Thisconfirmsthe

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: he
vapor
 from
110°C
to
100°C,
transforming
reversibly
the
water
vapor
to
water
 liquid
at
100°C
and
1
bar,
cooling
slowly
the
water
liquid
from
100°C
to
 44°C.
The
total
entropy
change
is
then
given
by:
 S 500 273 CPm 500 Δ fusH∅ 500 317 CL ΔS = dT + + ∫ ∫ Pm dT + 18 263 T 18 273 18 273 T ∅ V 100 373 CPm 100 Δ vapH 100 317 CL dT − + ∫ ∫ Pm dT 18 383 T 18 373 18 373 T ΔS = 
 ∅ 317 500 S 273 500 Δ fusH 500 L CPm ln + CPm ln + + 18 18 263 18 273 273 ∅ 317 100 V 373 100 Δ vapH 100 L CPm ln + CPm ln − 18 18 383 18 373 383 We
find
that
ΔS
=
233
J.K‐1.
 € 
 Example
#3
 Consider
the
adiabatic
irreversible
expansion
of
two
moles
of
an
ideal
 monoatomic
gas
under
a
constant
external
pressure,
Pext
=
1
bar.
In
the
 Marand’s
Notes:
Chapter
3
‐
The
Second
Law
of
Thermodynamics
 107
 initial
state,
the
gas
occupies
a
volume
VA
=
5
L
under
a
pressure
of

 PA
=
10
bars.
In
the
final
state
the
pressure
is
PB
=
1
bar.
 First,
we
need
to
define
initial
and
final
states
carefully.
 We
know
initial
pressure,
volume
and
number
of
moles,
therefore
we
can
 calculate
the
initial
temperature
TA
 TA
=
PA
VA
/
n
R
=
(106
x
5.10‐3)
/
(2
x
8.3145)
=
300.68
K
 The
final
state
is
given
by
writing
that
the
heat
is
equal
to
zero
for
an
 adiabatic
process:

dU
=
δw
+
δq
=
δw
 Since
the
gas
is
ideal,
dU
can
be
written
as:
dU
=
CV
dT
and
ΔU
=
CV
ΔT

 (CV
=
2
mol
x
3R/2

J/K.mol,
which
is
independent
of
temperature
for
an
 ideal
monoatomic
gas).
 Since
the
expansion
occurs
under
constant
external
pressure
then

 δw
=
‐Pext
dV
and
w...
View Full Document

Ask a homework question - tutors are online