1. At the beach in San Francisco (0 meters) the pressure of the atmosphere is 101.325 kPa

Question

1. At the beach in San Francisco (0 meters) the pressure of the atmosphere is 101.325 kPa (kilopascals) and in Denver, 1609.344 meters above sea level, the pressure of the atmosphere is about 83.437 kPa. Using this data, find a linear equation for pressure P in terms of altitude h. (Hint: write the pressure and altitude in each location as a point (h, P). Then use point-slope form to find the equation of the line.)

2. What is the rate of change of the pressure of the atmosphere as altitude increases in meters? Write a sentence answering this question using the phrase “rate of change.”

3. Mount Everest is 8848 meters high. What does your linear approximation predict for the pressure of the atmosphere on Mount Everest?

4. Compared to San Francisco, how much oxygen is available in a breath of air in Denver? Since 21% of the molecules in the air are oxygen molecules in either situation, we can just compute the ratio P(Denver)P(SF).

5. Using the same technique as question 3, express as a percentage how much oxygen is available in a breath on Mount Everest as compared to San Francisco.

6. Let’s do a very basic check of whether our linear model is good. What would it mean for the pressure of the atmosphere P(h) at an altitude h to equal zero?

7. Does our model indicate that P(h) will equal zero at some altitude? If so, P(h)=0 at what altitude?

8. The air pressure at the peak of Mount Everest is actually closer to 31.5 kPa. Draw some

conclusions about the validity of our model.

Answer 1

1. The two points are San Francisco (0, 101.325) and Denver (1609.344, 83.437). The slope is

m = (83.437 – 101.325) ÷ (1609.344 – 0) ≈ –0.01111.

So the linear equation is:

P(h) = 101.325 – 0.01111h

1. The rate of change of atmospheric pressure with respect to altitude is about –0.011 kPa per meter. This means that for every meter you ascend, the pressure decreases by about 0.011 kPa.
2. On Mount Everest (8848 m): P(8848) = 101.325 – 0.01111(8848) ≈ 3.07 kPa.
3. Compared to San Francisco, the oxygen in Denver is given by the ratio P(Denver)/P(SF) = 83.437 ÷ 101.325 ≈ 0.823. This means Denver has about 82.3% as much oxygen per breath as San Francisco.
4. On Mount Everest, the ratio is P(Everest)/P(SF) = 3.07 ÷ 101.325 ≈ 0.0303. This means only about 3.0% as much oxygen per breath as San Francisco, according to the linear model.
5. If P(h) = 0, that would mean there is no atmosphere at all, or a complete vacuum.
6. Solving for P(h) = 0 gives h = 101.325 ÷ 0.01111 ≈ 9119 m. So our model predicts zero pressure at about 9119 meters.
7. In reality, the pressure on Mount Everest is about 31.5 kPa, not 3.07 kPa. This shows that the linear model is not valid at very high altitudes. Atmospheric pressure actually decreases exponentially, so the linear model only works for small altitude changes like between San Francisco and Denver.