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CHAPTER 12
THE GASEOUS STATE OF MATTER
SOLUTIONS TO REVIEW QUESTIONS
1.
In Figure 12.1, color is the evidence of diffusion; bromine is colored and air is colorless.
If hydrogen and oxygen had been the two gases, this would not work because both gases
are colorless. Two ways could be used to show the diffusion. The change of density
would be one method. Before diffusion the gas in the flask containing hydrogen would be
much less dense. After diffusion, the gas densities in both flasks would be equal. A
second method would require the introduction of spark gaps into both flasks. Before
diffusion, neither gas would show a reaction when sparked. After diffusion, the gases in
both flasks would explode because of the mixture of hydrogen and oxygen.
2.
The pressure of a gas is the force that gas particles exert on the walls of a container. It
depends on the temperature, the number of molecules of the gas and the volume of the
container.
3.
The air pressure inside the balloon is greater than the air pressure outside the balloon. The
pressure inside must equal the sum of the outside air pressure plus the pressure exerted by
the stretched rubber of the balloon.
4.
The major components of dry air are nitrogen and oxygen.
5.
1 torr = 1 mm Hg
6.
The molecules of H 2 at 100°C are moving faster. Temperature is a measure of average
kinetic energy. At higher temperatures, the molecules will have more kinetic energy.
7.
1 atm corresponds to 4 L.
8.
The pressure times the volume at any point on the curve is equal to the same value. This
is an inverse relationship as is Boyle’s law. (PV = k)
9.
If T2 6 T1 , the volume of the cylinder would decrease (the piston would move
downward).
10.
The pressure inside the bottle is less than atmospheric pressure. We come to this
conclusion because the water inside the bottle is higher than the water in the trough
(outside the bottle).
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11.
The density of air is 1.29 g/L. Any gas listed below air in Table 12.3 has a density greater
than air. For example: O2 , H 2S, HCl, F2 , CO2.
12.
Basic assumptions of Kinetic Molecular Theory include:
(a)
(b)
(c)
(d)
(e)
13.
Gases consist of tiny particles.
The distance between particles is great compared to the size of the particles.
Gas particles move in straight lines. They collide with one another and with the
walls of the container with no loss of energy.
Gas particles have no attraction for each other.
The average kinetic energy of all gases is the same at any given temperature. It
varies directly with temperature.
The order of increasing molecular velocities is the order of decreasing molar masses.
increasing molecular velocity
;99999999999999
Rn, F2 , N2 CH4 , He, H2
99999999999999:
decreasing molar mass
At the same temperature the kinetic energies of the gases are the same and equal to
1
2 mv 2. For the kinetic energies to be the same, the velocities must increase as the molar
masses decrease.
14.
Average kinetic energies of all these gases are the same, since the gases are all at the
same temperature.
15.
Gases are described by the following parameters:
(a)
(b)
pressure
volume
(c)
(d)
temperature
number of moles
16.
An ideal gas is one which follows the described gas laws at all P, V and T and whose
behavior is described exactly by the Kinetic Molecular Theory.
17.
Boyle’s law: P1V1 = P2V2 , ideal gas equation: PV = nRT
If you have an equal number of moles of two gases at the same temperature the right side
of the ideal gas equation will be the same for both gases. You can then set PV for the first
gas equal to PV for the second gas (Boyle’s law) because the right side of both equations
will cancel.
18.
Charles’ law: V1>T1 = V2>T2 , ideal gas equation: PV = nRT
Rearrange the ideal gas equation to: V>T = nR>P
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If you have an equal number of moles of two gases at the same pressure the right side of
the rearranged ideal gas equation will be the same for both. You can set V>T for the first
gas equal to V>T for the second gas (Charles’ law) because the right side of both
equations will cancel.
19.
A gas is least likely to behave ideally at low temperatures. Under this condition, the
velocities of the molecules decrease and attractive forces between the molecules begin to
play a significant role.
20.
A gas is least likely to behave ideally at high pressures. Under this condition, the
molecules are forced close enough to each other so that their volume is no longer small
compared to the volume of the container. Attractive forces may also occur here and
sooner or later, the gas will liquefy.
21.
Equal volumes of H 2 and O2 at the same T and P:
(a)
have equal number of molecules (Avogadro’s law)
(b)
mass O2 = 16 times mass of H 2
(c)
moles O2 = moles H 2
(d)
average kinetic energies are the same (T same)
(e)
rate H 2 = 4 times the rate of O2 (Graham’s Law of Effusion)
(f)
density O2 = 16 times the density of H 2
density O2 = ¢
mass O2
≤
volume O2
density H 2 = ¢
mass H 2
≤
volume H 2
density O2 = ¢
mass O2
≤ (density H 2)
mass H 2
volume O2 = volume H 2
¢
mass O2
mass H 2
≤ = ¢
≤
density O2
density H 2
density O2 = a
22.
