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Kinetic theory questions

  1. May 14, 2004 #1
    by what factor does
    i. the mean square speed
    ii. the root mean square speed

    ... of molecules of a gas increase when its temperature is doubled. I can tell that it would be 2 and 2^0.5 for the answer but is there a way to use the expressions to find that?
  2. jcsd
  3. May 14, 2004 #2


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    I wish you had explained HOW you "can tell that it would be 2 and 2^0.5 for the answer " or, for that matter, told us which "expressions" you would want to use.

    I suppose you mean the fact that the temperature of a gas is proportional to the average speed of the molecules making up that gas. Of course, the "average speed", at least the mean, is the "root mean square speed" so the root means square speed (answer (ii)) would be multiplied by 2, the multiplier of the temperature, while the "mean square speed" (answer (i)) would be multiplied by the square of that: 4.
  4. May 14, 2004 #3
    the expression that i would use would be 3/2kT
  5. May 14, 2004 #4

    Dr Transport

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    mean square speed [tex] <v^{2}> =\int f(\vec{v}) v^{2} d\vec{v} [/tex]

    mean square speed [tex] <v> = \int f(\vec{v}) v d\vec{v} [/tex]

    root mean square speed [tex] \sqrt{<v^{2}> - <v>^{2} [/tex]

    where [tex] f(\vec{v}) [/tex] is the distribution function of the particles, How does the velocity or consequently the energy vary as the temperature varies.

    Now using the equipartition cipher is partially correct, but I'd use the above equations to find out for sure, that is if you know the distribution function and I believe that the equipartitio theorem assumes the Boltzmann distribution, so work out the integrals and see.
  6. May 14, 2004 #5
    thanx very much everyone
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