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Homework Help: Kinetic Theory of Gases Average question

  1. Mar 16, 2010 #1
    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;
    [tex]change \ in \ momentum \ = \ (mv) \ - \ (-mv) \ = \ 2mv[/tex]

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

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

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

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

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

    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 :yuck: . Where did the [(2m)/l] fraction go?

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

    (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.
  2. jcsd
  3. Mar 16, 2010 #2

    Andrew Mason

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    Homework Helper

    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 [itex]mv^2/2[/itex]. 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.

  4. Mar 16, 2010 #3
    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 [Broken]
    Last edited by a moderator: May 4, 2017
  5. Mar 16, 2010 #4
    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.
  6. Mar 16, 2010 #5

    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 wont 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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