# Ratio of oxygen in atmosphere

1. Mar 2, 2012

### SoggyBottoms

1. The problem statement, all variables and given/known data
Air at sealevel is composed of 80% nitrogen and 20% oxygen, but at the top of Mount Everest the composition of air is different. Give an expression for the ratio ($f_O$) of oxygen in the air as a function of height z above sealevel, assuming the temperature T is constant and the air is an ideal gas. The gravitational acceleration is g and the mass of a molecule of nitrogen is $m_N$ and the mass of a molecule of oxygen is $m_O$.

3. The attempt at a solution

The potential energy of a molecule of oxygen is $m_O g z$, so the probability of finding it at height z is: $P_{mO}(z) \propto e^{-\frac{m_O g z}{k_B T}}$. Similarly for nitrogen: $P_{mN}(z) \propto e^{-\frac{m_N g z}{k_B T}}$. So the ratio is simply $f_O = \frac{e^{-\frac{m_O g z}{k_B T}}}{e^{-\frac{m_N g z}{k_B T}}} = e^{-\frac{(m_O - m_N)gz}{k_B T}}?$

2. Mar 3, 2012

### D H

Staff Emeritus
No, it isn't. The troposphere is well mixed. That's why it's called the troposphere. Is this a problem from a physics text, or one that your teacher made up?

I'll ignore this minor problem.

That gives a ratio of one at sea level.

Hint: Assume each gas is individually in hydrostatic equilibrium.

3. Mar 3, 2012

### SoggyBottoms

Just a question my teacher came up with.

The hydrostatic equation is $dp = -\rho g dz$ with $dp = k_B T dn$ and $\rho = nm$:

$$\frac{dn}{n} = -\frac{m_O g}{k_B T}dz, \Leftrightarrow \\ \ln{\frac{n}{n_0}} = -\frac{m_O g }{k_B T}z \Leftrightarrow \\ n = n(0) e^{-\frac{m_O g }{k_B T}z}$$

If I do the same thing for nitrogen and divide again I get the same answer, so I'm not sure what to do...

4. Mar 6, 2012

Anyone?

5. Mar 6, 2012

### D H

Staff Emeritus
Correct.

You do not get the same answer. Oxygen and nitrogen have different molar masses.

Try computing nO(z)/ nN(z), the ratio of the oxygen density to nitrogen density. What does that tell you?

6. Mar 6, 2012

### SoggyBottoms

Since $n_n = n(0) e^{-\frac{m_N g }{k_B T}z}$ I get $\frac{n_O}{n_N} = \frac{n_0 e^{-\frac{m_O g }{k_B T}z}}{n_0 e^{-\frac{m_N g }{k_B T}z}} = e^{-\frac{(m_O - m_N) g }{k_B T}z}$...or am I missing something? Because that's the same thing I got in my original post.

7. Mar 7, 2012

### D H

Staff Emeritus
You are assuming that oxygen and nitrogen have the same density at sea level (i.e., the same n0. They obviously don't since the the atmosphere at sea level is 80% nitrogen, 20% oxygen.

8. Mar 7, 2012

### SoggyBottoms

You're right, so I should get $\frac{0.8}{0.2}e^{-\frac{(m_O - m_N)g}{k_B T}z} = 4 e^{-\frac{(m_O - m_N)g}{k_B T}z}$.

9. Mar 7, 2012

### D H

Staff Emeritus
Correct. So now it's just a matter of plugging in the numbers with z=height of Everest and then converting the resulting nitrogenxygen ratio to the percentage of the air at the top of Everest that is oxygen.

The result will be wrong (wrong in reality that is; the result will be correct for this homework problem). The makeup is still more or less 80%/20% ratio at the top of Everest. The troposphere is just too well-mixed for any disparities to show up. The upper reaches of the atmosphere (above the stratosphere) do exhibit this differentiation.