C H A P T E R 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 ¼ 1 mm Hg
6. The molecules of H2 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 ¼ k)
9. If T2 < 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).
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, H2S, HCl, F2, CO2.
- 124 -
- Chapter 12 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.
increasing molecular velocity
!
Rn; F2 ; N2 CH4 ; He; H2
!
decreasing molar mass
At the same temperature the kinetic energies of the gases are the same and equal to ½ mv2 . 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:
(c) temperature
(a) pressure
(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 whose behavior is
described exactly by the Kinetic Molecular Theory.
17. Boyle’s law: P1 V1 ¼ P2 V2 , ideal gas equation: PV ¼ 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 ¼ V2 =T2 , ideal gas equation: PV ¼ nRT
Rearrange the ideal gas equation to: V=T ¼ nR=P
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.
- 125 -
- Chapter 12 21. Equal volumes of H2 and O2 at the same T and P:
(a) have equal number of molecules (Avogadro’s law)
(b) mass O2 ¼ 16 times mass of H2
(c) moles O2 ¼ moles H2
(d) average kinetic energies are the same (T same)
(e) rate H2 ¼ 4 times the rate of O2 (Graham’s Law of Effusion)
(f) density O2 ¼ 16 times the density of H2
mass O2
mass H2
density H2 ¼
density O2 ¼
volume O2
volume H2
volume O2 ¼ volume H2
mass O2
mass H2
mass O2
ðdensity H2 Þ
density O2 ¼
density O2
density H2
mass H2
32
density O2 ¼
ðdensity H2 Þ ¼ 16 ðdensity H2 Þ
2
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 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.
(b) 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.
(c) 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.
23. N2 ðgÞ þ O2 ðgÞ ! 2 NOðgÞ
1 vol þ 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 of 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.
- 126 -
- Chapter 12 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, CC13F, and CCl2F2), 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.
(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.
28.
Oxygen atom ¼ O Oxygen molecule ¼ O2
An oxygen molecule contains 16 electrons.
Ozone molecule ¼ O3
- 127 -
- Chapter 12 SOLUTIONS TO EXERCISES
1. Pressure conversions:
a.
b.
c.
(a)
(b)
(c)
torr
inches Hg
768
752
745
30.2
29.6
29.3
kilopascals
102
100
99.3
760 torr
¼ 768 torr
in: Hg ! torr ð30:2 in: HgÞ
29:9 in: Hg
101:325 kPa
in: Hg ! kPa ð30:2 in: HgÞ
¼ 102 kPa
29:9 in: Hg
29:9 in: Hg
torr ! in: Hg ð752 torrÞ
¼ 29:6 in: Hg
760 torr
101:325 kPa
torr ! kPa ð752 torrÞ
¼ 100 kPa
760 torr
760 torr
kPa ! torr ð99:3 kPaÞ
¼ 745 torr
101:325 kPa
29:9 in: Hg
kPa ! in: Hg ð99:3 kPaÞ
¼ 29:3 in: Hg
101:325 kPa
2. Pressure conversions:
a.
b.
c.
(a)
(b)
(c)
mm Hg
lb/in.2
atmospheres
789
1700
1100
15.3
32
21
1.04
2.2
1.4
14:7 lb=in:2
¼ 15:3 lb=in:2
mm Hg ! lb=in ð789 mm HgÞ
760 mm Hg
1 atm
mm Hg ! atm ð789 mm HgÞ
¼ 1:04 atm
760 mm Hg
760 mm Hg
¼ 1700 mm Hg
lb=in:2 ! mm Hg ð32 lb=in:2 Þ
14:7 lb=in:2
1 atm
2
2
lb=in: ! atm ð32 lb=in: Þ
¼ 2:2 atm
14:7 lb=in:2
760 mm Hg
atm ! mm Hg ð1:4 atmÞ
¼ 1100 mm Hg
1 atm
14:7 lb=in:2
2
¼ 21 lb=in:2
atm ! lb=in: ð1:4 atmÞ
1 atm
2
- 128 -
- Chapter 12 3. (a)
(b)
(c)
(d)
1 atm
ð28 mm HgÞ
¼ 0:037 atm
760 mm Hg
1 atm
ð6000 cm HgÞ
¼ 78:95 atm
76 cm Hg
1 atm
ð795 torrÞ
¼ 1:05 atm
760 torr
1 atm
ð5:00 kPaÞ
¼ 0:0493 atm
101:325 kPa
4. (a)
(b)
(c)
(d)
1 atm
¼ 0:082 atm
ð62 mm HgÞ
760 mm Hg
1 atm
ð4250 cm HgÞ
¼ 55:92 atm
76 cm Hg
1 atm
ð225 torrÞ
¼ 0:296 atm
760 torr
1 atm
ð0:67 kPaÞ
¼ 0:0066 atm
101:325 kPa
5. P1 V1 ¼ P2 V2
or V2 ¼
P 1 V1
. Change 625 torr to atmospheres and mm Hg.
