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. 6. The molecules of 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. 9. If 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). T 2 6 T 1 , (PV = k) H 2 1 torr = 1 mm Hg - 140 - HEINS12-140-163v4.qxd 12/30/06 2:32 PM Page 140
Transcript
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 gasesare colorless. Two ways could be used to show the diffusion. The change of densitywould be one method. Before diffusion the gas in the flask containing hydrogen would bemuch less dense. After diffusion, the gas densities in both flasks would be equal. Asecond method would require the introduction of spark gaps into both flasks. Beforediffusion, neither gas would show a reaction when sparked. After diffusion, the gases inboth 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. Itdepends on the temperature, the number of molecules of the gas and the volume of thecontainer.
3. The air pressure inside the balloon is greater than the air pressure outside the balloon. Thepressure inside must equal the sum of the outside air pressure plus the pressure exerted bythe stretched rubber of the balloon.
4. The major components of dry air are nitrogen and oxygen.
5.
6. The molecules of at 100°C are moving faster. Temperature is a measure of averagekinetic 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. Thisis an inverse relationship as is Boyle’s law.
9. If the volume of the cylinder would decrease (the piston would movedownward).
10. The pressure inside the bottle is less than atmospheric pressure. We come to thisconclusion because the water inside the bottle is higher than the water in the trough(outside the bottle).
T2 6 T1,
(PV = k)
H2
1 torr = 1 mm Hg
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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 greaterthan air. For example:
12. Basic assumptions of Kinetic Molecular Theory include:
(a) Gases consist of tiny particles.(b) The distance between particles is great compared to the size of the particles.(c) Gas particles move in straight lines. They collide with one another and with the
walls of the container with no loss of energy.(d) Gas particles have no attraction for each other.(e) The average kinetic energy of all gases is the same at any given temperature. It
varies directly with temperature.
13. The order of increasing molecular velocities is the order of decreasing molar masses.
At the same temperature the kinetic energies of the gases are the same and equal toFor 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 thesame temperature.
15. Gases are described by the following parameters:
(a) pressure (c) temperature(b) volume (d) number of moles
16. An ideal gas is one which follows the described gas laws at all P, V and T and whosebehavior is described exactly by the Kinetic Molecular Theory.
17. Boyle’s law: ideal gas equation:If you have an equal number of moles of two gases at the same temperature the right sideof the ideal gas equation will be the same for both gases. You can then set PV for the firstgas equal to PV for the second gas (Boyle’s law) because the right side of both equationswill cancel.
18. Charles’ law: ideal gas equation:Rearrange the ideal gas equation to: V>T = nR>P PV = nRTV1>T1 = V2>T2,
PV = nRTP1V1 = P2V2,
1�2 mv2.
increasing molecular velocity
;99999999999999Rn, F2 , N2 CH4, He, H2
99999999999999:decreasing molar mass
O2, H2S, HCl, F2 , CO2.
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If you have an equal number of moles of two gases at the same pressure the right side ofthe rearranged ideal gas equation will be the same for both. You can set for the firstgas equal to for the second gas (Charles’ law) because the right side of bothequations will cancel.
19. A gas is least likely to behave ideally at low temperatures. Under this condition, thevelocities of the molecules decrease and attractive forces between the molecules begin toplay a significant role.
20. A gas is least likely to behave ideally at high pressures. Under this condition, themolecules are forced close enough to each other so that their volume is no longer smallcompared to the volume of the container. Attractive forces may also occur here andsooner or later, the gas will liquefy.
