Test Bank for Pilbeams Mechanical Ventilation 7th Edition by Cairo

Test Bank for Pilbeams Mechanical Ventilation 7th Edition by Cairo Chapter 01: Basic Terms and Concepts of Mechanical Ventilation Cairo: Pilbeam’s Mechanical Ventilation: Physiological and Clinical Applications, 7th Edition MULTIPLE CHOICE 1. The body’s mechanism for conducting air in and out of the lungs is known as which of the following? a. External respiration b. Internal respiration c. Spontaneous ventilation d. Mechanical ventilation ANS: C The conduction of air in and out of the body is known as ventilation. Since the question asks for the body’s mechanism, this would be spontaneous ventilation. External respiration involves the exchange of oxygen (O 2 ) and carbon dioxide (CO ) between the alveoli and the pulmonary capillaries. Internal respiration occurs at the cellular level and involves movement of oxygen from the systemic blood into the cells. 2. Which of the following are involved in external respiration? a. Red blood cells and body cells b. Scalenes and trapezius muscles c. Alveoli and pulmonary capillaries d. External oblique and transverse abdominal muscles ANS: C External respiration involves the exchange of oxygen and carbon dioxide (CO ) between the alveoli and the pulmonary capillaries. Internal respiration occurs at the cellular level and involves movement of oxygen from the systemic blood into the cells. Scalene 2 and trapezius muscles are accessory muscles of inspiration. External oblique and transverse abdominal muscles are accessory muscles of expiration. 3. The graph that shows intrapleural pressure changes during normal spontaneous breathing is depicted by which of the following? a. b. c. d. ANS: B During spontaneous breathing, the intrapleural pressure drops from about −5 cm H 2 . 1 2 O at end-expiration to about −10 cm H O at end-inspiration. The graph depicted for answer B shows that change from −5 cm H 4. During spontaneous inspiration alveolar pressure (P a. −1 cm H b. +1 cm H c. 0 cm H d. 5 cm H ANS: A 2 2 2 O 2 O O O A ) is about ________________. During normal spontaneous ventilation alveolar pressure will become −1 cm H 2 2 O to −10 cm H 2 O. O which is the lowest. During the exhalation of a normal spontaneous breath the alveolar pressure will become +1 cm H 2 O. 2 5. The pressure required to maintain alveolar inflation is known as which of the following? a. Transairway pressure (P b. Transthoracic pressure (P c. Transrespiratory pressure (P d. Transpulmonary pressure (P ANS: D TA ) TT ) TR L ) ) The definition of transpulmonary pressure (P L ) is the pressure required to maintain alveolar inflation. Transairway pressure (P ) is the pressure gradient required to produce airflow in the conducting tubes. Transrespiratory pressure (P ) is the pressure to inflate the lungs and airways during positive-pressure ventilation. Transthoracic pressure (P TT TR ) represents the pressure required to expand or contract the lungs and the chest wall at the same time. 6. Calculate the pressure needed to overcome airway resistance during positive-pressure ventilation when the proximal airway pressure (P a. 7 cm H b. 30 cm H c. 40 cm H awo d. 175 cm H ANS: B 2 O 2 2 ) is 35 cm H O O 2 O The transairway pressure (P 2 O and the alveolar pressure (P TA A ) is 5 cm H 2 O. ) is used to calculate the pressure required to overcome airway resistance during mechanical ventilation. This formula is P TA = P awo − P A. 7. The term used to describe the tendency of a structure to return to its original form after being stretched or acted on by an outside force is which of the following? a. Elastance b. Compliance c. Viscous resistance d. Distending pressure 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. 8. Calculate the pressure required to achieve a tidal volume of 400 mL for an intubated patient with a respiratory system compliance of 15 mL/cm H a. 6 cm H 2 b. 26.7 cm H c. 37.5 cm H d. 41.5 cm H ANS: B O 2 2 2 2 O. O O O ΔC = ΔV/ΔP then ΔP = ΔV/ΔC 9. Which of the following conditions causes pulmonary compliance to increase? 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. 10. Calculate the effective static compliance (C ) given the following information about a patient receiving mechanical ventilation: peak inspiratory pressure (PIP) is 56 cm H 2 s O, plateau pressure (P plat ) is 40 cm H 2 O, exhaled tidal volume (V ) is 650 mL, and positive end-expiratory pressure (PEEP) is 10 cm H a. 14.1 mL/cm H b. 16.3 mL/cm H c. 21.7 mL/cm H d. 40.6 mL/cm H ANS: C 2 2 2 2 O O O O The formula for calculating effective static compliance is C 2 O. s = V T /(P plat − EEP). 