Variation of air density with height

In summary, the ideal gas equation can be used to find the density of a gas at a specific point based on its pressure and temperature. However, in order to determine the variation of air density with height, the pressure variation with height is needed. This can be obtained from the barometric formula, which assumes an isothermal atmosphere. For a more realistic formula, empirical data or equations of state are needed. The temperature dependence of air density can be derived using the hydrostatic equation and assuming either isothermal or isentropic processes. However, these equations are only valid within certain regions of the atmosphere.
  • #1
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TL;DR Summary
Is there a way to determine how air density varies with height? We all know that it falls, but what is the relation. I am looking for a function like ##\rho_{\text{air}}(x)## where ##x## is the height from the earth's surface.
Using the ideal gas equation ##PV = nRT\Rightarrow PV = \frac{m}{M} RT## where ##m,M## are the mass and molecular weights of the gas respectively.

This yields ##\frac{m}{V} = \frac{PM}{RT} = \rho##, the density of the gas at a point with pressure ##P##.

If only we can obtain the variation of pressure with height from the Earth's surface : ##P(x) = ?##, we could use it to find ##\rho(x)##.

Does anyone know?
 
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  • #4
Caveat: That's for an isothermal atmosphere!
 
  • #5
vanhees71 said:
Caveat: That's for an isothermal atmosphere!

Yes, I noticed, thank you. This means that the variation of density of air with height, ##\rho (x)## is dependent both on pressure and temperature of air with height. Is there a resource besides wikipedia on how to derive the temperature dependence? Wikipedia only gives the results.
 
  • #6
For the derivation of the isothermal barometric formula assuming an ideal gas you have two posibilities.

(1) Thermodynamics

The phase-space distribution function is the Maxwell-Boltzmann function
$$f(\vec{x},\vec{p})=\exp\{[-\beta (H(\vec{x},\vec{p})-\mu \}.$$
The Hamiltonian reads
$$H=\frac{\vec{p}^2}{2m} + m g z.$$
Integrating over ##\vec{p}## gives the density. We need it not with the absolute norm but just relative to ##z=0## and thus get
$$\rho(\vec{x})=\rho_0 \exp(-\beta m g z)$$
with ##\beta=1/(k_{\text{T}} T)## and ##m## is the mass of the gas molecules.

(2) Hydrostatic equation

$$\vec{\nabla} P=\rho \vec{g}.$$
For an isothermal change of state you have
$$P=\rho k_{\text{B}} T/m$$
and thus the equation gets
$$\vec{\nabla} \rho = -\frac{\rho}{k_{\text{B}} T} m g \vec{e}_z$$
and thus through integration
$$\rho=\rho_{0} \exp \left (-\frac{m g z}{k_{\text{B}} t} \right).$$
Of course the isothermal atmosphere is only a good approximation for not too large changes of height.

For a realistic barometric formula you need an empirical equation of state. E.g., the pressure as function of height. For details see

https://en.wikipedia.org/wiki/Barometric_formula
 
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  • #8
From what I recall, the atmosphere is split into 3 or 4 regions, each region with it's own curve, for ballistic missile and space craft.
 
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  • #9
brotherbobby said:
This means that the variation of density of air with height, ##\rho (x)## is dependent both on pressure and temperature of air with height. Is there a resource besides wikipedia on how to derive the temperature dependence?

The derivation is similar to the derivation for the isothermal case as demonstrated by @vanhees71 in #6. It also starts with the hydrostatic equation

##p' = - \rho \cdot g##

but assuming isententropic processes instead of isothermal changes:

##p \cdot V^\kappa = const.##

Together with the density

##\rho = \frac{m}{V} = \frac{{M \cdot n}}{V}##

and the ideal gas equation

##p \cdot V = n \cdot R \cdot T##

this results in the differential equation

##p' = - \frac{{M \cdot g \cdot p_0^{\frac{{\kappa - 1}}{\kappa }} }}{{R \cdot T_0 }}p^{\frac{1}{\kappa }} ##

with the solution

##p = p_0 \cdot \left[ {1 - \left( {1 - \frac{1}{\kappa }} \right) \cdot \frac{{M \cdot g}}{{R \cdot T_0 }} \cdot z} \right]^{\frac{\kappa }{{\kappa - 1}}} ##

and

##T = T_0 - \left( {1 - \frac{1}{\kappa }} \right) \cdot \frac{{M \cdot g}}{R} \cdot z##

But that only works within the troposphere because it is convection dominated. It fails in the stratosphere (which is radiation dominated).
 
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