Entropy of translation and rotation of a molecules

  • Thread starter LmdL
  • Start date
  • #1
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Homework Statement


Hi all,
There is a question from the course book:
abyysfQ.png

X1kSDKo.png



Homework Equations


[tex]S=k_B ln W[/tex]

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!
 

Answers and Replies

  • #2
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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?
 
  • #3
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Already submitted my version of the HW, so it's not relevant.
 

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