21. Equal volumes of and at the same T and P:
(a) have equal number of molecules (Avogadro’s law)
(b)
(c)
(d) average kinetic energies are the same (T same)
(e) rate the rate of (Graham’s Law of Effusion)
(f)
22. 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 inverselyproportional to the pressure, at constant temperature. The Kinetic Molecular Theoryassumes the volume occupied by gases is mostly empty space. Decreasing thevolume of a gas by compressing it, increases the concentration of gas molecules,resulting in more collisions of the molecules and thus increased pressure upon thewalls of the container.
density O2 = a32
2b (density H2) = 16(density H2)
density O2 = ¢mass O2
mass H2≤ (density H2) ¢ mass O2
density O2≤ = ¢ mass H2
density H2≤
volume O2 = volume H2
density H2 = ¢ mass H2
volume H2≤ density O2 = ¢ mass O2
volume O2≤
density O2 = 16 times the density of H2
O2H2 = 4 times
moles O2 = moles H2
mass O2 = 16 times mass of H2
O2H2
V>T V>T
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(b) Charles’ law. Charles’ law states that the volume of a fixed mass of gas is directlyproportional to the absolute temperature, at constant pressure. According to KineticMolecular Theory, the kinetic energies of gas molecules are proportional to theabsolute temperature. Increasing the temperature of a gas causes the molecules tomove faster, and in order for the pressure not to increase, the volume of the gasmust increase.
(c) Dalton’s law. Dalton’s law states that the pressure of a mixture of gases is the sumof the pressures exerted by the individual gases. According to the Kinetic MolecularTheory, there are no attractive forces between gas molecules; therefore, in a mixtureof gases, each gas acts independently and the total pressure exerted will be the sumof the pressures exerted by the individual gases.
23.
According to Avogadro’s Law, equal volumes of nitrogen and oxygen at the sametemperature and pressure contain the same number of molecules. In the reaction, nitrogenand oxygen molecules react in a 1:1 ratio. Since two volumes of nitrogen monoxide areproduced, one molecule of nitrogen and one molecule of oxygen must produce twomolecules of nitrogen monoxide. Therefore each nitrogen and oxygen molecule must bemade 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 givenvolume, and 0°C and 760 torr are convenient reference points.
25. Conversion of oxygen to ozone is an endothermic reaction. Evidence for this statement isthat energy is required to convert to
26. Chlorofluorocarbons, (Freons, and ), are responsible for damaging theozone layer. When these compounds are carried up to the stratosphere (the outer part ofthe atmosphere) they absorb ultraviolet radiation and produce chlorine free radicals thatreact with ozone and destroy it.
27. Heating a mole of 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 theKinetic Molecular Theory.
N2
(O3)
CCl2F2CCl3F,
O3.O2(286 kJ>3 mol O2)
1 vol + 1 vol ¡ 2 vol
N2(g) + O21g2 ¡ 2 NO1g2
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(d) Average velocity of the molecules will increase. Increasing the temperatureincreases the average kinetic energies of the molecules; hence, the average velocityof the molecules will increase also.
(e) Number of molecules remains unchanged. Heating does not alter the number ofmolecules present, except if extremely high temperatures were attained. Then, the
molecules might dissociate into N atoms resulting in fewer molecules.
28.An oxygen molecule contains 16 electrons.
Ozone molecule = O3Oxygen molecule = O2Oxygen atom = O
N2N2
N2
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CHAPTER 12
SOLUTIONS TO EXERCISES
1. (a)
(b)
(c)
2. (a)
(b)
(c)
3. (a)
(b)
(c)
(d)
4. (a)
(b)
(c)
(d) 10.67 kPa2a 1 atm
101.325 kPab = 0.0066 atm
1225 torr2a 1 atm
760 torrb = 0.296 atm
14250. cm Hg2a 1 atm
76 cm Hgb = 55.92 atm
162 mm Hg2a 1 atm
760 mm Hgb = 0.082 atm
15.00 kPa2a 1 atm
101.325 kPab = 0.0493 atm
1795 torr2a 1 atm
760 torrb = 1.05 atm
16000. cm Hg2a 1 atm
76 cm Hgb = 78.95 atm
128 mm Hg2a 1 atm
760 mm Hgb = 0.037 atm
1715 mm Hg2a101.325 kPa
760 mm Hgb = 95.3 kPa
1715 mm Hg2a 1013 mbar
760 mm Hgb = 953 mbar
1715 mm Hg2a 1 torr
1 mm Hgb = 715 torr
1715 mm Hg2¢ 14.7 lb/in.2
760 mm Hg≤ = 13.8 lb>in.2
1715 mm Hg2a 1 in. Hg
25.4 mm Hgb = 28.1 in. Hg
1715 mm Hg2a 1 atm
760 mm Hgb = 0.941 atm
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5. Change 625 torr to atmospheres and mm Hg.
