Question

III. Diffusion in Capillaries Consider a capillary in an alveolus in the lung. The density of oxygen inside the capillary is 0.30 million molecules per cubic micrometer, and in the air outside the density is 5.0 million molecules per cubic micrometer. A) Will oxygen molecules flow in both directions ncross the membrane? In which direction will there be a net flow of oxygen. Explain. B) If the flow rate of oxygen molecules across the capillary membrane is about 4.0 million molecules per square micrometer per second, and the membrane itself is 7.0 nanometers thick, determine the diffusion coefficient.

Answer 1

a)

Given the density of oxygen inside the capillary = 0.3 x 10⁶ molecules/($\mu$m)³

Density of oxygen outside the capillary = 5 x 10⁶ molecules/($\mu$m)³

From ideal gas equation, we know that

PV = nRT

But n = m/M where m is the mass of the gas and M is the molecular weight of the gas

RT т P х V М

pRT Р М

Where $\rho$ is the density of the gas

According to Fick's law of diffustion, the net diffusion rate of gas across a fluid membrane is proportional to the difference in partial pressure, proportional to the area of the membrane and inversely proportional to the thickness of the membrane

From the above equation, it is clear that partial pressure of oxygen outside is higher than that of inside the capillary, therefore oxygen will diffuse from outside to inside.

b)

ДС F = -D Дz

Where F = flow rate of oxygen molecules across the capillary membrane = 4 x 10⁶/(µm)²/s. Dc = Difference in concentration densities = (0.3x10⁶ – 5 x 10⁶) = -4.7 x 10⁶ molecules/(µm)³, Dz = thickness of membrane = 7 x 10^-3$\mu$m

Substituting these values in the above equation we get

Diffusion coefficient, D = 4 x10⁶ x 7x10^-3/4.7x10⁶

= 5.96 x 10^{-3 (}$\mu$m)²S^-1