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Chapter 7
States of Matter and Changes in State
71
1.
A weather report indicates a barometric pressure of 29.2 in. What is this pressure expressed in torr?
29.2 in
x
25.4 mm
1 in
=
742 mm Hg = 742 torr
3.
If the gas in Exercise 2 is heated to 35
o
C, what would be the separation between the two mercury levels?
11
2 2
2
Write the gas law for the two experiments:
P V = nRT
&
P V = nRT
V,n, and R are not subscripted because they are unchanged.
PT
Dividing the two equations eliminates V, n, and R to yield
=
2
21
1
2
T
, so P = P
T
T = 273 + 35 = 308, T = 273 + 20 = 293, and P = 758 + 423 = 1181 mm from Exercise 2, so
308
P =
1183 mm
= 1241 mm, and the height of the column = 1241  758 = 483 mm
293
×
×
5.
Convert the following temperatures to the Celsius scale:
Use K =
°
C + 273.15
a)
4 K =
269
°
C
b)
350 K =
77
°
C
c)
186.4 K =
86.8
°
C
d) 657 K =
384
°
C
7.
What is the pressure inside a 7.20L container filled with 0.254 moles of CO
2
at 35
o
C?
P
=
nRT
V
=
(0.254)(0.0821)(308)
7.20
=
0.892 atm
9.
The pressure of air in a tire at 10
o
C is 22 psi. After several miles of driving at high speed, the pressure is 28 psi.
Assume the volume of the tire is unchanged and calculate the temperature of the air in the tire.
The number of moles of gas, its volume and the ideal gas law constant are all unchanged so we can write
nR
V
=
P
1
T
1
=
P
2
T
2
or T
2
= T
1
x
P
2
P
1
,
T
1
= 10 + 273 = 283 K, P
1
= 22 psi, and P
2
= 28 psi.
Note we can use any
units for the two pressures as long as they are the same because the ratio of pressures is unitless.
T
2
= 283
x
28
22
= 360 K which is 360  273 = 87
o
C
11. Determine the concentrations of the following gases in moles/L:
a)
3.0 moles of gas at 900 K and 2.6 atm
First, determine the volume:
V =
nRT
P
=
(3.0 mol)(0.0821 L
⋅
atm
⋅
K
1
⋅
mol
1
)(900 K)
2.6 atm
= 85 L
Then determine the concentration:
[gas] =
n
V
=
3.0 mol
85 L
= 0.035 mol/L = 0.035 M
b)
6.2 L of a gas at 250
o
C and 800 torr
First, determine the number of moles:
n =
PV
RT
=
(
800
760
atm)(6.2 L)
(0.0821 L
⋅
atm
⋅
K
1
⋅
mol
1
)(523 K)
= 0.15 mol
Then determine the concentration:
[gas] =
n
V
=
0.15 mol
6.2 L
= 0.025 mol/L = 0.025 M
c)
4.0 g of Ne in a 0.56 L container at 200
o
C.
First determine the number of moles:
4.0 g Ne
×
1 mole
20.18 g
= 0.20 mol Ne
Then determine the concentration:
[Ne] =
n
V
=
0.20 mol
0.56 L
= 0.35 mol/L = 0.35 M
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View Full DocumentStates of Matter and Changes in State
72
13. What are the partial pressures of O
2
and N
2
and the total pressure in a 3.5L flask at 77
o
C that contains 4.0 g O
2
and 7.0 g N
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 Fall '08
 BIGHAM

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