Identify which con gurations are identical this is

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Unformatted text preview: olecules in a after which you can remove these labels to e time we have added two in the text. identify which con gurations are identical. This is done chamber system. (which may sound like a uld only correspond to Let’s add a third molecule as shown in Figure 4.12. But nowthe chambers the f O2) the chance of all of 4.10 We begin this (important!) thought experiment with let’s designate evacung to the left or to the right molecules by A and by B, and the other gas-phase molecules that weexplicitly over a for two indistinguishable by C. And, in addition, let’s observe list !"#$%&'% zero. In fact, the possibil- ated except Hundred Molecules 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 r diminishes rapidly as we observe the available con gurations, which are (i) both)(½)(½) = ⅛ the le -hand !"'! 1. All in the le chamber: (ABC)–( ) (½ molecules in equal division of the mol- chamber, (ii) both molecules in the right-hand chamber, (iii) one molecule in the 2. All in the right chamber: ( )–(ABC) (½)(½)(½) = ⅛ trated that this process, as molecules throughout a le -hand chamber and one molecule in the right-hand chamber. We can draw a 3. Two in thatle and one in the right: the summarizes our ndings. e probability of nding one molecule !"!( neous process by a purely histogram ( as )–( C ) of the system. We did not in the le -hand chamber is ½, just ABthe probability of getting “heads” in a coin such as energy or en( AC )–( B ) toss is ½. e probability of nding both molecules in the le -hand chamber is an to say that the energy !"! ( of )–( A two (½)(½)**+*12 ), just as the probabilityBC tossing ) “heads in a row is (½)(½) = (¼). ) = (¼ )**+,-./0 not zero. So, we can o er the in the le a applies to the right-hand chamber; the probability of nding on of spontaneity by saying ButOne same logic nd two in the right: 4. ess moves from states of both molecules in the le -hand chamber is (½)(½) = (¼). e probability of nd( C )–( AB ) tes of high probability. In this example, we see that the lowest probability ing one molecule in the le chamber )–( AC ) molecule in the right chamber is and one ( that are relatively ordered, i.e. a large fraction of the gas molecules can B ½. and the states of high probability are )–( BC ) (A he right or the left of the container; us, we construct our histogram accordingly. Notice that for our case of two e the system is relatively disordered. identical molecules, there are three (n + 1) possible con gurations and that there But while each con g nding the molecules evenly has the same probability is a higher probability ofuration of the three moleculesdistributed between the two if we distinguish the individual molecules usion that we draw fromchambers. progression is of great prac- ((½)(½)(½) = ⅛), if the molecules are this simple indistinguishable we have a probability of cal) importance. It is importa...
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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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