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191
T
HE
F
IRST
L
AW OF
T
HERMODYNAMICS
19.1.
(a)
IDENTIFY
and
SET UP:
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
Wp
d
V
=
∫
Since
p
is constant,
2
1
21
()
V
V
Wpd
Vp
VV
==
−
∫
The problem gives
T
rather than
p
and
V
, so use the ideal gas law to rewrite the expression for
W
.
EXECUTE:
pV
nRT
=
so
11
1
,
pV
nRT
=
22
2
;
=
subtracting the two equations gives
pV V
nRT T
−=
−
Thus
Wn
R
TT
=−
is an alternative expression for the work in a constant pressure process for an ideal gas.
Then
(
)
(2.00 mol)(8.3145 J/mol K)(107 C
27 C)
1330 J
R
=
⋅
°
−
°
=
+
EVALUATE:
The gas expands when heated and does positive work.
19.2.
IDENTIFY:
At constant pressure,
.
Vn
R
T
=Δ = Δ
SET UP:
8.3145 J/mol K.
R
=⋅
T
Δ
has the same numerical value in kelvins and in C
°.
EXECUTE:
3
1.75 10 J
35.1 K.
(6 mol) (8.3145 J/mol K)
W
T
nR
×
Δ=
=
=
⋅
KC
Δ
and
2
27.0 C 35.1 C
62.1 C.
T
=°
+°
EVALUATE:
When
0
W
>
the gas expands. When
p
is constant and
V
increases,
T
increases.
19.3.
IDENTIFY:
Example 19.1 shows that for an isothermal process
12
ln(
/
).
R
T pp
=
pV
nRT
=
says
V
decreases
when
p
increases and
T
is constant.
SET UP:
358.15 K.
T
=
3.
pp
=
EXECUTE:
(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
⎛⎞
=
−
⎜⎟
⎝⎠
EVALUATE:
Since
V
decreases,
W
is negative.
Figure 19.3
19.4.
IDENTIFY:
Use the expression for
W
that is appropriate to this type of process.
SET UP:
The volume is constant.
EXECUTE:
(a)
The
pV
diagram is given in Figure 19.4.
(b)
Since
0,
0.
VW
=
19
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Chapter 19
EVALUATE:
For any constant volume process the work done is zero.
Figure 19.4
19.5.
IDENTIFY:
Example 19.1 shows that for an isothermal process
12
ln(
/
).
Wn
R
T pp
=
Solve for
1
.
p
SET UP:
For a compression (
V
decreases)
W
is negative, so
518 J.
W
=−
295.15 K.
T
=
EXECUTE:
(a)
1
2
ln
.
Wp
nRT
p
⎛⎞
=
⎜⎟
⎝⎠
/
1
2
.
R
T
p
e
p
=
518 J
0.692.
(0.305 mol)(8.314 J/mol K)(295.15 K)
W
nRT
−
==
−
⋅
/
0.692
(1.76 atm)
0.881 atm.
R
T
pp
e
e
−
=
(b)
In the process the pressure increases and the volume decreases. The
pV
diagram is sketched in Figure 19.5.
EVALUATE:
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)
IDENTIFY
and
SET UP:
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.
EXECUTE:
,
→
0,
V
Δ=
so
0
W
=
23
→
p
is constant; so
53
3 4
(5.00 10 Pa)(0.120 m
0.200 m )
4.00 10 J
WpV
=Δ=
×
−
=
−
×
(
W
is negative since the volume
decreases in the process.)
4
tot
1
2
2
3
WW
W
→→
=+=
−
×
EVALUATE:
The volume decreases so the total work done is negative.
19.7.
IDENTIFY:
Calculate
W
for each step using the appropriate expression for each type of process.
SET UP:
When
p
is constant,
.
V
=Δ
When
0,
V
0.
W
=
EXECUTE:
(a)
13
1
2
1
32
24
2
1
2
41
(
),
0,
(
) and
0.
W p
VVW
W pVV
W
=
=
The total work done by the system is
13
32
24
41
1
2
2
1
()
(
)
,
WWWW ppVV
+++=−
−
which is the area in the
pV
plane enclosed by the loop.
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 Spring '11
 Shaefer

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