Answer
The proportion of oxygen to nitrogen is lower, in a higher-altitude situation.
Work Step by Step
Example 18.4 calculates the variation of atmospheric pressure with elevation. The pressure p is given as the following:
$$p=p_o e^{\frac{-Mgy}{RT}}$$
Here, M represents the molar mass of the gas. Oxygen’s molar mass is greater than that of nitrogen’s, so the pressure dropoff will be steeper for oxygen, with increasing height y.