Sowefinallyget p t s v v s

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Unformatted text preview: 
 5
empty
spaces
and
31
molecules.
 
 
 
 
 
 SOLID
PHASE
 For
the
solid
phase
at
0K,
the
number
of

 configurations
is
given
by
W
=
36!
/
(
0!
x
36!
)

 since
there
are
36
sites,
0
empty
space
and

 36
molecules.
We
can
now
calculate
the
entropy

 for
each
of
the
“model”
phases.
 The
number
of
configurations
W(liq)
and
W(sol)
for
the
above
examples
 are:
376,992
for
the
liquid
phase
and
1
for
the
solid
phase.

 Therefore
ln
W(liq)
=
12.84
and
ln
W(sol)
=
0
 Marand’s
Notes:
Chapter
3
‐
The
Second
Law
of
Thermodynamics
 115
 We
must
account
for
the
fact
that
each
“model”
phase
examined
above
has
 a
different
number
of
molecules.
Therefore
we
must
calculate
ln(W)
on
a
 per
molecule
basis
and
we
obtain:
 ln
W(liq)
=
12.84
/
31
=
0.414
and

 ln
W(sol)
=
0
/
36
=
0
 We
can
now
calculate
the
molar
entropy
(for
one
mole
of
molecules)
as:

 Sm
=
Nav
k
ln
W
which
leads
to:
Sm
=
R
ln
W
 Sm
(liq)
=
8.3145
ln
W(liq)
=
3.4
J/(K.mol)
and
 Sm
(sol)
=
8.3145
ln
W(sol)
=
0
J/(K.mol)
at
0K
 Obviously,
the
above
values
should
not
be
taken
too
seriously,
because
our
 model
is
very
crude.
However,
these
calculations
show
that
the
molar
 entropy
of
the
liquid
is
larger
than
that
of
the
solid
at
0
K.

This
confirms
the
 notion
that
entropy
is
a
measure
of
the
disorder
present
in
the
system
and
 that
the
entropy
of
a
solid
at
0
K,
where
the
solid
is
in
perfect
crystalline
 order
is
0.
Note
that
at
temperatures
above
0
K,
the
entropy
of
the
solid
is
 small
but
non‐zero,
because
of
the
vibrations
present
in
molecules.
You
 could
visualize
that
because
of
the
vibrations
present,
molecules
may
 occupy
different
locations
within
the
same
lattice
site,
and
this
creates
 disorder.
At
0
K,
there
is
no
motion,
no
vibration,...
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This note was uploaded on 01/26/2014 for the course CHEM 3615 taught by Professor Aresker during the Spring '07 term at Virginia Tech.

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