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Entropy of translation and rotation of a molecules

  1. Dec 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Hi all,
    There is a question from the course book:
    abyysfQ.png
    X1kSDKo.png


    2. Relevant equations
    [tex]S=k_B ln W[/tex]

    3. The attempt at a solution
    My solution:
    So first of all, for each molecule, there are 2 motions: translational and rotational.
    For rotational I get:
    [tex]W_1 =\Omega \left ( \theta \right )[/tex]
    For translational, I should calculate the free (non forbidden volume) for molecule to move in. Since the overall volume is V and there are N molecules, the volume allowed for moving is:
    [tex]W_2 =V-NA \left ( \theta \right )[/tex]
    Since molecules are distinguishable, I need to divide by (N!)^(1/N) for each molecule, so:
    [tex]W_2 =\frac{V-NA \left ( \theta \right )}{\sqrt[N]{N!}}[/tex]
    Therefore, the entropy per molecule is:
    [tex]s= k_B ln \left ( \frac{V-NA \left ( \theta \right )}{\sqrt[N]{N!}}\Omega \left ( \theta \right ) \right )[/tex]
    And since there are N molecules:
    [tex]S= k_B N ln \left ( \frac{V-NA \left ( \theta \right )}{\sqrt[N]{N!}}\Omega \left ( \theta \right ) \right )=k_B ln \left (\frac{1}{N!} \left ( V-NA \left ( \theta \right ) \right )^N \left ( \Omega \left ( \theta \right ) \right )^N \right )[/tex]

    But according the book, the answer is:
    4XSXUoh.png

    So I have 2 questions:
    1. Do they mixed up between Omega and A? Or it's me?
    2. Where do a factor of 2 came from in the expression for the volume?

    Thanks!
     
  2. jcsd
  3. Dec 19, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Dec 19, 2014 #3
    Already submitted my version of the HW, so it's not relevant.
     
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