32
b(density H 2) = 16(density H 2)
2
Behavior of gases as described by the Kinetic Molecular Theory.
(a)
Boyle’s law. Boyle’s law states that the volume of a fixed mass of gas is inversely
proportional to the pressure, at constant temperature. The Kinetic Molecular Theory
assumes the volume occupied by gases is mostly empty space. Decreasing the
volume of a gas by compressing it, increases the concentration of gas molecules,
resulting in more collisions of the molecules and thus increased pressure upon the
walls of the container.
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(b)
(c)
23.
Charles’ law. Charles’ law states that the volume of a fixed mass of gas is directly
proportional to the absolute temperature, at constant pressure. According to Kinetic
Molecular Theory, the kinetic energies of gas molecules are proportional to the
absolute temperature. Increasing the temperature of a gas causes the molecules to
move faster, and in order for the pressure not to increase, the volume of the gas
must increase.
Dalton’s law. Dalton’s law states that the pressure of a mixture of gases is the sum
of the pressures exerted by the individual gases. According to the Kinetic Molecular
Theory, there are no attractive forces between gas molecules; therefore, in a mixture
of gases, each gas acts independently and the total pressure exerted will be the sum
of the pressures exerted by the individual gases.
N2(g) + O21g2 ¡ 2 NO1g2
1 vol + 1 vol ¡ 2 vol
According to Avogadro’s Law, equal volumes of nitrogen and oxygen at the same
temperature and pressure contain the same number of molecules. In the reaction, nitrogen
and oxygen molecules react in a 1:1 ratio. Since two volumes of nitrogen monoxide are
produced, one molecule of nitrogen and one molecule of oxygen must produce two
molecules of nitrogen monoxide. Therefore each nitrogen and oxygen molecule must be
made up on two atoms (diatomic).
24.
We refer gases to STP because some reference point is needed to relate volume to moles.
A temperature and pressure must be specified to determine the moles of gas in a given
volume, and 0°C and 760 torr are convenient reference points.
25.
Conversion of oxygen to ozone is an endothermic reaction. Evidence for this statement is
that energy (286 kJ>3 mol O2) is required to convert O2 to O3 .
26.
Chlorofluorocarbons, (Freons, CCl 3F, and CCl 2F2), are responsible for damaging the
ozone layer. When these compounds are carried up to the stratosphere (the outer part of
the atmosphere) they absorb ultraviolet radiation and produce chlorine free radicals that
react with ozone (O3) and destroy it.
27.
Heating a mole of N2 gas at constant pressure has the following effects:
(a)
Density will decrease. Heating the gas at constant pressure will increase its volume.
The mass does not change, so the increased volume results in a lower density.
(b)
Mass does not change. Heating a substance does not change its mass.
(c)
Average kinetic energy of the molecules increases. This is a basic assumption of the
Kinetic Molecular Theory.
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28.
(d)
Average velocity of the molecules will increase. Increasing the temperature
increases the average kinetic energies of the molecules; hence, the average velocity
of the molecules will increase also.
(e)
Number of N2 molecules remains unchanged. Heating does not alter the number of
molecules present, except if extremely high temperatures were attained. Then, the
N2 molecules might dissociate into N atoms resulting in fewer N2 molecules.
Oxygen atom = O
Oxygen molecule = O2
An oxygen molecule contains 16 electrons.
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CHAPTER 12
SOLUTIONS TO EXERCISES
1.
2.
3.
4.
(a)
1715 mm Hg2a
1 atm
b = 0.941 atm
760 mm Hg
(b)
1715 mm Hg2a
1 in. Hg
b = 28.1 in. Hg
25.4 mm Hg
(c)
1715 mm Hg2 ¢
14.7 lb/in.2
≤ = 13.8 lb>in.2
760 mm Hg
(a)
1715 mm Hg2a
1 torr
b = 715 torr
1 mm Hg
(b)
1715 mm Hg2a
1013 mbar
b = 953 mbar
760 mm Hg
(c)
1715 mm Hg2a
101.325 kPa
b = 95.3 kPa
760 mm Hg
(a)
128 mm Hg2a
(b)
16000. cm Hg2a
(c)
1795 torr2a
(d)
15.00 kPa2a
(a)
162 mm Hg2a
(b)
14250. cm Hg2a
(c)
1225 torr2a
(d)
10.67 kPa2a
1 atm
b = 0.037 atm
760 mm Hg
1 atm
b = 78.95 atm
76 cm Hg
1 atm
b = 1.05 atm
760 torr
1 atm
b = 0.0493 atm
101.325 kPa
1 atm
b = 0.082 atm
760 mm Hg
1 atm
b = 55.92 atm
76 cm Hg
1 atm
b = 0.296 atm
760 torr
1 atm
b = 0.0066 atm
101.325 kPa
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5.
6.
P1V1 = P2V2
or
V2 =
P1V1
. Change 625 torr to atmospheres and mm Hg.
