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Found in: Page 399

### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

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# The alveoli in emphysema victims are damaged and effectively form larger sacs. Construct a problem in which you calculate the loss of pressure due to surface tension in the alveoli because of their larger average diameters. (Part of the lung’s ability to expel air results from pressure created by surface tension in the alveoli.) Among the things to consider are the normal surface tension of the fluid lining the alveoli, the average alveolar radius in normal individuals and its average in emphysema sufferers.

The pressure loss due to surface tension is obtained as:$$2000 N/{m^2}$$.

See the step by step solution

## Step 1: Surface tension

The force acting per unit length on a surface film is called surface tension. It is a scalar quantity measures in newton per meter.

## Step 2: Construction of problem

The diameter of alveoli of an emphysema victim is increased twice of its initial value. Calculate the loss of pressure due to surface tension in the alveoli because of their larger average diameters, if the average diameter of alveoli is $$100 \mu m$$ and the normal surface tension of the fluid lining the alveoli is $$0.05 N/m$$.

## Step 3: Initial pressure inside the alveoli

The initial radius of the alveoli is:

$${r_i} = \frac{{{d_i}}}{2}$$

Here, $${d_i} = 100 \mu m$$ is the initial diameter of the alveoli.

$$\begin{array}{c}{r_i} = \frac{{100 \mu m}}{2}\\ = 50 \mu m\end{array}$$

The initial pressure inside the alveoli is:

$${P_i} = \frac{{4\gamma }}{{{r_i}}}$$

Here, $$\gamma = 0.05 N/m$$ is the surface tension of the fluid lining the alveoli, and $${r_i} = 50 \mu m$$ is the initial radius of the alveoli.

$$\begin{array}{c}{P_i} = \frac{{4 \times 0.05 N/m}}{{50 \mu m}}\\ = \frac{{4 \times 0.05 N/m}}{{\left( {50 \mu m} \right) \times \left( {\frac{{1{0^{ - 6}} m}}{{1 \mu m}}} \right)}}\\ = 4000 N/{m^2}\end{array}$$

## Step 4: Final pressure inside the alveoli

The final radius of the alveoli is,

$${r_f} = \frac{{{d_f}}}{2}$$

Here, $${d_f}$$ is the initial diameter of the alveoli.

Since, final diameter is twice of the initial diameter i.e., $${d_f} = 2{d_i}$$. Therefore,

$$\begin{array}{c}{r_f} = \frac{{2{d_i}}}{2}\\ = {d_i}\end{array}$$

Substitute $$100 \mu m$$for $${d_i}$$:

$${r_f} = 100 \mu m$$

The final pressure inside the alveoli is:

$${P_f} = \frac{{4\gamma }}{{{r_f}}}$$

Substitute $$0.05 N/m$$for $$\gamma$$, and $$100 \mu m$$ for $${r_f}$$,

$$\begin{array}{c}{P_f} = \frac{{4 \times 0.05 N/m}}{{100 \mu m}}\\ = \frac{{4 \times 0.05 N/m}}{{\left( {100 \mu m} \right) \times \left( {\frac{{1{0^{ - 6}} m}}{{1 \mu m}}} \right)}}\\ = 2000 N/{m^2}\end{array}$$

## Step 5: Pressure loss

The pressure loss is:

$$\Delta P = {P_i} - {P_f}$$

Substitute $${P_i} = 4000 N/{m^2}$$and $${P_f} = 2000 N/{m^2}$$

$$\begin{array}{c}\Delta P = \left( {4000 N/{m^2}} \right) - \left( {2000 N/{m^2}} \right)\\ = 2000 N/{m^2}\end{array}$$

Hence, the pressure loss due to surface tension is $$2000 N/{m^2}$$.

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