11. Based upon the following patient information, calculate the patient’s static lung compliance: exhaled tidal volume (V peak inspiratory pressure (PIP) is 28 cm H a. 0.02 L/cm H b. 0.03 L/cm H c. 0.22 L/cm H d. 0.34 L/cm H ANS: C 2 2 2 2 O O O O 2 O, plateau pressure (P The formula for calculating effective static compliance is C s = V T plat ) is 8 cm H /(P plat − EEP). 2 O, and PEEP is set at 5 cm H . 2 T 2 O. T TA ) is 675 mL, 12. A patient receiving mechanical ventilation has an exhaled tidal volume (V ) of 500 mL and a positive end-expiratory pressure setting (PEEP) of 5 cm H O. Patient-ventilator system checks reveal the following data: Time PIP (cm H 2 2 O) P plat (cm H O) 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. β 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 T ) by subtracting the plateau pressure from the peak inspiratory pressure. Analyzing the P will show any changes in the pressure needed to overcome airway resistance. Analyzing the P will demonstrate any changes in compliance. The P plateau plateau remained 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 to 14 cm H 2 O) and then a sharp increase to 23 cm H O at 1000. Increases in P signify increases in airway resistance. Airway resistance may be caused by secretion buildup, bronchospasm, mucosal edema, and mucosal inflammation. Tracheobronchial suctioning will remove any secretion buildup, and a β adrenergic bronchodilator will reverse bronchospasm. Increasing the tidal volume will add to the airway resistance according to TA Poiseuille’s law. Increasing the PEEP will not address the root of this patient’s problem; the patient’s compliance is normal. 13. The values below pertain to a patient who is being mechanically ventilated with a measured exhaled tidal volume (V ) of 700 mL. Time Peak Inspiratory Pressure (cm H 2 O) Plateau Pressure (cm H O) Analysis of this data points to which of the following conclusions? a. Airway resistance is 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 = 5 cm H 2 O, 1100 P TA = 6 cm H 2 O, and 1130 P TA = 6 cm H 2 TA . 3 = 5 cm H O. This data shows that there is no significant increase or decrease in this patient’s airway resistance. Analysis of the patient’s plateau pressure (P plat ) reveals an increase of 15 cm H O over the three and a half hour time period. This is directly related to a decrease in lung compliance. Calculation of the lung compliance (C S =V T /(P plat − EEP))at each time interval reveals a steady decrease from 20 mL/cm H 2 O to 14 mL/cm H 14. The respiratory therapist should expect which of the following findings while ventilating a patient with acute respiratory distress syndrome (ARDS)? a. An elevated plateau pressure (P b. A decreased elastic resistance c. A low peak inspiratory pressure (PIP) d. A large transairway pressure (P ANS: A plat TA ) ) gradient ARDS is a pathological condition that is associated with a reduction in lung compliance. The formula for static compliance (C ) utilizes the measured plateau pressure (P plat ) in its denominator (C S = V T /[P – EEP]). Therefore, with a consistent exhaled tidal volume (V T ), an elevated P plat will decrease C S . 15. The formula used for the calculation of static compliance (C a. (Peak inspiratory pressure [PIP] − EEP)/tidal volume (V b. (Plateau pressure [P plat ] − EEP)/tidal volume (V c. Tidal volume/(plateau pressure − EEP) d. Tidal volume/(peak pressure [PIP] − plateau pressure [P ANS: C C S =V T /(P plat − EEP) T ) S plat ) is which of the following? T ) plat ]) 2 2 O. 2 T O, 1000 P 2 TA TA S 16. Plateau pressure (P a. Inspiration b. End-inspiration c. Expiration d. End-expiration ANS: B plat ) is measured during which phase of the ventilatory cycle? The calculation of compliance requires the measurement of the plateau pressure. This pressure measurement is made during no-flow conditions. The airway pressure (P ) 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 awo 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). 17. The condition that is associated with an increase in airway resistance is which of the following? a. Pulmonary edema b. Bronchospasm c. Fibrosis d. Ascites ANS: B Airway resistance is determined by the gas viscosity, gas density, tubing length, airway diameter, and the flow rate of the gas through the tubing. The two factors that are most often subject to change are the airway diameter and the flow rate of the gas. The flow rate of the gas during mechanical ventilation is controlled. Pulmonary edema is fluid accumulating in the alveoli and will cause a drop in the patient’s lung compliance. Bronchospasm causes a narrowing of the airways and will, therefore, increase the airway resistance. Fibrosis causes an inability of the lungs to stretch, decreasing the patient’s lung compliance. Ascites causes fluid buildup in the peritoneal cavity and increases tissue resistance, not airway resistance. 