P2
(a)
1625 torr211 atm/760 torr2(525 mL)
= 288 mL
1.5 atm
(b)
1625 torr211mm Hg/1 torr2(525 mL)
= 721 mL
455 mm Hg
P1V1 = P2V2
(a)
or
V2 =
P1V1
. Change 722 torr Hg to atmospheres and mm Hg.
P2
1722 torr211 atm>760 torr21635 mL2
= 241 mL
2.5 atm
(b)
1722 torr2(1 mm Hg/1 torr21635 mL2
= 577 mL
795 mm Hg
7.
P2 =
P1V1
V2
10.75 atm21521 mL2
= 0.50 atm
776 mL
8.
P2 =
P1V1
V2
11.7 atm21225 mL2
= 3.3 atm
115 mL
9.
V1
V2
=
T1
T2
(a)
(b)
10.
or
V2 =
V1T2
;
T1
16.00 L21273 K2
= 6.60 L
248 K
16.00 L21100. K2
= 2.42 L
248 K
V1
V2
V1T2
=
or V2 =
;
T1
T2
T1
0.0°F = -18°C
(a)
(b)
Temperatures must be in Kelvin (°C + 273)
Temperatures must be in Kelvin (°C + 273)
16.00 L21255 K2
= 6.17 L
248 K
16.00 L21345 K2
= 8.35 L
248 K
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11.
Use the combined gas law
V2 =
12.
13.
14.
15.
or
P2V2
P1V1
=
T1
T2
or
P2V2
P1V1
=
T1
T2
or
P2V2
P1V1
=
T1
T2
P2V2
P1V1
=
T1
T2
V2 =
V2 =
P1V1T2
P2T1
P1V1T2
P2T1
V2 =
P1V1T2
P2T1
or
P2 =
P1V1T2
V2T1
or
T2 =
12.50 atm2122.4 L21268 K2
= 33.4 L
11.50 atm21300. K2
Use the combined gas law
11.0 atm21775 mL21298 K2
= 1.4 atm
1615 mL21273 K2
16.
P1V1T2
P2T1
10.950 atm211400. mL21275 K2
= 2.4 * 105 L
14.0 torr211 atm>760 torr21291 K2
Use the combined gas law
V2 =
P2V2
P1V1
=
T1
T2
V2 =
1740 mm Hg21410 mL21523 K2
= 7.8 * 102 mL
1680 mm Hg21300. K2
Use the combined gas law
V2 =
or
1740 mm Hg21410 mL21273 K2
= 3.6 * 102 mL
1760 mm Hg21300. K2
Use the combined gas law
V2 =
P2V2
P1V1
=
T1
T2
Use the combined gas law
Change 765 torr to atmospheres.
P2V2T1
P1V1
1765 torr211 atm/760 torr211.5 L2(292 K)
= 120 K (120 - 273) K = -150°C
11.5 atm212.5 L2
17.
Ptotal = PN2 + PH2O vapor = 721 torr
PH2O vapor = 19.8 torr
PN2 = 721 torr - 19.8 torr = 701 torr
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18.
Ptotal = PN2 + PH2O vapor = 705 torr
PH2O vapor = 23.8 torr
PN2 = 705 torr - 23.8 torr = 681 torr
19.
Ptotal = PN2 + PH2 + PO2
= 200. torr + 600. torr + 300. torr = 1100. torr = 1.100 * 103 torr
20.
Ptotal = PH2 + PN2 + PO2
= 325 torr + 475 torr + 650. torr = 1450. torr = 1.450 * 103 torr
21.
Ptotal = PCH4 + PH2O vapor (Solubility of methane is being ignored.)
PH2O vapor = 23.8 torr
PCH4 = 720. torr - 23.8 torr = 696 torr
To calculate the volume of dry methane, note that the temperature is constant, so
P1V1 = P2V2 can be used.
V2 =
22.
1696 torr212.50 L2
P1V1
=
= 2.29 L
P2
1760. torr2
Ptotal = PC3H8 + PH2O vapor
C3H 8 in propane
PH2O vapor = 20.5 torr
PC3H8 = 745 torr - 20.5 torr = 725 torr
To calculate the volume of dry propane, note that the temperature is constant, so
P1V1 = P2V2 can be used.
V2 =
23.
24.
1725 torr211.25 L2
P1V1
=
= 1.19 L C3H 8
P2
1760. torr2
1 mol of a gas occupies 22.4 L at STP
1 mol
11.75 L2a
b = 0.0781 mol O2
22.4 L
13.50 L2a
1 mol
b = 0.156 mol N2
22.4 L
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25.
(a)
16.02 * 1023 molecules2a
(b)
12.5 mol2a
(c)
26.
22.4 L
b = 56 L CH4
mol
22.4 L
b = 8.75 L O2
112.5 g2a
32.00 g
(a)
11.80 * 1024 molecules2a
(b)
17.5 mol2a
(c)
22.4 L
b = 22.4 L CO2
6.02 * 1023 molecules
22.4 L
b = 67.0 L SO3
6.02 * 1023 molecules
22.4 L
b = 170 L C2H6
mol
22.4 L
125.2 g2a
b = 7.96 L Cl2
70.90 g
27.
