Chapter 5 - Section B - Non-Numerical Solutions
5.1
Shown to the right is a
PV
diagram with two adi-
abatic lines 1
→
2 and 2
→
3, assumed to inter-
sect at point 2. A cycle is formed by an isothermal
line from 3
→
1. An engine traversing this cycle
would
produce
work. For the cycle
±
U
=
0, and
therefore by the ﬁrst law,
Q
+
W
=
0. Since
W
is negative,
Q
must be positive, indicating that
heat is absorbed by the system. The net result
is therefore a complete conversion of heat taken
in by a cyclic process into work, in violation of
Statement 1a
of the second law (Pg. 156). The
assumption of intersecting adiabatic lines is there-
fore false.
5.5
The energy balance for the over-all process is written:
Q
=
±
U
t
+
±
E
K
+
±
E
P
Assuming the egg is not scrambled in the process, its internal-energy change after it returns to its initial
temperature is zero. So too is its change in kinetic energy. The potential-energy change, however, is
negative, and by the preceding equation, so is
Q
. Thus heat is transferred to the surroundings.
The total entropy change of the process is:
±
S
total
=
±
S
t
+
±
S
t
surr
Just as
±
U
t
for the egg is zero, so is
±
S
t
. Therefore,
±
S
total
=
±
S
t
surr
=
Q
surr
T
σ
=
−
Q
T
σ
Since
Q
is negative,
±
S
total
is positive, and the process is irreversible.
5.6
By Eq. (5.8) the thermal efﬁciency of a Carnot engine is:
η
=
1
−
T
C
T
H
Differentiate:
±
∂η
∂
T
C
²
T
H
=−
1
T
H
and
±
∂
T
H
²
T
C
=
T
C
T
H
2
=
T
C
T
H
1
T
H
Since
T
C
/
T
H
is less unity, the efﬁciency changes more rapidly with
T
C
than with
T
H
. So in theory it is
more effective to decrease
T
C
. In practice, however,
T
C
is ﬁxed by the environment, and is not subject
to control. The
practical
way to increase
η
is to increase
T
H
. Of course, there are limits to this too.
5.11
For an ideal gas with constant heat capacities, and for the changes
T
1
→
T
2
and
P
1
→
P
2
, Eq. (5.14)
can be rewritten as:
±
S
=
C
P
ln
±
T
2
T
1
²
−
R
ln
±
P
2
P
1
²
(
a
)I
f
P
2
=
P
1
,
±
S
P
=
C
P
ln
±
T
2
T
1
²
If
V
2
=
V
1
,
P
2
P
1
=
T
2
T
1
Whence,
±
S
V
=
C
P
ln
±
T
2
T
1
²
−
R
ln
±
T
2
T
1
²
=
C
V
ln
±
T
2
T
1
²
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