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16.24.
Model:
Treat the oxygen gas in the cylinder as an ideal gas.
Solve: (a)
The number of moles of oxygen is
mol
50 g
1.563 mol
32 g mol
M
n
M
==
=
(b)
The number of molecules is
( )( )
23
1
23
A
1.563 mol
6.02
10
mol
9.41
10
Nn
N
−
×
=×
.
(c)
The volume of the cylinder
Thus,
()
2
22
0.10 m
0.40 m
1.257
10
m .
Vr
L
ππ
−
3
23
25
3
23
9.41
10
7.49
10
m
1.257
10
m
N
V
−
−
×
×
×
(d)
From the idealgas law
pV
=
nRT
we can calculate the absolute pressure to be
( )( )
1.563 mol
8.31 J mol K
293 K
303 kPa
1.257
10
m
nRT
p
V
−
=
×
where we used
T
=
20
°
C
=
293 K. But a pressure gauge reads
gauge pressure
:
g
1 atm
303 kPa
101 kPa
202 kPa
pp
=−
=
−
=
16.19.
Model:
Treat the gas in the sealed container as an ideal gas.
Solve: (a)
From the ideal gas law equation
pV
=
nRT
, the volume
V
of the container is
( ) ( )
3
5
2.0 mol
8.31 J mol K
273
30 K
0.0497 m
1.013
10 Pa
nRT
V
p
+
=
×
Note that pressure
must
be in Pa in the ideal gas law.
(b)
The beforeandafter relationship of an ideal gas in a sealed container (constant volume) is
( )
12
2
21
1
273
130 K
1.0 atm
1.33 atm
273
pV
T
TT
T
+
=⇒
= =
=
+
Note that gaslaw calculations
must
use T in kelvins
16.33.
Model:
Assume that the gas is an ideal gas.
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 Spring '10
 ZHOU

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