Kinetic Theory of Gases Average question

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Homework Help Overview

The discussion revolves around the Kinetic Theory of Gases, specifically focusing on the concept of root mean square velocity and its implications in understanding molecular behavior and energy in gases. The original poster expresses confusion regarding the mathematical definition and application of root mean square velocity in the context of momentum change and kinetic energy.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to understand the transition from individual particle momentum to the average behavior of gas molecules, questioning the mathematical steps involved in defining root mean square velocity. Some participants provide insights into the relationship between kinetic energy and average energy, while others express skepticism about the mathematical manipulation involved.

Discussion Status

Participants are exploring the mathematical definitions and physical interpretations of root mean square velocity. Some guidance has been offered regarding the relationship between average energy and temperature, while others are questioning the reasoning behind certain mathematical steps. The discussion reflects a mix of understanding and confusion, with no explicit consensus reached.

Contextual Notes

The original poster mentions a lack of familiarity with probability concepts, which may be influencing their understanding of the mathematical definitions presented in the problem. There is also an acknowledgment of the complexity of the topic as it relates to kinetic theory and gas behavior.

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Hey I'm studying the Kinetic Theory of gases and although it's in a chemistry
book it employs physics concepts ergo I have come here to ask you nice people :biggrin:

The 5th point is where I have a problem, the rest is just information for you.

I am assuming a particle is in a cube (ideal situations here) & it traverses a
distance l with a momentum p=mv.

1. As it travels forward it will hit the cube wall and the resultant change in
momentum will be;
change \ in \ momentum \ = \ (mv) \ - \ (-mv) \ = \ 2mv

2. The number of collisions per second will be;
No.\ of \ collisions \ per \ second \ = \ \frac{velocity}{distance} \ = \ \frac{v}{l}

3. The Change in momentum per second will be;
Change \ in \ momentum \ per \ second = 2mv \frac{v}{l} \ = \ \frac{2mv^2}{l}

4. I've assumed one particle in one direction. Assuming n particles with
a total change in momentum per second of;
total \ change \ in \ momentum \ per \ second \ = \ \frac{2m}{l} \ \displaystyle\sum_{i=1}^n v_i^2

5. Here is where I have the problem. The text says to do this next;

Define the root square mean velocity, \overline{v} such that;
\overline{v}^2 \ = \ \left( \ \frac{1}{n} \ \sum_{i=1}^n v_i^2 \right)

What does this mean? Is this some mathematical concept from probabilities? I haven't done probabilities in math in like 5 years & even then that was in school . Where did the [(2m)/l] fraction go?

Using this concept the book goes on to show that;
The \ total \ change \ in \ momentum \ = \ [2mn \overline{v}^2 ] / l

(that is a fraction with l in the denominator, all the rest in the numerator [latex issues] )
I can't understand this (or any of the further concepts on pressure etc...) without getting this concept.
 
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The root mean square velocity is useful because it is the speed of a molecule that has an energy equal to the average energy of all the molecules in the gas. This is because the kinetic energy of a molecule is mv^2/2. If you add up all the squares of all the velocities of the molecules. multiply by m/2 and divide by n, you will get the average energy of the gas molecules. This is useful because the temperature of the gas is directly related (proportional to) the average energy of the gas molecules.

AM
 
I don't see any logic nor reason for doing that mathematically.

I understand that you are converting to using kinetic energy but I mean, it seems as though you're just arbitrarily multiplying one side by m/2 then dividing by n.

What about multiplying both sides of the equation?
Or multiplying top and bottom by a clever choice of 1?

http://www.betz.lu/media/users/charel/math07.gif
 
Last edited by a moderator:
2mv^2/l gives the force for just a single molecule of velocity v.If there are n such molecules all moving with the same velocity the total force would be n*2mv^2/l.The chance of the molecules all moving with the same velocity is very remote so instead of using v we use v bar(the mean squared velocity)which leads to the final equation you wrote.The 2m/l fraction hasn't gone anywhere.
 
OH!

God it's so easy lol...

I must be so tired that it's obviously affecting my study @ this stage.

So the average velocity is just the sum of the velocities of all of the particles divided by the number of particles.

Great stuff, it makes even more sense conceptually to use this seeing as they won't all be hitting the walls at the same time. yes, thanks a lot. :)

------have to stop being scared away by sigma's :p-----
 

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