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Fraction of molecules have kinetic energy of 1/2mC^2

  1. Nov 1, 2016 #1
    1. The problem statement, all variables and given/known data
    Consider a gas of molecules of mass m at equilibrium at temperature T. Obtain an expression for the fraction of molecules having kinetic energy e = 1/2mC2 in the range e to e + de.

    This is problem 5.3, page 48 of Vincenti and Kruger's Intro to Physical Gas Dynamics

    2. Relevant equations
    Maxwellian speed distribution

    3. The attempt at a solution
    I'm not sure where to start really, so I don't have much of a solution.

    If I were finding the fraction of molecules with kinetic energy LESS than 1/2mC2, I would integrate the speed distribution function from 0 to 1/2mC2. But I'm not sure about finding the fraction at exactly 1/2mC2.

    My first thoughts were to change the speed distribution to an energy distribution by knowing that c2 = 2e/m, and then integrating the kinetic energy distribution from 0 to inf to find the total kinetic energy. Then divide 1/2mC2 by that total to find a fraction. Don't feel like this is right.

    Anyone care to nudge me in the right direction?
     
  2. jcsd
  3. Nov 1, 2016 #2

    NascentOxygen

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    Staff: Mentor

    Hi igowithit! :welcome:

    My reading of the task is that you are asked to:
    obtain an expression for the fraction of molecules having kinetic energy in the range e to e + ∆e.

    They have helpfully provided you with the formula to use, but wrote it where it has lead to confusion.
     
  4. Nov 1, 2016 #3

    vela

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    To elaborate on your reasoning here, what I believe you meant was:
    $$P(E < \frac 12 mC^2) = P(v < C) = \int_0^C f_v(v)\,dv$$
    You'd integrate the speed distribution from 0 to C.

    This fraction would be 0. The probability of picking one value out of the continuum of energies is 0. That's why the question is asking you to find the probability in a range of values, specifically you want to calculate
    $$P(e<E<e+de) = \int_e^{e+de} f_E(E)\,dE.$$

     
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