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

• blintaro
In summary, the density of the Earth's atmosphere decreases exponentially with height, according to the ideal gas law.
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

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!

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

Chet

Density is mass/volume?

blintaro said:
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

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?

blintaro said:

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

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?

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

Chet

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

Chet

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)

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

1 person
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. :)

## What is the density of Earth's atmosphere at sea level?

The density of Earth's atmosphere at sea level is approximately 1.2 kg/m^3. This means that for every cubic meter of air at sea level, it has a mass of 1.2 kilograms.

## How does the density of Earth's atmosphere change with increasing height?

The density of Earth's atmosphere decreases with increasing height. This is because the weight of the air above decreases as you move higher in the atmosphere, resulting in lower air pressure and therefore lower density.

## What is the average density of Earth's atmosphere?

The average density of Earth's atmosphere is approximately 0.9 kg/m^3. This takes into account the varying density at different heights in the atmosphere.

## How does temperature affect the density of Earth's atmosphere?

Temperature has a direct effect on the density of Earth's atmosphere. As the temperature of air increases, its density decreases. This is because the molecules in warmer air have more energy and are able to spread out, resulting in lower density.

## How is the density of Earth's atmosphere measured?

The density of Earth's atmosphere is typically measured using a device called a barometer. This measures the air pressure, which can then be used to calculate the density using the ideal gas law. Other methods include using satellites to measure the thickness of the atmosphere and its composition.

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