Pressure falls with an increase in altitude. This was first clarified by the Frenchman Florin Périer in 1648 on the instructions of Blaise Pascal. Mt. Puy de Dome, near where Périer lived, was 975 m high. Measurements showed that the mercury in a Torricellian tube falls by 8 mm when this mountain is climbed.

A fall in air pressure with an increase in altitude is quite natural, for a smaller column of air then presses down on the instrument.

If you have ever flown in an airplane, you should know that there is an instrument on the front wall of the cabin indicating the altitude of the airplane with an accuracy to within tens of metres. This instrument is called an altimeter. This is an ordinary barometer, but it has been calibrated to show heights above sea level.

Pressure falls with an increase in altitude; let us find a formula of this dependence. We single out a small layer of air with an area of 1 cm\(^2\) located between altitudes \(h_{1}\) and \(h_{2}\). The change of density with altitude is hardly noticeable within a layer which is not too large. Therefore, the weight of the volume of air we have singled out (it is a small cylinder of height \(h_{2} - h_{1}\) and base area of 1 cm\(^2\)) will be \[mg = \rho \, (h_{2} - h_{1}) \, g\] This weight is just what yields the fall in pressure caused by rising from altitude \(h_{1}\) to altitude \(h_{2}\), that is \[\dfrac{p_{1} - p_{2}}{\rho} = g \, (h_{2} - h_{1})\] But according to Boyle’s law, which should be known to the reader (and if not, he will find it in the second book, p. 33), the density of a gas is proportional to its pressure. Consequently, \[\dfrac{p_{1} - p_{2}}{p} \propto (h_{2} - h_{1})\]

On the left is the fraction by which the pressure grew when the altitude was lowered from \(h_{2}\) to \(h_{1}\). Hence, a growth in pressure by one and the same percent will correspond to identical drops of \(h_{2} - h_{1}\).

Measurements and calculations in complete agreement with each other show that the pressure will fall by 0.1 of its value for each kilometre rise above sea level. The same also holds for descents into deep shafts under sea level—the pressure will increase by 0.1 of its value when we descend by one kilometre.

We are talking about a change of 0.1 from the value at the previous altitude. This means that during an ascent of 1 km, the pressure decreases to 0.9 of the pressure at sea level; during an ascent through the next kilometre, it will become equal to 0.9 of 0.9 of the pressure at sea level; at an altitude of 3 km, the pressure will be equal to 0.9 of 0.9 of 0.9, i.e. \(0.9^{3}\) , of the pressure at sea level. It is not difficult to continue this reasoning further.

Denoting the pressure at sea level by \(p_{0}\), we can write out the pressure at altitude \(h\) (expressed in kilometres): \[p = p_{0} (0.87)^{h} = p_{0} \times 10^{-0.06h}\] A more precise number is written in parentheses: 0.9 is the rounded-off value. The formula presupposes the identical temperature at all altitudes. But as a matter of fact, the temperature of the atmosphere changes with altitude and does so, moreover, in accordance with a rather complicated law. Nevertheless, the formula yields fairly good results and may be used for altitudes up to hundreds of kilometres.

It is not hard to determine with the aid of this formula that on the top of the Elbrus—about 5.6 km— the pressure will fall by a factor of approximately two, while at an altitude of 22 km (the record height of a stratospheric balloon’s ascent with people), the pressure will fall to 50 mm Hg.

When we say that a pressure of 760 mm Hg is standard, we must not forget to add, “at sea level”. At an altitude of 5.6 km, the standard pressure will not be 760, but 380 mm Hg.

Along with pressure, air density also falls with an increase in altitude according to the same law. At an altitude of 160 km, not much air will remain.

In fact, \[(0.87)^{160} = 10^{-10}\]

The air density at the Earth’s surface is equal to about 1000 g/cm\(^3\), which means that according to our formula there should be \(10^{-7}\) g of air in 1 m\(^3\) at an altitude of 160 km. But in reality, as measurements performed with the aid of rockets show, the air density at this height is ten times as great.

Our formula gives us an even greater underestimation for heights of several hundreds of kilometres. The change of temperature with altitude and also a particular phenomenon—the decay of air molecules under the action of solar radiation—are responsible for the fact that the formula becomes useless at great heights. Here we shall not go into this.