1725 mL2a
17.03 g
1L
1 mol
ba
ba
b = 0.551 g NH3
1000 mL 22.4 L
mol
28.
1945 mL2a
42.08 g
1L
1 mol
ba
ba
b = 1.78 g C3H6
1000 mL 22.4 L
mol
29.
11.00 L NH 32a
1 mol
6.022 * 1023 molecules
b¢
≤ = 2.69 * 1022 molecules NH 3
22.4 L
mol
30.
11.00 L CH 42a
1 mol
6.022 * 1023 molecules
b¢
≤ = 2.69 * 1022 molecules CH 4
22.4 L
mol
31.
density of Cl 2 gas = 3.17 g>L (from table 12.3)
32.
density of CH 4 gas = 0.716 g>L (from table 12.3)
33.
(a)
110.0 g2>13.17 g>L2 = 3.15 liters
13.0 L210.716 g>L2 = 2.1 g CH 4
(b)
(c)
d = a
131.3 g Xe
1 mol
ba
b = 5.86 g>L Xe
mol
22.4 L
38.00 g F2
1 mol
d = ¢
b = 1.70 g>L F2
≤a
mol
22.4 L
d = ¢
30.07 g C2H6
1 mol
b = 1.34 g>L C2H6
≤a
mol
22.4 L
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34.
(a)
d = a
(b)
d = ¢
(c)
35.
44.01 g CO2
1 mol
b = 1.96 g>L CO2
≤a
mol
22.4 L
56.10 g C4H8
1 mol
d = ¢
b = 2.50 g>L C4H8
≤a
mol
22.4 L
38.00 g F2
1 mol
b = 1.696 g>L F2
≤a
mol
22.4 L
(a)
d = ¢
(b)
Assume 1.00 mol of F2 and determine the volume using the ideal gas equation,
PV = nRT.
11.00 mol210.0821 L atm>mol K21300. K2
nRT
V =
=
P
1.00 atm
= 24.6 L at 27°C and 1.00 atm
d =
36.
222.0 g Rn
1 mol
ba
b = 9.91 g>L Rn
mol
22.4 L
38.00 g
= 1.54 g>L F2
24.6 L
70.90 g Cl 2
1 mol
b = 3.165 g>L Cl 2
≤a
mol
22.4 L
(a)
d = ¢
(b)
Assume 1.00 mol of Cl 2 and determine the volume using the ideal gas equation,
PV = nRT.
V =
11.00 mol210.0821 L atm>mol K21295 K2
nRT
=
P
0.500 atm
= 48.4 L at 22°C and 0.500 atm
d =
37.
PV = nRT
70.90 g
= 1.46 g>L Cl 2
48.4 L
V =
nRT
P
V =
12.3 mol210.0821 L atm>mol K21300. K2
= 57 L Ne
750 torr
torr
760
atm
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nRT
P
10.75 mol210.0821 L atm>mol K21298 K2
= 19 L Kr
725 torr
torr
760
atm
38.
PV = nRT
39.
When working with gases, the identity of the gas does not affect the volume, as long as the
number of moles are known.
Total moles = mol H 2 + mol CO2 = 5.00 mol + 0.500 mol = 5.50 mol
V =
V = 15.50 mol2a
40.
V =
22.4 L
b = 123 L
mol
When working with gases, the identity of the gas does not affect the volume, as long as
the number of moles are known.
Total moles = mol N2 + mol HCl = 2.50 mol + 0.750 mol = 3.25 mol
V = 13.25 mol2a
22.4 L
b = 72.8 L
mol
41.
PV = nRT
14.15 atm210.250 L2
PV
T =
=
= 2.81 K
nR
14.50 mol210.0821 L atm>mol K2
42.
PV = nRT
10.500 atm215.20 L2
PV
=
= 0.13 mol N2
n =
RT
10.0821 L atm>mol K21250 K2
43.
The balanced equation is Zn(s) + H2SO4(aq) : H2(g) + ZnSO4(aq)
44.
(a)
(52.7 g Zn) a
(b)
(525 mL H2) a
1 mol H2 22.4 L 1000 mL
1 mol
ba
ba
ba
b = 1.81 * 104 mL H2
65.39 g 1 mol Zn
mol
1L
1 mol H2SO4
1L
1 mol
ba
ba
b = 0.0234 mol H2SO4
1000 mL 22.4 L
1 mol H2
The balanced equation is 2 H2O2(aq) : 2H2O(l) + O2(g)
(a)
(50.0 g H2O2)a
(b)
(225 mL O2)a
1 mol O2
1 mol
22.4 L 1000 mL
ba
ba
ba
b = 1.65 * 104 mL O2.