P2
(a)
ð625 torrÞð1 atm=760 torrÞð525 mLÞ
¼ 288 mL
1:5 atm
(b)
ð625 torrÞð1 atm Hg=1 torrÞð525 mLÞ
¼ 721 mL
455 mm Hg
6. P1 V1 ¼ P2 V2
(a)
(b)
or V2 ¼
P 1 V1
. Change 722 torr Hg to atmospheres and mm Hg.
P2
ð722 torrÞð1 atm=760 torrÞð635 mLÞ
¼ 241 mL
2:5 atm
ð722 torrÞð1 atm Hg=1 torrÞð635 mLÞ
¼ 577 mL
795 mm Hg
7.
P2 ¼
P 1 V1
V2
ð0:75 atmÞð521 mLÞ
¼ 0:50 atm
776 mL
8.
P2 ¼
P 1 V1
V2
ð1:7 atmÞð225 mLÞ
¼ 3:3 atm
115 mL
- 129 -
- Chapter 12 -
9.
V1 V2
¼
T1 T2
(a)
(b)
(c)
10.
(b)
(c)
V2 ¼
V1 T 2
; Temperatures must be in Kelvin ( C þ 273)
T1
ð125 mLÞð268 KÞ
¼ 114 mL
294 K
ð125 mLÞð308 KÞ
¼ 131 mL
294 K
ð125 mLÞð1095 KÞ
¼ 466 mL
294 K
V1 V2
¼
T1 T2
(a)
or
or
V2 ¼
V1 T 2
; Temperatures must be in Kelvin ( C þ 273)
T1
ð575 mLÞð298 KÞ
¼ 691 mL
248 K
ð575 mLÞð273 KÞ
¼ 633 mL ð32 F ¼ 0 C ¼ 273 KÞ
248 K
ð575 mLÞð318 KÞ
¼ 737 mL
248 K
11. Use the combined gas laws
V2 ¼
P1 V1 P2 V2
¼
T1
T2
or V2 ¼
P 1 V1 T 2
P2 T1
P1 V1 P2 V2
¼
T1
T2
or V2 ¼
P1 V1 T2
P2 T1
ð0:950 atmÞð1400: LÞð275 KÞ
¼ 2:4 105 L
ð4:0 torrÞð1 atm=760 torrÞð291 KÞ
14. Use the combined gas law
V2 ¼
P 1 V1 T 2
P2 T1
ð678 torrÞð25:6 LÞð308 KÞ
¼ 30:8 L
ð595 torrÞð292 KÞ
13. Use the combined gas law
V2 ¼
or V2 ¼
ð0:75 atmÞð1025 mLÞð308 KÞ
¼ 544 mL
ð1:25 atmÞð348 KÞ
12. Use the combined gas laws
V2 ¼
P1 V1 P2 V2
¼
T1
T2
P1 V1 P2 V2
¼
T1
T2
or V2 ¼
ð2:50 atmÞð22:4 LÞð268 KÞ
¼ 33:4 L
ð1:50 atmÞð300: KÞ
- 130 -
P1 V1 T2
P2 T1
- Chapter 12 -
15. Use the combined gas law
P 1 V1 P 2 V2
¼
T1
T2
or P2 ¼
P 1 V1 T 2
V2 T1
or T2 ¼
P 2 V2 T 1
P1 V1
ð1:0 atmÞð775 mLÞð298 KÞ
¼ 1:4 atm
ð615 mLÞð273 KÞ
16. Use the combined gas law
P 1 V1 P 2 V2
¼
T1
T2
Change 765 torr to atmospheres.
