191
T
HE
F
IRST
L
AW OF
T
HERMODYNAMICS
19.1.
(a)
I
DENTIFY
and
S
ET
U
P
:
The pressure is constant and the volume increases.
The
pV
diagram is
sketched in Figure 19.1
Figure 19.1
(b)
2
1
V
V
W
p dV
=
∫
Since
p
is constant,
2
1
2
1
(
)
V
V
W
p
dV
p V
V
=
=
−
∫
The problem gives
T
rather than
p
and
V
, so use the ideal gas law to rewrite the expression for
W
.
E
XECUTE
:
pV
nRT
=
so
1
1
1
,
pV
nRT
=
2
2
2
;
p V
nRT
=
subtracting the two equations gives
2
1
2
1
(
)
(
)
p V
V
nR T
T
−
=
−
Thus
2
1
(
)
W
nR T
T
=
−
is an alternative expression for the work in a constant pressure process for an ideal gas.
Then
2
1
(
)
(2.00 mol)(8.3145 J/mol
K)(107 C
27 C)
1330 J
W
nR T
T
=
−
=
⋅
°
−
°
= +
E
VALUATE
:
The gas expands when heated and does positive work.
19.2.
I
DENTIFY
:
At constant pressure,
.
W
p V
nR T
=
Δ
=
Δ
S
ET
U
P
:
8.3145 J/mol
K.
R
=
⋅
T
Δ
has the same numerical value in kelvins and in
C
°.
E
XECUTE
:
3
1.75
10
J
35.1 K.
(6 mol) (8.3145 J/mol
K)
W
T
nR
×
Δ
=
=
=
⋅
K
C
T
T
Δ
= Δ
and
2
27.0 C
35.1 C
62.1 C.
T
=
°
+
°
=
°
E
VALUATE
:
When
0
W
>
the gas expands. When
p
is constant and
V
increases,
T
increases.
19.3.
I
DENTIFY
:
Example 19.1 shows that for an isothermal process
1
2
ln(
/
).
W
nRT
p
p
=
pV
nRT
=
says
V
decreases
when
p
increases and
T
is constant.
S
ET
U
P
:
358.15 K.
T
=
2
1
3
.
p
p
=
E
XECUTE
:
(a)
The
pV
diagram is sketched in Figure 19.3.
(b)
1
1
(2.00 mol)(8.314 J/mol
K)(358.15 K)ln
6540 J.
3
p
W
p
⎛
⎞
=
⋅
= −
⎜
⎟
⎝
⎠
E
VALUATE
:
Since
V
decreases,
W
is negative.
Figure 19.3
19.4.
I
DENTIFY
:
Use the expression for
W
that is appropriate to this type of process.
S
ET
U
P
:
The volume is constant.
E
XECUTE
:
(a)
The
pV
diagram is given in Figure 19.4.
(b)
Since
0,
0.
V
W
Δ
=
=
19
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192
Chapter 19
E
VALUATE
:
For any constant volume process the work done is zero.
Figure 19.4
19.5.
I
DENTIFY
:
Example 19.1 shows that for an isothermal process
1
2
ln(
/
).
W
nRT
p
p
=
Solve for
1
.
p
S
ET
U
P
:
For a compression (
V
decreases)
W
is negative, so
518 J.
W
= −
295.15 K.
T
=
E
XECUTE
:
(a)
1
2
ln
.
W
p
nRT
p
⎛
⎞
=
⎜
⎟
⎝
⎠
/
1
2
.
W
nRT
p
e
p
=
518 J
0.692.
(0.305 mol)(8.314 J/mol
K)(295.15 K)
W
nRT
−
=
= −
⋅
/
0.692
1
2
(1.76 atm)
0.881 atm.
W
nRT
p
p e
e
−
=
=
=
(b)
In the process the pressure increases and the volume decreases. The
pV
diagram is sketched in Figure 19.5.
E
VALUATE
:
W
is the work done by the gas, so when the surroundings do work on the gas,
W
is negative.
Figure 19.5
19.6.
(a)
I
DENTIFY
and
S
ET
U
P
:
The
pV
diagram is sketched in Figure 19.6.
Figure 19.6
(b)
Calculate
W
for each process, using the expression for
W
that applies to the specific type of process.
E
XECUTE
:
1
2,
→
0,
V
Δ
=
so
0
W
=
2
3
→
p
is constant; so
5
3
3
4
(5.00
10
Pa)(0.120 m
0.200 m
)
4.00
10
J
W
p
V
=
Δ
=
×
−
= −
×
(
W
is negative since the volume
decreases in the process.)
4
tot
1
2
2
3
4.00
10
J
W
W
W
→
→
=
+
= −
×
E
VALUATE
:
The volume decreases so the total work done is negative.
19.7.
I
DENTIFY
:
Calculate
W
for each step using the appropriate expression for each type of process.
S
ET
U
P
:
When
p
is constant,
.
W
p V
=
Δ
When
0,
V
Δ
=
0.
W
=
E
XECUTE
:
(a)
13
1
2
1
32
24
2
1
2
41
(
),
0,
(
) and
0.
W
p V
V
W
W
p
V
V
W
=
−
=
=
−
=
The total work done by the system is
13
32
24
41
1
2
2
1
(
)(
),
W
W
W
W
p
p
V
V
+
+
+
=
−
−
which is the area in the
pV
plane enclosed by the loop.
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 Spring '11
 Shaefer

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