle Time (TCT) = 60 sec/f; TCT – Inspiratory Time (TI) = Expiratory Time (TE)

DIF: 2 REF: pg. 91

Question 12

What is the flow rate necessary to deliver a tidal volume (VT) of 600 mL, with a constant waveform, at a respiratory rate of 15 breaths/min with an I:E of 1:4?

a. 36 L/min
b. 40 L/min
c. 45 L/min
d. 60 L/min

Answer 12

ANS: C
Total Cycle Time (TCT) = 60 sec/f; Inspiratory Time (TI) = TCT/Inspired (I) + Expired (E); Flow rate = Tidal Volume (VT)/Inspiratory Time (TI)

DIF: 2 REF: pgs. 91, 92

Question 13

Setting flow rates high will cause which of the following to occur?

a. Improve gas exchange
b. Lengthen inspiratory time
c. Increase air trapping
d. Increase peak pressures

Answer 13

ANS: D
The flow setting on a mechanical ventilator determines how fast the inspired gas will be delivered to the patient. During continuous mandatory ventilation (CMV), high flows shorten inspiratory time (TI) and may result in higher peak pressures and poor gas distribution.

DIF: 1 REF: pg. 92

Question 14

Slow flow rates will cause which of the following to occur?

a. Poor gas exchange
b. Increase peak pressures
c. Shorten expiratory time
d. Decrease mean airway pressure

Answer 14

ANS: C
Slower flows may reduce peak pressures, improve gas distribution, and increase at the expense of increasing inspiratory time (TI). Unfortunately, shorter expiratory time (TE) can lead to air trapping, while using a longer TI may cause cardiovascular side effects.

DIF: 1 REF: pg. 92

Question 15

The flow wave form pattern that provides the shortest inspiratory time (TI) of all the available flow patterns with an equivalent peak flow rate setting is which of the following?

a. Sine
b. Rectangular
c. Ascending Ramp
d. Descending Ramp

Answer 15

ANS: B
Generally, a constant flow pattern provides the shortest inspiratory time (TI) of all the available flow patterns with an equivalent peak flow rate setting.

DIF: 1 REF: pg. 92

Question 16

The flow wave form pattern that is created during pressure targeted ventilation is which of the following?

a. Sine
b. Rectangular
c. Ascending Ramp
d. Descending Ramp

Answer 16

ANS: D
The descending wave form occurs naturally in pressure ventilation.

DIF: 2 REF: pg. 92

Question 17

The flow wave form pattern that will decrease peak pressure but at the same time may increase mean airway pressure is which of the following?

a. Sine
b. Rectangular
c. Ascending Ramp
d. Descending Ramp

Answer 17

ANS: D
In situations where plateau pressure (PPlateau) is critical, changing to a descending ramp in order to reduce peak pressures may increase the mean airway pressure.

DIF: 1 REF: pg. 93

Question 18

A patient having an acute, severe asthma exacerbation is intubated and set up on volume-controlled continuous mandatory ventilation (VC-CMV). To ensure volume delivery at the lowest peak pressure while providing for better air distribution, which flow wave form should be used?

a. Sine
b. Rectangular
c. Ascending Ramp
d. Descending Ramp

Answer 18

ANS: D
In patients with high airway resistance (Raw), the descending pattern is more likely to deliver a set tidal volume (VT) at a lower pressure and provide for better distribution of air through the lung than a constant or an accelerating flow.

DIF: 2 REF: pg. 92

Question 19

The most appropriate tidal volume setting for a 6’3” male ventilator patient with normal lungs is which of the following?

a. 300 mL
b. 500 mL
c. 700 mL
d. 900 mL

Answer 19

ANS: B
First calculate ideal body weight (IBW) = 106 + 6(ht – 60). Then using the range of 5 to 7 mL/kg IBW the tidal volume range for this patient is 445 mL to 623 mL.

