Laplace's Barometric Law

Starting with the Ideal Gas Law

$$P=nkT$$

Why does pressure change as you go up a mountain? Because $n$ goes down, this is due to gravity pulling the molecules down. So it would be relevant to know how the number density depends on gravity and height. Generally $n(z)\ne n(0)$ and $\frac{n(z)}{n(0)}<1$. We can look at an infinitesimally small portion of volume $dV$ where the number density becomes

$$n=\frac{dN}{dV}$$

with $dV=Adz$, where A is the area of the column of volume being analyzed. Assume the gas is at rest macroscopically, the net force is zero. If the force on each molecule was only $\mathbf{F}=-mg\hat{z}$ then the molecules would all fall down, there must be some force in the $\hat{z}$ direction. The gas below the column is providing some pressure upwards. The mechanical equilibrium condition,

$$(F_z{)}_{net}=0$$

The force pressing down from the molecules above the column is $F=PA=P(z+dz)A$ in the $-\hat{z}$ direction and the force pressing up from the molecules below the column is $F=P(z)A$ in the $+\hat{z}$ direction.

$$0=(F_z{)}_{net}=-mg\cdot dN+P(z)A-P(z+dz)A$$

The number of molecules $dN$ is $dN=n(z)dV=nAdz$. The expression becomes,

$$0=-mg\cdot n(z)dz+(P(z)-P(z+dz))$$

dividing by $dz$,

$$0=-mg\cdot n(z)+\frac{P(z)-P(z+dz)}{dz}$$ $$0=-mg\cdot n(z)-\frac{P(z+dz)-P(z)}{dz}$$

The second term is just the definition of the derivative in the limit that $dz$ goes to zero

$$0=-mg\cdot n(z)-\frac{dP(z)}{dz}$$

Using the ideal gas law

$$0=-mg\cdot n(z)-\frac{d}{dz}(n(z)\cdot kT)$$ $$0=-\frac{mg}{kT}\cdot n(z)-\frac{d}{dz}n(z)$$ $$\frac{dn(z)}{dz}=-\frac{mg}{kT}\cdot n(z)$$ $$\int \frac{1}{n(z)}dn(z)=\int \frac{mg}{kT}dz$$ $$\ln (n(z))=-\frac{mgz}{kT}+C$$ $$n(z)=e^{-mgz/kT}{e}^C=C{e}^{-mgz/kT}$$

Determining integration constant by setting $z=0$

$$n(0)=C$$ $$n(z)=n(0)e^{-mgz/kT}$$

The ratio becomes

$$\frac{n(z)}{n(0)}=e^{-mgz/kT}$$

Notice that the pressure has the same ratio

$$\frac{n(z)}{n(0)}=\frac{p(z)}{p(0)}=e^{-mgz/kT}$$