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

• SirCrayon
In summary, the conversation discusses finding the average height of molecules in a box filled with classical ideal gas molecules in a constant gravitational field. The distribution of molecular height is given by f(z) = C exp (-mgz/KT), where C is a normalization constant. The conversation then goes on to discuss finding the average height, as well as the limiting values of height as the height of the box approaches 0 and infinity. The solution involves solving integrals and making approximations.
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

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.

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

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##?

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

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

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]

SirCrayon said:
(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.

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

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?

## 1. What is the purpose of finding the average height of molecules in a box?

The average height of molecules in a box is a measure of the overall distribution of the molecules within the box. It can provide valuable information about the physical properties and behavior of the molecules, such as their density and movement.

## 2. How is the average height of molecules in a box calculated?

The average height, z, of molecules in a box is calculated by taking the sum of the individual heights of each molecule and dividing it by the total number of molecules in the box. This is represented by the formula: z = (z1 + z2 + ... + zn) / n, where z1, z2, ..., zn are the individual heights of each molecule and n is the total number of molecules.

## 3. Can the average height of molecules in a box change over time?

Yes, the average height of molecules in a box can change over time as the molecules move and interact with each other. This is especially true in dynamic systems, where the molecules are constantly undergoing reactions and changes in their positions.

## 4. How does the size of the box affect the average height of molecules?

The size of the box can affect the average height of molecules in several ways. A larger box may contain a higher number of molecules, resulting in a higher average height. Additionally, a larger box may also provide more space for the molecules to move and interact, potentially affecting their average height.

## 5. What factors can influence the accuracy of calculating the average height of molecules in a box?

Several factors can influence the accuracy of calculating the average height of molecules in a box. These include the precision of the measurements used to determine the individual heights of the molecules, the number of molecules in the box, and any external factors such as temperature or pressure that may affect the behavior of the molecules.

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