The average compliance of the lungs and chest wall combined is 0.1 L/cm H 2 O. 0.2 L/cm H 2 O 0.3 L/cm H 2 O 0.4 L/cm H 2 O

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Answer 1

Textbook Question

Chapter 2, Problem 1RQ

The average compliance of the lungs and chest wall combined is

0.1 L/cm H₂O.

0.2 L/cm H₂O

0.3 L/cm H₂O

0.4 L/cm H₂O

Expert Solution & Answer

Summary Introduction

Introduction:

The pulmonary ventilation process depends upon the elastic properties of the lungs and the chest wall. The presence of a pressure gradient allows the gas to move from a higher pressure to a lower pressure, and this is the primary principle of ventilation. The lungs and the chest wall expand when air is taken in, and collapse when air is expelled out, and therefore, they act as springs. The air applies an elastic force to the walls which are clinically evaluated to study the compliance of lungs.

Answer to Problem 1RQ

Correct answer:

0.1 L/cm H₂O.

Explanation of Solution

Justification/ Explanation for the correct answer:

Option (a) is given 0.1L/cm H₂O. Lung compliance refers to the ability of the lung’s wall to readily accept the elastic force. It is denoted as C_L and can be defined as the change in the volume of lungs per unit change in the pressure.

The compliance is expressed in liters per centimeter of water pressure (L/cm H₂O). A normal person generates a negative pleural pressure change during inspiration, which is equal to 5 cm H₂O. The lungs accept about 500 milliliters or 0.5 liters of volume at this change. Therefore, the compliance can be calculated by using the formula:

Compliance=change in volume in liters (ΔV)change in pressure, ΔP (cm H2O)=0.5 L of gas5 (cm H2O) = 0.1L/cm H2O

Hence, option (a) is the correct answer.

The explanations for the incorrect answers:

Option (b) states that the average compliance is 0.2L/cm of water, which is possible because of the opposing effect of the lungs. The above value is the unopposed compliance of the lungs and because the lungs are attached to the thoracic wall and the lung’ wall and thoracic wall both acts as springs and recoil each other. This decreases the compliance to half value. So, it is an incorrect option.

Option (c) states that the average compliance of the lung and the chest wall is 0.3 L/cm of water, and from the calculation shown above, it cannot be true. So, it is an incorrect option.

Option (d) states that the average compliance of the combined lung and chest wall is 0.4L/cm of water. The unopposed compliance is 0.2 L/cm of water, which is reduced to half, that is, 0.1L/cm of water because of the opposing effects of the lung and thoracic wall. So, it is an incorrect option.

Hence, options (b), (c), and (d) are incorrect.

Conclusion

The combined average compliance of the lung and chest wall is 0.1L/cm of water, which is half of the average compliance of the lung’s wall because the lung and thoracic wall show opposing effects and recoil.

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Answer 2

Textbook Question

Chapter 2, Problem 1RQ

The average compliance of the lungs and chest wall combined is

0.1 L/cm H₂O.

0.2 L/cm H₂O

0.3 L/cm H₂O

0.4 L/cm H₂O

Expert Solution & Answer

Summary Introduction

Introduction:

The pulmonary ventilation process depends upon the elastic properties of the lungs and the chest wall. The presence of a pressure gradient allows the gas to move from a higher pressure to a lower pressure, and this is the primary principle of ventilation. The lungs and the chest wall expand when air is taken in, and collapse when air is expelled out, and therefore, they act as springs. The air applies an elastic force to the walls which are clinically evaluated to study the compliance of lungs.

Answer to Problem 1RQ

Correct answer:

0.1 L/cm H₂O.

Explanation of Solution

Justification/ Explanation for the correct answer:

Option (a) is given 0.1L/cm H₂O. Lung compliance refers to the ability of the lung’s wall to readily accept the elastic force. It is denoted as C_L and can be defined as the change in the volume of lungs per unit change in the pressure.

The compliance is expressed in liters per centimeter of water pressure (L/cm H₂O). A normal person generates a negative pleural pressure change during inspiration, which is equal to 5 cm H₂O. The lungs accept about 500 milliliters or 0.5 liters of volume at this change. Therefore, the compliance can be calculated by using the formula:

Compliance=change in volume in liters (ΔV)change in pressure, ΔP (cm H2O)=0.5 L of gas5 (cm H2O) = 0.1L/cm H2O

Hence, option (a) is the correct answer.

The explanations for the incorrect answers:

Option (b) states that the average compliance is 0.2L/cm of water, which is possible because of the opposing effect of the lungs. The above value is the unopposed compliance of the lungs and because the lungs are attached to the thoracic wall and the lung’ wall and thoracic wall both acts as springs and recoil each other. This decreases the compliance to half value. So, it is an incorrect option.