34.02 g 2 mol H2O2
mol
1L
2 mol H2O2
1L
1 mol
ba
ba
b = 0.0201 mol H2O2
1000 mL 22.4 L
1 mol O2
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45.
The balanced equation is 4 NH3(g) + 5 O2(g) : 4 NO(g) + 6 H2O(g)
Remember that volume–volume relationships are the same as mole–mole relationships
when dealing with gases at the same T and P.
5 L O2
b = 3.1 L O2
4 L NH3
(a)
(2.5 L NH3)a
(b)
(25 L NH3)a
(c)
Limiting reactant problem.
(25 L O2)a
18.02 g
6 L H2O
1 mol
ba
ba
b = 30. g H2O
4 L NH3 22.4 L
1 mol
4 L NO
b = 20. L NO
5 L O2
(25 L NH3)a
4 L NO
b = 25 L NO
5 L NH3
Oxygen is the limiting reactant. 20. L NO us formed.
46.
The balanced equation is C3Hg(g) + 5 O2(g) : 3 CO2(g) + 4 H2O(g)
Remember that volume–volume relationships are the same as mole–mole relationships
when dealing with gases at the same T and P.
5 L O2
b = 36 L O2
(a) (7.2 L C3H8)a
1 L C3H8
44.01 g
3 L CO2
1 mol
ba
ba
b = 210 g CO2
1 L C3H8 22.4 L
1 mol
(b)
(35 L C3H8)a
(c)
Limiting reactant problem.
(15 L C3H8)a
4 L H2O
b = 60. L H2O
1 L C3H8
4 L H2O
b = 12 L H2O
5 L O2
Oxygen is the limiting reactant. 12 L H2O is formed.
(15 L O2)a
47.
The balanced equation is 2 KClO3(s) : 2 KCl(s) + 3 O2(g)
(0.525 kg KCl) a
1000 g
3 mol O2
1 mol
22.4 L
ba
ba
ba
b = 237 L O2
1 kg
74.55 g 2 mol KCl
1 mol
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48.
The balanced equation is C6H12O6(s) + 6 O2(g) : 6 CO2(g) + 6 H2O(l)
(1.50 kg C6H12O6)a
1000 g
6 mol CO2
22.4 L
1 mol
ba
ba
ba
b
1 kg
180.16 g 1 mol C6H12O6
1 mol
= 1.12 * 103 L CO2
49.
Like any other gas, water in the gaseous state occupies a much larger volume than in the
liquid state.
50.
During the winter the air in a car’s tires is colder, the molecules move slower and the
pressure decreases. In order to keep the pressure at the manufacturer’s recommended psi
air needs to be added to the tire. The opposite is true during the summer.
51.
(a)
the pressure will be cut in half
(b)
the pressure will double
(c)
the pressure will be cut in half
(d)
the pressure will increase to 3.7 atm or 2836 torr
nRT
PV = nRT
P =
V
11.5 mol210.0821 L atm>mol K21303 K2
P =
= 3.7 atm
10. L
760 torr
P = 3.7 atm a
b = 2.8 * 103 torr
1 atm
52.
(a)
(c)
P
T
V
(b)
P
(d)
T
n
V
V
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- Chapter 12 -
53.
The can is a sealed unit and very likely still contains some of the aerosol. As the can is
heated, pressure builds up in it eventually causing the can to explode and rupture with
possible harm from flying debris.
54.
One mole of an ideal gas occupies 22.4 liters at standard conditions. (0°C and 1 atm
pressure)
PV = nRT
11.00 atm21V2 = 11.00 mol210.0821 L atm>mol K21273 K2
V = 22.4 L
55.
Solve for volume using PV = nRT
(a)
(b)
(c)
56.
V =
V =
V =
10.2 mol Cl2210.0821 L atm>mol K21321 K2
= 5 L Cl2
180 cm>76 cm2 atm
14.2 g NH32a
1 mol
17.03 g
ba
0.0821 L atm
mol K
b(262 K)
0.65 atm
121 g SO32a
1 mol
80.07 g
ba
0.0821 L atm
mol K
b(328 K)
110 kPa
kPa
101.3
atm
4.2 g NH3 has the greatest volume
Assume 1 mol of each gas
(a)
SF6 = 146.1 g>mol
d = a
(b)
146.1 g
1 mol
ba
b = 6.52 g>L SF6
mol
22.4 L
Assume 25°C and 1 atm pressure
V1at 25°C2 = 122.4 L2a
298 K
b = 24.5 L
273 K
C2H 6 = 30.07 g>mol
d = a
30.07 g
1 mol
ba
b = 1.23 g>L C2H 6
mol
24.5 L
- 154 -
= 8.2 L NH3
= 6.5 L SO3
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Page 155
- Chapter 12 -
(c)
He at -80°C and 2.15 atm
11 mol210.0821 L atm>mol K21193 K2
= 7.37 L
2.15 atm
4.003 g
1 mol
d = a
ba
b = 0.543 g>L He
mol
7.37 L
SF6 has the greatest density
V =
57.