18. An increase in peak inspiratory pressure (PIP) without an increase in plateau pressure (P ) is associated with which of the following? a. Increase in static compliance (C b. Decrease in static compliance (C c. Increase in airway resistance d. Decrease in airway resistance ANS: C S ) S ) The PIP represents the amount of pressure needed to overcome both elastance and airway resistance. The P plat is the amount of pressure required to overcome elastance alone. Since the P has remained constant in this situation, the static compliance is unchanged. The difference between the PIP and the P plat plat is the transairway pressure (P ) and represents the pressure required to overcome the airway resistance. If P TA TA . 4 plat increases, the airway resistance is also increasing, when the gas flow rate remains the same. 19. The patient-ventilator data over the past few hours demonstrates an increased peak inspiratory pressure (PIP) with a constant transairway pressure (P a. Static compliance (C b. Static compliance (C c. Airway resistance (R d. Airway resistance (R ANS: B TA ). The respiratory therapist should conclude which of the following? S S ) has increased. ) has decreased. aw aw ) has increased. ) has decreased. The PIP represents the amount of pressure needed to overcome both elastance and airway resistance. The P is the amount of pressure required to overcome elastance alone, and is the pressure used to calculate the static compliance. Since P has stayed the same, it can be concluded that R has remained the same. Therefore, the reason the PIP has increased is because of an increase in the P plat aw . This correlates to a decrease in C 20. Calculate airway resistance (R aw S. ) for a ventilator patient, in cm H 2 plat TA O/(L/s), when the peak inspiratory pressure (PIP) is 50 cm H O, the plateau pressure (P a. 0.58 R b. 1.2 R c. 35 R d. 50 R ANS: C R aw =P TA aw aw aw aw /flow; or R aw plat ) is 15 cm H =(PIP − P 21. Calculate airway resistance (R aw plat 2 O, and the set flow rate is 60 L/min. )/flow ) for a ventilator patient, in cm H O/(L/s), with the following information: Peak inspiratory pressure (PIP) is 20 cm H a. 5 R b. 6 R c. 10 R d. 15 R ANS: B R aw aw aw aw aw =(PIP − P plat 2 O, plateau pressure (P )/flow and flow is in L/s. plat ) is 15 cm H 2 2 O, PEEP is 5 cm H 2 O, and set flow rate is 50 L/min. 2 22. Calculate the static compliance (C S ), in mL/cm H 2 O, when PIP is 47 cm H 2 O, plateau pressure (P plat ) is 27 cm H O, baseline pressure is 10 cm H a. 43 C b. 36 C c. 20 C S S S d. 0.065 C ANS: A C S = V T /(P S plat 2 O, and exhaled tidal volume (V − EEP)C S =V T /(P plat − EEP) T ) is 725 mL. 23. Calculate the inspiratory time necessary to ensure 98% of the volume is delivered to a patient with a C s . 5 2 = 40 mL/cm H O and the R aw = 1 cm H a. 0.04 s b. 0.16 s c. 1.6 s d. 4.0 s ANS: B 2 O/(L/s). Time constant = C (L/cmH 2 O) × R(cmH O/ L/s[]) . 98% of the volume will be delivered in 4 time constants. Therefore, multiply 4 times the time constant. 2 24. How many time constants are necessary for 95% of the tidal volume (V a. 1 b. 2 c. 3 d. 4 ANS: C T ) to be delivered from a mechanical ventilator? 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. 25. Compute the inspiratory time necessary to ensure 100% of the volume is delivered to an intubated patient with a C = 60 mL/cm H 2 O and the R a. 0.36 s b. 0.5 s c. 1.4 s d. 1.8 s ANS: D aw = 6 cm H Time constant (TC) = C (L/cm H 2 O/(L/s). 2 O) × R (cm H O/[L/s]). 100% of the volume will be delivered in 5 time constants. Therefore, multiply 5 times the time constant. 2 26. Evaluate the combinations of compliance and resistance and select the combination that will cause the lungs to fill fastest. a. C b. C c. C d. C s s s s ANS: C = 0.1 L/cmH = 0.1 L/cmH = 0.03 L/cmH = 0.103L/cmH 2 2 OR OR 2 aw aw OR 2 = 1cmH = 10 cmH aw OR aw = 1cmH 2 = 10 cmH O/(L/s) 2 2 O/(L/s) O/(L/s) 2 O/(L/s) Use the time constant formula, TC = C × R, to determine the time constant for each choice. The time constant for answer A is 0.1 s. The time constant for answer B is 1 s. The time constant for answer C is 0.03 s, and the time constant for answer D is 0.3 s. The product of multiplying the time constant by 5 is the inspiratory time needed to deliver 100% of the volume. 27. 