(a)
(b)
6. Change 722 torr Hg to atmospheres and mm Hg.
(a)
(b)
7.
8.
9. Temperatures must be in Kelvin
(a)
(b)
10. Temperatures must be in Kelvin
(a)
(b)16.00 L21345 K2
248 K= 8.35 L
16.00 L21255 K2248 K
= 6.17 L
0.0°F = -18°C
(°C + 273)V1
T1=
V2
T2 or V2 =
V1T2
T1;
16.00 L21100. K2248 K
= 2.42 L
16.00 L21273 K2248 K
= 6.60 L
(°C + 273)V1
T1=
V2
T2 or V2 =
V1T2
T1;
11.7 atm21225 mL2115 mL
= 3.3 atmP2 =P1V1
V2
10.75 atm21521 mL2776 mL
= 0.50 atmP2 =P1V1
V2
1722 torr2(1 mm Hg/1 torr21635 mL2795 mm Hg
= 577 mL
1722 torr211 atm>760 torr21635 mL22.5 atm
= 241 mL
P1V1 = P2V2 or V2 =P1V1
P2.
1625 torr211mm Hg/1 torr2(525 mL)
455 mm Hg= 721 mL
1625 torr211 atm/760 torr2(525 mL)
1.5 atm= 288 mL
P1V1 = P2V2 or V2 =P1V1
P2.
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11. Use the combined gas law
12. Use the combined gas law
13. Use the combined gas law
14. Use the combined gas law
15. Use the combined gas law
16. Use the combined gas law
Change 765 torr to atmospheres.
17.
PN2= 721 torr - 19.8 torr = 701 torr
PH2O vapor = 19.8 torr
Ptotal = PN2+ PH2O vapor = 721 torr
1765 torr211 atm/760 torr211.5 L2(292 K)
11.5 atm212.5 L2 = 120 K (120 - 273) K = -150°C
P1V1
T1=
P2V2
T2 or T2 =
P2V2T1
P1V1
11.0 atm21775 mL21298 K21615 mL21273 K2 = 1.4 atm
P1V1
T1=
P2V2
T2 or P2 =
P1V1T2
V2T1
V2 =12.50 atm2122.4 L21268 K211.50 atm21300. K2 = 33.4 L
Total moles = mol H2 + mol CO2 = 5.00 mol + 0.500 mol = 5.50 mol
V =10.75 mol210.0821 L atm>mol K21298 K2
725 torr
760
torr
atm
= 19 L KrV =nRT
PPV = nRT
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45. The balanced equation is Remember that volume–volume relationships are the same as mole–mole relationshipswhen dealing with gases at the same T and P.
(a)
(b)
(c) Limiting reactant problem.
Oxygen is the limiting reactant. 20. L NO us formed.
46. The balanced equation is Remember that volume–volume relationships are the same as mole–mole relationshipswhen dealing with gases at the same T and P.
(a)
(b)
(c) Limiting reactant problem.
Oxygen is the limiting reactant. 12 L H2O is formed.
47. The balanced equation is
(0.525 kg KCl)a1000 g
1 kgb a 1 mol
74.55 gb a 3 mol O2
2 mol KClb a 22.4 L
1 molb = 237 L O2
2 KClO3(s) : 2 KCl(s) + 3 O2(g)
(15 L O2)a4 L H2O
5 L O2b = 12 L H2O
(15 L C3H8)a 4 L H2O
1 L C3H8b = 60. L H2O
(35 L C3H8)a 3 L CO2
1 L C3H8b a 1 mol
22.4 Lb a44.01 g
1 molb = 210 g CO2
(7.2 L C3H8)a 5 L O2
1 L C3H8b = 36 L O2
C3Hg(g) + 5 O2(g) : 3 CO2(g) + 4 H2O(g)
(25 L NH3)a 4 L NO
5 L NH3b = 25 L NO
(25 L O2)a4 L NO
5 L O2b = 20. L NO
(25 L NH3)a6 L H2O
4 L NH3b a 1 mol
22.4 Lb a 18.02 g
1 molb = 30. g H2O
(2.5 L NH3)a 5 L O2
4 L NH3b = 3.1 L O2
4 NH3(g) + 5 O2(g) : 4 NO(g) + 6 H2O(g)
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48. The balanced equation is
