Finding the average height, z, of molecules in the box

[SirCrayon]

Homework Statement

Let a box of height h be filled with classical ideal gas molecules of mass m in a constant gravitational filed g. As will be shown later, the distribution molecular height z obeys:

f(z) = C exp (-mgz/KT)

Where C is the normalization constant
a) find the average height z, of molecules in the box
b) find the limiting value of z when h->0
c) find the limiting value of z when h->infinitiy

The Attempt at a Solution

I am a little confused as to where to begin, should i start with determining C?

I started with:

Area * Height * average_density = number of molecules
= Area * integral (z = 0, H [constant*exp(-mgz/(kT))] dz
= Area*constant*(kT/(mg))*integral(0,mgH)/(kT) [exp(-u)]du
= Area*constant*(kT/(mg))*(1-exp(-mgH/(kT))

Therefore,
number = N = Area*constant*(kT/(mg))*(1-e^(-mgH/(kT))

So:
Constant = N*(mg/(kT))/(Area*(1-e^(-mgH/(kT))

Am i on the right track? thanks in advance

 

[haruspex]

It doesn't ask you to find C, so I think you can just use C as given and not bother with determining it. You can only find it by knowing N, which is not given.

 

[SirCrayon]

Thanks,

I think i have figured out a) with the below:
<z> = ∫ z f(z) dz / ∫ f(z) dz

= ∫ z exp(-mgz/(kT)) dz / ∫ exp(-mgz/(kT)) dz

where the integrations are from z = 0 to z = h.

I am pretty stumped on b and c though

 

[haruspex]

b) and c) as posted make no sense. It would make sense to ask for limiting values of the average height. Perhaps the original question has $\bar z$?

 

[SirCrayon]

Yes sorry, all the z's in a,b,c are z¯

 

[haruspex]

Ok, so did you solve the integrals and get an answer for (a)?

 

[SirCrayon]

Yes I have,

<z> = ∫ z f(z) dz / ∫ f(z) dz

= ∫ z exp(-mgz/(kT)) dz / ∫ exp(-mgz/(kT)) dz

= (kT)^2/(mg)^2 * (e^(-mgh/kT)*(-mgh/kT - 1) - 1)) / -(kT/mg)[(e-mg/h/KT)-1]

 

[haruspex]

(kT)^2/(mg)^2 * (e^(-mgh/kT)*(-mgh/kT - 1) - 1)) / -(kT/mg)[(e-mg/h/KT)-1]

In LaTex:
$\frac{\left(\frac{kT}{mg}\right)^2\left(e^{-\frac{mgh}{kT}}\left(-\frac{mgh}{kT} - 1\right) - 1\right) }{ -\frac{kT}{mg}\left(e^{-\frac{mgh}{kT}}-1\right)}$
Right? But I think you have a sign wrong - pls double check.
Some cancellation would be in order.
So b asks for the limit of the above as h tends to 0.

 

[SirCrayon]

I can't seem to find the sign that's wrong, the only thing that I can reduce is the (KT/mg)^2 right? I can't figure out how to cancel anything else.

Setting h=0 should give me the term i need and same with inifinity for c). Thanks. But i can't seem to figure out the above

 

[haruspex]

I believe it should be $\frac{\left(\frac{kT}{mg}\right)^2\left(e^{-\frac{mgh}{kT}}\left(-\frac{mgh}{kT} - 1\right) +1\right) }{ -\frac{kT}{mg}\left(e^{-\frac{mgh}{kT}}-1\right)}$, i.e. $\frac{kT}{mg}\frac{1-e^{-\frac{mgh}{kT}}\left(1+\frac{mgh}{kT} \right) }{1 -e^{-\frac{mgh}{kT}}}$
What's an approximation for e^-x for small x?