Dobson unit

NASA uses a baseline value of 220 DU for ozone. This was chosen as the starting point for observations of the Antarctic ozone hole, since values of less than 220 Dobson units were not found before 1979. Also, from direct measurements over Antarctica, a column ozone level of less than 220 Dobson units is a result of the ozone loss from chlorine and bromine compounds.[5]

In addition, Dobson units are often used to describe total column densities of sulfur dioxide, which occurs in the atmosphere in small amounts due to the combustion of fossil fuels, from biological processes releasing dimethyl sulfide, or by natural combustion such as forest fires. Large amounts of sulfur dioxide may be released into the atmosphere as well by volcanic eruptions. The Dobson unit is used to describe total column amounts of sulfur dioxide because it appeared in the early days of ozone remote sensing on ultraviolet satellite instruments (such as TOMS).

The Dobson unit arises from the ideal gas law

${\displaystyle PV=nRT,}$

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 kelvins (K).

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

${\displaystyle n_{\text{air}}={\frac {A_{av}n}{V}},}$

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

${\displaystyle n_{\text{air}}={\frac {A_{av}P}{RT}}.}$

The number density (molecules/volume) of air at standard temperature and pressure (T = 273 K and P = 101325 Pa) is, by using this equation,

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

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

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

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, do we go from units of molecules per cubic meter, a volume, to molecules per square centimeter, an area? This must be done by integration. To get a column density, we must integrate the total column over a height. Per the definition of Dobson units, we see that 1 DU = 0.01 mm of trace gas when compressed down to sea level at standard temperature and pressure. So if we integrate our number density of air from 0 to 0.01 mm, we find the number density which is equal to 1 DU:

${\displaystyle \int _{0~{\text{mm}}}^{0.01~{\text{mm}}}(2.69\times 10^{25}~{\text{molecules}}\cdot {\text{m}}^{-3})\,\mathrm {d} x=2.69\times 10^{25}~{\text{molecules}}\cdot {\text{m}}^{-3}\cdot 0.01~{\text{mm}}-2.69\times 10^{25}~{\text{molecules}}\cdot {\text{m}}^{-3}\cdot 0~{\text{mm}}}$

${\displaystyle {}=2.69\times 10^{25}~{\text{molecules}}\cdot {\text{m}}^{-3}\cdot 10^{-5}~{\text{m}}=2.69\times 10^{20}~{\text{molecules}}\cdot {\text{m}}^{-2}.}$

And thus we come up with the value of 1 DU, which is 2.69×10²⁰ molecules per meter squared.