49. Like any other gas, water in the gaseous state occupies a much larger volume than in theliquid state.
50. During the winter the air in a car’s tires is colder, the molecules move slower and thepressure decreases. In order to keep the pressure at the manufacturer’s recommended psiair 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
52.
P
(a)
V
T
(c)
P
T
(b)
V
n
(d)
V
P = 3.7 atm a760 torr
1 atmb = 2.8 * 103 torr
P =11.5 mol210.0821 L atm>mol K21303 K2
10. L= 3.7 atm
PV = nRT P =nRT
V
= 1.12 * 103 L CO2
(1.50 kg C6H12O6)a1000 g
1 kgb a 1 mol
180.16 gb a 6 mol CO2
1 mol C6H12O6b a 22.4 L
1 molb
C6H12O6(s) + 6 O2(g) : 6 CO2(g) + 6 H2O(l)
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53. The can is a sealed unit and very likely still contains some of the aerosol. As the can isheated, pressure builds up in it eventually causing the can to explode and rupture withpossible harm from flying debris.
54. One mole of an ideal gas occupies 22.4 liters at standard conditions. (0°C and 1 atmpressure)
T =11.3 * 109 atm211.0 L212.0 g>mol210.0821 L atm>mol K211.4 * 103 g2 = 2.3 * 107 K
11.3 * 109 atm211.0 L2 = ¢ 1.4 * 103 g
2.0 g>mol≤10.0821 L atm>mol K21T2
¢1.4 g
cm3 ≤ ¢ 1000 cm3
L≤ = 1.4 * 103 g>L
PV = nRT or PV = a g
molar massbRT
10.019 mol CO2¢2 mol CO2
2 mol CO≤ a 22.4 L
molb = 0.43 L CO2 = 430 mL CO2
0.0095 mol O2
mol CO = 0.019 mol
a800 mm Hg * 1 atm
760 mm Hgb10.500 L2 = 1n210.0821 L atm>mol K21333 K2
mol O2 = 0.038 mol
11.8 atm210.500 L O22 = 1n210.0821 L atm>mol K21288 K2O2:
PV = nRTCOO2
2 CO + O2 ¡ 2 CO2
PN2= 1790 torr2¢ 0.051 mol N2
0.085 mol≤ = 4.7 * 102 torr
PCO2= 1790 torr2¢0.013 mol CO2
0.085 mol≤ = 1.2 * 102 torr
PO2= 1790 torr2¢ 0.020 mol O2
0.085 mol≤ = 1.9 * 102 torr
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(b) mass of air in the ball
(c) Actually the pressure changes when the temperature changes. Since pressure isdirectly proportional to moles we can calculate the change in moles required tokeep the pressure the same at 30°C as it was at 20°C.
of air required to keep the pressure the same at 30°C.
0.01 mol air must be allowed to escape from the ball.
or 0.3 g air must be allowed to escape.
62. Use the combined gas laws to calculate the bursting temperature
63. To double the volume of a gas, at constant pressure, the temperature (K) must be doubled.
64. at 22°C and 740 torr
after change in temperature (P constant)
after change in pressure (T constant)
Since temperature is constant,
torr (pressure to change 2 V to V)P2 = 1740 torr2a2 V
Vb = 1.5 * 103
P1V1 = P2V2 or P2 =P1V1
V2
V = volume
2 V = volume
V = volume
T2 = 21300. K2 = 600. K = 327°C
T2 = 2 T1 T2 =2 V1T1
V1 V1
T1=
2 V1
T2
V2 = 2 V1 V1
T1=
V2
T2
T2 =P2V2T1
P1V1=176 cm212.00 L21293 K2165 cm211.75 L2 = 392 K 1119°C2
T2 = T2 T1 = 20°C 1293 K2 V2 = 2.00 L V1 = 1.75 L
P2 = 1.00 atm (76 cm) P1 = 65 cmP1V1
T1=
P2V2
T2
(T2).
