# Density of Earth's atmosphere w.r.t height

1. Sep 7, 2014

### blintaro

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

"It's possible to use the ideal gas law to show that the density of the Earth's atmosphere decreases exponentially with height, that is
ρ=ρ0exp(-z/z0)
where z is the height above sea level, ρ0 is the density at sea level and z0 is called the "scale height" of the atmosphere.

a.) Determine value of z0:

b.) What is the density of the air in Denver, at an elevation 1600 m? What percent of sea-level density is this?

2. Relevant equations
Ideal gas law: Pv=nRT

possibly P=P0 + ρgh ? Seems unlikely as ρ varies?

3. The attempt at a solution

Pv=nRT
assumed we'd be comparing two volumes of gas (at height z0 and z) with equal number of molecules, at equal temperature, thus
P1v1=P2v2
Then substituted v=ρ(mass)
Again assuming equal mass would imply
P1ρ1=P2ρ2

Not really sure how to proceed to involve height instead of pressure or even if on right track... Somehow the latter seems more likely. Help would be appreciated!

2. Sep 7, 2014

### Staff: Mentor

What is your understanding of the definition of the word density?

Chet

3. Sep 7, 2014

### blintaro

Density is mass/volume?

4. Sep 7, 2014

### Staff: Mentor

Right, mass per unit volume. From the ideal gas law, if the molecular weight of the gas is M, can you write an equation for the density of the gas as a function of T, P, M, and R?

Chet

5. Sep 9, 2014

### blintaro

n=mass/molecular weight = m/M

PV=(m/M)RT
=>PM/(RT)=m/V=ρ

But how does ρ vary with height? Can I use P=P0 + ρgz? But then I would be plugging ρ into an expression for ρ right?

6. Sep 9, 2014

### Staff: Mentor

Correct!

No. You have to use the differential version of the equation:

dP/dz=-ρg =-(PMg)/(RT)

Do you know how to solve this differential equation for P?

Chet

7. Sep 9, 2014

### blintaro

dP/dz=-(PMg)/(RT)

∫(P)-1dP=∫(-Mg)/(RT)dz

ln(P/P0)=(-Mg)/(RT)(z-z0) + c

(ρ)/(ρ0) =C(ez/ez0)(-Mg)/(RT)

ρ=ρ0C(ez/ez0)(-Mg)/(RT)

Did I do that right?

8. Sep 9, 2014

### Staff: Mentor

Almost. The boundary condition should be P=P0 and ρ=P0M/(RT) at z = 0.

Chet

9. Sep 10, 2014

### blintaro

I'm not quite sure I follow...

pp0(P-1)dP = (-Mg)/(RT)∫z0dz

ln(P/P0)=(-Mg)/(RT)z

ρ0=(P0M)/(RT)
=>P0=(ρ0RT)/M

ρ=(PM)/(RT)
=>P=(ρRT)/M

Then would (RT)/M cancel on both top and bottom?
ln(P/P0) = ln((ρ)[STRIKE]((RT)/M)[/STRIKE]/((ρ0)[STRIKE](RT)/M))[/STRIKE]

leaving
ln(ρ/ρ0)=(-Mgz)/(RT)

10. Sep 10, 2014

### Staff: Mentor

Yes. This is the correct answer. From this result, you should be able to back out an equation for zo, by comparing with the original equation in the problem statement .

Chet

11. Sep 10, 2014

### blintaro

I just realized that I had read the original equation wrong, I thought ρ=ρ0exp(-z/z0) was ρ=ρ0-z/z, not ρ=ρ0(e-z/z) Whoops, lots of unnecessary headaches there!

So then

ρ=ρ0(e(-Mg)/(RT))-z

and z0 = (RT)/(Mg), great, thanks for helping me out with this. :)