# If a specific mass of air is polluted, will the pollution stick with it?

Let us say a car pollutes 1m3 of air. Is that pollution attached to that specific air mass or will it fall to the ground or take its own path through the air, independent of the air it first polluted?

And how far can it travel, typically?

• Pour yourself a cup o' tea, and add some drops of milk. What happens to the milk? Why would air and gaseous pollutants behave differently?
– Erik
Jul 12 '19 at 14:59
• There will be dispersion reducing the concentration of all the pollutants. The particulates will drop out depending on their size and type. Some will be scrubbed out of the air by rain, and some will react with components of the atmosphere. Overall you will see the pollutants follow along with the airmass, but as you move away from the source of the pollution the concentration will diminish. A atmospheric scientist should be able to give you a better answer.
– user824
Jul 12 '19 at 16:26
• Jul 12 '19 at 16:50

Like many of my answers on this Stack Exchange, the answer is yes and no. Part of it depends on the pollutant itself. Let's call the mass by it's technical term, a parcel. There are a couple things that can change the pollution level of the parcel, other than the source (a car in your example):

• Chemical reactions between the pollutant, the parcel, light, and water vapor.
• Diffusion (the parcel gets all stretched and mixes with surrounding parcels as well as the molecules slowly migrating)
• Deposition (if the pollutant is an aerosol, it could be falling from gravity or be taken up by rain. If it is a gas, it could be reacting with aerosols or the surface of the earth).

Let's put this into math! $$\frac{D\chi}{Dt}=S+R_{prod}-R_{sink}-D-F$$ where $$\chi$$ is the concentration of the pollutant, $$S$$ is the source, $$R_{prod}$$ is the chemical reactions that produce the pollutant, $$R_{sink}$$ is the chemical reactions that destroy the pollutant, $$D$$ is deposition, and $$F$$ is diffusion.

How far does this polluted mass travel? Well, that is complicated, as the parcel is immediately stretched and contorted. But an actual answer can be computed. Provided that there is no additional source, the pollutant should be reduced drastically by the equation $$\tau=\frac{\chi}{R_{sink}+D+F}$$, where $$\tau$$ stands for lifetime. How far will it travel? Well, let's say the wind speed is $$v$$, which has dimensions of length per time. $$\tau$$ is a time scale. So we can say that the pollutant will be drastically reduced by $$v\times\tau$$ distance downwind.

Carbon monoxide is a toxic gas that has a lifetime of 1-2 months. So given a wind speed of about 5 m s$$^{-1}$$, it may travel 26,297 kilometers. Carbon dioxide, which is also a pollutant from cars, has a lifetime of 35-95 years, so it can travel 14989539 kilometers. Given that the diameter of the earth is 8000 kilometers, you can see why the pollution just spreads over the face of the earth.

Nitrogen monoxide is also a pollutant emitted by cars. However, it is rapidly oxidized and has a lifetime of a few seconds. So it may only travel a few meters.

Air is a fluid, as such it tends to mix and concentrations, of anything, tend to disperse into the wider volume. Erik's cup of tea metaphor is pretty accurate, currents of air will pull apart your exemplar 1m3 and disperse the pollutants therein, the heavier particulate, which is mainly soot, water droplets from combustion, and unburned fuel, will full to the ground fairly quickly. Finer particulate and reactive gases like nitrogen and sulfur oxides will react with water to form acids and fall back to earth with the next rain storm. The exception to this general rule of mixing occurs when there is an inversion layer that puts a "lid" on vertical circulation in a particular area causing a build up of pollutants which often results in smog formation.