Unformatted text preview: the histogram can calculate a histogram de ning the probmolecules molecules and
distributed roughly equally
ability distribution for the two molecules in a
after which you can remove these labels to
between each of the the center.
ce time we have added a very narrow spike attwo sides is much more is done chamber system.
ule distribution is
identify which con gurations are identical. This two in the text. H..,532*0 L5/6763.308 )*+,-./01,234+,0*+,-5/6763.308,/9,4753/:1,;310536:</=1,/9,
probable than the distribution being skewed
(which may sound like a
to either the right or the left side.
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
We begin this (important!) thought experiment with let’s designate evacuB71C",K*70,31,0*+,-5/6763.308,0*70,7..,/9,0*+,>/.+?:.+1,
ng to the left or to the right
for object in contact with by
experience possibil- molecules by A and by B, and the other gas-phase molecules that weexplicitly over a
we except Hundred indistinguishable a C. And, in addition, let’s observe list
zero. In fact, thethat when atedplace a hot two Molecules
out the possible arrangements: we, at certain intervals, count the number of times we
al even be skewed w
period of time during which
n toenergy will oto spontaneously from the high temperature
r diminishes rapidly as we observe the available con
emperature body. But why All in the le ithamber: gurations,do
1. is!"'! Does c have anything to which H",'M#,F,!"( )(½) = ⅛ the le -hand
(ABC)–( are (i) both)(½
) (½ molecules in
equal order of the mol- chamber, (ii) both molecules in the right-hand chamber, (iii) one molecule in the
With divisionand disorder?
2. All in the right chamber:
trated that this process,
I",N'M#O ½)(½)(½) = ⅛
as molecules throughout a le -hand chamber and one molecule in the right-hand chamber. We can draw a
3. Two in the summarizes our ndings. e probability of nding one molecule
neous process by a purely histogram thatle and one in the right:
( as )–( C )
of the system. We did not in the le -hand chamber is ½, just ABthe probability of getting “heads” in a coin 4.11
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)(½H..,.+A ), just as the probabilityBC tossing ) “heads in a row is (½)(½) = (¼).
) = (¼
not zero. So, we can o er
on of spontaneity by saying ButOne same logic nd two in the right:
4. the in the le a applies to the right-hand chamber; the probability of nd...
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