A certain box contains 100 particles of an ideal gas. How many times more likely

Question

A certain box contains 100 particles of an ideal gas. How many times more likely is it to find the particles evenly split between the left and right halves of the box, than to find 90 particles on 1 side and 10 on the other. Given the multiplicity W of some configuration, the probability P for the same configuration is, P=W/N total where the total number of configuration is N total = 2^ N for N particles in the two-sided box. To find how many times one probability is more likely than the other, divide the first probability by the second probability.

Explanation / Answer

First probability= W/Ntotal which involves a 50:50 distribution.

P1= 1/ 2^100

Second probability involves a 9:1 ratio hence multiplicity is 9

P2= 9/2^100

HEnce the answer= P1/P2

= 1/9

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