2
$\begingroup$

Given the minimal temperature measured in the Earth's atmosphere is not $0\space K$, air molecules are moving. I see no reason for air molecule to move all at the same speed. The question is restricted to the absolute value of speed (I know that if there is no wind, the algebraic sum of air molecule velocity is zero).

My question is: what is the distribution of air molecule speed (at sea level)?

I'm looking for a density graph or histogram in which I can see for each speed the proportion of air molecule going at that speed. I imagine this distribution don't change with altitude (there is just less air molecule) but I may be mistaken.

Bonus question: Is this distribution specific to Earth's atmosphere? (I'm thinking of other gas such as other planet's atmosphere, a volume of gas in an hermetic box in a lab,...)?

Improve this question
$\endgroup$
1

1 Answer 1

Reset to default
4
$\begingroup$

The Maxwell-Boltzmann graph represents just this. $\hskip2in$Maxwell boltzmann
Source: (wewwchemistry.com)

It can represent different temperatures too. You see a probability density (simplified here to # of molecules) against speed or kinetic energy. The area underneath gives the proportion of molecules at a speed range.

With altitude changes, the pressure would change. The distribution is only dependant on temperature and the mass of molecules. Consider the ideal gas law $PV=nRT$ for this. Higher altitude = less pressure = less temperature, so the graph would look more like $T_1$. The atmosphere is not related to this in an ideal model, but would have an effect practically in the real world.

If you would like to know more, the root-mean-square (kind of like average) velocity of gas molecules is given by $v_{rms}=\sqrt{\frac{3kT}{m}}$. This only depends on the mass of a molecule ($m$), temperature ($T$), and the Boltzmann constant ($k$).

Plus: Graph is commonly used in chemistry to see if a reaction will happen on its own, which is what the $E_a$ (activation energy) is about.

Improve this answer
$\endgroup$
5
  • $\begingroup$ The graphs do not look as though the areas under them are equal? $\endgroup$
    – Farcher
    Jul 16, 2019 at 21:56
  • $\begingroup$ @Farcher they never claimed to represent the same gas, but if they did, the total area under should actually be the same. If you're talking about the shaded area, that's about the proportion of molecules with $E_k>E_a$ which is about the chemistry bit I mentioned and is meant to be different depending on temperature. $\endgroup$
    – KingMongolian
    Jul 16, 2019 at 21:59
  • $\begingroup$ The site from which the graphs came from indicates that it is the same gas sample at different temperatures. I think you could have found a better example? $\endgroup$
    – Farcher
    Jul 16, 2019 at 22:02
  • $\begingroup$ @Farcher They should be the same area and they look pretty close. These graphs are always pretty deceiving. In fact, I calculated the area with MATLAB out of curiosity and the difference is about $0.07\%$, negligible I believe. $\endgroup$
    – KingMongolian
    Jul 16, 2019 at 22:12
  • $\begingroup$ I was curious what the velocity of diatomic nitrogen at 20oC (293K) would be and came up with 416 m s-1 Does that sound right? Calculation is:((2*1.38e-23*293)/4.667e-26)**0.5 Where 4.667e-26 is the mass in kg of a molecule of nitrogen given by: 2*(14/6e23)/1000 $\endgroup$
    – AJP
    Jul 13 at 15:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.