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 only 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. s 2 28. Calculate the static compliance (C S ), in mL/cm H 2 O, when PIP is 26 cm H 2 O, plateau pressure (P plat ) is 19 cm H O, baseline pressure is 5 cm H a. 16 b. 20 c. 22 d. 30 ANS: D C S =V T /(P plat 2 O, and exhaled tidal volume (V − EEP) 29. What type of ventilator increases transpulmonary pressure (P 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 T ) is 425 mL. L ) by mimicking the normal mechanism for inspiration? 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. 30. Air accidentally trapped in the lungs due to mechanical ventilation is known as which of the following? a. Plateau pressure (P plat ) 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. 31. The transairway pressure (P a. 5 cm H b. 10 cm H c. 20 cm H d. 30 cm H ANS: B P TA 2 =PIP − P O 2 2 2 O O O plat TA ) shown in this figure is which of the following? , where the PIP is 30 cm H 32. Use this figure to compute the static compliance (C a. 14 mL/cm H b. 20 mL/cm H c. 33 mL/cm H d. 50 mL/cm H ANS: D C s =P plat 2 2 2 2 O O O O − EEP; the P plat 2 O and the P in the figure is 20 cm H 2 S plat is 20 cm H 2 O. The PEEP is 5 cm H 2 O. ) for an intubated patient with an exhaled tidal volume (V O and the PEEP is 10 cm H 2 O. T 2 ) of 500 mL. 33. Evaluate the combinations of compliance and resistance and select the combination that will cause the lungs to empty slowest. a. C b. C c. C d. C S S S S ANS: B = 0.05 L/cmH = 0.05 L/cmH = 0.03 L/cmH = 0.03 L/cmH 2 2 2 2 O R O R O R O R aw aw aw aw = 2cmH = 6cmH = 5cmH = 8cmH 2 2 2 2 O/(L/s) O/(L/s) O/(L/s) O/(L/s) Use the time constant formula, TC = C × 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 s, B is 0.3 s, C is 0.15 s, and D is 0.24 s. To find out how many seconds for emptying, multiply the time constant by 5. 34. Use this figure to compute the static compliance for an intubated patient with an inspiratory flow rate set at 70 L/min. a. 0.2 cm H b. 11.7 cm H c. 16.7 cm H d. 20 cm H ANS: B 2 2 O/(L/s) 2 2 O/(L/s) O/(L/s) O/(L/s) Use the graph to determine the PIP (34 cm H 2 O) and the P plat (20 cm H O). Convert the flow into L/s (70 L/min/60 = 1.2 L/s). Then, R aw = (PIP – P plat )/flow. 35. The ventilator that functions most physiologically uses which of the following? a. Open loop b. Double circuit c. Positive pressure d. Negative pressure ANS: D 2 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. 36. 1 mm Hg = a. 7.5 mm H b. 1.30 atm c. 1.36 cm H d. 1034 cm H ANS: C 2 O 2 O 2 O 1 mm Hg = 1.36 cm H 2 O 1 kPa = 7.5 mm Hg 1 Torr = 1 mm Hg 1 atm = 760 mm Hg = 1034 cm H 2 O 37. Which of the following statements best defines elastance? a. Ability of a structure to stretch b. Ability of a structure to return to its natural shape after stretching c. Ability of a structure to stretch and remain in that position d. Ability of a structure to fill and empty during static conditions ANS: B The opposite of compliance, elastance is the tendency of a structure to return to its original form after being stretched or acted on by an outside force. 38. The term used to describe the opposition to movement offered by adjacent structures such as the lungs and their adjacent organs is which of the following? a. Elastance b. Compliance c. Viscous resistance d. Distending pressure ANS: C 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. . 7 39. Using a sensor, where do current ICU mechanical ventilators measure airway pressure (P a. At the patient’s airway opening b. By placing a specially designed balloon in the esophagus c. Proximal to the expiratory valve d. Airway pressure cannot be measured ANS: C AW )? During mechanical ventilation, proximal airway pressure is not typically measured at the airway opening due to accumulation of secretions and technical errors can alter sensor measurements. Current generation ICU mechanical ventilators measure airway pressure (P ) using a sensor positioned proximal to the expiratory valve, which is closed during the inspiration. The ventilator manometer pressure displayed on the user interface of the ventilator is typically designated as airway pressure (P aw ). Intrapleural pressure (P pl ) is the pressure in the potential space between the parietal and visceral pleurae. P pl . 8 aw is normally about –5 cm H O at the end of expiration during spontaneous breathing. It is about –10 cm H 2 O at the end of inspiration. Because P is often difficult to measure in a patient, a related measurement is used, the esophageal pressure (P ), which is obtained by placing a specially designed balloon in the esophagus; changes in the balloon pressure are used to estimate pressure and pressure changes in the pleural space.