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Homework Help: Thermodynamics - Entropy&Heat dispersion

  1. Feb 24, 2008 #1
    Hey, just a small mathematical enquiry about why temperature tends to average out in, say a box of gases. Lets just use the standard "Partitioned vessel" view of it, with One side having 1 mole of Helium at..say 398k and the other side with 1 mole of helium at 298K. The heat capacity is (rounded up) 21 J/Mol K and I'm assuming it's the same over the temperature range.

    Now to calculate the final temperature when both gases are mixed I'm just going to use [tex]q=nC_{v}\deltaT[/tex] to find the total thermal energy then calculate for T:

    [tex]q_{298K} = 1*21*298 = 6258J[/tex]
    [tex]q_{398K} = 1*21*398 = 8358J[/tex]

    [tex]T= \frac{q_{298K}+q_{398K}}{2*21} = 348K[/tex]

    Now with the final temperature, and discounting any residual T = 0 crystal entropy the entropy of that mixture is calculated by:

    [tex]\int^{398}_{0} \frac{C_{v}}{T} dT = C_{p}ln(\frac{398}{0})[/tex]
    Stuck here, ln(0) is not possible I did have a grand plan to try and prove it to myself but how do you deal with this part in the integral?
  2. jcsd
  3. Feb 25, 2008 #2
    solve as an improper integral?

    maybe, I haven't taken calc II in about 3 years.
  4. Feb 25, 2008 #3
    This isnt a homework or coursework question by the way, it's just me enquiring (Directed at whomever moved it)
  5. Feb 26, 2008 #4
    You take into account that the heat capacity at temperatures near absolute zero is not a constant. In the Debye theory of solids, for example, it is
    proportional to the third power of the temperature, so C/T goes as the
    square of T. Then you can do the integral.
  6. Feb 26, 2008 #5
    So its:
    [tex]C_p ln(398) - 0^2[/tex]?
  7. Feb 27, 2008 #6
    No, I don't think so. My suggestion was just to break the range of the integration into several sections: 0 to T1, T2, ..., 398. and I only addressed my remarks to the range near zero (the part of your post in bold print). Many (solid) materials follow an empirical law of the type

    [tex] C= aT^3 + \gamma T [/tex]

    over a small range in the neighbourhood of absolute zero of temperature. (i.e. the range 0 to T1. In this range, you can do the integral, thus avoiding the problem of log(0).
    The values of the constants in the equation above as well as the temperature T1 will depend upon the material you are considering. For most materials, you will also have to divide the range from T1 to 398 into several smaller ranges as well because 1) the above equation loses its validity and 2) there may be phase transitions which will also contribute (i.e. the heat capacity is not a continuous function over the whole temperature range). For example, for oxygen, the ranges are

    0 to 14 K contributing 0.54 cal/deg
    14 to 23.66 K contributing 1.5
    phase change contributing 0.948
    23.66 to 43.76 contributing 4.661
    phase change contributing 4.058
    43.76 to 54.39 contributing 2.397
    fusion contributing 1.954
    54.39 to 90.13 (liquid phase) contributing 6.462
    Vaporization contributing 18.07
    Total: Entropy of gas (T=90.13) is 40.59

    For higher temperatures, you can use cp=7R/2, up to about T=298 K; at
    temperatures above this, you need another empirical equation.

    As you see, the calculation of entropy is not exactly trivial. For helium, it might be
    less complicated but I don't really know. A final remark: I have only addressed myself to the question you asked about the integral in neighbourhood of T=0, not to the rest of your post. Maybe you don't need to consider the absolute entropy but only the
    change in entropy from the boiling point to the higher temperature. In that case, you could simply reformulate the problem to avoid the lower temperature range(s)!
    Would that work in your case?
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