Ment times higher than the other twocan c lls o

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Unformatted text preview: assigning a probability probability that is the system and n out how and arrangements have a to each arrangemolecules being distributed equally three becomes states the probability of each con guration changes.ment times higher than the other twocan c lls o uniformly on both sides of the distribution, and there are everythingmore di leftult, but we !"#$%&'' !"#$%&') with For three identical molecules, eight di erent begin to!"#$%&'(general trend start to emerge. Two Molecules Three and everything on see a on the Molecules Ten Molecules 1 possible states. FIGURE we consider two molecules in the right. If you want fto from this to yourself, arrangements are possible, but only four of For example,4.11 Iseeprove the histograms we can !"% you an label each molecule (say A, B, and C) these we have not the but 100 molecules, canotherwise vacated bulb with of chambers d this to the case for which are distinct. For 10, arrangements of for ten molecules that the probabilitytwothe !"& and connected distributed roughly urations, two molecules on the left done by computer, write down all the possible con gwe calculate e 4.14. Now the individual counting is best and two molecules molecules!"$ by a passageway andequally being after which you can two sides is much more on the right, the states can be realized by ar- between each of theremove these labels to in the !"$ reside in the le or in the right the probability that both molecules are e probability that all molecules atoms in three di erent ways, so identify which conatgurations are identical. = ¼; is done in the text. left chamber the same time, (½)(½) This ranging the !"# lly zero; and, moreover, these the probability that the distribution then than the distribution being skewed that arrangements have a probability that is probable we calculate the probability that both to either the right or the left side. !"# molecules are in the right chamber at the !"# three times higher than the other two unlikely Let’s add a third molecule as shown in Figure 4.12. But now let’s designate the tly skewed to one chamber or the other is increasingly states !"' same time, (½)(½) = ¼; then we calculate the molecules is increased. with everything on the left and everything on molecules rmodynamics that statesthe righthen- And, in addition, let’s explicitly list by A and by B, and is in other w C. the that by place the foundation for the S yourself, e probability that one molecule the right. If you want to prove this to econd Law of !"! !"! !"! you can label each molecule (say A, B,H..,532*0 thechamber and arrangements:chamber, ½, we and C) out universe, the total the left ever a H..,.+A spontaneous event takes place in the possible one is in entropy ofH..,532*0 the universe H..,.+A H..,.+A H..,532*0 The Case for 100 Molecules the possible con gurations, can calculate a histogram de ning the proband write down allWhat is remarkable is t...
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This note was uploaded on 11/08/2012 for the course CHEMISTRY 109 taught by Professor Flemingcrim during the Fall '10 term at Wisconsin.

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