Option (c) states that the average compliance of the lung and the chest wall is 0.3 L/cm of water, and from the calculation shown above, it cannot be true. So, it is an incorrect option.

Option (d) states that the average compliance of the combined lung and chest wall is 0.4L/cm of water. The unopposed compliance is 0.2 L/cm of water, which is reduced to half, that is, 0.1L/cm of water because of the opposing effects of the lung and thoracic wall. So, it is an incorrect option.

Hence, options (b), (c), and (d) are incorrect.

Conclusion

The combined average compliance of the lung and chest wall is 0.1L/cm of water, which is half of the average compliance of the lung’s wall because the lung and thoracic wall show opposing effects and recoil.

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Answer 3

Textbook Question

Chapter 2, Problem 1RQ

The average compliance of the lungs and chest wall combined is

0.1 L/cm H₂O.

0.2 L/cm H₂O

0.3 L/cm H₂O

0.4 L/cm H₂O

Expert Solution & Answer

Summary Introduction

Introduction:

The pulmonary ventilation process depends upon the elastic properties of the lungs and the chest wall. The presence of a pressure gradient allows the gas to move from a higher pressure to a lower pressure, and this is the primary principle of ventilation. The lungs and the chest wall expand when air is taken in, and collapse when air is expelled out, and therefore, they act as springs. The air applies an elastic force to the walls which are clinically evaluated to study the compliance of lungs.

Answer to Problem 1RQ

Correct answer:

0.1 L/cm H₂O.

Explanation of Solution

Justification/ Explanation for the correct answer:

Option (a) is given 0.1L/cm H₂O. Lung compliance refers to the ability of the lung’s wall to readily accept the elastic force. It is denoted as C_L and can be defined as the change in the volume of lungs per unit change in the pressure.

The compliance is expressed in liters per centimeter of water pressure (L/cm H₂O). A normal person generates a negative pleural pressure change during inspiration, which is equal to 5 cm H₂O. The lungs accept about 500 milliliters or 0.5 liters of volume at this change. Therefore, the compliance can be calculated by using the formula:

Compliance=change in volume in liters (ΔV)change in pressure, ΔP (cm H2O)=0.5 L of gas5 (cm H2O) = 0.1L/cm H2O

Hence, option (a) is the correct answer.

The explanations for the incorrect answers:

Option (b) states that the average compliance is 0.2L/cm of water, which is possible because of the opposing effect of the lungs. The above value is the unopposed compliance of the lungs and because the lungs are attached to the thoracic wall and the lung’ wall and thoracic wall both acts as springs and recoil each other. This decreases the compliance to half value. So, it is an incorrect option.

Option (c) states that the average compliance of the lung and the chest wall is 0.3 L/cm of water, and from the calculation shown above, it cannot be true. So, it is an incorrect option.

Option (d) states that the average compliance of the combined lung and chest wall is 0.4L/cm of water. The unopposed compliance is 0.2 L/cm of water, which is reduced to half, that is, 0.1L/cm of water because of the opposing effects of the lung and thoracic wall. So, it is an incorrect option.

Hence, options (b), (c), and (d) are incorrect.

Conclusion

The combined average compliance of the lung and chest wall is 0.1L/cm of water, which is half of the average compliance of the lung’s wall because the lung and thoracic wall show opposing effects and recoil.

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Answer 4

Textbook Question

Chapter 2, Problem 1RQ

The average compliance of the lungs and chest wall combined is

0.1 L/cm H₂O.

0.2 L/cm H₂O

0.3 L/cm H₂O

0.4 L/cm H₂O

Expert Solution & Answer

Summary Introduction

Introduction:

The pulmonary ventilation process depends upon the elastic properties of the lungs and the chest wall. The presence of a pressure gradient allows the gas to move from a higher pressure to a lower pressure, and this is the primary principle of ventilation. The lungs and the chest wall expand when air is taken in, and collapse when air is expelled out, and therefore, they act as springs. The air applies an elastic force to the walls which are clinically evaluated to study the compliance of lungs.

Answer to Problem 1RQ

Correct answer:

0.1 L/cm H₂O.

Explanation of Solution

Justification/ Explanation for the correct answer:

Option (a) is given 0.1L/cm H₂O. Lung compliance refers to the ability of the lung’s wall to readily accept the elastic force. It is denoted as C_L and can be defined as the change in the volume of lungs per unit change in the pressure.