DIF: 2 REF: pgs. 92, 93

Question 20

A 5’10” male patient with normal lungs has been intubated and requires mechanical ventilation with volume-controlled continuous mandatory ventilation (VC-CMV). The tidal volume and ventilator rate settings that should be recommended for this patient are which of the following?

a. VT = 525 mL, rate = 14 bpm
b. VT = 750 mL, rate = 12 bpm
c. VT = 825 mL, rate = 10 bpm
d. VT = 950 mL, rate = 8 bpm

Answer 20

ANS: A
First calculate ideal body weight (IBW) for a male, which is 106 + 6(ht – 60) = 75 kg. Then using the range of 5 to 7 mL/kg IBW, the tidal volume range for this patient is 375 mL to 525 mL. Minute ventilation is about 100 mL/kg IBW. Therefore, minute ventilation should be approximately 7.5 L/min. Dividing the calculated minute ventilation by the tidal volume range for this patient provides a range of rates for the tidal volumes: 14 to 20 bpm depending on the set volumes. The most appropriate volume and rate combination for this patient is 525 mL × 14 bpm = 7.35 L/min.

DIF: 2 REF: pgs. 90, 91

Question 21

A 5’2” female patient with normal lungs has been intubated and requires mechanical ventilation with volume-controlled continuous mandatory ventilation (VC-CMV). The tidal volume (VT) and ventilator rate settings that should be recommended for this patient are which of the following?

a. VT = 315 mL, rate = 20 bpm
b. VT = 364 mL, rate = 14 bpm
c. VT = 468 mL, rate = 12 bpm
d. VT = 563 mL, rate = 10 bpm

Answer 21

ANS: B
First calculate ideal body weight (IBW) for a female, which is 105 + 5(ht – 60) = 52 kg. Then using the range of 5 to 7 mL/kg IBW the tidal volume range for this patient is 260 mL to 364 mL. Minute ventilation is about 100 mL/kg IBW. Therefore, minute ventilation should be approximately 5.2 L/min. Then dividing the calculated minute ventilation by the tidal volume range for this patient provides a range of rates for the tidal volumes: 14 to 20 bpm depending on the set volumes. The most appropriate volume and rate combination for this patient is 364 mL × 14 bpm = 5.1 L/min.

DIF: 2 REF: pg. 91

Question 22

A 47-year-old, 5’6”, 112 lb female patient, is still under the effects of anesthesia following a hysterectomy. Her body temperature is 37° C. She has no history of lung disease. The appropriate initial minute ventilation for this patient is which of the following?

a. 5.1 L/min
b. 6.1 L/min
c. 11.2 L/min
d. 13.5 L/min

Answer 22

ANS: B
First calculate ideal body weight (IBW) for a female, using the formula 105 + 5(ht – 60). This patient’s IBW is 51 kg. Minute ventilation is about 100 mL/kg IBW, which would be 51,000 mL/min or 5.1 L/min.

DIF: 2 REF: pg. 90

Question 23

A 26-year-old, 6’6”, 250 lb male patient, is still under the effects of anesthesia following knee surgery. His body temperature is 37° C. He has no history of lung disease. The appropriate initial minute ventilation for this patient is which of the following?

a. 8.9 L/min
b. 9.7 L/min
c. 11.4 L/min
d. 13.6 L/min

Answer 23

ANS: B
First calculate ideal body weight (IBW) for a male, using the formula 106 + 6(ht – 60). This patient’s IBW is 97 kg. Minute ventilation is about 100 mL/kg IBW, which would be 97,000 mL/min or 9.7 L/min.

DIF: 2 REF: pgs. 90, 91

Question 24

A 39-year-old, 5’4”, 138 lb female patient requires intubation and mechanical ventilation. Her body temperature is 39° C. She has no history of lung disease. The appropriate initial minute ventilation for this patient is which of the following?

a. 5.7 L/min
b. 6.8 L/min
c. 7.6 L/min
d. 13.8 L/min

Answer 24

ANS: B
First calculate ideal body weight (IBW) for a female, using the formula 105 + 5(ht – 60). This patient’s IBW is 57 kg. Minute ventilation is about 100 mL/kg IBW, which would be 57,000 mL/min or 5.7 L/min. VE would have to be increased by 10% for each degree above 37° C: a total increase of 20% of 5.7 = 1.14; therefore, the new minute ventilation (VE) would be 5.7 + 1.14 = 6.8 L/min.

