Dobson unit - caclulation problem

  • Thread starter dorin1993
  • Start date
  • Tags
    Unit
In summary, the number of molecules in the atmosphere at a uniform concentration of 3*10^12 molecules/cm^3 between 15 km and 30 km is 4.5*10^18 molecules.
  • #1
dorin1993
11
0
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
 
Physics news on Phys.org
  • #2
How many molecules in a 15 km column with a 1 square cm base?
 
  • #3
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:
  • #4
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
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
You already did. Think it over. What is the base of the column?

Yes, it was that simple from the very beginning.
 
  • #7
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:
  • #8
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
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:
 

1. What is a Dobson unit?

A Dobson unit is a unit of measurement used to describe the concentration of ozone in the Earth's atmosphere. It is equal to the number of molecules of ozone that would be required to create a layer of pure ozone 0.01 millimeters thick at a temperature of 0 degrees Celsius and a pressure of 1 atmosphere.

2. How is a Dobson unit calculated?

A Dobson unit is calculated by measuring the total amount of ozone in a vertical column of the Earth's atmosphere, from the surface to the edge of the stratosphere. This measurement is then divided by the surface area of the column and expressed in units of DU (Dobson units).

3. What is the purpose of calculating Dobson units?

Calculating Dobson units is important for monitoring the health of the Earth's ozone layer. It allows scientists to track changes in ozone levels over time and assess the effectiveness of measures taken to reduce ozone depletion.

4. How do scientists measure Dobson units?

To measure Dobson units, scientists use instruments called Dobson spectrophotometers. These instruments use ultraviolet light to measure the amount of ozone in a vertical column of the atmosphere. The data collected by these instruments is then used to calculate the total amount of ozone in the column.

5. Can Dobson units be converted to other units of measurement?

Yes, Dobson units can be converted to other units of measurement, such as parts per million (ppm) or molecules per square centimeter. However, these conversions are not exact and may vary depending on factors such as temperature and pressure. It is important to use consistent units when comparing ozone measurements.

Similar threads

  • Biology and Chemistry Homework Help
Replies
3
Views
2K
  • Biology and Chemistry Homework Help
Replies
8
Views
2K
  • Biology and Chemistry Homework Help
Replies
2
Views
11K
  • Introductory Physics Homework Help
Replies
12
Views
664
  • Introductory Physics Homework Help
Replies
6
Views
5K
  • Biology and Chemistry Homework Help
Replies
4
Views
13K
  • Calculus and Beyond Homework Help
Replies
30
Views
3K
  • Introductory Physics Homework Help
Replies
15
Views
1K
  • Biology and Chemistry Homework Help
Replies
6
Views
6K
Replies
3
Views
5K
Back
Top