Is Entropy Defined by the Gibbs Entropy Formula Extensive?

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SUMMARY

The Gibbs Entropy Formula demonstrates that entropy is extensive for two independent systems A and B, expressed as S(A,B) = S(A) + S(B). The formula S = -k ∑ pi ln(pi) is utilized, where pi represents the probability of each microstate. The successful derivation shows that S(A,B) can be calculated as k ln(Ω(A)Ω(B)), confirming the extensiveness of entropy through the relationship between the number of microstates for noninteracting systems.

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Homework Statement



Show explicitly that Entropy as defined by the Gibbs Entropy Formula is extensive. That is, for two independent (noninteracting) systems A and B,

S(A,B) = S(A) + S(B)

where S(A,B) is the entropy of A and B considered as part of a larger system.

Homework Equations



S = -k \sum pi ln(pi)

The Attempt at a Solution



I honestly have no idea where to start! I tried letting pi = 1/Ω, to obtain,

S = k \sum (1/Ω)ln(Ω), and then tried summing S(A) and S(B) together to obtain S(A,B), but it didn't work out. I also tried just summing up S(A) and S(B) without writing in terms of Ω...didn't work either. I then tried,

S = -k \sum pi ln(pi) ==>
S = k \sum (1/Ω) ln(Ω) ==>
S = k (1/Ω) ln(Ω)\sum 1, \sum 1 = Ω
S = k (1/Ω) ln(Ω)Ω
S = k ln(Ω)
and then I summed up S(A) and S(B) which WORKED,
S(A,B) = k ln(Ω(A))+k ln(Ω(B)) = k ln(Ω(A)Ω(B)) = k ln (Ω(A,B)), but I don't think this argument works. Plus the prof derived the Gibbs Entropy Formula from k ln Ω... so I don't think I'm even on the right track! Any ideas or suggestions? Thanks!
 
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Tsar_183 said:
S(A,B) = k ln(Ω(A))+k ln(Ω(B)) = k ln(Ω(A)Ω(B)) = k ln (Ω(A,B)), but I don't think this argument works.
This looks fine to me. The number of microstates for two noninteracting systems Ω(A,B) is the product of the number of microstates for each system individually Ω(A)Ω(B).
 

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