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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 gasphase 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 righthand 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 righthand 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 righthand 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.
 Fall '10
 FlemingCrim
 Chemistry, Mole

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