Most2 two chamber system done inthe thermodynamic

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Unformatted text preview: hat this disarmingly simple statement leads must increase. 1. ability distributionhamber: molecules in a All in the le c for the two (ABC)–( ) (½)(½)(½) = ⅛ after which you can remove these labels to e time we have added directly to con gurations are of the This is importanttext. the identify whichthe formulationidentical.most2. two chamber system. done inthe thermodynamic variable –)–(ABC) All in the right chamber: ( (½)(½)(½) = ⅛ )*+,-./01,234+,0*+,-5/6763.308,/9,4753/:1,;310536:</=1,/9, (which may sound like a Gibbs Free Energy. e remainder of the chapter and signi cant parts of chapters uld only correspond to 3. Two 4.12. lereactions in electrochem6, 7, and 8 that treat chemical equilibrium, >/.+?:.+1,6+0@++=,0*+,.+A,7=;,532*0,-/5</=,/9,0*+, acid-base nowone designate Let’s add a third molecule as shown in Figurein the But and the andthe right: f O2) the chance of all of 4.10 We begin this (important!) thought experiment with let’schambers the C ) evacuAB )–( B71C",D/@,>7=8,@781,75+,0*+5+,0/,-.7?+,EE,/9,0*+, developing ng to the left or to the right istry, are by A and by B, and the a thorough understanding of Gibbs Free( Energy. molecules built upon indistinguishable gas-phase molecules that weexplicitly over a ated except Hundred Molecules other by C. And, in addition, let’s observe list B ) for two !"#$%&'% zero. In fact, the possibilAC )–( >/.+?:.+1,3=,0*+,.+A,13;+,9/5,0*+,=F'!!,>/.+?:.+,?71+G out the possible arrangements: we, at certain intervals, count the number of(times we period of time during which n to even be skewed to ( BC )–( A ) r diminishes rapidly as we observe the available con gurations, which H",' both)(½)(½) = ⅛ the le -hand !"'! 1. All in the le chamber: (ABC)–( are (i) (½ molecules in ) equal division of the mol- chamber, (ii) both molecules in the right-hand chamber,a(iii)two in the right: the 4. One in the le nd one molecule in 2. All in the right chamber: ( )–(ABC) (½)(½)(½) = ⅛ trated that this process, ( C )–( AB ) I",'!! K+,?/:.;,?*//1+,7=8,/9,0*+,'!!, as molecules throughout a le -hand chamber and one molecule in the right-hand chamber. We can draw a B )–( AC 3. Two in thatle and one in the right: the summarizes our ndings. e probability of >/.+?:.+1,0/,6+,0*+,/=+,/=,0*+,532*0", !"!( nding one (molecule ) neous process by a purely histogram J",#'!! ( ABtheatC ) molecular level from two Aa )–( BC ) )–( probability of getting “heads”(in coin of the system. We did not in the leexamine the issueis ½, just as We -hand chamber of probability the complesuch as energy or en( AC )–( B ) toss is ½ e probability of nding both molecules incon leuration of the three molecules has the same probability the g -hand chamber is mentary .perspectives. But an to say that the energy !"! ( of tossing ) “heads in a row is (½)(½) = ¼). )–( Awhile each (½)(½H..,.+A ), just as the probabilityBCwe distinguish the individual molecules(((½)(½)(½) = ⅛), if th...
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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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