(a)
Empirical formula. Assume 100 g starting material
80.0 g C
6.66
= 6.66 mol C
= 1
12.01 g>mol
6.66
20.0 g H
19.8
= 19.8 mol H
= 2.97
1.008 g>mol
6.66
Empirical formula = CH 3
Empirical mass = 12.01 g + 3.024 g = 15.03 g>mol
(b)
Molecular formula.
a
2.01 g 22.4 L
ba
b = 30. g>mol (molar mass)
1.5 L
mol
30. g>mol
= 2; Molecular formula is C2H 6
15.03 g>mol
(c)
Valence electrons = 2142 + 6 = 14
H
58.
H
H
C
C
H
H
H
Lewis structure
PV = nRT
(a)
(b)
1790 torr211 atm2
b 12.0 L2 = 1n210.0821 L atm>mol K21298 K2
760 torr
n = 0.085 mol (total moles)
a
mol N2 = total moles - mol O2 - mol CO2
0.65 g O2
0.58 g CO2
= 0.085 mol 32.00 g>mol
44.01 g>mol
mol N2 = 0.085 mol - 0.020 mol O2 - 0.013 mol CO2 = 0.052 mol
10.052 mol N22 ¢
28.02 g N2
≤ = 1.5 g N2
mol
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- Chapter 12 -
PO2 = 1790 torr2 ¢
(c)
0.020 mol O2
≤ = 1.9 * 102 torr
0.085 mol
PCO2 = 1790 torr2 ¢
PN2 = 1790 torr2 ¢
59.
0.013 mol CO2
≤ = 1.2 * 102 torr
0.085 mol
0.051 mol N2
≤ = 4.7 * 102 torr
0.085 mol
2 CO + O2 ¡ 2 CO2
Calculate the moles of O2 and CO to find the limiting reactant.
PV = nRT
O2: 11.8 atm210.500 L O22 = 1n210.0821 L atm>mol K21288 K2
mol O2 = 0.038 mol
CO: a
800 mm Hg * 1 atm
b 10.500 L2 = 1n210.0821 L atm>mol K21333 K2
760 mm Hg
Limiting reactant is CO
mol CO = 0.019 mol
0.0095 mol O2 will react with 0.019 mol CO.
10.019 mol CO2 ¢
60.
PV = nRT
¢
1.4 g
3 ≤¢
cm
or
2 mol CO2 22.4 L
b = 0.43 L CO2 = 430 mL CO2
≤a
2 mol CO
mol
PV = a
g
b RT
molar mass
1000 cm3
≤ = 1.4 * 103 g>L
L
11.3 * 109 atm211.0 L2 = ¢
61.
1.4 * 103 g
≤ 10.0821 L atm>mol K21T2
2.0 g>mol
11.3 * 109 atm211.0 L212.0 g>mol2
= 2.3 * 107 K
T =
10.0821 L atm>mol K211.4 * 103 g2
(a)
Assume atmospheric pressure of 14.7 lb>in.2 to begin with.
Total pressure in the ball = 14.7 lb>in.2 + 13 lb>in.2 = 28 lb>in.2
PV = nRT
128 lb>in.22 ¢
1 atm
≤ 12.24 L2 = 1n210.0821 L atm>mol K21293 K2
14.7 lb>in.2
n = 0.18 mol air
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- Chapter 12 -
(b)
mass of air in the ball
m = 10.18 mol2a
(c)
29 g
b = 5.2 g air
mol
Actually the pressure changes when the temperature changes. Since pressure is
directly proportional to moles we can calculate the change in moles required to
keep the pressure the same at 30°C as it was at 20°C.
PV = nRT
128 lb>in.22 ¢
1 atm
≤ 12.24 L2 = 1n210.0821 L atm>mol K21303 K2
14.7 lb>in.2
n = 0.17 mol of air required to keep the pressure the same at 30°C.
0.01 mol air (0.18 - 0.17) must be allowed to escape from the ball.
10.01 mol air2a
62.
29 g
b = 0.29 g or 0.3 g air must be allowed to escape.
mol
Use the combined gas laws to calculate the bursting temperature (T2).
P1V1
P2V2
P2 = 1.00 atm (76 cm)
=
P1 = 65 cm
T1
T2
V2 = 2.00 L
V1 = 1.75 L
T1 = 20°C 1293 K2
T2 =
T2 = T2
176 cm212.00 L21293 K2
P2V2T1
=
= 392 K 1119°C2
P1V1
165 cm211.75 L2
63.
To double the volume of a gas, at constant pressure, the temperature (K) must be doubled.
V1
V2
=
V2 = 2 V1
T1
T2
V1
2 V1
2 V1T1
T2 = 2 T1
=
T2 =
T1
T2
V1
T2 = 21300. K2 = 600. K = 327°C
64.