DIF: 2 REF: pg. 88

Question 25

A patient has a body temperature of 40° C. How should the initial minute ventilation setting be adjusted?

a. Increase it by 15%
b. Decrease it by 18%
c. Decrease it by 25%
d. Increase it by 30%

Answer 25

ANS:	D
Minute ventilation (VE) would have to be increased by 10% for each degree above 37° C.

DIF: 1 REF: pg. 88

Question 26

The pattern that has been shown to improve the distribution of gas in the lungs for an intubated patient on volume-controlled continuous mandatory ventilation (VC-CMV) is which of the following?

a. Sine waveform
b. Ascending ramp
c. Descending ramp
d. Square waveform

Answer 26

ANS: C
Studies comparing the descending flow pattern with the constant flow pattern suggest that the descending flow pattern improves the distribution of gas in the lungs.

DIF: 1 REF: pg. 93

Question 27

A 47-year-old, 6’1” male patient is admitted to the hospital due to trauma from a motor vehicle accident. Forty-eight hours post admission, the patient is suffering from respiratory distress with severe hypoxemia and is intubated. A chest x-ray, done prior to intubation , reveals a ground glass appearance bilaterally. The physician requests the volume-controlled continuous mandatory ventilation (VC-CMV) mode for this patient. The initial settings for the ventilator should be which of the following?

a. VT = 450 mL, rate = 18 bpm, PEEP = 8 cm H2O
b. VT = 600 mL, rate = 10 bpm, PEEP = 5 cm H2O
c. VT = 750 mL, rate = 15 bpm, PEEP = 10 cm H2O
d. VT = 900 mL, rate = 12 bpm, PEEP = 5 cm H2O

Answer 27

ANS: A
First calculate ideal body weight (IBW) for a male, using the formula 106 + 6(ht – 60). This patient’s IBW is 84 kg. Minute ventilation is about 100 mL/kg IBW, which would be 8.4 L/min. Since the patient appears to have Acute Respiratory Distress Syndrome (ARDS) the tidal volume should be set to between 4 and 6 mL/kg. This would make the appropriate tidal volume range 336 mL to 504 mL. This eliminates all of the choices except “A.” Dividing the tidal volume range into 8.4 L/min gives the set rate range at 17 to 25 bpm. This also eliminates all but choice “A.”

DIF: 3 REF: pgs. 90, 91

Question 28

A 65-year-old, 73-inch-tall, 195 lb male patient was admitted 2 days ago for renal failure. The patient has a history of Chronic Obstructive Pulmonary Disease (COPD) and has a pulse of 122 bpm, BP 153/88, and temperature 37° C. The patient is intubated for acute-on-chronic respiratory failure with hypoxemia. The physician requests volume-controlled continuous mandatory ventilation (VC-CMV). The initial settings for the ventilator should be which of the following?

a. VT = 700 mL, rate = 12 bpm, PEEP = 3 cm H2O
b. VT = 900 mL, rate = 10 bpm, PEEP = 5 cm H2O
c. VT = 450 mL, rate = 20 bpm, PEEP = 8 cm H2O
d. VT = 800 mL, rate = 15 bpm, PEEP = 10 cm H2O

Answer 28

ANS: A
First calculate ideal body weight (IBW) for a male, using the formula 106 + 6(ht – 60). This patient’s IBW is 84 kg. Minute ventilation is about 100 mL/kg IBW, which would be 8.4 L/min. Since the patient has a history of Chronic Obstructive Pulmonary Disease (COPD), the tidal volume should be set to between 8 and 10 mL/kg with rates of between 8 and 12 bpm.. This would make the appropriate tidal volume range 672 mL to 840 mL. This eliminates the two choices where the volume is outside of this range, leaving choices “A” and “D.” Choice “D” can be eliminated because the rate of 15 is outside the suggested rates for COPD patients. Rapid rates will lead to air trapping in patients with high airway resistance.

