Changes

For ideal gases, the density is given by

For ideal gases, the density is given by

:<math>\rho =\frac{PM}{RT}</math>

:<math>\rho =\frac{PM}{RT}</math>

Where M is the molar mass, R is the ideal gas constant and T is the [[temperature]] (Absolute temperature, in Kelvin or Rankine) of the gas.

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:

we can then set up the equation as follows:

:<math>\frac{dP}{dy}=-\rho g=-\frac{PMg}{RT}</math>

:<math>\frac{dP}{dy}=-\rho g=-\frac{PMg}{RT}</math>

Note that we are using the scalar value of gravity (g), so the minus sign is included due to gravity is downwards, in the negative direction of the y-axis.

 

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

using the method of [[separation of variables]], we can rearrange the equation so

:<math>\frac{dP}{P}=-\frac{Mg}{RT}dy</math>

:<math>\frac{dP}{P}=-\frac{Mg}{RT}dy</math>

Integrating both sides gives

Integrating both sides gives

:<math>\ln\frac{P}{P_0}=-\frac{Mg\left(y-y_0\right)}{RT}</math>

:<math>\ln\frac{P}{P_0}=-\frac{Mg\left(y-y_0\right)}{RT}</math>

Where P₀ is the reference pressure at point y₀ (often taken at the point which P₀ is the atmospheric pressure).

 

Where <math>P_0</math> is the reference pressure at point <math>y_0</math> (often taken at the point which <math>P_0</math> is the atmospheric pressure).

Rearranging gives

Rearranging gives

:<math>P=P_0 \exp{\left(-\frac{Mg\left(y-y_0\right)}{RT}\right)}</math>

:<math>P=P_0 \exp{\left(-\frac{Mg\left(y-y_0\right)}{RT}\right)}</math>

==References==

==References==

{{Reflist}}

{{Reflist}}