Tounderstandtheconceptofconfigurations

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vapor,
liquid
and
solids
are
33.58,
 75.29
and
36.8
J.K‐1.mol‐1,
respectively.

Again,
we
set
the
total
heat
to
be
 zero
(use
of
a
thermally
insulated
container).
Also,
the
heat
is
equal
to
the
 enthalpy
change
(constant
pressure
process).
Therefore:
 T 500 273 S 500 500 F L 0= CPmdT + Δ fusH∅ + ∫ ∫ CPmdT + 18 263 18 18 273 373 TF 100 100 100 V ∫ CPmdT − 18 Δ vapH∅ + 18 ∫ CL dT Pm 18 383 373 
 The
first
integral
corresponds
to
the
heating
of
(500/18)
moles
of
ice
from

 € ‐10°C
to
0°C,
the
second
term
to
the
fusion
of
(500/18)
moles
of
ice,
the
 third
term
to
the
heating
of
(500/18)
moles
of
liquid
from
0°C
to
TF.
The
 fourth
term
corresponds
to
the
cooling
of
(100/18)
moles
of
water
vapor
 from
110°C
to
100°C,
the
fifth
term
to
the
condensation
of
(100/18)
moles
 of
water
vapor
and
the
sixth
term
to
the
cooling
of
(100/18)
moles
of
liquid
 H2O
from
100°C
to
TF.
Note
that
we
equate
the
standard
condensation
 enthalpy
change
with
minus
the
standard
vaporization
enthalpy
change,
 since
condensation
is
the
opposite
of
vaporization.
 We
find
that
TF
=
317
K
(44°C),
a
temperature
consistent
with
the
 assumption
that
the
final
state
of
the
system
is
the
liquid
state.
 Marand’s
Notes:
Chapter
3
‐
The
Second
Law
of
Thermodynamics
 106
 We
can
now
calculate
the
entropy
change
for
the
system
by
following
the
 reversible
path
consisting
of:
Heating
slowly
the
ice
from
‐10°C
to
0°C,
 transforming
the
ice
reversibly
to
water
at
0°C
and
1
bar,
heating
up
slowly
 the
molten
ice
(water
liquid)
from
0°C
to
44°C.
Cooling
slowly
t...
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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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