Atmospheric pressure
First consider the simple idea with the
height of the atmosphere being reduced so that the density if uniform and equal to that at sea
level (1.293 kgm
-3) so 10
5 = hρg and this gives the height
of the atmosphere (h) = 10
5/ρg = 7.88 km
PV = nRT
but P = 10
5 Pa, R = 8.3 and T = 293
oC and so
Volume of
the atmosphere = 4/3πR
3 – 4/3πr
3 =
4/3π(R
3 – r
3)
R = 6.4x10
3 km + 7.88 km =
6.4079x10
3 km = 6.4079x10
6 m
r = 6.40x10
3 km =
6.40x10
6 m
So volume of the Earth's atmosphere = 4/3π(2.63116x10
20 – 2.62144x10
20) = 4/3π9.72x10
18 m
3 = 4.07x10
18
m
3
However the density of this "uniform atmosphere" = 1.293 kgm
-3
and from this you can find a value for the mass of the Earth's atmosphere.
Mass of
Earth's atmosphere = density x volume = 1.293 x 4.07x10
18 = 5.26x10
18
kg.
The mass of a "molecule of air" is about 10
-26 kg and so the atmosphere would
contain some 5.26x10
44 molecules = 8.73x10
20 moles (n).
Now
using PV = nRT Pressure (P) = 8.73x10
20 x 8.3 x 273/4.12x10
18 =
4.80x10
5 Pa
Which is of the right order for the accepted value for the
pressure of the atmosphere.
The Earth only has an atmosphere because of the
gravitational attraction of the Earth's field. Without that the molecules would simply disperse
into space. The escape velocity for a given planet is [2Rg]1/2 where R is the radius of the
planet and g the gravitational field strength at its surface. So you see that if g = 0 the escape
velocity would also be zero and the planet would lose its atmosphere.
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