Chapter 11 - Section B - Non-Numerical Solutions
11.6
Apply Eq. (11.7):
¯
T
i
≡
∂(
nT
)
∂
n
i
P
,
T
,
n
j
=
T
∂
n
∂
n
i
T
,
P
,
n
j
=
T
¯
P
i
≡
∂(
nP
)
∂
n
i
P
,
T
,
n
j
=
P
∂
n
∂
n
i
T
,
P
,
n
j
=
P
11.7
(
a
) Let
m
be the mass of the solution, and define the partial molar mass by:
¯
m
i
≡
∂
m
∂
n
i
T
,
P
,
n
j
Let
M
k
be the molar mass of species
k
. Then
m
=
k
n
k
M
k
=
n
i
M
i
+
j
n
j
M
j
(
j
=
i
)
and
∂
m
∂
n
i
T
,
P
,
n
j
=
∂(
n
i
M
i
)
∂
n
i
T
,
P
,
n
j
=
M
i
Whence,
¯
m
i
=
M
i
(
b
) Define a partial
specific
property as:
˜
M
i
≡
∂
M
t
∂
m
i
T
,
P
,
m
j
=
∂
M
t
∂
n
i
T
,
P
,
m
j
∂
n
i
∂
m
i
T
,
P
,
m
j
If
M
i
is the molar mass of species
i
,
n
i
=
m
i
M
i
and
∂
n
i
∂
m
i
T
,
P
,
m
j
=
1
M
i
Because constant
m
j
implies constant
n
j
, the initial equation may be written:
˜
M
i
=
¯
M
i
M
i
11.8
By Eqs. (10.15) and (10.16),
¯
V
1
=
V
+
x
2
dV
dx
1
and
¯
V
2
=
V
−
x
1
dV
dx
1
Because
V
=
ρ
−
1
then
dV
dx
1
=
−
1
ρ
2
d
ρ
dx
1
whence
¯
V
1
=
1
ρ
−
x
2
ρ
2
d
ρ
dx
1
=
1
ρ
1
−
x
2
ρ
d
ρ
dx
1
=
1
ρ
2
ρ
−
x
2
d
ρ
dx
1
¯
V
2
=
1
ρ
+
x
1
ρ
2
d
ρ
dx
1
=
1
ρ
1
+
x
1
ρ
d
ρ
dx
1
=
1
ρ
2
ρ
+
x
1
d
ρ
dx
1
With
ρ
=
a
0
+
a
1
x
1
+
a
2
x
2
1
and
d
ρ
dx
1
=
a
1
+
2
a
2
x
1
these become:
¯
V
1
=
1
ρ
2
[
a
0
−
a
1
+
2
(
a
1
−
a
2
)
x
1
+
3
a
2
x
2
1
]
and
¯
V
2
=
1
ρ
2
(
a
0
+
2
a
1
x
1
+
3
a
2
x
2
1
)
679