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 = [tex]200\mu m = 200\times 10^{- 6}\ m[/tex]
Gauge Pressure inside, [tex]P_{in} = 25\ mmHg[/tex]
Blood Pressure outside, [tex]P_{o} = 10\ mmHg[/tex]
Now,
Change in pressure, [tex]\Delta P = 25 - 10 = 15\ mmHg = 1.99\times 10^{3}\ Pa[/tex]
Since the alveolus is considered to be a spherical shell
The surface tension can be calculated as:
[tex]\Delta P = \frac{4\pi T}{R}[/tex]
[tex]T = \frac{1.99\times 10^{3}\times 200\times 10^{- 6}}{4\pi} = 0.0318\ N/m = 0.318\ mN/m[/tex]
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.