# Atmospheric Pressure Above the Sea Level

1. Nov 26, 2012

### anomalocaris

1. The problem statement, all variables and given/known data

The atmospheric pressure deceases exponentially with height. At 5.5 km, the pressure is half that at sea level. At what height is the pressure one eighth that of sea level?

2. Relevant equations
All I can think of is Pat=ρgh

I know 1 atm=101325 Pa=101.325 kPa and that this is the atmospheric pressure at sea level.

3. The attempt at a solution
$\frac{101325 Pa}{2}$=ρ(9.8 m/s2)(5.5 km)
50662.5 Pa=ρ(9.8 m/s2)(5500 m)
50662.5 Pa=ρ(53900 m2s-1)
ρ=0.93993 kg/m3

I thought that in solving for density, I could change the equation to make
$\frac{101325 Pa}{8}$=(0.93993 kg/m3)(9.8 m/s2)h
and then solve for h.
h=(12665.625 Pa)/((0.93993 kg/m3)(9.8 m/s2))
h=1380 m=1.38 km

This was my homework and for the life of me I do not know how I got this right, because the answer I found above is "incorrect." The correct answer should be 16.5 km. Does anyone have any ideas as to where I am going wrong? I'm wondering if I used the wrong formula? I do not think I understand the concept so well.

Thank you!

2. Nov 26, 2012

### Staff: Mentor

The key to this problem statement is the words deceases exponentially with height. So

P = P0e-h/H

where H is the so called scale height, and is equal to the height interval over which the pressure decreases by a factor of e.

3. Nov 27, 2012

### anomalocaris

Oh okay! We didn't go over this in lecture and it is not in my text book. I'm not familiar with this formula. From what equation does one derive this?

For this particular problem, would I set it up like this:

$101325 Pa/8$=($101325 Pa/2$)e-5.5 km/H

Thank you so much for your help!

Last edited: Nov 27, 2012
4. Nov 27, 2012

### haruspex

It's what is meant by the information you were given: atmospheric pressure deceases exponentially with height. That is, there exist some constants P0 and H s.t. the pressure at height h is P0e-h/H.
No, you have no unknowns in there, and you have the relationship backwards.
Given that P(h) = P0e-h/H, what equations do you get for h=0 (sea level) and h = 5.5km?

5. Nov 27, 2012

### Staff: Mentor

I agree with haruspex on his answer to the first part of your question. As to the second part, where does the relation come from: It comes from a combination of the hydrostatic equation and the ideal gas law:

dp/dz = -ρg

where ρ is the density of air and g is the acceleration of gravity. From the ideal gas law,

ρ = (pM)/(RT)

where M is the molecular weight of air.

Combining the two equations, we get

dp/dz = -p / H

where H is called the scale height:

H = (RT)/(Mg)

In reality, since T varies with altitude z, H does also. But, often people use an average value over the troposphere to estimate how the pressure varies with altitude. Often, the scale height is taken to be constant at ~ 7 km.