# Lung compliance study Part /: If a patient generates a negative pleural pressure change of -8 cm H 2 O during inspiration, and the lungs accept a new volume of 630 mL, what is the compliance of the lungs? Part II: If the same patient, 6 hours later, generates a pleural pressure of -12 cm H 2 O during inspiration, and the lungs accept a new volume of 850 mL, what is the compliance of the lungs? Part III: In comparing Part II to Part I, the patient's lung compliance is A. increasing. B. decreasing.

### 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 15RQ
Textbook Problem

## Lung compliance studyPart /: If a patient generates a negative pleural pressure change of -8 cm H 2 O during inspiration, and the lungs accept a new volume of 630 mL, what is the compliance of the lungs?Part II: If the same patient, 6 hours later, generates a pleural pressure of -12 cm H 2 O during inspiration, and the lungs accept a new volume of 850 mL, what is the compliance of the lungs?Part III: In comparing Part II to Part I, the patient's lung compliance isA. increasing.B. decreasing.

Expert Solution
Summary Introduction

To review:

Part I: The compliance of the lungs if a patient generates a negative pleural pressure change of -8 cm H2O during inspiration, and the lungs accept a new volume of 630 ml.

Part II: The compliance of the lungs of the same patient, 6 hours later, if he generates a pleural pressure of -12 cm H2O during inspiration, and the lungs accept a new volume of 850 ml.

Part III: Comparing Part II to Part I, is the patient’s lung compliance increasing or decreasing.

Introduction:

The curve between the pressure and volume, which describes the elastic strength of the lungs, allowing the inhaling of air, is called lung compliance (CL). CL is described as the alteration in the ration in the volume of the lung (ΔV) for per unit change in the pressure (ΔP). It is mathematically expressed in milliliters per centimeter of pressure of water (mL/cmH2O).

### Explanation of Solution

Part I: In the given question, the change in negative pleural pressure is 8cmH2O during inspiration, and the new volume of air accepted by the lungs is 630mL.

Now, the compliance of the lung will be:

CL=ΔV(L)ΔP(cmH2O)=0.63L8(cmH2O)=0.07875L/cmH2O=78.75ml/cmH2O

Hence, the lung compliance of the patient is 78.75mL/cmH2O.

Part II: In the given question, the change in negative pleural pressure is 12cmH2O during inspiration, and the new volume of air accepted by the lungs is 850 ml

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