Chapter 6  Section B  NonNumerical Solutions
6.1
By Eq. (6.8),
∂
H
∂
S
P
=
T
and isobars have positive slope
Differentiate the preceding equation:
∂
2
H
∂
S
2
P
=
∂
T
∂
S
P
Combine with Eq. (6.17):
∂
2
H
∂
S
2
P
=
T
C
P
and isobars have positive curvature.
6.2
(
a
) Application of Eq. (6.12) to Eq. (6.20) yields:
∂
C
P
∂
P
T
=
∂
{
V
−
T
(∂
V
/∂
T
)
P
}
∂
T
P
or
∂
C
P
∂
P
T
=
∂
V
∂
T
P
−
T
∂
2
V
∂
T
2
P
−
∂
V
∂
T
P
Whence,
∂
C
P
∂
P
T
= −
T
∂
2
V
∂
T
2
P
For an ideal gas:
∂
V
∂
T
P
=
R
P
and
∂
2
V
∂
T
2
P
=
0
(
b
) Equations (6.21) and (6.33) are both general expressions for
dS
, and for a given change of state
both must give the same value of
dS
. They may therefore be equated to yield:
(
C
P
−
C
V
)
dT
T
=
∂
P
∂
T
V
dV
+
∂
V
∂
T
P
d P
Restrict to constant
P
:
C
P
=
C
V
+
T
∂
P
∂
T
V
∂
V
∂
T
P
By Eqs. (3.2) and (6.34):
∂
V
∂
T
P
=
β
V
and
∂
P
∂
T
V
=
β
κ
Combine with the boxed equation:
C
P
−
C
V
=
β
T V
β
κ
6.3
By the definition of
H
,
U
=
H
−
PV
.
Differentiate:
∂
U
∂
T
P
=
∂
H
∂
T
P
−
P
∂
V
∂
T
P
or
∂
U
∂
T
P
=
C
P
−
P
∂
V
∂
T
P
566
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Substitute for the final derivative by Eq. (3.2), the definition of
β
:
∂
U
∂
T
P
=
C
P
−
β
PV
Divide Eq. (6.32) by
dT
and restrict to constant
P
. The immediate result is:
∂
U
∂
T
P
=
C
V
+
T
∂
P
∂
T
V
−
P
∂
V
∂
T
P
Solve for the two derivatives by Eqs. (6.34) and (3.2); substitution gives:
∂
U
∂
T
P
=
C
V
+
β
κ
(β
T
−
κ
P
)
V
6.4
(
a
) In general,
dU
=
C
V
dT
+
T
∂
P
∂
T
V
−
P
dV
(6.32)
By the equation of state,
P
=
RT
V
−
b
whence
∂
P
∂
T
V
=
R
V
−
b
=
P
T
Substituting this derivative into Eq. (6.32) yields
dU
=
C
V
dT
, indicating that
U
=
f
(
T
)
only.
(
b
) From the definition of
H
,
d H
=
dU
+
d
(
PV
)
From the equation of state,
d
(
PV
)
=
R dT
+
b d P
Combining these two equations and the definition of part (
a
) gives:
d H
=
C
V
dT
+
R dT
+
b d P
=
(
C
V
+
R
)
dT
+
b d P
Then,
∂
H
∂
T
P
=
C
V
+
R
By definition, this derivative is
C
P
. Therefore
C
P
=
C
V
+
R
.
Given that
C
V
is constant, then
so is
C
P
and so is
γ
≡
C
P
/
C
V
.
(
c
) For a mechanically reversible adiabatic process,
dU
=
dW
. Whence, by the equation of state,
C
V
dT
= −
P dV
= −
RT
V
−
b
dV
= −
RT
d
(
V
−
b
)
V
−
b
or
dT
T
= −
R
C
V
d
ln
(
V
−
b
)
But from part (
b
),
R
/
C
V
=
(
C
P
−
C
V
)/
C
V
=
γ
−
1.
Then
d
ln
T
= −
(γ
−
1
)
d
ln
(
V
−
b
)
or
d
ln
T
+
d
ln
(
V
−
b
)
γ
−
1
=
0
From which:
T
(
V
−
b
)
γ
−
1
=
const
.
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 brown
 Equations, Trigraph, dt, ∂t, Eqs.

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