- #1

Ghost Repeater

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## Homework Statement

Consider a cylindrical parcel of air of area A and infinitesimal height dz. If this air parcel is to remain stationary, the difference between the total pressure forces exerted on its top and bottom faces must be equal to its weight. Use this information and the ideal gas law to show that

$$\frac{dP}{dz} = - \frac{mg}{kT} P$$

where m is the mass of an air molecule, g is the gravitational field strength, k is Boltzmann's constant, T is the absolute temperature, and P is the air pressure at the vertical position z corresponding to the center of the particle.[/B]

## Homework Equations

$$PV = NkT$$

$$\rho = \frac{M}{V}$$[/B]

## The Attempt at a Solution

I'm calling the pressure difference between top and bottom dP. Since this has to equal the weight of the cylinder, I have

$$dP = -W_{cyl} = -Mg$$

where M is the weight of all the air in the cylinder. I can express this in terms of the density of air as

$$dP = -\rho_{air} V g$$

where V is the volume of the cylinder.

This in turn can be written in terms of the given mass m of a single molecule:

$$dP = -m \frac{N}{V} Vg$$

I can write the volume of the cylinder differentially as V = Adz, so now I have:

**$$\frac{dP}{dz} = -m \frac{N}{V} A g$$**

Finally, using the IGL, I can write

$$\frac{N}{V} = \frac{P}{kT}$$.

So I have

$$\frac{dP}{dz} = -m\frac{P}{kT}A g$$[/B]

So you see my problem: I have this factor of A still hanging around and I can't find a way to get rid of it. I don't see the flaw in my algebra, but physically it's obvious that the area of the cylinder can't make any difference to the pressure gradient!

So you see my problem: I have this factor of A still hanging around and I can't find a way to get rid of it. I don't see the flaw in my algebra, but physically it's obvious that the area of the cylinder can't make any difference to the pressure gradient!