DIF: 3 REF: pgs. 90, 91

Question 29

A 57-year-old, 5’3”, 165 lb female patient arrives in the open heart unit following coronary artery bypass surgery. The patient has a history of diabetes and no history of pulmonary disease. The most appropriate initial volume-controlled continuous mandatory ventilation (VC-CMV) settings are which of the following?

a. VT = 220 mL, rate = 25 bpm, PEEP = 10 cm H2O
b. VT = 360 mL, rate = 15 bpm, PEEP = 5 cm H2O
c. VT = 550 mL, rate = 12 bpm, PEEP = 12 cm H2O
d. VT = 750 mL, rate = 10 bpm, PEEP = 8 cm H2O

Answer 29

ANS: B
First calculate ideal body weight (IBW) for a female, using the formula 105 + 5(ht – 60). This patient’s IBW is 55 kg. Minute ventilation is about 100 mL/kg IBW, which would be 5.54 L/min. Since the patient is post-op, the tidal volume should be set to between 5 and 8 mL/kg with rates of between 10 and 20 bpm. This would make the appropriate tidal volume range 275 mL to 440 mL. This fact eliminates choices “A,” “C,” and “D.” The positive-end-expiratory pressure (PEEP) may also be too high for a post-op open heart patient in choices “A,” “C,” and “D.”

DIF: 3 REF: pg. 88

Question 30

With which flow waveform pattern will the mean airway pressure be the highest?

a. Sine
b. Square
c. Ascending ramp
d. Descending ramp

Answer 30

ANS: D
The descending ramp flow pattern keeps mean airway pressure high and may improve gas distribution.

DIF: 1 REF: pg. 93

Question 31

A mechanically ventilated patient is going to be placed on pressure support ventilation following an acceptable spontaneous weaning trial. The patient is a 5’9” male who weighs 185 lbs. During volume-controlled continuous mandatory ventilation (VC-CMV) his average peak inspiratory pressure (PIP) was about 26 cm H2O and the plateau pressure (PPlateau) was 16 cm H2O. What initial pressure support level should be set?

a. 5 cm H2O
b. 10 cm H2O
c. 15 cm H2O
d. 20 cm H2O

Answer 31

ANS: B
The easiest way to establish the initial setting for pressure support ventilation (PSV) is to set it equal to the transairway pressure (Peak Inspiratory Pressure (PIP) – Plateau Pressure (Pplateau)).

DIF: 2 REF: pg. 93

Question 32

When changing the control variable from volume-control (VC) to pressure control (PC), the initial inspiratory pressure should be set based on which of the following methods?

a. Body surface area multiplied by 4
b. Plateau pressure measurement taken during VC ventilation
c. Maximum peak inspiratory pressure during VC ventilation
d. Maximum peak inspiratory pressure minus plateau pressure during VC ventilation

Answer 32

ANS: B
Initial pressure is set at the plateau pressure (Pplateau) value during volume-controlled continuous mandatory ventilation (VC-CMV) and must be adjusted as necessary to achieve tidal volume (VT).

DIF: 1 REF: pg. 93

Question 33

A 63-year-old, 5’11”, 185 lb male patient with a history of Chronic Obstructive Pulmonary Disease (COPD) is admitted to the hospital due to liver failure. Over the course of the 48 hours he has developed respiratory distress. The respiratory therapist performs a physical assessment and finds the following: heart rate 135 bpm, respiratory rate 28 with accessory muscle use. Breath sounds are decreased bilaterally with coarse crackles in the right base. A chest x-ray from 24 hours ago shows bilateral lower lobe infiltrates. The patient has a nonproductive cough. The respiratory therapist draws an arterial blood gas which reveals: pH 7.31; partial pressure of carbon dioxide (PaCO2) 57 mm Hg; partial pressure of oxygen (PaO2) 58 mm Hg; arterial oxygen saturation (SaO2) 87%; bicarbonate (HCO3-) 27 mEq/L while receiving oxygen via nasal cannula 3 L/min. The respiratory therapist should recommend which of the following for this patient?

a. Continue with current therapy and monitor the patient closely.
b. Place the patient on a nonrebreather mask with 15 L/min oxygen.
c. Intubate and place on pressure-controlled continuous mandatory ventilation (PC-CMV), peal inspiratory pressure (PIP) 40 cm H2O, positive-end-expiratory pressure (PEEP) 8 cm water (H2O), fractional inspired oxygen (FIO2) 1.0.
d. Use BiPAP with IPAP 10 cm H2O, EPAP 5cm H2O, bleed in 4 L/min oxygen.