The compliance is expressed in liters per centimeter of water pressure (L/cm H₂O). A normal person generates a negative pleural pressure change during inspiration, which is equal to 5 cm H₂O. The lungs accept about 500 milliliters or 0.5 liters of volume at this change. Therefore, the compliance can be calculated by using the formula:

Compliance=change in volume in liters (ΔV)change in pressure, ΔP (cm H2O)=0.5 L of gas5 (cm H2O) = 0.1L/cm H2O

Hence, option (a) is the correct answer.

The explanations for the incorrect answers:

Option (b) states that the average compliance is 0.2L/cm of water, which is possible because of the opposing effect of the lungs. The above value is the unopposed compliance of the lungs and because the lungs are attached to the thoracic wall and the lung’ wall and thoracic wall both acts as springs and recoil each other. This decreases the compliance to half value. So, it is an incorrect option.

Option (c) states that the average compliance of the lung and the chest wall is 0.3 L/cm of water, and from the calculation shown above, it cannot be true. So, it is an incorrect option.

Option (d) states that the average compliance of the combined lung and chest wall is 0.4L/cm of water. The unopposed compliance is 0.2 L/cm of water, which is reduced to half, that is, 0.1L/cm of water because of the opposing effects of the lung and thoracic wall. So, it is an incorrect option.

Hence, options (b), (c), and (d) are incorrect.

Conclusion

The combined average compliance of the lung and chest wall is 0.1L/cm of water, which is half of the average compliance of the lung’s wall because the lung and thoracic wall show opposing effects and recoil.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

Summary Introduction

Introduction:

The pulmonary ventilation process depends upon the elastic properties of the lungs and the chest wall. The presence of a pressure gradient allows the gas to move from a higher pressure to a lower pressure, and this is the primary principle of ventilation. The lungs and the chest wall expand when air is taken in, and collapse when air is expelled out, and therefore, they act as springs. The air applies an elastic force to the walls which are clinically evaluated to study the compliance of lungs.

Answer to Problem 1RQ

Correct answer:

0.1 L/cm H₂O.

Explanation of Solution

Justification/ Explanation for the correct answer:

Option (a) is given 0.1L/cm H₂O. Lung compliance refers to the ability of the lung’s wall to readily accept the elastic force. It is denoted as C_L and can be defined as the change in the volume of lungs per unit change in the pressure.

The compliance is expressed in liters per centimeter of water pressure (L/cm H₂O). A normal person generates a negative pleural pressure change during inspiration, which is equal to 5 cm H₂O. The lungs accept about 500 milliliters or 0.5 liters of volume at this change. Therefore, the compliance can be calculated by using the formula:

Compliance=change in volume in liters (ΔV)change in pressure, ΔP (cm H2O)=0.5 L of gas5 (cm H2O) = 0.1L/cm H2O

Hence, option (a) is the correct answer.

The explanations for the incorrect answers:

Option (b) states that the average compliance is 0.2L/cm of water, which is possible because of the opposing effect of the lungs. The above value is the unopposed compliance of the lungs and because the lungs are attached to the thoracic wall and the lung’ wall and thoracic wall both acts as springs and recoil each other. This decreases the compliance to half value. So, it is an incorrect option.

Option (c) states that the average compliance of the lung and the chest wall is 0.3 L/cm of water, and from the calculation shown above, it cannot be true. So, it is an incorrect option.

Option (d) states that the average compliance of the combined lung and chest wall is 0.4L/cm of water. The unopposed compliance is 0.2 L/cm of water, which is reduced to half, that is, 0.1L/cm of water because of the opposing effects of the lung and thoracic wall. So, it is an incorrect option.

Hence, options (b), (c), and (d) are incorrect.

Conclusion

The combined average compliance of the lung and chest wall is 0.1L/cm of water, which is half of the average compliance of the lung’s wall because the lung and thoracic wall show opposing effects and recoil.

Answer 8

Expert Solution & Answer

Summary Introduction

Introduction:

The pulmonary ventilation process depends upon the elastic properties of the lungs and the chest wall. The presence of a pressure gradient allows the gas to move from a higher pressure to a lower pressure, and this is the primary principle of ventilation. The lungs and the chest wall expand when air is taken in, and collapse when air is expelled out, and therefore, they act as springs. The air applies an elastic force to the walls which are clinically evaluated to study the compliance of lungs.

Answer to Problem 1RQ

Correct answer:

0.1 L/cm H₂O.

Explanation of Solution

Justification/ Explanation for the correct answer:

Option (a) is given 0.1L/cm H₂O. Lung compliance refers to the ability of the lung’s wall to readily accept the elastic force. It is denoted as C_L and can be defined as the change in the volume of lungs per unit change in the pressure.

