Worked exercise — The column of air pressing on our feet

Problem

Atmospheric pressure at sea level is $P_\text{atm} \approx 101\,325\;\text{Pa}$. How much air mass presses on a surface of $1\;\text{m}^2$? And, assuming the atmosphere has constant density $\rho_\text{aria} = 1{,}2\;\text{kg/m}^3$, what is its equivalent height?

Solution

Step 1 — Force on the surface. Pressure is force per unit area, and the force is the weight of the overlying column of air:

$F = P_\text{atm}\cdot A = 101\,325\cdot 1 = 101\,325\;\text{N}$

Step 2 — Mass of the column. From $F = m g$:

$m_\text{colonna} = \frac{F}{g} = \frac{101\,325}{9{,}81} \approx 10\,329\;\text{kg}$

$\ev{m_\text{colonna} \approx 1{,}0\cdot 10^{4}\;\text{kg}}$

It’s like having $10$ tonnes above every square metre; we don’t notice it because the same pressure acts inside us.

Step 3 — Equivalent height. If air had constant density, the mass would be $m = \rho\, A\, h$, from which

$h = \frac{m_\text{colonna}}{\rho\cdot A} = \frac{10\,329}{1{,}2\cdot 1} \approx 8600\;\text{m}$

$\ev{h \approx 8{,}6\;\text{km}}$

A little more than Everest. In reality the atmosphere extends much higher, but its density decreases exponentially with altitude: the value found here is the “equivalent height” of an ideal atmosphere of constant density.

Linked atoms

Argomenti: Fluid statics and dynamics Concetti: Pressure · Stevin’s law · Weight force Competenze: Unit conversion