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Effusion and speed of molecules

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data

    2mfd8jn.png

    Part(a): Find rate of incidence of gas molecules.
    Part(b): Find v1 and v2.
    Part (c): Find mean free path for faster molecules.
    Part (d): Find mean free path of 10m/s molecules.

    2. Relevant equations



    3. The attempt at a solution

    Part (a)

    Letting ##\phi## be flux, and ##\Delta p## be change in momentum of rebounding molecule,
    [tex]\phi \Delta p = \left(\frac{4}{3}\pi r^3\right)\rho g[/tex]
    [tex]\phi (2m\overline c) = \left(\frac{4}{3}\pi r^3\right)\rho g[/tex]
    [tex]R = \frac{2}{3}\frac{r\rho g}{m \overline c}[/tex]

    Funny how the mass of the molecule is not given. But this make sense as the more massive the atom travelling at a certain speed, the less flux you need to support the weight.

    Part (b)

    t5lhs3.png

    The speed distribution of molecules coming out of the oven is given by ##f = 2\alpha^2 v^3 e^{-\alpha v^2}##.

    [tex]v_2 = \int_0^{\infty} v f dv[/tex]
    [tex]= \frac{3}{4}\sqrt{\frac{\pi}{\alpha}}[/tex]
    [tex]= \sqrt{\frac{9}{8}\frac{\pi kT_1}{m}}[/tex]

    For interest, let's calculate the temperature of the Chamber:

    [tex]v_2 = \sqrt{\frac{8kT_2}{\pi m}} = \sqrt{\frac{9}{8}\frac{\pi kT_1}{m}}[/tex]
    [tex]T_2 = \left(\frac{3}{8}\pi\right)^2 T_1\approx 972K[/tex]

    To calculate ##v_1##, we use the speed distribution of the chamber (not the one coming out of the oven!)

    [tex]g = 2\alpha_2^2 v^3 exp(-\alpha_2 v^2)[/tex]

    [tex]v_1 = 2\alpha_2^2 \int_0^{\infty} v^4 exp(-\alpha_2 v^2)[/tex]
    [tex] = \sqrt{\frac{9}{8}\frac{\pi kT_2}{m}}[/tex]
    [tex] = \pi \sqrt{\frac{81}{512}\frac{\pi k T_1}{m}} \approx 1112 ms^{-1}[/tex]

    Part (c)
    I'm not sure whether the mean free path calculated here is correct.

    The relative speed is ##v_r = |v_2 - v_1| = 1112 - 944 = 168 ms^{-1}## and ##n = \frac{P}{kT} = 2.47*10^{17}## and ##\sigma = 1.52*10^{-19}##.

    Mean free path = (Distance travelled)/(No. of collisions during that distance)

    [tex]\lambda = \frac{v_2t}{v_r n \sigma t} = 176m [/tex]

    Part (d)
    In this case, ##v_r = |1112 - 10| = 1102##.

    [tex] \lambda = \frac{10t}{v_r n \sigma t} = 0.24 m[/tex]

    This is roughly 700 times less than above.

    This means that the atoms in the beam have a much larger mean free path than the atoms in the chamber. Thus velocity of the beam reduces much only after 176m into its flight path.
     
  2. jcsd
  3. May 2, 2014 #2
    bumpp , anyone mind checking my answers?
     
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