PHYS 132, WEEK 9
NOVEMBER 1518
18.16.
Solve:
The average translational kinetic energy per molecule is
e
avg
=
1
2
mv
rms
2
=
3
2
k
B
T
v
rms
=
3
k
B
T
m
Since we want the
v
rms
for H
2
and N
2
to be equal,
3
k
B
T
H
2
m
H
2
=
3
k
B
T
m
N
2
T
H
2
=
m
H
2
m
N
2
T
N
2
=
2 u
28 u
373 K
(
)
=
26.6 K
18.20.
Solve:
(a)
The average translational kinetic energy per molecule is
e
avg
=
1
2
mv
rms
2
=
3
2
k
B
T
This means
e
avg
doubles if the temperature
T
doubles.
(b)
The rootmeansquare speed
v
rms
increases by a factor of
2
as the temperature doubles.
(c)
The mean free path is
l
=
1
4
2
p
N V
(
)
r
2
Because
N
/
V
and
r
do not depend on
T
, doubling temperature has no effect on
l
.
18.25.
Solve:
The volume of the air is
V
=
6.0 m
¥
8.0 m
¥
3.0 m
=
144.0 m
3
, the pressure
p
=
1 atm
=
1.013
¥
10
5
Pa, and the
temperature
T
=
20
°
C
=
293 K. The number of moles of the gas is
n
=
pV
RT
=
5991 mols
This means the number of molecules is
N
=
nN
A
=
5991 mols
(
)
6.022
10
23
mol

1
(
)
=
3.61
10
27
molecules.
Since air is a diatomic gas, the room’s thermal energy is
E
th
=
N
e
avg
=
N
5
2
k
B
T
(
)
=
3.65
10
7
J
Assess:
The room’s thermal energy can also be obtained as follows:
E
th
=
nC
V
T
=
(5991 mols)(20.8 J/mol K)(293 K)
=
3.65
¥
10
7
J
18.29.
Visualize:
Refer to Figure 18.13. At low temperatures,
C
V
=
3
2
R
=
12.5 J / mol K.
At room temperature and modestly hot
temperatures,
C
V
=
5
2
R
=
20.8 J / mol K.
At very hot temperatures,
C
V
=
7
2
R
=
29.1 J / mol K.
Solve:
(a)
The number of moles of diatomic hydrogen gas in the rigid container is
0.20 g
2 g /
mol
=
0.1 mol
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 Spring '08
 ALL
 Energy, Kinetic Energy, avg, fi vrms, fi vesc

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