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Simple problem (thermodynamics): is it just too simple?

  1. Apr 25, 2009 #1
    1. The problem statement, all variables and given/known data
    Each molecule of a substance can exist in either of two states; these states differ in energy by 25 eV.

    a. What is the entropy of this sample at room temperature?

    b. What is the internal energy of this sample at room temperature?

    2. Relevant equations
    dU = TdS
    U =(f/2)nRT
    (f = degrees of freedom)

    3. The attempt at a solution
    I don't understand how i am supposed to derive the total entropy of the sample from the difference in energy "states" (the DIFFERENCE in entropy between states would be easy): the problem talks about molecules, then atoms.. are these states referring to electronic excitement, or what?.. i need some opinions about this, please..
     
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  3. Apr 25, 2009 #2

    nicksauce

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    U = (f/2)nRT only applied to quadratic degrees of freedom, not a number of different energy states. Presumably the way to go about this would be to calculate the partition function, and derive the entropy and energy from that.
     
  4. Apr 25, 2009 #3
    Yes, that was my first attempt but i got stuck:
    Boltzmann's statistics show that all particles should be in the ground-state, but then i don't seem to go anywhere without at least knowing the ground-state energy... (Which cannot be deduced from the information given, to my knowing)
    This is why i went back to basics, but at this point i just wonder if the problem is well stated!
     
  5. Apr 25, 2009 #4

    Mapes

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    The zero level of energy is arbitrary. Why not take the energy levels to be 0 and 25 eV?
     
  6. Apr 26, 2009 #5
    Right... but then it seems i end-up with 0 internal energy for the system?
     
  7. Apr 26, 2009 #6

    Mapes

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    I don't think that would maximize entropy. What partition function are you using?
     
  8. Apr 26, 2009 #7
    Z = Z1N/N!
    Where Z1 is the sum of Boltzmann factors, which reduce to 1 if the gound-state energy is zero.. Here N is actually NA, Avogadro's number, so after getting 0 total internal energy i end up with a negative entropy... that's a problem!
     
  9. Apr 26, 2009 #8
    oh, this is the partition function for non-interacting, indistinguishable particles, to answer your question in words..
     
  10. Apr 26, 2009 #9

    Mapes

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    OK, got it. But this partition function assumes an infinite number of possible states; you only have two. You need to identify your ensemble and construct a partition function for a two-state system.
     
  11. Apr 26, 2009 #10
    got you.. but given that, i think all i know is that all particles should be in the same state so it doesn't give much of a partition.. i'm gonna think about that
     
  12. Apr 26, 2009 #11

    Mapes

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    Well, I'd dispute that. At absolute zero they'll all be in the same lower energy state, and at very high temperatures they'll all be in the same higher energy state. But at intermediate temperatures there should be a mix to give a per-molecule average of (25 eV)(x), where x is a fraction.
     
  13. Apr 26, 2009 #12
    yes, but the only relevant statistics i can think of is Boltzmann's and it gives a chance of e to the minus 900 and something.. which is zero, for all practical purposes
     
  14. Apr 26, 2009 #13

    Mapes

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    OK, just to confirm, your partition function is [1+exp(-e/kT)]N/N!, where e is 25 eV? Did you calculate the entropy and internal energy? (And e is definitely 25 eV, not 25 meV? As you've noticed, 25 eV is quite high.)
     
  15. Apr 26, 2009 #14
    right. internal energy comes down to zero, entropy becomes –133 J or something (i'm not at home), which is nonsense; i use: F = –kT ln(Z) = (U-TS)/T, then S = – dF/dT
     
  16. Apr 26, 2009 #15

    Mapes

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    Do you get S/N = k ln[1+exp(-e/kT)] - k ln N + k + (e/T)/[1+exp(e/kT)]?
     
  17. Apr 26, 2009 #16
    sorry, i had to get home (thanks for your time, by the way..)
    so yes, i obtain that (with Stirling's approximation) and everything is just too small to be considered, except for the term
    –K ln(N) which is just –F/T. (with F = 133 kJ)
     
  18. Apr 26, 2009 #17

    Mapes

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    McQuarrie in Statistical Mechanics recommends ignoring the (- k ln N + k) part because it's not a function of temperature. The k ln[1+exp(-e/kT)] + (e/T)/[1+exp(e/kT)] part captures the key features of the two-state system: transition in energy per molecule from 0 eV at low temperatures to 25/2 eV at high temperatures, and zero heat capacity at very low and very high temperatures. Also see http://books.google.com/books?id=z6...aX7EI3WzATuyK37Cg&client=firefox-a#PPA215,M1".

    I'm not sure where the 133 kJ comes from?

    (Also, I made an error in my post 11 above. At high temperatures the system will be split evenly between the energy levels rather than entirely at the higher energy level.)
     
    Last edited by a moderator: Apr 24, 2017
  19. Apr 26, 2009 #18
    this link is gold, thank you for that and for the time spent... i'm gonna look at it in details tomorrow; the 133kJ comes from computing F = NkT[ln(N) – 1], since N is just one mole in the problem and T is 298K, but i just obtained 138kJ by checking it.. anyway,
    thanks again, i should be good with that!
     
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