Density of air

To calculate the density of air as a function of altitude, one requires additional parameters. For the troposphere, the lowest part of the atmosphere, they are listed below, along with their values according to the International Standard Atmosphere, using for calculation the universal gas constant instead of the air specific constant:

${\displaystyle p_{0}=}$ sea level standard atmospheric pressure, 101325 Pa

${\displaystyle T_{0}=}$ sea level standard temperature, 288.15 K

${\displaystyle g=}$ earth-surface gravitational acceleration, 9.80665 m/s²

${\displaystyle L=}$ temperature lapse rate, 0.0065 K/m

${\displaystyle R=}$ ideal (universal) gas constant, 8.31447 J/(mol·K)

${\displaystyle M=}$ molar mass of dry air, 0.0289654 kg/mol

Temperature at altitude ${\displaystyle h}$ meters above sea level is approximated by the following formula (only valid inside the troposphere, no more than ~18 km above Earth's surface (and lower away from Equator)):

${\displaystyle T=T_{0}-Lh\,}$

The pressure at altitude ${\displaystyle h}$ is given by:

${\displaystyle p=p_{0}\left(1-{\frac {Lh}{T_{0}}}\right)^{gM/RL}}$

Density can then be calculated according to a molar form of the ideal gas law:

${\displaystyle \rho ={\frac {pM}{RT}}\,={\frac {pM}{RT_{0}(1-Lh/T_{0})}}={\frac {p_{0}M}{RT_{0}}}\left(1-{\frac {Lh}{T_{0}}}\right)^{gM/RL-1}\,}$

where:

${\displaystyle M=}$ molar mass

${\displaystyle R=}$ ideal gas constant

${\displaystyle T=}$ absolute temperature

${\displaystyle p=}$ absolute pressure

Note that the density close to the ground is ${\displaystyle \rho _{0}={\frac {p_{0}M}{RT_{0}}}}$

It can be easily verified that the hydrostatic equation holds:

${\displaystyle {\frac {dp}{dh}}=-g\rho }$.

As the temperature varies with height inside the troposphere by less than 25%, ${\displaystyle {\frac {Lh}{T_{0}}}<0.25}$ and one may approximate:

${\displaystyle \rho =\rho _{0}e^{({\frac {gM}{RL}}-1)\cdot ln(1-{\frac {Lh}{T_{0}}})}\approx \rho _{0}e^{-({\frac {gM}{RL}}-1){\frac {Lh}{T_{0}}}}=\rho _{0}e^{-({\frac {gMh}{RT_{0}}}-{\frac {Lh}{T_{0}}})}}$

Thus:

${\displaystyle \rho \approx \rho _{0}e^{-h/H_{n}}}$

Which is identical to the isothermal solution, except that H_n, the height scale of the exponential fall for density (as well as for number density n), is not equal to RT₀/g M as one would expect for an isothermal atmosphere, but rather:

${\displaystyle {\frac {1}{H_{n}}}={\frac {gM}{RT_{0}}}-{\frac {L}{T_{0}}}}$

Which gives H_n = 10.4 km.

Note that for different gasses, the value of H_n differs, according to the molar mass M: It is 10.9 for nitrogen, 9.2 for oxygen and 6.3 for carbon dioxide. The theoretical value for water vapor is 19.6, but due to vapor condensation the water vapor density dependence is highly variable and is not well approximated by this formula.

The pressure can be approximated by another exponent:

${\displaystyle p=p_{0}e^{({\frac {gM}{RL}})\cdot ln(1-{\frac {Lh}{T_{0}}})}\approx p_{0}e^{-{\frac {gM}{RL}}{\frac {Lh}{T_{0}}}}=p_{0}e^{-{\frac {gMh}{RT_{0}}}}}$

Which is identical to the isothermal solution, with the same height scale H_p = RT₀/gM. Note that the hydrostatic equation no longer holds for the exponential approximation (unless L is neglected).

H_p is 8.4 km, but for different gasses (measuring their partial pressure), it is again different and depends upon molar mass, giving 8.7 for nitrogen, 7.6 for oxygen and 5.6 for carbon dioxide.

Further note that since g, Earth's gravitational acceleration, is approximately constant with altitude in the atmosphere, the pressure at height h is proportional to the integral of the density in the column above h, and therefore to the mass in the atmosphere above height h. Therefore the mass fraction of the troposphere out of all the atmosphere is given using the approximated formula for p:

${\displaystyle 1-{\frac {p(h=11km)}{p_{0}}}=1-\left({\frac {T(11km)}{T_{0}}}\right)^{gM/RL}=76\%}$

For nitrogen, it is 75% , while for oxygen this is 79%, and for carbon dioxide - 88%.

Higher than the troposphere, at the tropopause, the temperature is approximately constant with altitude (up to ~20 km) and is 220 K. This means that at this layer L = 0 and T= 220K, so that the exponential drop is faster, with H_TP = 6.3 km for air (6.5 for nitrogen, 5.7 for oxygen and 4.2 for carbon dioxide). Both the pressure and density obey this law, so, denoting the height of the border between the troposphere and the tropopause as U:

${\displaystyle p=p(U)\cdot e^{-(h-U)/H_{TP}}=p_{0}\left(1-{\frac {LU}{T_{0}}}\right)^{gM/RL}\cdot e^{-(h-U)/H_{TP}}}$

${\displaystyle \rho =\rho (U)e^{-(h-U)/H_{TP}}=\rho _{0}\left(1-{\frac {LU}{T_{0}}}\right)^{gM/RL-1}e^{-(h-U)/H_{TP}}}$