P21_063 - Thus, W = 3! 2! 1! = 3 . (d) The most likely...

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63. We adapt the discussion of § 21-7 to 3 and 5 particles (as opposed to the 6 particle situation treated in that section). (a) The least multiplicity conFguration is when all the particles are in the same half of the box. In this case, using Eq. 21-18, we have W = 3! 3! 0! =1 . (b) Similarly for box B , W =5! / (5! 0!) = 1 in the “least” case. (c) The most likely conFguration in the 3 particle case is to have 2 on one side and 1 on the other.
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Unformatted text preview: Thus, W = 3! 2! 1! = 3 . (d) The most likely conFguration in the 5 particle case is to have 3 on one side and 2 on the other. Thus, W = 5! 3! 2! = 10 . (e) We use Eq. 21-19 with our result in part (c) to obtain S = k ln W = ( 1 . 38 10 23 ) ln 3 = 1 . 5 10 23 J / K . (f) Similarly for the 5 particle case (using the result from part (d)), we Fnd S = k ln 10 = 3 . 2 10 23 J/K....
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