ð765 torrÞð1 atm=760 torrÞð1:5 LÞð292 KÞ
¼ 120 K ð120 K 273Þ; K ¼ 153 C
ð1:5 atmÞð2:5 LÞ
17. Ptotal ¼ PO2 þ PH2 O vapor ¼ 772 torr
PH2 O vapor ¼ 21:2 torr
PO2 ¼ 772 torr 21:2 torr ¼ 751 torr
18. Ptotal ¼ PCH4 þ PH2 O vapor ¼ 749 mm Hg
PH2 O ¼ 30:0 torr ¼ 30:0 mm Hg
PCH4 ¼ 749 mm Hg 30:0 mm Hg ¼ 719 mm Hg
19.
Ptotal ¼ PN2 þ PH2 þ PO2
¼ 200: torr þ 600: torr þ 300: torr ¼ 1100: torr ¼ 1:100 103 torr
20.
Ptotal ¼ PH2 þ PN2 þ PO2
¼ 325 torr þ 475 torr þ 650: torr ¼ 1450: torr ¼ 1:450 103 torr
21. Ptotal ¼ PCH4 þ PH2 O vapor (Solubility of methane is being ignored.)
PH2 O vapor ¼ 23:8 torr
PCH4 ¼ 720: torr 23:8 torr ¼ 696 torr
To calculate the volume of dry methane, note that the temperature is constant, so P1 V1 ¼ P2 V2 can be
used.
V2 ¼
22.
P1 V1 ð696 torrÞð2:50 LÞ
¼
¼ 2:29 L
P2
ð760: torrÞ
Ptotal ¼ PC3 H8 þ PH2 O vapor
C3 H8 in propane
PH2 O vapor ¼ 20:5 torr
PC3 H8 ¼ 745 torr 20:5 torr ¼ 725 torr
- 131 -
- Chapter 12 To calculate the volume of dry propane, note that the temperature is constant, so P1 V1 ¼ P2 V2
can be used.
V2 ¼
P1 V1 ð725 torrÞð1:25 LÞ
¼
¼ 1:19 L C3 H8
ð760: torrÞ
P2
23. 1 mol of a gas occupies 22.4 L at STP
1 mol
ð1:75 LÞ
¼ 0:0781 mol O2
22:4 L
1 mol
24. ð3:50 LÞ
¼ 0:156 mol N2
22:4 L
22:4 L
25. (a)
6:02 1023 molecules
¼ 22:4 L CO2
23
molecules
6:02
10
22:4 L
(b) ð2:5 molÞ
¼ 56 L CH4
mol
22:4 L
(c) ð12:5 gÞ
¼ 8:75 L O2
32:00 g
22:4 L
24
26. (a)
1:80 10 molecules
¼ 67:0 L SO3
6:02 1023 molecules
22:4 L
(b) ð7:5 molÞ
¼ 170 L C2 H6
mol
22:4 L
(c) ð25:2 gÞ
¼ 7:96 L Cl2
70:90 g
27. ð725 mLÞ
28. ð945 mLÞ
1L
1000 mL
1L
1000 mL
1 mol
22:4 L
1 mol
22:4 L
17:03 g
mol
42:08 g
mol
¼ 0:551 g NH3
¼ 1:78 g C3 H6
1 mol
22:4 L
29. ð1025 molecules CO2 Þ
¼ 3:813 1020 L CO2
mol
6:022 1023 molecules
1 mol
30. ð10:5 L CO2 Þ
22:4 L
6:022 1023 molecules
¼ 2:82 1023 molecules CO2
mol
31. density of Cl2 gas ¼ 3:17 g=L (from table 12.3)
1L
ð10:0 gÞ
¼ 3:15 L
3:17 g
32. density of CH4 gas ¼ 0:716 g=L (from table 12.3)
ð3:0 LÞð0:716 g=LÞ ¼ 2:1 g CH4
- 132 -
- Chapter 12 33. (a)
(b)
(c)
(d)
34. (a)
(b)
(c)
(d)
35. (a)
Density
of Gases 4:003 g He
1 mol
¼ 0:179 g=L He
d¼
mol
22:4 L
20:01 g HF
1 mol
d¼
¼ 0:893 g=L HF
mol
22:4 L
42:08 g C3 H6
1 mol
d¼
¼ 1:89 g=L C3 H6
mol
22:4 L
120:9 g CCl2 F2
1 mol
d¼
¼ 5:40 g=L CCl2 F2
mol
22:4 L
222 g Rn
1 mol
d¼
¼ 9:91 g=L Rn
mol
22:4 L
46:01 g NO2
1 mol
d¼
¼ 2:054 g=L NO2
mol
22:4 L
80:07 g SO3
1 mol
¼ 3:575 g=L SO3
mol
22:4 L
28:05 g C2 H4
1 mol
¼ 1:252 g=L C2 H4
mol
22:4 L
Assume 1.00 mol of NH3 and determine the volume using the ideal gas equation, PV ¼ nRT.
