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18.88:
a) Following Example 18.4,
,
RT
pM
dy
dP

=
which in this case becomes
,
0
αy
T
dy
R
Mg
p
dp


=
which integrates to
.
1
or
,
1
ln
ln
0
0
0
0
α

=

=
R
Mg
T
αy
p
p
T
αy
Rα
Mg
p
p
b)
Using the first equation above, for sufficiently small
,
)
1
ln(
,
0
0
T
αy
T
αy
α

≈

and this
gives the expression derived in Example 18.4.
c)
,
8154
.
0
K)
288
(
m)
m)(8863
C
10
6
.
0
(
1
2
=
°
×


6576
.
5
m)
C
10
K)(0.6
mol
J
3145
.
8
(
)
s
m
80
.
9
)(
10
8
.
28
(
2
2
3
=
°
×
⋅
×
=


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