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Problems On Temperature

  1. Aug 16, 2006 #1
    [URGENT] Problems On Temperature

    Hi all,

    I was just release from Chemistry lesson and here is what happen.

    My chem teacher said that " At 0 Kelvin, which is absolute zero, there is no kinetic energy in the molecules anymore."

    Being Very Well Read( in physic), i know that it is not true, thus i argue with him.

    In the end he used a formula K.E. = 3/2KT ( or 1/2 K T i can't remember what is the formula)

    where K = bottzman's constant.
    And T = Thermodynamic Temperrature ( kelvin)

    It works out that when T=0, KE = 0 hence he conclude that at absolute zero KE = 0

    However, although i know that KE is not zero because of Quantum mechanics effects, i am unable to prove him wrong.

    I know about the extrapolation of the graph( energy against temperature), however since it is a " extrapolation" he refuse to accept it as evidence for KE not equals to Zero.

    Can someone help me to show that KE is not zero when absolute zero is reached.( pls work it in such a way that even non-quantum Mechanic student can understand.)

    Thx :)
    Last edited: Aug 16, 2006
  2. jcsd
  3. Aug 16, 2006 #2


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    Try this page for a discussion on the kinetic theory of gases

    more topics - http://hyperphysics.phy-astr.gsu.edu/hbase/kinetic/ktcon.html

    Discussion of temperature - http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/temper.html

    Discussion on liquid helium

    Discussion of particle spin and Bose-Einstein Condensate

    Discussion of absolute zero

    I hope ZapperZ, marlon and others will contribute to the discussion, and perhaps provide additional references.
  4. Aug 16, 2006 #3
    Thx for the help!
  5. Aug 16, 2006 #4
    after reading all that has been said.

    Can i conclude that The presence of " zero-point motion/energy" and " Bose-Einstein condensate" is because it need to agree with Heisenberg's uncertainty principle?
  6. Aug 16, 2006 #5
    And that the Formula KE = 3/2 K T only accounts for translational motion and not rotation and vibration motion?

    If so, shouldn't all gas be able to solidfy at 0 K since no translational motion equals to all atoms drop down on the floor of the container forming a solid?
  7. Aug 16, 2006 #6


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    What you want to do here is look at the fundamental assumption of thermodynamics laws that are used in standard textbooks. You'll see that most of them assume the validity of kinetic theory. Now look at the assumptions based on kinetic theory. You'll see that in many cases, there's a limit to where kinetic theory can and cannot work. At low enough temperatures, three things occur:

    1. a possible phase transition. Here, kinetic theory fails miserably in predicting when something will undergo a phase transition from gas to liquid, or liquid to solid. In such a case, various thermodynamics properties undergo an abrupt discontinuity, including temperature variation. What this means is that one simply cannot extrapolate the trend one sees at higher temperatures and assumes that this is what will happen at very low temperatures. So the usage of E = 3/2 kT, while valid at higher temperatures, would not surve at very low temperatures, and certainly would not be valid across a phase transition (example: T doesn't change at the ice-water interface even though you're adding energy (heat) to it.)

    2. When a material has solidified, for most practical application, the description of the material's property can be accurately described via the quantum simple harmonic oscillator (SHO). When this occurs, then any undergraduate physics student than tell you that the LOWEST possible energy that a SHO system can have is not zero, but 1/2 [itex]\hbar \omega[/itex]. So essentially, even at T=0 in such a system, there is still a small, but non-zero vibration.

    3. The deBoer effect that are more obvious in the nobel gasses at very low temperatures, is the correction one needs to account for in the specific heats of these gasses[1]. So this is clearly something that is present experimentally and one of the clearest indication of zero-point energy, no matter how low the temperature gets.

    If your instructor do not buy any of these, then there's nothing that can be done. However, I have a feeling that he/she was simply trying to illustrate the properties of a classical system and did not want to confuse a student by delving into areas which aren't covered in the course.


    [1] Ashcroft and Mermin, "Solid State Physics", pg. 412 (1976).
  8. Aug 16, 2006 #7
    if youd have read it all more carefully you'd know helium will never solidify under normal pressure.

    the theory that states T = 3/2K.E. doesn't take into account the attraction and repulsion between atoms.
    it's good only for gas much above it's boiling temperature.

    and yes, you could say it's heisenberg's uncertainty principle - you can't have something that stands still.

    [EDIT] haven't seen ZapperZ already covered it =)
  9. Aug 16, 2006 #8
    Nicely done! THX guys!
  10. Aug 16, 2006 #9
    BTW, i just checked Wiki and google "deBoer effect" i couldn't find anything so can someone tell me what it means?
  11. Aug 16, 2006 #10


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  12. Aug 16, 2006 #11
    one more thing, it is stated that we can't have zero entropy because this will mean that Kelvin will tend to zero.

    "As a system asymptotically approaches absolute zero of temperature all processes virtually cease and the entropy of the system asymptotically approaches a minimum value."

    However using Equation S = K Ln w, when we plot the graph(S against w) out, there should be a X-intercept shouldn't it?

    so is it possible to have zero entropy?
  13. Aug 17, 2006 #12
    Hi again, today i had chem lesson again.

    I presented what u guys said, my chem teacher "sort of agree" with it.

    HOWEVER, he ask, the 3rd law of thermadyanics states that when Kelvin reaches 0, entropy will also be 0. if there were to be motion, the entorpy will not be zero anymore since there is more than one way in which the particles can arrange itself.

    can anyone help me account for this?

    EDIT 1: BTW we are currently in the chapter " entropy" for our chem lesson
  14. Aug 17, 2006 #13


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    This might help, but I would wish others to provide comments.

    Please refer back to ZapperZ's comments in post #6.

    Much of our understanding of the world comes from observations in the macroscopic world (large collections of atoms/molecules), as opposed to the microscopic (quantum) world. At the atomic and subatomic level, our generalized understanding does not apply.

    This might be of interest
  15. Aug 17, 2006 #14
    So are u trying to say that the 3rd law of thermodyanics(3LOT) is being defied? since there are residual entropy in absolute zero, which is forbidden by 3LOT
  16. Aug 17, 2006 #15


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    Not quite.

    The third law of thermodynamics can be stated as:
    The key is 'minimum value' which is not necessarily zero.
    from http://en.wikipedia.org/wiki/Third_law_of_thermodynamics

    If the 3rd Law of Thermodynamics is stated as "entropy is zero at absolute zero", then that would be incorrect - actually it would be incorrectly stated.
  17. Aug 17, 2006 #16
    But how do u explain the equation S = K Ln w

    if i use Equation S = K Ln w and plot a graph(S against w) out, there should be a X-intercept shouldn't it(since Ln has a x-intercept and at x-intercept s=0)?
  18. Aug 21, 2006 #17
    erh.....anyone can ans my qns pls?
  19. Aug 21, 2006 #18


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    What does one mean by explaining the equation?

    Taking the derivative,

    dS/dw = K/w, so at w=0, dS/dw 'blows up', goes to infinity.

    ln w |w=0 is similarly undefined, or 'blows up'.

    Does 'w' have to go to zero?
  20. Aug 21, 2006 #19
    oh sry, i get it now, the only possible way to get S=0 is have w=1, which in this case alrdy show that there is only one way to arrange the particles. thx! :)

    p.s. hope u are not irritated by me
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