V = volume at 22°C and 740 torr
2 V = volume after change in temperature (P constant)
V = volume after change in pressure (T constant)
Since temperature is constant, P1V1 = P2V2
P2 = 1740 torr2a
or
P2 =
P1V1
V2
2V
b = 1.5 * 103 torr (pressure to change 2 V to V)
V
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Page 158
- Chapter 12 -
65.
Volume is constant, so
T2 =
66.
P1
P2
=
T1
T2
or
T2 =
T1P2
;
P1
1500. torr21295 K2
= 211 K = -62°C
700. torr
The volume of the tires remains constant (until they burst), so
P1
P2
T1P2
=
;
or T2 =
T1
T2
P1
71.0°F = 21.7°C = 295 K
T2 =
67.
Use the combined gas laws.
P1V1
P2V2
P1V1T2
=
or P2 =
T1
T2
V2T1
P2 =
68.
144 psi21295 K2
= 433 K = 160°C = 320°F
30. psi
P1 and T1 are at STP
11.00 atm21800. mL21303 K2
= 3.55 atm
1250. mL21273 K2
Use the combined gas law
P1V1
P2V2
=
T1
T2
or
V2 =
P1V1T2
P2T1
First calculate the volume at STP.
V2 =
1400. torr21600. mL21273 K2
= 275 mL = 0.275 L
1760. torr21313 K2
At STP, a mole of any gas has a volume of 22.4 L
10.275 L2a
1 mol
6.022 * 1023 molecules
b¢
≤ = 7.39 * 1021 molecules
22.4 L
1 mol
Each molecule of N2O contains 3 atoms, so:
17.39 * 1021 molecules2a
69.
3 atoms
b = 2.22 * 1022 atoms
1 molecule
Pressure varies directly with absolute temperature.
P1
P2
=
T1
T2
P2 =
P1T2
T1
T1 = 25°C + 273 = 298 K
T2 = 212°F = 100°C = 373 K
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Page 159
- Chapter 12 -
132 lb>in.221373 K2
= 40. lb>in.2
298 K
At 212°F the tire pressure is 40. lb>in.2
The tire will not burst.
P2 =
70.
A column of mercury at 1 atm pressure is 760 mm Hg high. The density of mercury is
13.6 times that of water, so a column of water at 1 atm pressure should be 13.6 times as
high as that for mercury.
1760 mm2113.62 = 1.03 * 104 mm 133.8 ft2
71.
Use the ideal gas equation
PV = nRT
n =
RT
PV
Change 2.20 * 103 lb>in.2 to atmosphere
12.20 * 103 lb>in.22 ¢
n =
72.
(150. atm)(55 L)
= 3.3 * 102 mol O2
0.0821 L # atm
a
b(300. K)
mol # K
The conversion is: m3 ¡ cm3 ¡ mL ¡ L ¡ mol
11.00 m32a
73.
1 atm
≤ = 150. atm
14.7 lb>in.2
100 cm 3 1 mL
1L
1 mol
b ¢
ba
b = 44.6 mol Cl 2
≤a
3
1m
1000 mL 22.4 L
1 cm
First calculate the moles of gas and then convert moles to molar mass.
10.560 L2a
1 mol
b = 0.0250 mol
22.4 L
1.08 g
= 43.2 g>mol (molar mass)
0.0250 mol
74.
The conversion is: g>L ¡ g>mol
a
1.78 g 22.4 L
ba
b = 39.9 g>mol (molar mass)
L
mol
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Page 160
- Chapter 12 -
75.
PV = nRT
(a)
V =
(b)
n =
10.510 mol210.0821 L atm>mol K21320. K2
nRT
=
= 8.4 L H 2
P
1.6 atm
10.789 atm2116.0 L2
PV
=
= 0.513 mol CH 4
RT
10.0821 L atm>mol K21300. K2
The molar mass for CH 4 is 16.04 g/mol
116.04 g>mol210.513 mol2 = 8.23 g CH 4
(c)
PV = nRT, but n =
g
where M is the molar mass and g is the grams of the gas.
M
gRT
. To determine density, d = g>V.
M
g
gRT
g
PM
=
.
Solving PV =
for
produces
M
V
V
RT
14.00 atm2144.01 g>mol2
g
=
= 8.48 g>L CO2
d =
V
10.0821 L atm>mol K21253 K2
Thus, PV =
(d)
76.
g
PM
=
from part (c), solve for M (molar mass)
V
RT
12.58 g>L210.0821 L atm>mol K21300. K2
dRT
=
= 63.5 g>mol (molar mass)
M =
P
1.00 atm
Since d =
C2H 21g2 + 2HF1g2 ¡ C2H 4F21g2
1.0 mol C2H 2 ¡ 1.0 mol C2H 4F2
15.0 mol HF2 ¢
1 mol C2H 4F2
≤ = 2.5 mol C2H 4F2
2 mol HF
C2H 2 is the limiting reactant. 1.0 mol C2H 4F2 forms, no moles C2H 2 remain.
