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Shkiper50 [21]
3 years ago
13

The gauge pressure inside an alveolus with a 200 µm radius is 25 mmHg, while the blood pressure outside is only 10 mmHg. Assumin

g 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?
Physics
1 answer:
alekssr [168]3 years ago
3 0

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.

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