# Relative concentration vs. altitude

Discussion in 'Physics' started by boks, Jan 30, 2009.

1. ### boks Thread Starter Active Member

Oct 10, 2008
218
0
Earth's atmosphere has roughly four molecules of nitrogen for every oxygen molecule at sea level;
more precisely, the ratio is 78:21. Assuming a constant temperature at all altitudes (not really very
accurate), what is the ratio at an altitude of 10km? Explain why your result is qualitative reasonable.

Hint; This problem concerns the number density of oxygen molecules as a function of height. The
density is related in a simple way to the probability that a given oxygen molecule will be found at a
particular height. You know how to calculate such probabilities.

Any idea how to solve this?

Last edited: Jan 31, 2009
2. ### BillO Distinguished Member

Nov 24, 2008
985
136
At first glance, I'd be tempted to treat them like ideal gases, then use the ideal gas law and partial pressures to calulate the change in concentration with altitude.

3. ### boks Thread Starter Active Member

Oct 10, 2008
218
0
$PV = N k_B T$, right?

4. ### Alexei Smirnov Active Member

Jan 7, 2009
43
1
Use barometric formula at constant T for each gas separately:

PN2(h)=PN2(0)*exp(-MN2*h*A)
PO2(h)=PO2(0)*exp(-MO2*h*A)

where A = g/kT (constant).

Pressure ratio at height h is:
PN2(h) / PO2(h) = PN2(0) / PO2(0) * exp( -(MN2-MO2)*h*A )

The exponent with mass difference is responsible for pressure (or number of molecules) ratio change with altitude.

Thanks,
Alexei

5. ### triggernum5 Active Member

May 4, 2008
216
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Does the ratio simply vary with the specific gravities?

6. ### boks Thread Starter Active Member

Oct 10, 2008
218
0
I found this formula, describing nanoparticles in air. I guess it then also can be used for gas molecules?

Last edited: Feb 1, 2009

Jan 7, 2009
43
1
and height