10.01 mol air2a29 g
molb = 0.29 g
(0.18 - 0.17)
n = 0.17 mol
128 lb>in.22¢ 1 atm
14.7 lb>in.2≤12.24 L2 = 1n210.0821 L atm>mol K21303 K2
PV = nRT
m = 10.18 mol2a29 g
molb = 5.2 g air
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65. Volume is constant, so or
66. The volume of the tires remains constant (until they burst), so
or
67. Use the combined gas laws.
or
68. Use the combined gas law or
First calculate the volume at STP.
At STP, a mole of any gas has a volume of 22.4 L
Each molecule of contains 3 atoms, so:
69. Pressure varies directly with absolute temperature.
T2 = 212°F = 100°C = 373 K
T1 = 25°C + 273 = 298 KP2 =P1T2
T1
P1
T1=
P2
T2
17.39 * 1021 molecules2a 3 atoms
1 moleculeb = 2.22 * 1022 atoms
N2O
10.275 L2a 1 mol
22.4 Lb ¢6.022 * 1023 molecules
1 mol≤ = 7.39 * 1021 molecules
V2 =1400. torr21600. mL21273 K2
1760. torr21313 K2 = 275 mL = 0.275 L
V2 =P1V1T2
P2T1
P1V1
T1=
P2V2
T2
P2 =11.00 atm21800. mL21303 K2
1250. mL21273 K2 = 3.55 atm
P1 and T1 are at STPP2 =P1V1T2
V2T1
P1V1
T1=
P2V2
T2
T2 =144 psi21295 K2
30. psi= 433 K = 160°C = 320°F
71.0°F = 21.7°C = 295 K
T2 =T1P2
P1;
P1
T1=
P2
T2
T2 =1500. torr21295 K2
700. torr= 211 K = -62°C
T2 =T1P2
P1;
P1
T1=
P2
T2
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At 212°F the tire pressure is
The tire will not burst.
70. A column of mercury at 1 atm pressure is 760 mm Hg high. The density of mercury is13.6 times that of water, so a column of water at 1 atm pressure should be 13.6 times ashigh as that for mercury.
71. Use the ideal gas equation
Change to atmosphere
72. The conversion is:
73. First calculate the moles of gas and then convert moles to molar mass.
74. The conversion is:
a1.78 g
Lb a22.4 L
molb = 39.9 g>mol (molar mass)
g>L ¡ g>mol
1.08 g
0.0250 mol= 43.2 g>mol (molar mass)
10.560 L2a 1 mol
22.4 Lb = 0.0250 mol
11.00 m32a100 cm
1 mb3¢ 1 mL
1 cm3 ≤ a 1 L
1000 mLb a 1 mol
22.4 Lb = 44.6 mol Cl2
m3 ¡ cm3 ¡ mL ¡ L ¡ mol
n =(150. atm)(55 L)
a0.0821 L # atm
mol # Kb(300. K)
= 3.3 * 102 mol O2
12.20 * 103 lb>in.22¢ 1 atm
14.7 lb>in.2≤ = 150. atm
2.20 * 103 lb>in.2
n =RT
PVPV = nRT
1760 mm2113.62 = 1.03 * 104 mm 133.8 ft2
40. lb>in.2
P2 =132 lb>in.221373 K2
298 K= 40. lb>in.2
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75.
(a)
(b)
The molar mass for is 16.04 g/mol
(c) but where M is the molar mass and g is the grams of the gas.
Thus, To determine density,
Solving for produces
(d) Since from part (c), solve for M (molar mass)
76.
is the limiting reactant. forms, no moles remain. According to the equation, 2.0 mol HF yields Therefore,
The flask contains and 3.0 mol HF when the reaction is complete.
