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Heat Capacity Power Series Approximation

  1. Oct 31, 2011 #1
    1. The problem statement, all variables and given/known data

    "Derive a more accurate approximation for the heat capacity at high temperatures, by keeping terms through [itex]x^{3}[/itex] in the expansions of the exponentials and then carefully expanding the denominator and multiplying everything out. Throw away terms that will be smaller than [itex](\frac{ε}{kT})^{2}[/itex] in the final answer. When the smoke clears, you should find [itex]C = Nk(1 - \frac{1}{12}(\frac{ε}{kT})^{2})[/itex]"

    2. Relevant equations

    In addition to the above:
    The "exact" formula for the heat capacity was found in an earlier part to the question:
    [itex]C = \frac{Nε^{2}e^{\frac{ε}{kT}}}{kT^{2}(e^{\frac{ε}{kT}} - 1)^{2}}[/itex]

    3. The attempt at a solution

    I used the Power Series expansion for small x:

    [itex]e^{x} ≈ 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6}[/itex]

    where [itex]x = \frac{ε}{kT}[/itex]

    I expanded all this out, factoring out [itex]x^{2}[/itex] on bottom to cancel the same on top, removed powers of x greater than 2:

    [itex]C = N k \frac{x^{2}(1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6})}{(x + \frac{x^{2}}{2} + \frac{x^{3}}{6})^{2}}[/itex]

    [itex]C = N k \frac{x^{2}(1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6})}{x^{2} + x^{3} + \frac{7 x^{4}}{12} + \frac{x^{5}}{6} + \frac{x^{6}}{36}}[/itex]

    [itex]C = Nk\frac{1 + x + \frac{x^{2}}{2}}{1 + x + \frac{7 x^{2}}{12}}[/itex]

    I'm not sure how to take this any further. I have gotten something that almost resembles what I want:

    [itex]C (1 + x + \frac{7x^{2}}{12}) = N k (1 + x +\frac{7x^{2}}{12} - \frac{x^{2}}{12})[/itex]

    But clearly, I can't just divide by the part attached to C and call it a day. Any tips to help get to the target answer?
     
  2. jcsd
  3. Oct 31, 2011 #2

    I like Serena

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    Welcome to PF, Fuzzletop! :smile:

    You can use: [itex]{1 \over 1 - x}=1+x+x^2+x^3+...[/itex]

    But before you do so, you should expand your denominator with one extra term in the first step.
    And then you need to keep all powers up to power 3 (power 5 in the denominator since that reduces to power 3).
     
  4. Oct 31, 2011 #3
    Thanks!

    I did manage to get something of an answer. I did so while still ignoring the powers of 3 (I tried to use them at first, but I couldn't get it to work, so I switched back to just up to powers of 2, and I tried this below; I'm sure including the powers of 3 would work equally as well):

    [itex]C = Nk\frac{(1 + x + x^{2} - \frac{5x^{2}}{12} - \frac{x^{2}}{12})}{(1 + x + x^{2} - \frac{5x^{2}}{12})}

    = Nk\frac{(\frac{1}{1 - x} - \frac{5x^{2}}{12} - \frac{x^{2}}{12})}{(\frac{1}{1 - x} - \frac{5x^{2}}{12})}[/itex]

    Which eventually comes very close to the goal equation, but for a term [itex]\frac{5x^{2}}{12}[/itex] which I removed as it's "smaller than [itex](\frac{ε}{kT})^{2}[/itex]" (a bit less than half), and I got the answer, [itex]C = Nk(1 - \frac{1}{12}(\frac{ε}{kT})^{2})[/itex]

    Thanks a lot! I'm not usually one to think of using series expansions unless I've been told to do so explicitly, so that was a great help!
     
  5. Oct 31, 2011 #4

    I like Serena

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    You're welcome! :smile:

    Actually, I intended (but with terms up to the 3rd power):
    [tex]C = Nk\frac{1 + x + \frac{x^{2}}{2}}{1 + x + \frac{7 x^{2}}{12}}
    = Nk(1 + x + \frac{x^{2}}{2})(1 - (x + \frac{7 x^{2}}{12}) + (x + \frac{7 x^{2}}{12})^2 - ...)[/tex]
     
  6. Oct 31, 2011 #5

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