Chapter 1: 93PP (page 32)

CALCULATING LUNGS VALUE IN HUMANS.In humans, oxygen and carbon dioxide are exchanged in the blood within many small sacs called alveoli in the lungs. Alveoli provide a large surface area for gas exchange. Recent careful measurements show that the total number of alveoli in a typical pair of lungs is about $480 \times 10^{6}$ and that the average volume of a single alveolus is $4.2 \times 10^{6} {μm}^{3}$ . (The volume of a sphere is $\frac{4}{3} {πr}^{3}$ and the area of a sphere isdata-custom-editor="chemistry" $A = 4 {πr}^{2}$ .)
1.93If we assume that alveoli are spherical, what is the diameter of a typical alveolus? (a)0.20 mm; (b) 2 mm; (c) 20 mm;(d) 200 mm.

Short Answer

Expert verified

The diameter of a typical alveolus is 0.20 mm.

Step by step solution

Given data

Total numbers of alveoli, $N = 480 \times 10^{6}$

Average volume of alveoli, $V = 4.2 \times 10^{6} {μm}^{3}$

Introduction

Gas exchanging region of the lungs is called alveoli. The product of the total number of alveoli and the average volume of a single alveolus will give the total volume of the alveoli.

Step 3: Find the diameter of a typical alveolus

The volume of a sphere in terms of its radius r is,

$V = \frac{4}{3} {πr}^{3}$

And its diameter, $d = 2 r$ .

Use the above formula to find the diameter of a typical alveolus by assuming that it is a sphere.

$V = \frac{4}{3} {πr}^{3} 4.2 \times 10^{6} {(μm)}^{3} = \frac{4}{3} \times 3.14 \times r^{3} r^{3} = 1.003 \times 10^{6} {(μm)}^{3} r = 100 μm$

As known that,

$1 μm = 0.001 mm$

Therefore, the radius will be,

$\begin{array}{rcl} r & = & 100 \times 0.001 mm \\ = & 0.1 mm \end{array}$

Diameter is define as below.

$\begin{array}{rcl} d & = & 2 r \\ = & 2 \times 0.1 mm \\ = & 0.20 mm \end{array}$

Hence, the option (a) is correct.

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Short Answer

Step by step solution

Given data

Introduction

Step 3: Find the diameter of a typical alveolus

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