Answer 33

ANS: D
This patient is showing signs of ventilatory failure as evidenced by his acute-on-chronic respiratory acidosis with uncorrected hypoxemia. This patient should be tried on noninvasive positive pressure ventilation (NPPV) prior to intubation to try to avoid it if possible. Using a nonrebreathing mask would not address the patient’s ventilatory problem and may cause oxygen induced hypoventilation. Continuing with current therapy would not address the problem of impending ventilatory failure. If intubated and mechanically ventilated with pressure-controlled continuous mandatory ventilation (PC-CMV), starting off at 40 cm H2O is too high. When the peak inspiratory pressure (PIP) or plateau pressure (Pplateau) from volume ventilation are not available, an initial pressure of 10 – 15 cm H2O should be set followed by volume measurements and pressure adjustments when appropriate.

DIF: 3 REF: pg. 93

Question 34

The mode of ventilation that provides pressure-limited, time-cycled breaths that use a set tidal volume as a feedback control is which of the following?

a. Pressure Support Ventilation (PSV)
b. Pressure Regulated Volume Control (PRVC)
c. Pressure-Controlled Continuous Mandatory Ventilation (PC-CMV)
d. Bilevel Positive Airway Pressure (Bilevel PAP)

Answer 34

ANS: B
Pressure support ventilation (PSV) is pressure-limited and flow cycled. Pressure regulated volume control (PRVC) is a dual control mode of ventilation that provides the benefits of pressure breathing along with targeting a set volume. These breaths are pressure-limited, time-cycled and use tidal volume as a feedback. The pressure is adjusted by the ventilator to meet the set volume. Pressure-controlled continuous mandatory ventilation (PC-CMV) is pressure-limited and time-cycled, but does not have a volume feedback system. BiLevel PAP is similar to PC-CMV in that it is pressure-limited and time-cycled with no volume feedback.

DIF: 1 REF: pgs. 99, 100

Question 35

The mode of pressure ventilation that is patient- or time-triggered and flow-cycled is which of the following?

a. Volume Support (VS)
b. Pressure Support Ventilation (PSV)
c. Pressure-Controlled Continuous Mandatory Ventilation (PC-CMV)
d. Bileval Positive Airway Pressure (Bilevel PAP)

Answer 35

ANS:	D
Volume support (VS) and pressure support ventilation (PSV) are both patient-triggered and flow-cycled. Pressure-controlled continuous mandatory ventilation (PC-CMV) is patient- or time-triggered and time-cycled. Bilevel PAP is patient- or time-triggered and flow-cycled.

DIF: 1 REF: pgs. 98, 99

Question 36

A patient receiving mechanical ventilation via pressure regulated volume control (PRVC) has a set target volume of 500 mL with an upper pressure limit setting of 35 cm H2O. During the respiratory therapist’s first patient ventilator system check, 25 cm H2O was needed to deliver the set volume. Several hours later the pressure to deliver the set volume is 15 cm H2O. The respiratory therapist should do which of the following?

a. Switch the patient to pressure support ventilation (PSV).
b. No action should be taken.
c. The set volume should be reduced.
d. The upper pressure limit should be reduced.

Answer 36

ANS: D
The upper pressure limit should be reduced so that the lungs are protected from pressures that would be too high for the recovering lungs. There is nothing in this scenario that gives any clues to the fact that the patient should be switched to PSV. Not taking any action in this situation could cause harm to the patient if the pressure were to rise suddenly, as with a cough or forcible exhalation.

DIF: 3 REF: pgs. 99, 100