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Homework Help: Kinetic-Molecular Model of Ideal Gas: vrms/vav

  1. Feb 13, 2006 #1

    This should be a pretty simple problem to answer... I'm just a bit confused on this, and want to make sure I'm right. It's an easy problem:

    Molecules in a gas can only move in the x direction (i.e., [tex]v_{y}=v_{z}=0)[/tex]. You set up an experiment in which you measure the velocity of a few molecules and the result that you obtain is the following (expressed in m/s):

    2, -4, 6, 1, -3, -2, -5, 2, -1, 4, 3, -5

    Calculate: a) the average x-component of the velocity [tex](v_{x})_{av}[/tex], b) the average speed [tex](v)_{av}[/tex], and c) the root mean square of the velocity [tex]v_{rms}[/tex]

    For a), the x-component of velocity is literally just the average, right? No absolute values b/c we're not talking about speed here.

    For b) because I'm being asked for the average speed, here is where I take the absolute values of all of these and average them together, right?

    For c) This is where I'm most confused... Here, wouldn't I just square what I got for b) and then take the square root of it? That seems to make absolutley no sense. Would I then use the formula below?

    I noticed something in the book: [tex](v_{x}^2)_{av}, (v_{y}^2)_{av}, (v_{z}^2)_{av}[/tex] must all be equal. Hence: [tex](v_{x}^2)_{av} = \displaystyle{\frac{1}{3}}(v^2)_{av}[/tex]

    This wouldn't apply for this situation, correct? As the y and the z components are 0, right?
    Last edited: Feb 13, 2006
  2. jcsd
  3. Feb 13, 2006 #2
    Its been a while since I did some gas theory, but isn't the rms of the velocities, simply [tex]v_{rms} = \frac{A}{\sqrt{2}} [/tex] (A = Ans to part b.)
    Don't quote me on that, I'm not sure, but I think its right since rms voltage in electronics is defined in a similar way.
    Last edited by a moderator: Feb 13, 2006
  4. Feb 13, 2006 #3
    Thanks for responding, heh

    I don't quite understand why you get that, could you explain it further?
    (I'm not doubting you, I just don't understand)

    Thanks again
  5. Feb 13, 2006 #4
    Bump, please...?
  6. Feb 13, 2006 #5
    Would this be the same as the root-mean-square in mathematics?


    So my problem would look like this:

    Is this correct?
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