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. 2119 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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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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