Question

The gauge pressure inside an alveolus with a 200 µm radius is 25 mmHg, while the blood pressure outside is only 10 mmHg. Assuming the alveolus acts like a spherical bubble, what is the surface tension of the fluid and membrane around the outside of the alveolus? How does this compare to the surface tension of water?

Answer 1

Answer:

The surface tension is 0.0318 N/m and is sufficiently less than the surface tension of the water.

Solution:

As per the question:

Radius of an alveolus, R = $200\mu m = 200\times 10^{- 6}\ m$

Gauge Pressure inside, $P_{in} = 25\ mmHg$

Blood Pressure outside, $P_{o} = 10\ mmHg$

Now,

Change in pressure, $\Delta P = 25 - 10 = 15\ mmHg = 1.99\times 10^{3}\ Pa$

Since the alveolus is considered to be a spherical shell

The surface tension can be calculated as:

$\Delta P = \frac{4\pi T}{R}$

$T = \frac{1.99\times 10^{3}\times 200\times 10^{- 6}}{4\pi} = 0.0318\ N/m = 0.318\ mN/m$

And we know that the surface tension of water is 72.8 mN/m

Thus the surface tension of the alveolus is much lesser as compared to the surface tension of water.