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Homework Help: Ideal gas

  1. Mar 8, 2008 #1
    1. The problem statement, all variables and given/known data
    The following is a probability density distribution for an ideal gas:
    [tex] p(v_x,v_y,v_z) = (2 \pi \sigma ^2)^{-3/2} \exp\left({\frac{v_x^2 +v_y^2+v_z^2}{2 \sigma^2}}\right) [/tex]

    [tex]E = 0.5 m |\vec{v}|^2[/tex]

    Find p(E).

    2. Relevant equations



    3. The attempt at a solution
    I want to just plug in what E but that gives the wrong answer and I cannot figure out why.
    [tex] p(E) = (2 \pi \sigma ^2)^{-3/2} \exp\left({\frac{E}{m \sigma^2}}\right) [/tex]
    Why is that wrong? It makes no sense, but I just plugged in what E was!! This is really bothering me.
     
  2. jcsd
  3. Mar 8, 2008 #2

    kdv

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    What is the correct answer? It's a probability *density* so there is also a factor [tex] dv_x dv_y dv_z [/tex] that must be multiplied to get a probability. If you want the probability density in terms of the energy you wil have to relate the above factors to the infinitesimal [tex] dE [/tex]

    EDIT: it's a density in speed so the factor to convert is probably dv
     
  4. Mar 8, 2008 #3
    Interesting. BTW, is it generally true that if y=y(x) and x=x(z), then y=y(x(z))? That is, you can write y as a function of z just by plugging in x as a function of z. That rule seems to fail here. But in general, when is it true?
     
    Last edited: Mar 8, 2008
  5. Mar 8, 2008 #4

    kdv

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    It is true in general, yes.

    The problem here is that what you are looking for is *not* simply y(x(z))!

    what you are looking for is the probability density in energy so the probability of measuring an energy between E and E+dE.. If you just plug in v in terms of energy, you are still giving the probability density in *speed*, so you are giving th eprobability of measuring the speed between v and v+dv. This is not what you are asked to find!
     
  6. Mar 8, 2008 #5
    So, back to the original problem. [itex]p(v_x, v_y, v_z)dv_x dv_y dv_z[/tex] gives the probability of finding the particle's velocity in the box with side lengths [itex]dv_x, dv_y, dv_z[/itex] with the one corner at [itex](v_x, v_y, v_z)[/itex].

    And I want to find p(E) such that [itex]p(E)dE[/tex] gives the probability of finding the particle's energy between E and E+dE.

    We know that [itex]dE = 1/2m(2v_x dv_x + 2v_y dv_y+2v_z dv_z)[/itex], but I still don't really know how to proceed. Do express each of dv_x dv_y dv_z as a function of dE somehow?

    You said something about just using dv, but I am not sure what you meant. Can you elaborate?
     
  7. Mar 8, 2008 #6

    kdv

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    You can't directly get from the distribution in v_x, v_y an dv_z to the density in Energy since there is more information in the first than in the second (since E depends only on v^2). So you need to find the probability density in the speed first. So you integrate over all directions using [tex] dv_x dv_y dv_z = v^2 dv d\Omega [/tex]. But the result of the angular integral is simply [tex] 4 \pi v^2 dv [/tex] times the probability density in v_x, v_y and v_z so

    [tex] \rho(v) dv = 4 \pi \rho(v_x,v_y,v_z) v^2 dv [/tex]
    where it is understood that in the sceon term you rewrite everything in terms of the speed v.
     
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