According to the equation, 2.0 mol HF yields 1.0 mol C2H 4F2 . Therefore,
5.0 mol HF - 2.0 mol HF = 3.0 mol HF unreacted
The flask contains 1.0 mol C2H 4F2 and 3.0 mol HF when the reaction is complete.
The flask contains 4.0 mol of gas.
14.0 mol210.0821 L atm>mol K21273 K2
nRT
P =
=
= 9.0 atm
V
10.0 L
77.
18.30 mol Al2 ¢
3 mol H 2 22.4 L
b = 279 L H 2 at STP
≤a
2 mol Al
mol
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- Chapter 12 -
78.
According to Graham’s Law of Effusion, the rates of effusion are inversely proportional
to the molar mass.
molar mass N2
28.02
rate He
=
=
= 27.000 = 2.646
rate N2
A molar mass He A 4.003
Helium effuses 2.646 times faster than nitrogen.
79.
(a)
According to Graham’s Law of Effusion, the rates of effusion are inversely
proportional to the molar mass.
16.04
rate He
=
= 24.007 = 2.002
rate CH 4
A 4.003
(b)
80.
Helium effuses twice as fast as CH 4 .
x = distance He travels
100 - x = distance CH 4 travels
D = distance traveled
DHe = 2 DCH4
x = 21100 - x2
3x = 200
x = 66.7 cm
The gases meet 66.7 cm from the helium end.
Assume 100. g of material to start with. Calculate the empirical formula.
C
185.7 g2a
1 mol
b = 7.14 mol
12.01 g
7.14
= 1.00 mol
7.14
H
114.3 g2a
1 mol
b = 14.2 mol
1.008 g
14.2
= 1.99 mol
7.14
The empirical formula is CH 2 . To determine the molecular formula, the molar mass
must be known.
a
2.50 g 22.4 L
ba
b = 56.0 g>mol (molar mass)
L
mol
56.0
The empirical formula mass is 14.0
= 4
14.0
Therefore, the molecular formula is (CH 2)4 = C4H 8
81.
2 CO1g2 + O21g2 : 2 CO21g2
110.0 mol CO2 ¢
Determine the limiting reactant
2 mol CO2
≤ = 10.0 mol CO2 (from CO)
2 mol CO
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- Chapter 12 -
18.0 mol O22 ¢
2 mol CO2
≤ = 16 mol CO2 (from O2)
1 mol O2
CO: the limiting reactant,
O2: in excess, 3.0 mol O2 unreacted.
(a) 10.0 mol CO react with 5.0 mol O2
10.0 mol CO2 and 3.0 mol O2 are present, no CO will be present.
(b)
82.
P =
113 mol210.0821 L atm>mol K21273 K2
nRT
=
= 29 atm
V
10. L
2 KClO31s2
¢
" 2 KCl1s2 + 3O 1g2
2
First calculate the moles of O2 produced. Then calculate the grams of KClO3 required to
produce the O2 . Then calculate the % KClO3 .
10.25 L O22a
1 mol
b = 0.011 mol O2
22.4 L
10.011 mol O22 ¢
a
83.
0.90 g
b11002 = 75% KClO3 in the mixture
1.20 g
Assume 1.00 L of air. The mass of 1.00 L of air is 1.29 g.
P2V2
P1V1
=
T1
T2
V2 =
d =
84.
2 mol KClO3 122.6 g
b = 0.90 g KClO3 in the sample
≤a
3 mol O2
mol
1760 torr211.00 L21290. K2
P1V1T2
=
= 1.8 L
P2T1
1450 torr21273 K2
1.29 g
m
=
= 0.72 g>L
V
1.8 L
Each gas behaves as though it were alone in a 4.0 L system.
(a) After expansion: P1V1 = P2V2
For CO2
For H 2
(b)
1150. torr213.0 L2
P1V1
=
= 1.1 * 102 torr
V2
4.0 L
150. torr211.0 L2
P1V1
=
= 13 torr
P2 =
V2
4.0 L
P2 =
Ptotal = PH2 + PCO2 = 110 torr + 13 torr = 120 torr (2 sig. figures)
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- Chapter 12 -
85.
P1 = 40.0 atm
P2 = P2
V1 = 50.0 L
V2 = 50.0 L
T1 = 25°C = 298 K
T2 = 25°C + 152°C = 177°C = 450. K
Gas cylinders have constant volume, so pressure varies directly with temperature.
140.0 atm21450. K2
P1T2
=
= 60.4 atm
P2 =
T1
298 K
86.
You can identify the gas by determining its density.
mass of gas = 1.700 g - 0.500 g = 1.200 g
volume of gas: Charles law problem. Correct volume to 273 K
10.4478 L21273 K2
V1
V2
V1T2
V2 =
=
=
= 0.3785 L
T1
T2
T1
323 K
1.200 g
m
=
= 3.170 g>L
d =
V
0.3785 L
gas is chlorine (see Table 12.3)
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