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Thermal physics - partition function

  1. Feb 18, 2005 #1
    Hi,
    i'm having trouble with a thermal physics problem relating to the partition function and i was wondering if anyone could help me out. the problem is as follows:

    (a) Consider a molecule which has energy levels En=c|n| , where n is a vector with integer components. Compute the partition function for a
    one-dimensional ideal gas of such molecules. Show that the average energy (assuming large temperature compared to the difference between adjacent energy levels) gives U=N(kB)T.

    (b) Take c=0.5eV. For a lab-sized box of gas of our molecules, estimate the temperature (in Kelvins) at which the assumption of part (a) breaks down.

    (c) Find the entropy. Figure out if this expression breaks down at low temperature. Are the two breakdown temperatures of (b) and (c) similar? Why or why not?

    ----------------------------------
    For part b), I guess the assumption is that the temperature, t (in fundamental units), is much greater than the change in energy levels, (E(n+1) - E(n)). As i understand, this assumption was used so that the summation in the partition function can be approximated by an integral.

    Let T be temperature in kelvins and kB be the boltzman constant. I got,

    t >> E(n+1) - E(n)
    t >> c|n+1| - c|n|
    t >> c
    t >> 0.5eV
    (kB)T >> 0.5eV
    T >> [0.5eV * 1.602 *10^-19 J/eV] / kB
    T >> 5802.26 Kelvins

    Is my approach for doing this problem correct?

    ----------------------------------------

    I'm mainly stuck on part c). let s = entropy (in fundamental units), F = free energy, t = temperature (in fundamental units) and Z = partition function. let N = # of gas molecules in the box.

    I found the entropy as follows:

    from part a), i found Z to be 1/N! * (t/c)^N

    s = - dF/dt
    = - d/dt [-t ln(z)]
    = d/dt [t ln(1/N! * (t/c)^N)]
    = d/dt t[ln(t^N) - ln(c^N) - N ln(N) + N] --> by stirling's approx.
    .
    .
    .
    = N [ln(t/cN) +2]

    I don't know where to go from here. how does this expression break down at low temperature? It seems like the only temperature that doesn't work is 0 and also negative temperatures, since you can't take the log of 0 or negatives. but t can't be negative anyway because we're using fundamental units (eg. NOT the celsius scale). how would i find the breakdown temperature in this case? is it supposed to be different than the answer in part b)? I understand why we must assume a large temperature in part a) (t/change in energy >> 1) - it's so we could approximate the summation by an integral when we're finding the partition function. But I don't understand why we need to assume a large temperature when finding the entropy. any help would be appreciated. thanks.
     
  2. jcsd
  3. Feb 18, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Well,what is the limit
    [tex] \lim_{t\rightarrow 0} S [/tex]

    and compare it with the limit which should be,according to the third principle of thermodynamics.

    Daniel.
     
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