nRT ð1:00 mol NH3 Þð0:0821 L atm=mol KÞð298 KÞ
¼
¼ 20: L at 25 C and 1:2 atm
P
1:2 atm
17:03 g
d¼
¼ 0:85 g=L NH3
20: L
V¼
(b)
Assume 1.00 mol of Ar and determine the volume using the ideal gas equation, PV ¼ nRT.
nRT ð1:00 mol ArÞð0:0821 L atm=mol KÞð348 KÞ
¼
¼ 29:1 L Ar at 75 C and 745 torr
745 torr
P
torr
760
39:95 g
atm
¼ 1:37 g=L Ar
d¼
29:1 L
V¼
36. (a)
Assume 1.00 mol C2H4 and determine the volume using the ideal gas equation, PV ¼ nRT.
nRT ð1:00 mol C2 H4 Þð0:0821 L atm=mol KÞð305 KÞ
¼
¼ 33 L C2 H4 at 32 C and 0:75 atm
P
0:75 atm
28:05 g
d¼
¼ 0:85 g=L C2 H4
33 L
Assume 1.00 mol of He and determine the volume using the ideal gas equation, PV ¼ nRT.
V¼
(b)
nRT ð1:00 mol HeÞð0:0821 L atm=mol KÞð330: KÞ
¼ 26:0 L He at 57 C and 791 torr
¼
791
torr
P
torr
4:003 g
d¼
¼ 0:154 g=L He 760 atm
26:0 L
V¼
- 133 -
- Chapter 12 -
37. PV ¼ nRT V ¼
nRT
P
V¼
38. PV ¼ nRT V ¼
nRT
P
V¼
39. PV ¼ nRT V ¼
RV
RT
n¼
ð75 mol NH3 Þð0:0821 L atm=mol KÞð295 KÞ
¼ 1:9 103 L NH3
729 torr
torr
760
atm
ð105 mol CH4 Þð0:0821 L atm=mol KÞð312 KÞ
¼ 1:8 103 L CH4
1:5 atm
ð1:2 atmÞð5:25 LÞ
¼ 0:26 mol O2
ð0:0821 L atm=mol KÞð299 KÞ
0
40. PV ¼ nRT n ¼
PV
RT
1
B752 torrC
@
torrAð9:55 LÞ
760
atm
¼ 0:362 mol CO2
n¼
ð0:0821 L atm=mol KÞð318 KÞ
0
41. PV ¼ nRT T ¼
PV
nR
B732 torrC
@
torrAð645 LÞ
760
atm
¼ 300: K
T¼
ð25:2 molÞð0:0821 L atm=mol KÞ
0
42. PV ¼ nRT T ¼
PV
nR
1
1
B675 torrC
@
torrAð725 LÞ
760
atm
¼ 209 K
T¼
ð37:5 molÞð0:0821 L atm=mol KÞ
43. The balanced equation is ZnðsÞ þ H2 SO4 ðaqÞ ! H2 ðgÞ þ ZnSO4 ðaqÞ
1 mol
1 mol H2
22:4 L 1000 mL
(a) ð52:7 g ZnÞ
¼ 1:81 104 mL H2
65:39 g 1 mol Zn
mol
1L
1L
1 mol
1 mol H2 SO4
¼ 0:0234 mol H2 SO4
(b) ð525 mL H2 Þ
1 mol H2
1000 mL 22:4 L
44. The balanced equation is 2 H2 O2 ðaqÞ ! 2 H2OðlÞ þ O2 ðgÞ
1 mol
1 mol O2
22:4 L 1000 mL
(a) ð50:0 g H2 O2 Þ
¼ 1:65 104 mL O2
34:02 g 2 mol H2 O2
mol
1L
1L
1 mol
2 mol H2 O2
¼ 0:0201 mol H2 O2
(b) ð225 mL O2 Þ
1 mol O2
1000 mL 22:4 L
- 134 -
- Chapter 12 45. The balanced equation is 4 NH3 ðgÞ þ 5 O2 ðgÞ ! 4 NOðgÞ þ 6 H2 OðgÞ
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
¼ 3:1 L O2
(a) ð2:5 L NH3 Þ
4 L NH3
6 L H2 O
1 mol
18:02 g
(b) ð25 L NH3 Þ
¼ 30: g H2 O
4 L NH3
22:4 L
1 mol
(c)
Limiting reactant problem.
