The First Law of Thermodynamics
1911
process
d
b
→
d
b
db
db
U
Q
W
→
Δ
=
−
0,
db
W
=
as already noted.
30.0 J
8.0 J
22.0 J.
d
b
b
d
U
U
U
→
Δ
=
−
=
−
= +
Then
22.0 J
db
d
b
db
Q
U
W
→
= Δ
+
= +
(positive; heat absorbed).
E
VALUATE
:
The signs of our calculated
ad
Q
and
db
Q
agree with the problem statement that heat is absorbed in
these processes.
19.44.
I
DENTIFY
:
.
U
Q
W
Δ
=
−
S
ET
U
P
:
0
W
=
when
0.
V
Δ
=
E
XECUTE
:
For each process,
.
Q
U
W
= Δ
+
No work is done in the processes
ab
and
dc
, and so
450 J
bc
abc
W
W
=
=
and
120 J.
ad
adc
W
W
=
=
The heat flow for each process is: for
,
90 J.
ab Q
=
For
,
bc
440 J
450 J
890 J.
Q
=
+
=
For
,
180 J
120 J
300 J.
ad
Q
=
+
=
For
,
350 J.
dc Q
=
Heat is absorbed in each process. Note that the arrows
representing the processes all point in the direction of increasing temperature (increasing
U
).
E
VALUATE
:
U
Δ
is path independent so is the same for paths
adc
and
abc
.
300 J
350 J
650 J.
adc
Q
=
+
=
90 J
890 J
980 J.
abc
Q
=
+
=
Q
and
W
are path dependent and are different for these two paths.
19.45.
I
DENTIFY
:
Use
pV
nRT
=
to calculate
/
.
c
a
T
T
Calculate
U
Δ
and
W
and use
U
Q
W
Δ
=
−
to obtain
Q
.
S
ET
U
P
:
For path
ac
, the work done is the area under the line representing the process in the
pV
diagram.
E
XECUTE
:
(a)
5
3
5
3
(1.0
10
J)(0.060 m
)
1.00.
(3.0
10
J)(0.020 m
)
c
c
c
a
a
a
T
p V
T
p V
×
=
=
=
×
.
c
a
T
T
=
(b)
Since
,
c
a
T
T
=
0
U
Δ
=
for process
abc
. For
ab
,
0
V
Δ
=
and
0.
ab
W
=
For
bc
,
p
is constant and
5
3
3
(1.0
10
Pa)(0.040 m
)
4.0
10
J.
bc
W
p V
=
Δ
=
×
=
×
Therefore,
3
4.0
10
J.
abc
W
= +
×
Since
0,
U
Δ
=
3
4.0
10
J.
Q
W
=
= +
×
3
4.0
10
J
×
of heat flows into the gas during process
abc
.
(c)
5
5
3
3
1
2
(3.0
10
Pa
1.0
10
Pa)(0.040 m
)
8.0
10
J.
W
=
×
+
×
= +
×
3
8.0
10
J.
ac
ac
Q
W
=
= +
×
E
VALUATE
:
The work done is path dependent and is greater for process
ac
than for process
abc
, even though the
initial and final states are the same.
19.46.
I
DENTIFY
:
For a cycle,
0
U
Δ
=
and
.
Q
W
=
Calculate
W
.
S
ET
U
P
:
The magnitude of the work done by the gas during the cycle equals the area enclosed by the cycle in the
pV
diagram.
E
XECUTE
:
(a)
The cycle is sketched in Figure 19.46.
(b)
4
4
3
3
(3.50
10
Pa
1.50
10
Pa)(0.0435 m
0.0280 m
)
310 J.
W
=
×
−
×
−
= +
More negative work is done for
cd
than
positive work for
ab
and the net work is negative.
310 J.
W
= −
(c)
310 J.
Q
W
=
= −
Since
0,
Q
<
the net heat flow is out of the gas.
E
VALUATE
:
During each constant pressure process
W
p V
=
Δ
and during the constant volume process
0.
W
=
Figure 19.46
19.47.
I
DENTIFY
:
Use the 1st law to relate
tot
Q
to
tot
W
for the cycle.
Calculate
ab
W
and
bc
W
and use what we know about
tot
W
to deduce
ca
W
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1912
Chapter 19
(a)
S
ET
U
P
:
We aren°t told whether the pressure increases or decreases in process
bc
. The two possibilities for
the cycle are sketched in Figure 19.47.
Figure 19.47
In cycle I, the total work is negative and in cycle II the total work is positive. For a cycle,
0,
U
Δ
=
so
tot
tot
Q
W
=
The net heat flow for the cycle is out of the gas, so heat
tot
0
Q
<
and
tot
0.
W
<
Sketch I is correct.
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 Spring '06
 Buchler
 Physics, Thermodynamics, First Law Of Thermodynamics, Heat, ΔU

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