The Second Law of Thermodynamics
2011
E
VALUATE
:
For a cycle
0,
U
Δ
=
so by
U
Q
W
Δ
=
−
it must be that
net
net
Q
W
=
for a cycle. We can also check
that
net
0:
U
Δ
=
net
1
2
2
3
3
1
2180 J
1050 J
1130 J
0
U
U
U
U
→
→
→
Δ
= Δ
+ Δ
+ Δ
=
−
−
=
(Carnot),
e
e
<
as it must.
20.41.
I
DENTIFY
:
,
pV
nRT
=
so
pV
is constant when
T
is constant. Use the appropriate expression to calculate
Q
and
W
for each process in the cycle.
H
.
W
e
Q
=
S
ET
U
P
:
For an ideal diatomic gas,
5
2
V
C
R
=
and
7
2
.
p
C
R
=
E
XECUTE
:
(a)
3
2.0
10
J.
a
a
p V
=
×
3
2.0
10
J.
b
b
p V
=
×
pV
nRT
=
so
a
a
b
b
p V
p V
=
says
.
a
b
T
T
=
(b)
For an isothermal process,
2
1
ln(
/
).
Q
W
nRT
V
V
=
=
ab
is a compression, with
,
b
a
V
V
<
so
0
Q
<
and heat is
rejected.
bc
is at constant pressure, so
.
p
p
C
Q
nC
T
p V
R
=
Δ
=
Δ
V
Δ
is positive, so
0
Q
>
and heat is absorbed.
cd
is
at constant volume, so
.
V
V
C
Q
nC
T
V
p
R
=
Δ
=
Δ
p
Δ
is negative, so
0
Q
<
and heat is rejected.
(c)
3
2.0
10
J
241 K.
(1.00)(8.314 J/mol
K)
a
a
a
p V
T
nR
×
=
=
=
⋅
241 K.
b
b
b
a
p V
T
T
nR
=
=
=
3
4.0
10
J
481 K.
(1.00)(8.314 J/mol
K)
c
c
c
p V
T
nR
×
=
=
=
⋅
(d)
3
3
3
0.0050 m
ln
(1.00 mol)(8.314 J/mol
K)(241 K)ln
1.39
10
J.
0.010 m
b
ab
a
V
Q
nRT
V
⎛
⎞
⎛
⎞
=
=
⋅
= −
×
⎜
⎟
⎜
⎟
⎝
⎠
⎝
⎠
3
7
2
(1.00)(
)(8.314 J/mol
K)(241 K)
7.01
10
J.
bc
p
Q
nC
T
=
Δ
=
⋅
=
×
3
5
2
(1.00)(
)(8.314 J/mol
K)(
241 K)
5.01
10
J.
ca
V
Q
nC
T
=
Δ
=
⋅
−
= −
×
net
610 J.
ab
bc
ca
Q
Q
Q
Q
=
+
+
=
net
net
610 J.
W
Q
=
=
(e)
3
H
610 J
0.087
8.7%
7.01
10
J
W
e
Q
=
=
=
=
×
E
VALUATE
:
We can calculate
W
for each process in the cycle.
3
1.39
10
J.
ab
ab
W
Q
=
= −
×
5
3
3
(4.0
10
Pa)(0.0050 m
)
2.00
10
J.
bc
W
p V
=
Δ
=
×
=
×
0.
ca
W
=
net
610 J,
ab
bc
ca
W
W
W
W
=
+
+
=
which does equal
net
.
Q
20.42.
(a)
I
DENTIFY
and
S
ET
U
P
:
Combine Eqs.(20.13) and (20.2) to eliminate
C
Q
and obtain an expression for
H
Q
in
terms of
W
,
C
,
T
and
H
.
T
1.00 J,
W
=
C
268.15 K,
T
=
H
290.15 K
T
=
For the heat pump
C
0
Q
>
and
H
0
Q
<
E
XECUTE
:
C
H
;
W
Q
Q
=
+
combining this with
C
C
H
H
Q
T
Q
T
= −
gives
H
C
H
1.00 J
13.2 J
1
/
1
(268.15/290.15)
W
Q
T
T
=
=
=
−
−
(b)
Electrical energy is converted directly into heat, so an electrical energy input of 13.2 J would be required.
(c)
E
VALUATE
:
From part (a),
H
C
H
.
1
/
W
Q
T
T
=
−
H
Q
decreases as
C
T
decreases. The heat pump is less efficient as
the temperature difference through which the heat has to be °pumped± increases. In an engine, heat flows from
H
T
to
C
T
and work is extracted. The engine is more efficient the larger the temperature difference through which the
heat flows.
20.43.
I
DENTIFY
:
b
c
T
T
=
and is equal to the maximum temperature. Use the ideal gas law to calculate
.
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 Spring '06
 Buchler
 Physics, Second Law Of Thermodynamics, Heat engine, tc, Qh, nCV ΔT

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