# 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

### Cardiopulmonary Anatomy & Physiolo...

7th Edition
Des Jardins + 1 other
Publisher: Cengage Learning,
ISBN: 9781337794909

### Cardiopulmonary Anatomy & Physiolo...

7th Edition
Des Jardins + 1 other
Publisher: Cengage Learning,
ISBN: 9781337794909

#### Solutions

Chapter
Section
Chapter 2, Problem 1RQ
Textbook Problem

## The average compliance of the lungs and chest wall combined is0.1 L/cm H2O.0.2 L/cm H2O0.3 L/cm H2O0.4 L/cm H2O

Expert Solution
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.

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 CL 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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