Chapter 19
1. Each atom has a mass of
m = M
/
N
A
, where
M
is the molar mass and
N
A
is the
Avogadro constant. The molar mass of arsenic is 74.9 g/mol or 74.9
10
–3
kg/mol. 7.50
10
24
arsenic atoms have a total mass of (7.50
10
24
) (74.9
10
–3
kg/mol)/(6.02
10
23
mol
–1
) = 0.933 kg.
2. (a) Eq. 193 yields
n = M
sam
/
M
= 2.5/197 = 0.0127 mol.
(b) The number of atoms is found from Eq. 192:
N = nN
A
= (0.0127)(6.02
10
23
) = 7.64
10
21
.
3. (a) We solve the ideal gas law
pV = nRT
for
n
:
6
3
8
100Pa
1.0 10
m
5.47
10
mol.
8.31J/mol K
220K
pV
n
RT
(b) Using Eq. 192, the number of molecules
N
is
6
23
1
16
A
5.47
10
mol
6.02
10
mol
3.29
10
molecules.
N
nN
4. With
V
= 1.0
10
–6
m
3
,
p
= 1.01
10
–13
Pa, and
T
= 293 K, the ideal gas law gives
13
6
3
23
1.01
10
Pa
1.0
10
m
4.1
10
mole.
8.31 J/mol
K
293 K
pV
n
RT
ﾴ
ﾴ
ﾴ
ﾴ
Consequently, Eq. 192 yields
N = nN
A
= 25 molecules. We can express this as a ratio
(with
V
now written as 1 cm
3
)
N
/
V
= 25 molecules/cm
3
.
5. (a) In solving
pV = nRT
for
n
, we first convert the temperature to the Kelvin scale:
(40.0
273.15)K
313.15 K
T
+
. And we convert the volume to SI units: 1000 cm
3
=
1000
10
–6
m
3
. Now, according to the ideal gas law,
5
6
3
2
1.01
10 Pa
1000
10
m
3.88
10
mol.
8.31J/mol K
313.15K
pV
n
RT
(b) The ideal gas law
pV = nRT
leads to
819
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CHAPTER 19
5
6
3
2
1.06
10 Pa
1500
10
m
493K.
3.88
10
mol
8.31J/mol K
pV
T
nR
We note that the final temperature may be expressed in degrees Celsius as 220°C.
6. Since (standard) air pressure is 101 kPa, then the initial (absolute) pressure of the air
is
p
i
= 266 kPa. Setting up the gas law in ratio form (where
n
i
=
n
f
and thus cancels out
— see Sample Problem 191), we have
2
3
2
3
1.64
10
m
300K
266kPa
1.67
10
m
273K
f
f
f
f
i
i
i
p V
T
p
pV
T
ﾴ
ﾴ ﾴ
ﾴ
ﾴ
ﾴ ﾴ
ﾴ
ﾴ
ﾴ
ﾴ
ﾴ
which yields
p
f
= 287 kPa. Expressed as a gauge pressure, we subtract 101 kPa and
obtain 186 kPa.
7. (a) With
T
= 283 K, we obtain
(
) (
)
(
) (
)
3
3
100
10 Pa
2.50m
106mol.
8.31J/mol K
283K
pV
n
RT
ﾴ
=
=
=
ﾴ
(b) We can use the answer to part (a) with the new values of pressure and temperature,
and solve the ideal gas law for the new volume, or we could set up the gas law in ratio
form as in Sample Problem 191 (where
n
i
=
n
f
and thus cancels out):
3
100kPa
303K
2.50m
300kPa
283K
f
f
f
f
i
i
i
p V
T
V
pV
T
ﾴ
ﾴ ﾴ
ﾴ
ﾴ
ﾴ ﾴ
ﾴ
ﾴ
ﾴ ﾴ
ﾴ
which yields a final volume of
V
f
= 0.892 m
3
.
8. (a)
Eq. 1945 (which gives 0) implies
Q = W
.
Then Eq. 1914, with
T
= (273 +
30.0)K leads to gives
Q
= –3.14
10
3
J, or 
Q
 = 3.14
10
3
J.
(b) That negative sign in the result of part (a) implies the transfer of heat is
from
the gas.
9. The pressure
p
1
due to the first gas is
p
1
=
n
1
RT
/
V
, and the pressure
p
2
due to the
second gas is
p
2
=
n
2
RT
/
V
. So the total pressure on the container wall is
1
2
1
2
1
2
.
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 Fall '08
 Many
 Physics, Mass

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