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Unformatted text preview: 57.43
J/K
and
 ΔSD‐B

=
n
CPm
ln(TB
/
TD)
=
2
(5R/2)
ln
(192.44/30.07)
=
77.16
J/K.
 Marand’s
Notes:
Chapter
3
‐
The
Second
Law
of
Thermodynamics
 109
 The
change
in
entropy
for
the
A‐B
process
is
therefore
equal
to
19.7
J/K.
 You
could
envision
any
other
reversible
path
involving
combinations
of
 adiabatic
reversible,
isothermal
reversible,
etc…
and
get
the
same
answer.

 You
may
want
to
practice
this
type
of
calculation
on
your
own
following
the
 path
A

E

B
where
the
transformation
A

E
is
isothermal
(TA
=
TE)
and
 reversible
to
a
volume
VE
=
VB
and
the
process
E

B
is
a
slow
isochoric
 cooling
from
TE
to
TB.
(See
figure
below).
You
should
again
find
that
ΔSA‐>B
is
 identical
to
that
found
by
the
other
paths.
 
 
 
 
 Marand’s
Notes:
Chapter
3
‐
The
Second
Law
of
Thermodynamics
 110
 The
Third
Law
of
Thermodynamics
and
Temperature
Dependence
of
 Standard
Molar
Entropies
 In
the
previous
chapter,
we
discussed
the
concept
of
standard
state
for
the
 enthalpy
of
formation
of
substances.

Such
physical
quantities
will
allow
us
 to
examine,
for
example,
the
composition
of
a
reactive
mixture
once
it
 reaches
chemical
equilibrium.
Along
the
same
lines,
we
can
define
standard
 molar
entropies.
It
is
not
necessary
in
this
case
to
define
the
molar
entropy
 of
formation,
as
was
done
for
enthalpies,
because
through
the
third
law
of
 thermodynamics,
we
can
specify
exactly
the
entropy
of
a
substance
at
any
 T
and
P.
 We
have
seen
earlier
that
if
P
is
constant,
then
heating
or
cooling
of
one
 mole
of
a
substan...
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