Relative concentration vs. altitude

Thread Starter

boks

Joined Oct 10, 2008
218
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?
 
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BillO

Joined Nov 24, 2008
999
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.
 
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
 

Thread Starter

boks

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

 
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