Barometric Formula

In contrast to incompressible fluids, the density of gases changes as a function of the applied pressure. The formula for the hydrostatic pressure has therefore to be modified according to $$\mathrm{d}p = \varrho g\,\mathrm{d}h$$ In the next step, the density is going to be replaced with the help of the ideal gas equation $$pV = nRT$$ For the following derivation, a constant temperature throughout the whole atmosphere is assumed which is not true in reality and sets therefore some limitations to its validity. Using the formula $m = \varrho V$ and solving the resulting equation for $p$ leads to $$\varrho = \frac{pM}{RT}$$ where the fraction $m/n$ has been replaced by the molar mass $M$ of the gas. Inserting this into the first equation results gives the following differential equation: $$\mathrm{d}p = \frac{pMg}{RT}\,\mathrm{d}h$$ This can be solved by separating the variables and integrated according to $$\int_{p(h_0)}^{p(h_1)}\frac{\mathrm{d}p}{p} =\int_{h_0}^{h_1}\frac{Mg}{RT}\,\mathrm{d}h$$ Solving both integrals leads to $$\ln\left(\frac{p(h_1)}{p(h_0)}\right) = \frac{Mg}{RT}\Delta h$$ which can finally be solved for the pressure in the height $h_1$: $$\boxed{p(h_1) = p(h_0)e^{\frac{Mg}{RT}\Delta h}}$$ This equation is called barometric formula. If the pressure at a given height $h_0$ is known, then it can be calculated in a height $h_1 = h_0 + \Delta h$.