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Understanding Kinetic Molecular Theory and Graham's Law

  1. Mar 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Ok, firstly, I apologise for posting something which is probably trivial to any physics student, but my understanding of physics is pretty poor, so baby steps would be appreciated!

    2. Relevant equations

    PV=nRT
    E = 1/2 mv2

    3. The attempt at a solution

    My understanding of it so far:

    At the molecular level, a point particle of an ideal gas (ignoring rotational/vibrational components and forces between particles) has kinetic energy equal to:
    E = 1/2 mv2

    But particles are constantly colliding with each other as well as the walls of the container. Hence an expression for average kinetic energy tells us more information. My course notes give this as:
    Ebar = 1/2 mv(bar)-2

    an explanation as to what that means and how they got there would be nice :S

    And then there's another jump to average molar kinetic energy of an ideal gas, which is given in a different form by different sources. If I have reasoned correctly, the form in my notes for Eaverage, molar = 3/2 RT = 3/2 PV when n=1 in the ideal gas equation.

    but how did they jump from 1/2 mv2 to this? No doubt avogadro's constant comes into play, but its clearly not as simple as taking the second expression and multiplying.

    And finally, the jump to graham's law. I understand and accept that kinetic energy is only dependant on the temperature; hence two gases at equal temperatures have equal kinetic energy.

    If the rate of diffusion/effusion is a velocity, then how did we get to ratex = constant/sqrt(molar massx)?

    thanks!
     
    Last edited: Mar 25, 2010
  2. jcsd
  3. Mar 25, 2010 #2

    Char. Limit

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    Gold Member

    There's a property of the electron called the Root-mean-square speed. The formula for root-mean-square speed is something like this:

    [tex]v_{rms} = \sqrt{\frac{3RT}{M_m}}[/tex]

    Where v is the rms speed, M_m is the molar mass of the gas, T is temperature, R is 8.314, and 3 is 3.

    Plug the rms speed into the kinetic energy of a particle equation and you should get the average kinetic energy of an ideal gas. But don't trust me. (seriously, don't.) Try it.
     
  4. Mar 27, 2010 #3
    Ok i figured half of it out, it was meant to be a v with a bar instead of v^-2 for the second equation. gah, i feel stupid now.

    And with the rms equation, is the constant always 8.314? Would we ever need to use the other versions of the gas constant?
     
  5. Mar 28, 2010 #4

    Char. Limit

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    Well, the constant is always 8.314, and here's the reason...

    The units of the radicand are ...

    [tex](\frac{J}{mol-K})(K)(\frac{mol}{g}) =
    \frac{J-mol-K}{mol-K-g} = \frac{J}{g}[/tex]

    And J/g is near m^2/s^2, the square of speed, and then the square root takes care of everything.

    But, you ask, Joules aren't g-m^2/s^2, they're kg-m^2/s^2. Well, I'm pretty sure that you can ignore this for the same reason that you can throw a 3 in the equation... it's all about proportionality.
     
  6. Mar 28, 2010 #5

    Borek

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    Staff: Mentor

    This is not property of the electron :bugeye:

    Depends on what units are other values expressed in. If you are given pressure in PSI and volume in cubic feet you may prefer other R value. It is all about convenience. But it is different just because it is expressed in different units, physical sense it still the same.

    --
     
  7. Mar 28, 2010 #6

    Char. Limit

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    I meant to say gas...
     
  8. Mar 28, 2010 #7

    Borek

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    Staff: Mentor

    Last edited: Mar 28, 2010
  9. Mar 30, 2010 #8
    okay, thanks for the input, i think i get this concept better now :)
     
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