Editing Dobson unit (section)

==Derivation==

The Dobson unit arises from the ideal gas law

: <math>PV = nRT,</math>

where ''P'' and ''V'' are pressure and volume respectively, and ''n'', ''R'' and ''T'' are the number of moles of gas, the gas constant (8.314 J/(mol·K)), and ''T'' is temperature in [[kelvin]]s<!-- lower-case and pluralized – see Kelvin#Usage conventions --> (K).

The number density of air is the number of molecules or atoms per unit volume:

: <math>n_\text{air} = \frac{A_{av} n}{V},</math>

and when plugged into the real gas law, the [[number density]] of air is found by using pressure, temperature and the real gas constant:

: <math>n_\text{air} = \frac{A_{av} P}{RT}.</math>

The number density (molecules/volume) of air at standard temperature and pressure (''T'' = 273&nbsp;K and ''P'' = 101325&nbsp;Pa) is, by using this equation,

: <math>n_\text{air} = \frac{A_{av} P}{RT} = \frac{(6.02 \times 10^{23}\,\frac{\text{molecules}}{\text{mol}}) \cdot (101325~\text{Pa})}{8.314 \frac{\text{J}}{\text{mol K}} \cdot 273~\text{K}}.</math>

With some unit conversions of joules to pascal cubic meters, the equation for molecules/volume is

:<math>\frac{(6.02 \times 10^{23}~\frac{\text{molecules}}{\text{mol}}) \cdot (101325~\text{Pa})}{8.314~\frac{\text{Pa}\,\text{m}^3}{\text{mol K}} \cdot 273~\text{K}} = 2.69 \times 10^{25}~\text{molecules}\cdot\text{m}^{-3}.</math>

A Dobson unit is the total amount of a trace gas per unit area. In atmospheric sciences, this is referred to as a column density. How, though, are units of molecules per ''cubic'' meter, a volume, to be converted to molecules per ''square centimeter'', an area? This must be done by integration. To get the column density, integrate the total column over a height. Per the definition of Dobson units, 1 DU = 0.01&nbsp;mm of trace gas when compressed down to sea level at standard temperature and pressure. So integrating the number density of air from 0 to 0.01&nbsp;mm, it becomes equal to 1&nbsp;DU:

:<math>\int_{0~\text{mm}}^{0.01~\text{mm}} (2.69 \times 10^{25}~\text{molecules}\cdot\text{m}^{-3})\,\mathrm{d}x = 2.69 \times 10^{25}~\text{molecules}\cdot\text{m}^{-3} \cdot 0.01~\text{mm} - 2.69 \times 10^{25}~\text{molecules}\cdot\text{m}^{-3} \cdot 0~\text{mm}</math>
::<math>{} = 2.69 \times 10^{25}~\text{molecules}\cdot\text{m}^{-3} \cdot 10^{-5}~\text{m} = 2.69 \times 10^{20}~\text{molecules}\cdot\text{m}^{-2}.</math>

Thus the value of 1 DU is 2.69{{e|20}} molecules per meter squared.