Consider simple models for the earth's atmosphere. Neglect winds, convection, etc, and neglect variation in gravity.

Assume that the atmosphere is isothermal $\left(\right.$ at $\left.0^{\circ} \mathrm{C}\right)$. By using an expression for the distribution of molecules with height, estimate roughly the height below which half the molecules lie.
The molecular number density at height $h$ is denoted by $n(h)$. From the condition of mechanical equilibrium $d p=-n m g d h$ and the equation of state $p=n k T$, we find

$$
\frac{1}{p} d p=-\frac{m g}{k T} d h .
$$

Thus $n(h)=n_{0} \exp (-m g h / k T)$. Let $\int_{0}^{H} n(h) d h / \int_{0}^{\infty} n(h) d h=\frac{1}{2}$, then

$$
H=\frac{k T}{m g} \ln 2=\frac{R T}{N_{0} m g} \ln 2 .
$$

The average molecular weight of the atmosphere is 30 . We have

$$
H=\frac{8.31 \times 10^{7} \times 273}{30 \times 980} \times \ln 2 \approx 8 \times 10^{5} \mathrm{~cm}=8 \mathrm{~km} .
$$