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Why molecules exhibit classical behaviour at room temp

  1. Nov 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Explain why molecules in water at room temperature behave like classical particles.


    2. Relevant equations

    [tex]E = \frac{3}{2}kT = \frac{p^{2}}{2m}[/tex]

    [tex]p = \frac{h}{p}[/tex]

    [tex]\lambda = \frac{h}{\sqrt{3mkT}}[/tex]

    for room temp, T = 300K.
    m is mass of water molecule
    k is boltzmanns constant

    3. The attempt at a solution
    For classical behaviour the de broglie wavelength should be negligible in comparison to the interparticle spacing, [tex](\frac{V}{N})^{1/3}[/tex]. I remember this being told in the lectures.

    I work out the mass m of a molecule by doing 18 x 10^-3 kg / avagadros number, and plug it in to find the wavelength.

    I work out the interparticle spacing by taking V = 1 litre = 10^-3 m^3, and N = 1 kg / mass of a molecule, and I find that the order of magnitude different is only a factor of 10. What have I done wrong?
     
    Last edited: Nov 13, 2009
  2. jcsd
  3. Nov 13, 2009 #2
    So you take:

    [tex]
    \frac{18\times10^{-3}kg}{1mol}\cdot\frac{1mol}{6.02\times10^{23}molecule}=2.99\times10^{-26}kg/molecule
    [/tex]

    I take this value and put it into the wavelength equation:

    [tex]
    \lambda=\frac{6.626\times10^{-34}}{\sqrt{3\cdot2.99\times10^{-26}\cdot1.38\times10^{-23}\cdot300}}\approx1.1\times10^{12}
    [/tex]

    This should be several orders of magnitude smaller than [itex](V/N)^{1/3}[/itex].
     
  4. Nov 13, 2009 #3
    I'm getting 3.438 x 10^-11 when plugging those numbers into my calculate (for wavelength).

    EDIT: evidently im screwing something up putting the numbers in my calculater :|

    EDIT2: no im not... you put it in your calculator wrong :) missed the square root maybe?
     
    Last edited: Nov 13, 2009
  5. Nov 13, 2009 #4
    You're right, I must've slipped a finger and gotten that. But you are right, you should get [itex]3.44\times10^{-11}[/itex]. But either answer is much less than [itex](1\times10^{-3})^{1/3}=0.10[/itex]
     
  6. Nov 13, 2009 #5
    For the interparticle spacing I get 3.1 x 10^-10 m. Maybe you confused my metres units and mass m label in my original post, ill touch it up a little bit now.

    So, whats the crack? Where's the mistake? This was the route my lecturer went down, he just didn't plug in numbers and didn't say anything about water being 1kg for 1litre etc.
     
  7. Nov 13, 2009 #6
    N stands for the number of molecules in the problem, correct? If this is so V/N=V where we are looking at only one molecule.
     
  8. Nov 14, 2009 #7
    Yes, N is the number of molecules in the problem. But we are not told how many molecules there are in a set amount of water volume. So I took 1kg to occupy 1 litre (I think that's right). I then took 1 kg / mass of the molecules to find the number of molecules in 1kg, and then (V/N)^(1/3) to find the inter particle spacing? anyway, if we even took one molecule, V would be ~ 10^-10 anyway, making the interparticle spacing of the same order again...
     
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