Barometric formula

Example — Barometric formula

The atmospheric pressure $P$ varies with altitude $h$ according to: $\frac{dP}{dh} = -\frac{0{,}028\cdot g}{R\,T}\,P = -\alpha\,P,$ where $g=9{,}81$ m/s², $R=8{,}314$ J/(mol$\cdot$K), $T$ is the absolute temperature.

We separate: $\dfrac{dP}{P}=-\alpha\,dh$, hence $P(h) = P_0\,e^{-\alpha h}$.

With $T=293$ K: $\alpha = \dfrac{0{,}028\times 9{,}81}{8{,}314\times 293} \approx 1{,}13\times 10^{-4}$ m$^{-1}$. Therefore: $\boxed{P(h) = 101\,325\cdot e^{-0{,}000113\,h}}$ (with $h$ in metres, $P$ in Pascals).

Numerical example. At what altitude is the pressure half that at sea level? $e^{-0{,}000113\,h} = 0{,}5 \implies h = \frac{\ln 2}{0{,}000113} \approx 6\,135\text{ m}.$ It is a reasonable result: at the summit of Everest, at about $8\,848$ m, the pressure is about a third of that at sea level.

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Topics: Differential equations
Concepts: Decay · Barometric formula · Separable variables
Functions: Exponential function
Skills: Calculating · Integrating · Modelling