93PP - University Physics with Modern Physics

Question

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 $\mathbf{480}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}$ and that the average volume of a single alveolus is $\mathbf{4}\mathbf{.}\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{6}}\mathbf{ }{\mathbf{\mu m}}^{\mathbf{3}}$. (The volume of a sphere is $\frac{\mathbf{4}}{\mathbf{3}}{\mathbf{\pi r}}^{\mathbf{3}}$ and the area of a sphere is data-custom-editor="chemistry" $\mathbf{A}\mathbf{=}\mathbf{4}{\mathbf{\pi r}}^{\mathbf{2}}$.) 

 

1.93 If 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.

Step-by-Step Solution

Answer

The diameter of a typical alveolus is 0.20 mm. 

Step 1: Given data

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

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

Step 2: 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,

$\mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^3$ 

 

And its diameter, $\mathrm{d}=2\mathrm{r}$. 

 

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

$$\begin{gathered} \mathrm{V}=\frac{4}{3}{\mathrm{\pi r}}^3 \\ 4.2\times {10}^6 {\left(\mathrm{\mu m}\right)}^3=\frac{4}{3}\times 3.14\times {\mathrm{r}}^3 \\ {\mathrm{r}}^3=1.003\times {10}^6 {\left(\mathrm{\mu m}\right)}^3 \\ \mathrm{r}=100 \mathrm{\mu m} \end{gathered}$$ 

 

As known that,

$1 \mathrm{\mu m}=0.001 \mathrm{mm}$ 

 

Therefore, the radius will be,

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

 

Diameter is define as below.

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

 

Hence, the option (a) is correct.