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Entropy of translation and rotation of a molecules
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[QUOTE="LmdL, post: 4945777, member: 232683"] [h2]Homework Statement [/h2] Hi all, There is a question from the course book: [ATTACH=full]174809[/ATTACH] [ATTACH=full]174810[/ATTACH] [h2]Homework Equations[/h2] [tex]S=k_B ln W[/tex] [h2]The Attempt at a Solution[/h2] 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: [ATTACH=full]174811[/ATTACH] 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! [/QUOTE]
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Entropy of translation and rotation of a molecules
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