Answer:
(a) the approximate mass of a column of air is: 1225 (Kg), (b) the weight of this amount of air is: 12017.25 (N) and (c) the pressure at the bottom is: 120172.5 (KPa)
Explanation:
We need to remember that the equation that related mass and volume is the density as:[tex]p=\frac{m}{V}[/tex], where p is the density, m is the mass and V is the volume and assuming that air density is 1.225 (Kg/m3) and knowing that exosphere is the upper atmosphere and its height is 10000 (Km). We now can get the volume as:[tex]Volume=A*h_{upper atmosphere} =0.0001*10000000=1000(metre^{3})[/tex], so that the approximate mass of a column of air is: [tex]m=p*V=1.225*1000=1225 (Kg)[/tex]. Then the weight of this amount of air is:[tex]W=m*g[/tex], where W is the weight, m is the mass and g the gravity acceleration, so:[tex]W=1225*9.81=12017.25 (N)[/tex]. Finally to find the pressure we know that gauge pressure equation [tex]P=p*g*h[/tex] where P is the pressure, p is the density of the column and h is the height of the column, so [tex]P=1.225*9.81*10000000=120172.5 (KPa)[/tex].
