For ideal gases, the density is given by
:<math>\rho =\frac{PM}{RT}</math>
Where <math>M </math> is the molar mass, <math>R </math> is the ideal gas constant and <math>T </math> is the [[temperature]] (Absolute temperature, in Kelvin or Rankine) of the gas.
we can then set up the equation as follows:
:<math>\frac{dP}{dy}=-\rho g=-\frac{PMg}{RT}</math>
Note that we are using the scalar value of gravity (<math>g</math>), so the minus sign is included due to gravity is downwards, in the negative direction of the y-axis.
using the method of [[separation of variables]], we can rearrange the equation so
:<math>\frac{dP}{P}=-\frac{Mg}{RT}dy</math>
Integrating both sides gives
:<math>\ln\frac{P}{P_0}=-\frac{Mg\left(y-y_0\right)}{RT}</math>
Where P<submath>0P_0</submath> is the reference pressure at point y<submath>0y_0</submath> (often taken at the point which P<submath>0P_0</submath> is the atmospheric pressure).
Rearranging gives
:<math>P=P_0 \exp{\left(-\frac{Mg\left(y-y_0\right)}{RT}\right)}</math>
==References==
{{Reflist}}