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Calculating Entropy

  1. Oct 27, 2016 #1
    1. The problem statement, all variables and given/known data
    52 distinguishable particles have been in a box long enough to reach equilibrium. The box is divided into two equal-volume cells. Let's say that there are 103 sub-states (s1 through s1000) available to each particle on each side, regardless of how many other particles are around. (In a more realistic case that number would be much greater and depend on the temperature, but our key results would not change.) So a microstate is specified by giving the position (left half or right half) of every particle and its sub-state within this half.

    Edit: forgot to add the question

    What is the entropy of this configuration? (26 particles in left side of box)

    2. Relevant equations
    1. Use the binomial distribution to find the number of possibilities 26 particles are in the left side of the box
    2. ln(omega) for entropy, where omega is the distribution size (ie the number of accessible microstates)

    3. The attempt at a solution
    Using the binomial formula 52 choose 26 yields 495918532948104 possibilities for 26 particles to be in the left side of the box. Among those 26 particles, they each have an additional 1000 different states, which means there are a total of 100026 different states on the left side for a particular combination.

    Therefore, the total number of microstates is (52 choose 26) * 100026. Taking the natural log of that yields 213.439, which isn't the entropy of this particular configuration.

    What am I doing wrong?

    Thanks all!
     
  2. jcsd
  3. Oct 28, 2016 #2

    BvU

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    Hello Fruit,

    What about the other side of the divider ?

    (and how do you know your answer isn't right ?)
     
  4. Oct 28, 2016 #3
    Hi BvU,

    I know my answer isn't right because it's an online homework assignment with real time feedback.

    If I consider the right side as well and add it on top of the left side, then I should have my original answer doubled. But that is still wrong.
     
  5. Oct 29, 2016 #4

    BvU

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    No. Do you understand why not ?
     
  6. Oct 29, 2016 #5
    Aha! I had the right idea.

    There are 52 choose 26 different combinations for 26 particles to be in the left side. Then there are 100026 for both sides of box. Apparently I was a factor of 100026 off in the natural log. Thanks!
     
  7. Oct 30, 2016 #6

    BvU

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    Well done. The factor 495918532948104 only appears once.
     
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