# Homework Help: Dobson unit - caclulation problem

1. Nov 14, 2012

### dorin1993

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

2. Nov 14, 2012

### Staff: Mentor

How many molecules in a 15 km column with a 1 square cm base?

3. Nov 14, 2012

### dorin1993

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 [Broken]

http://imageshack.us/a/img29/5307/13079304.jpg [Broken]

Last edited by a moderator: May 6, 2017
4. Nov 14, 2012

### Staff: Mentor

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?

5. Nov 14, 2012

### dorin1993

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?

6. Nov 14, 2012

### Staff: Mentor

You already did. Think it over. What is the base of the column?

Yes, it was that simple from the very beginning.

7. Nov 14, 2012

### dorin1993

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: Nov 14, 2012
8. Nov 15, 2012

### Staff: Mentor

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.

9. Nov 15, 2012

### dorin1993

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. Nov 15, 2012

### Staff: Mentor

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.

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. Nov 15, 2012

### dorin1993

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. Nov 15, 2012

Good