PROBLEM SET 6.1: AIRWAY RESISTANCE AND ALVEOLAR GAS EXCHANGE
ANSW ER KEY
1. The density of Hg is 13.6 g cm3
and the acceleration due to gravity is 981
... [Show More] cm s-2
. Convert the
pressure exerted by a column of Hg 1 mm high to Pascals, N m-2
. If the atmospheric pressure on
top of a 3000 m peak is 527 mm Hg, what is it in Pascals?
The pressure exerted by a column of Hg is its gravitational force divided by its area. For a column
of Hg 1 mm high with a cross-sectional area A, the gravitational force on the column downward is
F = m x g
where g is the acceleration due to gravity = 981 cm s-2. The mass of the column of Hg is just its
density times the volume. Thus m = DV = D x A x h, where D = 13.6 g cm-3 is the density, A is the
area and h is the height, which is 1 m m in this case. The pressure is
P = F/A = (m x g) / A = (D x A x h x g)/A = D x g x h
So the pressure of a 1 mm colum n of Hg is
P = 13.6 g cm-3 x 981 cm s-2 x 0.1 cm = 1334.2 g cm s-2 cm-2 = 1334.2 dyne cm-2
The unit of dyne in the cgs system is not part of ISI; It is converted to N (kg m s-2) by multiplying by
10-5 N dyne-1 ; converting from cm-2 to m-2 requires multiplying by 104
cm2
m-2 ; This gives
P = 133.4 N m-2 = 133.4 Pa
As the equivalent measure of 1 mm Hg: 1 mm Hg = 133.4 Pa
If atmospheric pressure at the top of a 3000 m peak is 527 mm Hg, then this is equivalent to
527 mm Hg x 133.4 N m -2 mm Hg-1 = 70302 N m-2 = 70302 Pa
2. A. The typical diameter of an alveolus is about 0.3 mm . If there are 300 x 106
alveoli in the
lungs calculate the approximate area for gas exchange, assuming that the alveoli are
spheres.
The surface area of a single alveolus given that it approximates a sphere and its diameter is 0.3 mm
= 0.3 x 10
-3 m, is
A = 4 x B x r2
= 4 x B x [0.15 x 10-3 m]2
= 2.82 x 10-7 m2
The total area is the area per alveolus times the number of alveoli, which is 300 x 106
; thus the total
area is
Total Area = 2.82 x 10-7 m -2 x 300 x 106
= 84.8 m2 [Show Less]