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Calculating temperature from molecular speed distribution

  1. Jan 12, 2015 #1
    1. The problem statement, all variables and given/known data
    A sealed container of 0.10 m3 holds a sample of 3.0x1024 atoms of helium gas in equilibrium. The distribution of speeds of the helium atoms shows a peak at 1100 m s-1.
    Take the mass of a helium atom to be 4.0 amu.

    I, calculate the temperature and pressure of the helium gas.
    ii, what is the average kinetic energy of the helium atoms?
    iii, what is the position of the maximum in the energy distribution?


    2. Relevant equations
    ,<E> = (3/2)KT
    delta vsd = sqrt(<v2> - <v>2) (?)
    PV = NKT
    f(v) = Bv2e-mv^2 / 2KT
    <v> = sqrt((8KT) / (pi m))
    vmp = sqrt((2KT) / m)


    3. The attempt at a solution
    I believe that the peak in the Maxwell-Boltzmann speed distribution corresponds to the most probable speed (vmp), hence I think the temperature can be found by rearranging the equation for vmp to T = (vmp2 * m) / (2K) and then finding pressure from P = NKT / V. One thing i'm not clear on is whether the mass needed is that of one atom, or all of them.

    Part ii, seems reasonably straight forward once temperature is known (ie just plugging values in to the equation).

    I do not know the definition of the maximum in this context but assume it is the highest energy of any of the helium molecules. I understand that the speed and energy distributions are linked somehow but I cant understand how you obtain information about one from the other. im wondering whether standard deviation of molecular speeds has something to do with it or even the square root rule for averages? I am very unsure of this one.

    Thanks in advance guys, any help would be hugely appreciated!
     
  2. jcsd
  3. Jan 12, 2015 #2

    Dick

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    You are pretty much right to begin with. The maximum probable speed is the speed at which the maximum number of atoms can be found to be travelling close to. It's the maximum of the probability distribution for speed. From that you can get the temperature. And, of course, m is just the mass is just of one atom. Why would you think differently? Think intensive versus extensive properties.
     
    Last edited: Jan 12, 2015
  4. Jan 12, 2015 #3

    ehild

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    You need the maximum of the energy-distribution function f(E).
     
  5. Jan 13, 2015 #4
    Thanks for the replies guys! its great to know that I was on the right track with the first part. I wondered about what mass was needed because the number of atoms was given so I thought maybe that was why (to calculate total mass).

    It makes sense to me that you can find the maximum of the energy distribution from the speed distribution, but I am struggling to think how to go about this. I understand that kinetic energy is related to the square of speed but I am unsure exactly of how the distribution functions relate to each other.
    Could it be that the f(E)/g(E) is simply ewqual to (1/2)m f(v)^2?
     
  6. Jan 13, 2015 #5

    ehild

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    E=1/2 mv^2 for ideal gases. You can find the energy distribution function using that the number of molecules having speed between v and v+dv is equal to the number of those which have corresponding energy between E and E+dE: f(v)dv = fE(E)dE. You have to include dE. Differentiate E=1/2 mv2 to get the relation between dv and dE. Express v in terms of E and substitute for v in f(v)dv.
     
  7. Jan 13, 2015 #6
    Sorry to keep bothering you, im just having real trouble visulising what to do for this one!
    So dE/dv = mv. When you say express v in terms of E do you just mean v=dE / dv*m? Is the value of dv just an arbitrary choice? surely dv is needed in order to do this calculation.
     
  8. Jan 13, 2015 #7

    ehild

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    Use E=1/2 mv2 to express v in terms of E: ##v=\sqrt{2mE}##
    As for dv, you got that dv=dE/(mv). Substitute the previous expression for v.
    Write out f(v)dv in terms of E.
     
  9. Jan 13, 2015 #8
    I'm sure I must be frustrating you, I know im frustrating myself aha. I do appreciate this help greatly though, thankyou so much! I didn't know what to expect as this was my first post, I have been very pleasantly surprised.

    dv=dE / m* sqrt(2mE)
    So f(v)dv= f(v)* sqrt(2mE) = f(E)dE to
    Ie. f(v)dsqrt(2mE)= dE / m* sqrt(2mE)


    That's as far as I've got, my mind really had deserted me.
    Maybe dE =f(v)dv*m*sqrt(2mE) and that's how to find the maximum?
     
  10. Jan 13, 2015 #9

    ehild

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    You should substitute ##\sqrt{2mE}## for v into f(v) = Bv2e-mv^2 / 2KT
    Have you not learnt it at the classes?
    You find the distribution function for energy here: http://en.wikipedia.org/wiki/Maxwell–Boltzmann_distribution.
    Scroll down to eq (7).
    The maximum is at that energy where the derivative of f(E) is zero.
     
    Last edited: Jan 13, 2015
  11. Jan 13, 2015 #10
    I have been having to self teach this bit from a text book...I think ill need to start from the beginning again.
    Thanks for everything.
     
  12. Jan 13, 2015 #11

    ehild

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    You need to study Probability Theory, too, to understand the concept of distribution functions. It is very difficult if you do not have a teacher!
     
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