The compliance is expressed in liters per centimeter of water pressure (L/cm H₂O). A normal person generates a negative pleural pressure change during inspiration, which is equal to 5 cm H₂O. The lungs accept about 500 milliliters or 0.5 liters of volume at this change. Therefore, the compliance can be calculated by using the formula:

Compliance=change in volume in liters (ΔV)change in pressure, ΔP (cm H2O)=0.5 L of gas5 (cm H2O) = 0.1L/cm H2O

Hence, option (a) is the correct answer.

The explanations for the incorrect answers:

Option (b) states that the average compliance is 0.2L/cm of water, which is possible because of the opposing effect of the lungs. The above value is the unopposed compliance of the lungs and because the lungs are attached to the thoracic wall and the lung’ wall and thoracic wall both acts as springs and recoil each other. This decreases the compliance to half value. So, it is an incorrect option.

Option (c) states that the average compliance of the lung and the chest wall is 0.3 L/cm of water, and from the calculation shown above, it cannot be true. So, it is an incorrect option.

Option (d) states that the average compliance of the combined lung and chest wall is 0.4L/cm of water. The unopposed compliance is 0.2 L/cm of water, which is reduced to half, that is, 0.1L/cm of water because of the opposing effects of the lung and thoracic wall. So, it is an incorrect option.

Hence, options (b), (c), and (d) are incorrect.

Conclusion

The combined average compliance of the lung and chest wall is 0.1L/cm of water, which is half of the average compliance of the lung’s wall because the lung and thoracic wall show opposing effects and recoil.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Summary Introduction

Introduction:

The pulmonary ventilation process depends upon the elastic properties of the lungs and the chest wall. The presence of a pressure gradient allows the gas to move from a higher pressure to a lower pressure, and this is the primary principle of ventilation. The lungs and the chest wall expand when air is taken in, and collapse when air is expelled out, and therefore, they act as springs. The air applies an elastic force to the walls which are clinically evaluated to study the compliance of lungs.

Answer 12

Answer to Problem 1RQ

Correct answer:

0.1 L/cm H₂O.

Answer 13

Explanation of Solution

Justification/ Explanation for the correct answer:

Option (a) is given 0.1L/cm H₂O. Lung compliance refers to the ability of the lung’s wall to readily accept the elastic force. It is denoted as C_L and can be defined as the change in the volume of lungs per unit change in the pressure.

The compliance is expressed in liters per centimeter of water pressure (L/cm H₂O). A normal person generates a negative pleural pressure change during inspiration, which is equal to 5 cm H₂O. The lungs accept about 500 milliliters or 0.5 liters of volume at this change. Therefore, the compliance can be calculated by using the formula:

Compliance=change in volume in liters (ΔV)change in pressure, ΔP (cm H2O)=0.5 L of gas5 (cm H2O) = 0.1L/cm H2O

Hence, option (a) is the correct answer.

The explanations for the incorrect answers:

Option (b) states that the average compliance is 0.2L/cm of water, which is possible because of the opposing effect of the lungs. The above value is the unopposed compliance of the lungs and because the lungs are attached to the thoracic wall and the lung’ wall and thoracic wall both acts as springs and recoil each other. This decreases the compliance to half value. So, it is an incorrect option.

Option (c) states that the average compliance of the lung and the chest wall is 0.3 L/cm of water, and from the calculation shown above, it cannot be true. So, it is an incorrect option.

Option (d) states that the average compliance of the combined lung and chest wall is 0.4L/cm of water. The unopposed compliance is 0.2 L/cm of water, which is reduced to half, that is, 0.1L/cm of water because of the opposing effects of the lung and thoracic wall. So, it is an incorrect option.

Hence, options (b), (c), and (d) are incorrect.

Answer 14

Conclusion

The combined average compliance of the lung and chest wall is 0.1L/cm of water, which is half of the average compliance of the lung’s wall because the lung and thoracic wall show opposing effects and recoil.

Answer 15

Ch. 2 - The average compliance of the lungs and chest wall...Ch. 2 - Normally, the airway resistance in the...Ch. 2 - 3. In the normal individual in the upright...Ch. 2 - Prob. 4RQ Ch. 2 - 5. When arranged for flow Poiseuille’s law states...Ch. 2 - 6. During a normal exhalation, the 1....Ch. 2 - At rest, the normal pleural pressure change during...Ch. 2 - Prob. 8RQ Ch. 2 - Prob. 9RQ Ch. 2 - Assuming that pressure remains constant, if the...

The average compliance of the lungs and chest wall combined is 0.1 L/cm H 2 O. 0.2 L/cm H 2 O 0.3 L/cm H 2 O 0.4 L/cm H 2 O

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Answer to Problem 1RQ

Explanation of Solution

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