4 L NO
¼ 20: L NO
ð25 L O2 Þ
5 L O2
4 L NO
¼ 25 L NO
ð25 L NH3 Þ
4 L NH3
Oxygen is the limiting reactant. 20. L NO us formed.
46. The balanced equation is C3 HgðgÞ þ 5 O2 ðgÞ ! 3 CO2 ðgÞ þ 4 H2 OðgÞ
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
(a) ð7:2 L C3 H8 Þ
¼ 36 L O2
1 L C3 H8
3 L CO2
1 mol
44:01 g
(b) ð35 L C3 H8 Þ
¼ 210 g CO2
1 L C3 H8
22:4 L
1 mol
(c)
Limiting reactant problem.
4 L H2 O
¼ 60: L H2 O
ð15 L C3 H8 Þ
1 L C3 H8
4 L H2 O
¼ 12 L H2 O
ð15 L O2 Þ
5 L O2
Oxygen is the limiting reactant. 12 L H2O is formed.
47. The balanced equation is 2 KClO3 ðsÞ ! 2 KClðsÞ þ 3 O2 ðgÞ
1000 g
1 mol
3 mol O2
22:4 L
ð0:525 kg KClÞ
¼ 237 L O2
1 kg
74:55 g 2 mol KCl
1 mol
48. The balanced equation is C6 H12 O6 ðsÞ þ 6 O2 ðgÞ ! 6 CO2 ðgÞ þ 6 H2 Oðl Þ
1000 g
1 mol
6 mol CO2
22:4 L
ð1:50 kg C6 H12 O6 Þ
¼ 1:12 103 L CO2
1 kg
180:2 g 1 mol C6 H12 O6
1 mol
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.
- 135 -
- Chapter 12 51. (a)
(b)
(c)
(d)
the pressure will be cut in half
the pressure will double
the pressure will be cut in half
the pressure will increase to 3.7 atm or 2836 torr
nRT
V
ð1:5 molÞð0:0821 L atm=mol KÞð303 KÞ
P¼
¼ 3:7 atm
10: L
760 torr
P ¼ 3:7 atm
¼ 2:8 103 torr
1 atm
PV ¼ nRT P ¼
52. (a)
(c)
P
T
V
(b)
P
(d)
T
n
V
V
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 ¼ nRT
ð1:00 atmÞðVÞ ¼ ð1:00 molÞð0:0821 L atm=mol KÞð273 KÞ
V ¼ 22:4 L
55. Solve for volume using PV ¼ nRT
(a)
(b)
V¼
ð0:2 mol Cl2 Þð0:0821 L atm=mol KÞð321 KÞ
¼ 5 L Cl2
ð80 cm=76 cmÞ atm
1 mol
0:0821 L atm
ð4:2 g NH3 Þ
ð262 KÞ
17:03 g
mol K
V¼
¼ 8:2 L NH3
0:65 atm
- 136 -
- Chapter 12 -
(c)
1 mol
0:0821 L atm
ð21 g SO3 Þ
ð328 KÞ
80:07 g
mol K
V¼
¼ 6:5 L SO3
110 kPa
kPa
101:3
atm
4.2 g NH3 has the greatest volume
56. Assume 1 mol of each gas
(a)
SF6 ¼ 146:1 g=mol
146:1 g
1 mol
d¼
¼ 6:52 g=L SF6
mol
22:4 L
(b)
Assume 25 C and 1 atm pressure
298 K
Vðat 25 CÞ ¼ ð22:4 LÞ
¼ 24:5 L
273 K
C2 H6 ¼ 30:07 g=mol
30:07 g
1 mol
¼ 1:23 g=L C2 H6
d¼
mol
24:5 L
(c)
He at 80 C and 2.15 atm
ð1 molÞð0:0821 L atm=mol KÞð193 KÞ
¼ 7:37 L
2:15 atm
4:003 g
1 mol
d¼
¼ 0:543 g=L He
mol
7:37 L
V¼
SF6 has the greatest density
57. (a)
Empirical formula. Assume 100 g starting material
80:0 g C
¼ 6:66 mol C
12:01 g=mol
20:0 g H
¼ 19:8 mol H
1:008 g=mol
