# Dobson units?

1. Homework Statement
If the ozone concentration in a particular sample of air were 25 dobson units, what sould this concentration be in:
a)ppm
b) %(v/v)

2. Homework Equations
1 dob = 2.69*10^20 ozone molecules / m^2
http://en.wikipedia.org/wiki/Ozone

3. The Attempt at a Solution
The question is asking the concentration whereas dobsons are measured molecules per surface area. How do I go about doing this since the units don't seem to agree? Also for concentration problems, I will need to know the solution which is a mixture of air but what compositions are they? The solute is ozone.

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## Answers and Replies

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25 dobson units! Don't forget your sun glasses! Normal is about 300 DU of ozone.

Seriously, a dobson unit IS a volume measure. It is like you look a column of air all the way up, removed the ozone and see how large a column the of ozone would be at 1 atm and 0 C. The column of the air compressed down at 1 atm and 0 C would be about 8 kilometers. 25 dobson units would be about .25 mm. From that you can get your concontrations. Not much ozone is there? See:

http://ozonewatch.gsfc.nasa.gov/facts/dobson.html

for a good discussion.

25 dobson units! Don't forget your sun glasses! Normal is about 300 DU of ozone.

Seriously, a dobson unit IS a volume measure. It is like you look a column of air all the way up, removed the ozone and see how large a column the of ozone would be at 1 atm and 0 C. The column of the air compressed down at 1 atm and 0 C would be about 8 kilometers. 25 dobson units would be about .25 mm. From that you can get your concontrations. Not much ozone is there? See:

http://ozonewatch.gsfc.nasa.gov/facts/dobson.html

for a good discussion.
Are we always considering the surface area of the earth? So 25 Dob is .25mm thick of ozone on the surface of the earth? How do you work out the volume? Is it done by integrating from the surface of the earth to 0.25mm in the radial direction? It seems complicated and imprecise as the radius of the earth has to be used which could take on many values.

You are making it too hard. Read the article I sent you. You take all the ozone in the atmosphere in a column one meter square and compress standard temp and pressure and it comes out to be .25 mm high in your case. Then you take all the rest of the air in the column and compress it at standard pressure and temp and it comes out to be 8 kilometers high. The ppm is simply .25 mm/8 kilometers.... Ok, you are right, the dobson units are one dimensional, but you assign it to an arbitrary column area which makes it a volume.

The dobson units are measured with optics. One looks at the attenuation of light at four frequencies. Two are attenuated by ozone, two that are not. The air column is basically a constant around the World. Yea, I know it varies a bit from place to place, but not by much.

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I've seen the rest of the article. One possible error is specifying that the rest of the air composes a column with a base of 1cm^2 that is 8km high. How exact is this 8km?

ppm is atoms of solute per million of atoms in the solution. Let's consider the solute to be .25mm high with a base of 1cm^2. The number of ozone molecules is 6.725*10^17. The solution is 8km high on a base of 1cm^2 but how many molecules in total is in this column? If we assume the concentration to be the same as ozone than all can be worked out as you suggested, 25/(8*10^8) which when converted to ppm is 0.03125.
The answer suggested ppm=0.025

When calculating (%v/v), again I assumed all air were O3 and calculated according its molar mass and got 3.127*10^-6. The answer suggested 2.5*10^-6. So I am a little high and as expected because I have assumed the air to contain all O3 which in reality contain smaller molecules and lighter molecules so my answers were all a bit high. There is also that inaccurate 8km figure. I wonder how they did it.