Unformatted text preview: ( C. ½)(½ explicitly list
byThis frommoleculesby½)(And, in=addition, let’s explicitly list
moves done the
states
aother identical. of
move these labelsThis(is done)(½) = ⅛
ons )
ol. ( are –(ABC) to ½(½)(in the text. ⅛
)–(ABC)
rgements: in the This)(½ ½ in on=
ofight:probability. processdone)(½) see thatN</'%0<,%=2%#).1'$O0)%='$(<.'%'),&012%P'$&'(.'.%
spontaneous In this example, the text. the lowest probability
rlaThis isare identical. text. is based we a purely statistical treatment
ations done
.ehigh
right:
tpoint )–( C rdered, i.e. or involve the internal
are
molecules can
(right:did
ABin
ight:relatively o) discuss a large fraction of the gastheenergy, the eneamber:Figure thein Figure and thedesignate ) = ⅛
ownAB )–(we (ABC)–(now let’s states#)%*0#)*%,0%(%+0&'%0&P'&'P%(&&()*'+'),5
)
½ of let’s
((or the shown
(½
olecule asleft)of 4.12. But)(½)(½) But now high designate the
ightABC)–( C ) ) container; 4.12. =(⅛)(½)(½ probability are
( ( in)–(C )
ABAB )–(And,
)–( C.
ityACby)–(BCshownan energylet’s explicitlynow have another
scale. So we let’s
ownassociated 4.12.isordered. let’saddition, let’s explicitly listthe
therACtheasB with addition, 4.12. But nowlist
emoleculerelativelyin ButFigurein designate the designate
s(
B,ystemFigure )) by in now
and is other d C. And, chamber:BA) ) (
=⅛
½ A spontaneous
( ( by )–(ABC) ()–(ABC)
BC and
AC BC )–( A
)–(
oncept of the ))in addition, ½) =(½)(½)(½)let’s explicitly list
therAC )–(spontaneity: C.½)(let’sin ⅛ process moves from
by((B, )–( B other by )( And, explicitly list
addition,
ements:C. And,
BCBC )–( the )
A A right:
)
ility)–(we draw higher probability. We also notice something
to states of from this simple progression is of great pracandhat in
ngements:
e(right:
right:
n(t one
)
heABC)–( of (ABC)–()(½)() ) =a⅛)(characterized by a more
right:
mber:tatesAB )lowest(½ C ½ (½ ½)(½) = ⅛
eCs )–(
)–(
)
(
ight:AB )–( C It((ABC)–()(½ because we )(½) illustrated that
)((ABC)–( AB))is important ) = ⅛ (½)(½have = ⅛
importance. ) AB probability re
e(right:)–(ABC) if ((½)(½)(½))= ⅛ are sequestered in one
hamber:
½)(½
( Cexample,( AC )–( Bmolecules
)–(
—(BAC AB )B) ) )–(ABC)a (½)(½)(½) = ⅛
)–(
(Cfor)–( molecules seeking
)–( )–( AC
amber:)–( AB )) all the ) particular distribution between
dividual AC (
C )–(ABC)
B
( hamber: ) (BC )–(½A ½) = ⅛ (½)(½)(½) = ⅛
c ordered
)–(ABC)
)( )
re(( BC AC con
)–(
r spontaneous ) (½)(based on purely s
a(B(AB )–( theA) ))( guration; just asais the case of all the clothes
(( )–()–( placed
dight: and AC processsingle drawer! is tatistical treatment
one in BC right:in a
A )–(
olded )–(in BCdiscuss or involve the internal energy, the enlink between probright:oneinC) ( AB )–(
and)–(
the
AB
point )–( the
and two BCwe) ) right: C )
e(A A did BC right:
right:
nd(order will)loom large in our understanding of processes in
f((the three B )with an)–( C ) same probability have another
AB )–( Cmolecules has ) ) scale. So we now
( C )–(B the
AB AB the
AC )–( AB A...
View
Full
Document
This note was uploaded on 11/08/2012 for the course CHEMISTRY 109 taught by Professor Flemingcrim during the Fall '10 term at University of Wisconsin.
 Fall '10
 FlemingCrim
 Chemistry, Mole

Click to edit the document details