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What is the change in entropy?

  1. Oct 13, 2013 #1
    1. The problem statement, all variables and given/known data

    A system with 100,000 molecules has 2 energy levels, A and B. At first, the 2 energy levels are populated equally. After a reversible process, energy level A is populated by 65% of the molecules and the system is at 293K.

    a) What is the difference in energy between the two levels?
    b) How much heat was added or removed from the system?
    c) What is the change in entropy?


    2. Relevant equations

    Boltzmann distribution
    Probabilistic definition of entropy


    3. The attempt at a solution

    I computed the Boltzmann factor for the two energy levels and got the correct answer for (a) = 2.5*10^-21 J.

    Then, when I looked at (b) and (c), I thought they wouldn't ask for the change in entropy after asking for the heat, if the change in entropy was required for the calculations of the heat. So I pondered, and got this:

    Since in the process, 50%-(100%-65%)=15% of the molecules were transferred from B to A, so the energy change of the process would be= 0.15(100,000)(-ΔUab) = 0.15(100,000)(-2.5*10^-21 J) = -3.75*10^-17 J.

    But then I was stuck, because I only had the energy change, and just couldn't find the heat without knowing whether there was work and how much.
    And then the solution did what exactly I thought they wouldn't do:

    1. Find ΔS using the probabilistic definition:
    Initial S= -(100,000)(kb)[2(0.5ln0.5)]
    Final S= -(100,000)(kb)[0.65ln0.65 + 0.35ln0.35)]
    ΔS= -4570kb

    2. ΔS= q/T (reversible)
    ∴ q= TΔS = -1.85*10^-17 J

    Then for (c), they put "see above."

    --------

    My questions/confusions are:

    1). Does my attempt (multiplying the energy change for one molecule by the number of molecules) really get me to the energy change?
    2). I'm confused about how they use ΔS= q/T to get the heat here. The problem doesn't say the process is isothermal?

    According to the Boltzmann factor:
    Nb/Na = exp(-ΔUab/kbT)

    The process changes Nb/Na, and ΔUab has to be constant (does it?), T has to change.

    -------

    I hope the question isn't too long! Thanks for any help!
     
  2. jcsd
  3. Oct 14, 2013 #2

    BruceW

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    Homework Helper

    Welcome to physicsforums!
    yeah, it's a tricky problem, more like statistical physics than thermodynamics really.

    hmm. I get an energy difference of 1.5*10^-21 J ... Maybe our answers are different because we used different significant figures for our value of the Boltzmann constant. Anyway, you got it correct, so it doesn't matter.

    Yep, that is the correct total energy change for the system.

    the problem does not say the process is isothermal. The problem only says that the process is reversible, which ensures that the equation ΔS= q/T is true.

    yeah. It is meant to be implied that ΔUab is constant. And generally in these kinds of problems, the different possible energy levels are fixed. It is just the number of molecules in each energy level changes.
     
  4. Oct 15, 2013 #3
    Thanks for clarifying some of my confusions!
    But for ΔS= q/T, I thought it was defined as dS=dq(rev.)/T, so you would have to integrate the function to get the entropy change.
    If T is not constant, why can you take it out of the integral and simply integrate dq to get q?
    Thanks again!
     
  5. Oct 16, 2013 #4

    BruceW

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    Homework Helper

    yeah, that is strange. I'm not sure about that. Maybe they assume there is a path through phase-space where first, the temperature changes while entropy is constant, then the entropy changes while temperature is constant. And since the heat input is path-independent for a reversible process, if we take any other path through phase space, we still get the same heat input. They don't really say how the system gets from the first part to the second part (apart from saying the process is reversible), so it is hard to say exactly what they had in mind. I guess it is also possible that the difference in the two energy levels does change... It is just quite rare in most problems.
     
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