Question

Assume the alveolar radii are 0.05 and 0.07 mm towards the end of exhalation. Knowing that air flows from high to low pressures, calculate the pressure difference between the alveoli and justify why the smaller alveolus does not collapse. (Assume surface tension of alveolar wall due to surfactant is 25 mN/m)

Answer 1

Answer:

$0.286 mN/m^2$

Explanation:

Using Laplace's law of surface tension:

$P=\frac{2T}{r}$

where:

P = pressure

T = surface tension

r = radius

In the longer alveoli with radius of  0.07 mm = $0.07*10^{-2}m$; we have:

P =  $\frac{2*25mN/m}{0.07*10^{-2}m}$

P = $0.714 mN/m^2$

In the smaller alveoli with radius 0.05 mm = $0.05 * 10^{-2} m$; we have:

P = $\frac{2*25mN/m}{0.05*10^-2m}$

P = $1.00 mN/m^2$

The pressure difference can now be calculate as follows:

Pressure difference = $(1.00 - 0.714)mN/m^2$

Pressure difference = $0.286 mN/m^2$