# 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 105 = hρg and this gives the height of the atmosphere (h) = 105/ρg = 7.88 km

PV = nRT but P = 105 Pa, R = 8.3 and T = 293 oC and so

Volume of the atmosphere = 4/3πR3 – 4/3πr3 = 4/3π(R3 – r3)

R = 6.4x103 km + 7.88 km = 6.4079x103 km = 6.4079x106 m
r = 6.40x103 km = 6.40x106 m

So volume of the Earth's atmosphere = 4/3π(2.63116x1020 – 2.62144x1020) = 4/3π9.72x1018 m3 = 4.07x1018 m3

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 4.07x1018 = 5.26x1018 kg.

The mass of a "molecule of air" is about 10-26 kg and so the atmosphere would contain some 5.26x1044 molecules = 8.73x1020 moles (n).

Now using PV = nRT Pressure (P) = 8.73x1020 x 8.3 x 273/4.12x1018 = 4.80x105 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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© Keith Gibbs 2020