Is Entropy Defined by the Gibbs Entropy Formula Extensive?

  • Thread starter Thread starter Tsar_183
  • Start date Start date
  • Tags Tags
    Entropy
Click For Summary
The discussion centers on demonstrating that entropy defined by the Gibbs Entropy Formula is extensive, specifically for two independent systems A and B. The key equation to prove is S(A,B) = S(A) + S(B), where S represents entropy. The initial attempts involved manipulating the formula using the number of microstates, Ω, but the user expressed confusion about the validity of their approach. A crucial point made is that the number of microstates for two noninteracting systems is the product of their individual microstates, which supports the extensive property of entropy. Ultimately, the argument aligns with the Gibbs Entropy Formula, confirming its extensiveness.
Tsar_183
Messages
6
Reaction score
0

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!
 
Last edited:
Physics news on Phys.org
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).
 

Similar threads

Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K