High Temperature Limit of Entropy in a Two Level System

AI Thread Summary
In a two-level system at high temperature, the entropy is calculated using the formula S = k ln(Ω), where Ω represents the number of configurations of particles. The correct expression for Ω is Ω = N! / ((N/2)!(N/2)!), leading to a calculation involving Stirling's approximation. The resulting entropy expression simplifies to S = Nk ln(2) after resolving the logarithmic terms. The discussion highlights a common mistake with sign errors during the calculation process. The final goal is to confirm that the high temperature limit of entropy is indeed Nk ln(2).
Kara386
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Homework Statement


Sounds like a physics problem but I'm sure of the physics, stuck on the maths. At high T a two level system has ##\frac{N}{2}## particles in each level. If entropy is given by ##S = k\ln(\Omega)##, where ##\Omega## is the number of ways of getting ##\frac{N}{2}## particles per level, show the high temperature limit is ##Nk\ln(2)##.

Homework Equations

The Attempt at a Solution


To the best of my knowledge, ##Omega = \frac{N!}{(\frac{N}{2})!(\frac{N}{2})!}##. Taking ##ln## of this and using the Stirling approximation:

##N\ln(N) - N - [\frac{N}{2}\ln(\frac{N}{2}) - \frac{N}{2}] - [\frac{N}{2}\ln(\frac{N}{2})-\frac{N}{2}]##
##= N\ln(N) - N\ln(\frac{N}{2})##
## = -Nln(2)##
I've gone wrong with a minus sign somewhere but I really can't see where! Thanks for any help!
 
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How do you write ##-N \ln(N/2)## as a sum of two logarithms?
 
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DrClaude said:
How do you write ##-N \ln(N/2)## as a sum of two logarithms?
Oh yes, there's the missing minus sign. Thank you! :)
 
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