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

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* 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* 10^(3)\ Pa`

Since the alveolus is considered to be a spherical shell

The surface tension can be calculated as:

`\Delta P = (4\pi T)/(R)`

`T = (1.99* 10^(3)* 200* 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.