Marandsnoteschapter3thesecondlawofthermodynamics 118

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Unformatted text preview: 
=
‐Pext
ΔV
 Therefore:

3R
(TB
‐
TA)
=
‐Pext
(VB
‐
VA)
 Since
VB
=
n
R
TB
/
PB
and
PB
=
Pext
we
get:
 TB
=
(TA
+
Pext
VA
/
3R)
/
(1
+
n
/
3)
which
leads
to
TB
=
192.44
K
and

 VB
=
32
L.
 Now
that
we
know
the
final
and
the
initial
states,
we
can
design
any
 number
of
reversible
processes
between
A
and
B
(see
figure
below).
 Marand’s
Notes:
Chapter
3
‐
The
Second
Law
of
Thermodynamics
 108
 1)
 We
can
envision,
an
isobaric
slow
heating
from
state
A
to
a
state
C,
 characterized
by:
PC
=
PA
and
VC
=
VB.
The
temperature
at
state
C
is
defined
 by
TC
=
PC
VC
/
nR
.


 Therefore,
TC
=
PA
VB
/
nR
=
1924.3
K.
Then,
we
carry
out
an
isochoric
slow
 cooling
from
state
C
to
state
B.
 The
change
in
entropy
between
A
and
B,
ΔSA‐B
=
ΔSA‐C
+
ΔSC‐B
is
given
by:
 ΔSA‐C

=
n
CPm
ln(TC
/
TA)
=
2
(5R/2)
ln
(1924.3/300.68)
=
77.16
J/K
and
 ΔSC‐B

=
n
Cvm
ln(TB
/
TC)
=
2
(3R/2)
ln
(192.44/1924.3)
=
‐
57.43
J/K.
 The
change
in
entropy
for
the
A‐B
process
is
therefore
equal
to
19.73
J/K
 
 2)
 We
could
have
also
envisioned
the
path
A

D

B,
which
consists
 in
a
slow
isochoric
cooling
from
A
to
D
followed
by
a
slow
isobaric
heating
 from
D
to
B
(see
figure
next
page).
 State
D
is
characterized
by
PD
=
PB
and
VD
=
VA.
In
this
case
TD
=
PD
VD
/
n
R
 which
is
to
say
TD
=
PB
VA
/
n
R
=
30.07
K
 The
entropy
change
between
A
and
B
is
therefore
the
sum
of
the
entropy
 changes
between
A
and
D
and
between
D
and
B.
Therefore:
 ΔSA‐D

=
n
CVm
ln(TD
/
TA)
=
2
(3R/2)
ln
(30.07/300.68)
=
...
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