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Topic: Enrichment
Replies: 38   Last Post: Aug 24, 2012 1:33 PM

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Peter Duveen

Posts: 163
From: New York
Registered: 4/11/12
Re: Enrichment
Posted: Aug 21, 2012 11:15 PM
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Well, Bob, I was hoping you would ask about the temperature, and how it is related to the formula .

First let me derive the formula.

Imagine a tiny cube of breadth d filled with a gas at pressure P and density p.

Divide the molecules of the gas, considered equal in mass, into three groups, each of which bounces back and forth between the two parallel sides of the cube, each traveling at the mean velocity v.

Let's focus on one molecule going between one pair of the three pairs of parallel sides of the cube.

The mass of the molecule is m. It hits one side and bounces off, resulting in a change of momentum of 2 mv. We also have the time between collisions of the molecule on that one side, because it has to go back and forth each time. we have vt = 2d, or t = 2d/v.

Now if we divide the change in momentum by the change in time, we get the average force exerted on the one face by this single molecule. It is mv^2/d. If this is in turn divided by the area of a face of the cube, we have the pressure exerted on one face by that molecule, or


Now if there are N molecules in the cube, we have divided them all into three groups of N/3 molecules, each group bouncing between parallel sides of the cube. Thus, the pressure on one of the faces of the cube is N/3 times the pressure of one molecule, which we have already calculated, so it is N/3 mv^2/d^3 = P. But Nm is just the entire mass of gas in the cube, and that divided by d^3 is merely the density, which we shall call "p". So we finally have 1/3 pv^2 = P, or v^2 = 3P/p.

You will notice that we have arrived at a rather important result using simple algebra. It is important because it says the microscopic characteristic of the molecular world, namely, the velocity of molecules, is related to the macroscopic quantities of the gas, P and p, which are fairly easily measured.

Now where does the temperature come in? We use the conventional formula for the state of an ideal gas, PV = KT, where P is the pressure, V is the volume, K is a constant of proportionality, and T is the absolute temperature, of a sample of gas. K is specific to the sample.

Simply solve for P, and put the value in our first formula. That is, P = KT/V

Replacing P in our first formula, we have:

v^2 = 3 KT/Vp

Now note that Vp is nothing more than the entire mass of the sample of gas. Moving that to the left side of the equation, we have:

mv^2 = 3KT, or

1/2mv^2 = 3/2 KT

>From this equation, we can conclude that the kinetic theory of gases implies that the absolute temperature is directly proportional to the kinetic energy of the molecules.

I erred in saying "speed" in my previous post. I should have said "kinetic energy."

So, using simple algebra, and starting from first principles, we have shown that the molecular theory of gases implies that the mean velocity of the molecules is given by v = sqrt 3P/p, and that the temperature is proportional to the kinetic energy of the molecules. We have thus connected our microscopic theory to important macroscopic properties--tangible properties that can be measured.

Bob, I hope this clarifies what I was referring to in my previous reply.

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