How to Calculate Dobson Units for Atmospheric Ozone Concentration?

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Discussion Overview

The discussion revolves around calculating Dobson Units (DU) for atmospheric ozone concentration, specifically focusing on how to convert a given concentration of ozone molecules into a measure of DU based on a defined column of the atmosphere. Participants explore the mathematical relationships and conversions necessary for this calculation.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Dorin seeks help in calculating the number of Dobson Units from a uniform concentration of ozone molecules given in molecules/cm3 over a specified altitude range.
  • One participant asks how many molecules are present in a 15 km column with a 1 cm2 base.
  • Another participant emphasizes that the problem can be simplified by focusing on the number of molecules in the specified volume rather than the complexities of ozone and Dobson Units.
  • Calculations are presented, showing that the volume of the column is 15*105 cm3 and that this leads to a total of 4.5*1018 molecules over the 1 cm2 base.
  • Confusion arises regarding how to convert the total number of molecules into a surface density (molecules/cm2) for the purpose of calculating Dobson Units.
  • One participant proposes a method involving roots and squares of the concentration, which is met with skepticism by others who question the validity of this approach.
  • Another participant suggests using the ideal gas law to conceptualize the problem, indicating that the number of molecules can be converted to moles and then related to volume.
  • Clarification is provided that the previously calculated number of molecules over the 1 cm2 area is the key value needed for the DU calculation.
  • Ultimately, a participant expresses realization about the simplicity of the problem once the focus is shifted to the base area rather than the volume of the column.

Areas of Agreement / Disagreement

While there is some agreement on the calculations leading to the total number of molecules, there remains uncertainty and confusion regarding the conversion to Dobson Units and the methods used to arrive at that conversion. No consensus is reached on the best approach to the problem.

Contextual Notes

Participants express varying levels of understanding regarding the mathematical steps involved in the conversion process, leading to multiple proposed methods and some confusion about the relevance of certain calculations.

dorin1993
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calculate the number of DU assuming that the entire atmospheric O3 column is at a uniform concentration of 3*10^12 molecules/cm^3 between 15 km and 30 km and zero elsewhere.

I have no idea how to solve it. I don't know how to convert the 3*10^12 molecules/cm^3 to units of molecules/cm^2 that I can calculate by dividing in 2.69*10^16 molecules/cm^2 and give the DU.

Is anyone know how to solve this?

thank you,
Dorin
 
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How many molecules in a 15 km column with a 1 square cm base?
 
This is all the given information...
I attach the 2 pages in the book that explain about the dobson unit
the question is from this book.

http://imageshack.us/a/img824/1039/77622876.jpg http://imageshack.us/a/img29/5307/13079304.jpg
 
Last edited by a moderator:
You missed the point. You can easily calculate the answer to my question. Just forget about ozone, DU, Dobson spectrometers, whatever.

It is a pretty simple question - if there are n molecules of something in each cubic centimeter, how many molecules are present in the column 15 km high and with a 1 square cm base?
 
OK - so i have volume of 15km * 1cm * 1cm = 15*10^5 cm^3
then - 3*10^12 [molecules/cm^3] * 15*10^5 [cm^3] = 4.5*10^18 molecules

now - how can i calculate the molecule of the base? meaning the number of molecules for cm^2?
 
You already did. Think it over. What is the base of the column?

Yes, it was that simple from the very beginning.
 
Sorry, i still don't get it :\
I know the volume and i know how much molecules i have on it. Also i know that the base is 1 cm^2.
I even drown it and thinking of it over and over again and I don't know... I keep thinking about the diameter of one molecule and how much is the distans between them... but you said it's simple.


Edit:
Or maybe i can take 3*10^12 [molecules/cm^3] and do third root and then 2nd power

3√(3*10^12) = 1.44*10^3 molecules/cm
(1.44*10^3)^2 = 2.08*10^8 molecules/cm^2

Is it?
 
Last edited:
You know the number of molecules hanging over a 1 cm2 surface. They are dispersed over 15 km column, but now imagine combining them all together in the one place at Earth surface (that's what the DU is about) - what would the volume be? This is just a matter of converting number of molecules to number of moles and plugging it into PV=nRT.
 
Ok - so V = 7.47*10^-6 [mole] * 8.314 * 273K / 1.013*10^5 = 0.167 cm^3
again, I get volume - and I need a surface (molec/cm^2) that i can divide in 2.69*10^16 molecules/cm^2 and get DU.

But I don't understand why. The concentration of the molecule is given, and it's around the Earth from 15 km to 30 km high. It's not that I have only 4.5*10^18 molecules (on column 1cm * 1cm * 15 km )

Also, why is my calculation not correct?
3√(3*10^12) = 1.44*10^3 molecules/cm
(1.44*10^3)^2 = 2.08*10^8 molecules/cm^2
and then I can:
DU = 2.08*10^8 [molecules/cm^2] / 2.69*10^16 [molecules/cm^2] ?
 
  • #10
No idea what you are squaring and rooting and what for, so it is hard to say what you are doing. But the result you would get (around 10-8 DU) seems off.

dorin1993 said:
The concentration of the molecule is given, and it's around the Earth from 15 km to 30 km high. It's not that I have only 4.5*10^18 molecules (on column 1cm * 1cm * 15 km )

There are many more molecules in the whole atmosphere, but it doesn't matter. You have calculated that you have exactly 4.5*1018 molecules over 1 square cm. You have calculated that long ago, that is already the number you are looking for and it was from the very beginning. It can be expressed in different ways (text you posted mentions thickness of a layer, which is why I asked you to use ideal equation to calculate the volume in hope it will push you in the right direction), but you already have your answer - you just need to convert number of molecules over 1 cm2 to Dobson Unit - which is a simple division.
 
  • #11
Ohh! now I get it. All the time I referred the column as a volume and no thinking at all about the base 1cm^2.
Now I realize how easy it is.
Thank you so much for halping and explaining!
 
  • #12
dorin1993 said:
Ohh! now I get it.

Good :biggrin:
 

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