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

[blintaro]

Homework Statement

"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?

Homework Equations

Ideal gas law: Pv=nRT

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

The Attempt at a Solution

Not quite sure how to go about this one. Started with ideal gas law
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!

 

[Chestermiller]

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

Chet

 

[blintaro]

Density is mass/volume?

 

[Chestermiller]

Density is mass/volume?

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

 

[blintaro]

Hi, sorry for late reply:

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?

 

[Chestermiller]

Hi, sorry for late reply:

n=mass/molecular weight = m/M

PV=(m/M)RT
=>PM/(RT)=m/V=ρ

Correct!

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?

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

 

[blintaro]

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

∫(P)^-1dP=∫(-Mg)/(RT)dz

ln(P/P₀)=(-Mg)/(RT)(z-z₀) + c

(ρ)/(ρ₀) =C(e^z/e^z₀)^(-Mg)/(RT)

ρ=ρ₀C(e^z/e^z₀)^(-Mg)/(RT)

Did I do that right?

 

[Chestermiller]

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

∫(P)^-1dP=∫(-Mg)/(RT)dz

ln(P/P₀)=(-Mg)/(RT)(z-z₀) + c

(ρ)/(ρ₀) =C(e^z/e^z₀)^(-Mg)/(RT)

ρ=ρ₀C(e^z/e^z₀)^(-Mg)/(RT)

Did I do that right?

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

Chet

 

[blintaro]

almost. The boundary condition should be p=p₀ and ρ=p₀m/(rt) at z = 0.

Chet

I'm not quite sure I follow...

∫^p_p0(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)

 

[Chestermiller]

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

 

[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. :)