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Particle Statistics Help

  1. May 14, 2014 #1
    1. The problem statement, all variables and given/known data

    A container holds N molecules of nitrogen gas at T = 280K. Find the number of molecules with kinetic energies between .0300 eV and .0312 eV.

    dE = .0012 eV
    E = .0306 eV
    kT = 3.8668*10^-21 J or .02413 eV


    2. Relevant equations

    N(E) = (2N)(E^1/2)(e^(-E/kT)(π^-1/2)((kT)^-3/2) (defines the energy distribution of the gas)
    dN = N(E)dE
    dN = (2N)(E^1/2)(e^(-E/kT)(π^-1/2)((kT)^-3/2)dE

    3. The attempt at a solution

    I assumed that N(E) can be approximated to be linear therefore the above equation can be used. E was found to be the average energy between the two given values and dE was found by taking the difference between the two given energy values. kT was found by multiplying Boltzmann's constant (1.381*10^-23) and multiplying by the given temperature.

    The answer is supposed to be in terms of N (the overall number of molecules in the volume).

    I calculated the number to be
    (2)(N)(.175 ev^1/2)(.281)(.5642)(266.79 ev^3/2)(.0012 eV) = .0178 N

    However, the actual answer turns out to be 6.1 * 10^-6 N.

    I can't figure out what I've done incorrectly and any help would be appreciated!
     
    Last edited: May 14, 2014
  2. jcsd
  3. May 14, 2014 #2

    tms

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    To solve the problem you just have to integrate the distribution between the stated energy limits.
     
  4. May 14, 2014 #3
    Since the interval was small, a Riemann sum approximation appeared to provide a result close enough to an integration.

    I get that there would be some error from this approximation, but I don't think it would account for the four degrees of magnitude by which I'm off. For ease, we've been allowed to use this approximation by representing E as the midpoint and dE as the ΔE.

    EDIT: After doing the integral I got a value of .222 which still seems incorrect.
     
    Last edited: May 14, 2014
  5. May 14, 2014 #4

    tms

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    Check your units, especially for Boltzmann's constant.
     
  6. May 14, 2014 #5
    I used 8.62*10^-5 evK for Boltzmann, but it still seemed to not work. Everything else is either unitless or in eV.
     
  7. May 14, 2014 #6

    Simon Bridge

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    How many particles between energy E and E+dE
    How did you decide which distribution to use?
     
  8. May 14, 2014 #7
    The number of molecules between E and E+dE is some fraction of N and what we're trying to solve. The distribution equation was given earlier in the chapter and describes the likelihood of certain energies occuring in the system.
     
  9. May 14, 2014 #8

    Simon Bridge

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    You need to be able to make a concrete statement about what the energy distribution is telling you.
    What do you mean by "likelihood of energies occurring"? What does that mean in relation to the equation you wrote down?
     
  10. May 14, 2014 #9

    tms

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    I'm sorry, but I don't see what is wrong. The method you are using seems right, and gives reasonable answers in other cases. The answer you got (and I got) seems a little big, but that's just an eyeball estimate. The "correct" answer you give seems a lot too small.

    Are you sure you have stated the problem correctly?
     
  11. May 15, 2014 #10
    I just double checked the problem and all the information I've written appears to be correct. Maybe the book answer is wrong, though this seems unlikely.
     
  12. May 15, 2014 #11
    N(E) refers to the distribution of molecules at respective energies, so the expected number of molecules to be found at a certain interval (between E and E+dE).
     
  13. May 15, 2014 #12

    Simon Bridge

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    You are saying: the number of moecules between E and E+dE is given by
    ... that reads to me like:$$n(E) = 2N_{0}E^{1/2}e^{-E/kT}π^{-1/2}(kT)^{-3/2} = \frac{2N_0 E^{1/2}}{\sqrt{π (kT)^{3}}}e^{-E/kT}$$You say you chose that distribution because it was the only one given earlier rather than because you thought it through and decided that was the appropriate one for the job?

    What are the units of n(E) (or your N(E)) as written above?

    ...
    Anyway, if n(E) is supposed to be the density of states,
    then the number of molecules between ##E## and ##E+\text dE## would be ##n(E)\text dE##
    and the total number of molecules between ##E_a## and ##E_b## would be:$$N=\int_{E_a}^{E_b}n(E)\;\text{d}E$$

    refs.
    http://hep.ph.liv.ac.uk/~hock/Teaching/StatisticalPhysics-Part3-Handout.pdf
    http://www.physics.udel.edu/~glyde/PHYS813/Lectures/chapter_4.pdf
     
  14. May 15, 2014 #13
    The distribution equation describes the distribution of molecular energies for a gas in thermal equilibrium. The derivation of the distribution equation is not done and N(E) is instead attributed to being given by the Maxwell-Boltzmann distribution.

    What you wrote is the same as the given distribution for n(E).

    dN = n(E)dE

    So the number of molecules between two energy levels E and E+dE, would be the integral you wrote above. I did a midpoint Riemann Approximation as that was the advised method of computing the answer in the example. Additionally, after not getting the correct answer I did the integral and got another incorrect answer. All my energies and constants were in terms of eV and my temperature was in terms of K. I still can't figure out where I went wrong.
     
  15. May 15, 2014 #14

    Simon Bridge

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    I copied down what you wrote for N(E), but called it n(E) instead.
    It that the same n(E) in dN=n(E)dE that you wrote just now?

    I don't think so. Lets check:

    The number of states between E and E+dE would be n(E)dE

    ##n(E)dE=g(E)f_EdE## where ##g(E)## is the density of states and ##f_E## is the probability density there is a particle with energy E. For Maxwell statistics: $$f_EdE=2\sqrt{\frac{E}{\pi}}\left(\frac{1}{kT}\right)^{3/2}e^{-E/kT}dE$$... looks a bit like yours right?

    If the states are equally spaced, then ##n(E)dE=N_{tot}f_EdE## is the number of particles with energy between E and E+dE.

    Does it make a difference that Nitrogen is a diatomic gas?

    Lastly:
    If you are certain that you have followed the correct procedure etc then you have two options. Either you have misunderstood the question or the model answer is wrong.
    Check with fellow classmates, and with the person who set the question, to settle this matter.

    Good luck.
     
  16. May 15, 2014 #15

    tms

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    Wouldn't that just add more factors of ##1/kT##? Which would make the result larger.
     
  17. May 16, 2014 #16

    Simon Bridge

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    Certainly affects how the energy is partitioned off, but the question may only be concerned with translational states anyway.

    I have a statistical mechanics crash course someplace... oh yes:
    http://home.comcast.net/~szemengtan/StatisticalMechanics/IdealGas.pdf [Broken]
    Whole thing - but see 4.7, as relevant to above.
    Well it may help.
     
    Last edited by a moderator: May 6, 2017
  18. May 16, 2014 #17
    Thank you both! I'll watch the video, re-read the section in the book dealing with this and talk with some friends. The help has been much appreciated.
     
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