Entropy of translation and rotation of a molecules

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SUMMARY

The discussion centers on the calculation of entropy for molecules considering both translational and rotational motions. The user derives the entropy formula using the equation S=k_B ln W, where W is defined for translational and rotational movements. The user questions the distinction between the symbols Omega and A in the context of volume calculations and seeks clarification on the origin of a factor of 2 in the volume expression. The correct entropy expression is confirmed to be S= k_B N ln \left ( \frac{V-NA \left ( \theta \right )}{\sqrt[N]{N!}}\Omega \left ( \theta \right ) \right ).

PREREQUISITES
  • Understanding of statistical mechanics concepts, particularly entropy.
  • Familiarity with the Boltzmann constant (k_B) and its application in thermodynamics.
  • Knowledge of distinguishable and indistinguishable particles in statistical physics.
  • Basic grasp of rotational and translational motion in molecular dynamics.
NEXT STEPS
  • Study the derivation of the Boltzmann entropy formula in detail.
  • Learn about the implications of distinguishable versus indistinguishable particles in statistical mechanics.
  • Explore the concept of phase space and its relation to entropy calculations.
  • Investigate the role of rotational and translational motions in thermodynamic systems.
USEFUL FOR

This discussion is beneficial for students of thermodynamics, physicists studying statistical mechanics, and anyone interested in molecular entropy calculations.

LmdL
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Homework Statement


Hi all,
There is a question from the course book:
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Homework Equations


S=k_B ln W

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:
W_1 =\Omega \left ( \theta \right )
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:
W_2 =V-NA \left ( \theta \right )
Since molecules are distinguishable, I need to divide by (N!)^(1/N) for each molecule, so:
W_2 =\frac{V-NA \left ( \theta \right )}{\sqrt[N]{N!}}
Therefore, the entropy per molecule is:
s= k_B ln \left ( \frac{V-NA \left ( \theta \right )}{\sqrt[N]{N!}}\Omega \left ( \theta \right ) \right )
And since there are N molecules:
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 )

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!
 
Already submitted my version of the HW, so it's not relevant.
 

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