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Classical/quantum transition for an ideal gas in 1m^3 box

  1. Feb 18, 2014 #1
    1. The problem statement, all variables and given/known data
    Hey guys,

    So I have to estimate the temperature at which the classical / quantum transition happens for a 1m^3 box of air. This is done by comparing the de Broglie wavelength with the average distance between the particles - so basically the transition happens when they are comparable.


    2. Relevant equations
    Average distance between the particles: [itex](\frac{V}{N})^{\frac{1}{3}}=n^{-\frac{1}{3}}[/itex] where n is the number density

    de Broglie wavelength: [itex]\lambda = \frac{h}{\sqrt{2\pi m k T}}[/itex] where k is the boltzmann constant.


    3. The attempt at a solution

    Right, so here's what I did. Since the volume V = 1, we can say that
    [itex]N=n[/itex], using the distance between the particles equation.

    We also know that [itex]pV=NkT=nkT[/itex] since n = N, so assuming standard pressure (since the question asks me to estimate), we can say that [itex]n=\frac{10^{5}}{kT}[/itex].

    The transition happens when [itex]\lambda ≈ n^{-\frac{1}{3}}[/itex]. So replacing λ with our expression for n, we get this
    [itex]\frac{h}{\sqrt{2\pi m k T}}=(\frac{10^{5}}{kT})^{-\frac{1}{3}}[/itex]

    Using the approximation that the mass m of air is around 30 kg / mol, the mass of one air molecule is around 5 x 10-23kg. Plugging that in and solving for T gives me

    [itex]T≈5.6\times 10^{-4}K[/itex], which seems wrong for some reason..

    Can you guys help me out? thanks!
     
  2. jcsd
  3. Feb 18, 2014 #2

    TSny

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    kg??
     
  4. Feb 18, 2014 #3
    is that wrong? i just divided 30 by avogadro's number o.o
     
  5. Feb 18, 2014 #4
    oh wait i think its 5 x 10^-26 kg...its 30 g/mol not 30kg/mol XD
     
  6. Feb 18, 2014 #5
    So after that correction i get 0.035 K...is that better?
     
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