We would like to estimate how quickly the non-uniformities in gas composition in the alveoli are damped out. Consider an alveolus to be spherical with a diameter of 0.1mm. Let the sphere have an initial uniform concentration of oxygen, ci, and at a certain instant, the walls of the alveolus are raised to an oxygen concentration of c¥ and maintained at this value. If the oxygen diffusivity in the alveolus is approximated as that of water, 2.4x10-9 m2 /s, how long does it take for the concentration change (c-ci) at the center to be 90% of the final concentration change?

It will take 11.775 seconds.

Explanation: As a sphere with a diameter of 0.1 mm, the area of an alveolus is

A = 4.π.r²

r for an alveolus would be: r = 0.00005m or r = 5.m

Finding the area:

A = 4.3.14.(5.)²

A = 3.14.m²

The concentration change is to be 90% of the final, so

c = 0.9.3.14.

c = 28.26.

The oxygen diffusivity is 2.4.m²/s, that means in 1 second 2.4. of oxygen spread in one alveolus area. So:

1 second = 2.4.m²

t seconds = 28.26.m²

t =

t = 11.775s

For a concentration change at the center to be 90%, it will take 11.775s.

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.