(b)
6:66
¼1
6:66
19:8
¼ 2:97
6:66
Empirical formula ¼ CH3
Empirical mass ¼ 12:01 g þ 3:024 g ¼ 15:03 g=mol
2:01 g 22:4 L
Molecular formula:
¼ 30: g=molðmolar massÞ
1:5 L
mol
30: g=mol
¼ 2; Molecular formula is C2 H6
15:03 g=mol
(c)
Valence electrons ¼ 2ð4Þ þ 6 ¼ 14
H H
H C C H
H H
- 137 -
- Chapter 12 58. PV ¼ nRT
ð790 torrÞð1 atmÞ
(a)
ð2:0 LÞ ¼ ðnÞð0:0821 L atm=mol KÞð298 KÞ
760 torr
n ¼ 0:085 molðtotal molesÞ
(b)
mol N2 ¼ total moles mol O2 mol CO2
¼ 0:085 mol 0:65 g O2
0:58 g CO2
32:00 g=mol 44:01 g=mol
mol N2 ¼ 0:085 mol 0:020 mol O2 0:013 mol CO2 ¼ 0:052 mol
28:02 g N2
ð0:052 mol N2 Þ
¼ 1:5 g N2
mol
(c)
0:020 mol O2
PO2 ¼ ð790 torrÞ
¼ 1:9 102 torr
0:085 mol
0:013 mol CO2
PCO2 ¼ ð790 torrÞ
¼ 1:2 102 torr
0:085 mol
0:051 mol N2
PN2 ¼ ð790 torrÞ
¼ 4:7 102 torr
0:085 mol
! 2 CO2
59. 2 CO þ O2 Calculate the moles of O2 and CO to find the limiting reactant.
PV ¼ nRT
O2 : ð1:8 atmÞð0:500 L O2 Þ ¼ ðnÞð0:0821 L atm=mol KÞð288 KÞ
mol O2 ¼ 0:038 mol
800 mm Hg 1 atm
CO :
ð0:500 LÞ ¼ ðnÞð0:0821 L atm=mol KÞð333 KÞ
760 mm Hg
mol CO ¼ 0:019 mol
Limiting reactant is CO
0:0095 mol O2 will react with 0:019 mol CO:
2 mol CO2
22:4 L
¼ 0:43 L CO2 ¼ 430 mL CO2
ð0:019 mol COÞ
2 mol CO
mol
g
60. PV ¼ nRT or PV ¼
RT
molar mass
1:4 g 1000 cm3
¼ 1:4 103 g=L
L
cm3
1:4 103 g
9
1:3 10 atm ð1:0 LÞ ¼
ð0:0821 L atm=mol KÞðTÞ
2:0 g=mol
T¼
ð1:3 109 atmÞð1:0 LÞð2:0 g=molÞ
¼ 2:3 107 K
ð0:0821 L atm=mol KÞð1:4 103 gÞ
- 138 -
- Chapter 12 61. (a)
Assume atmospheric pressure of 14.7 lb=in.2 to begin with.
Total pressure in the ball ¼ 14:7 lb=in:2 þ 13 lb=in:2 ¼ 28 lb=in:2
PV ¼ nRT
1 atm
2
ð2:24 LÞ ¼ ðnÞð0:0821 L atm=mol KÞð293 KÞ
ð28 lb=in: Þ
14:7 lb=in:2
n ¼ 0.18 mol air
(b)
(c)
mass of air in the ball
molar mass of air is about 29 g/mol
29 g
m ¼ ð0:18 molÞ
¼ 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 ¼ nRT
1 atm
ð2:24 LÞ ¼ ðnÞð0:0821 L atm=mol KÞð303 KÞ
ð28 lb=in:2 Þ
14:7 lb=in:2
n ¼ 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.
29 g
ð0:01 mol airÞ
¼ 0:29 g or 0:3 g air must be allowed to escape.
mol
62. Use the combined gas laws to calculate the bursting temperature (T2).
P 1 V1 P 2 V2
¼
T1
T2
T2 ¼
P1 ¼ 65 cm
P2 ¼ 100 atmð76 atmÞ
V1 ¼ 1:75 L
T1 ¼ 20 Cð293 KÞ
V2 ¼ 2:00 L
T2 ¼ T2
P2 V2 T1 ð76 cmÞð2:00 LÞð293 KÞ
¼
¼ 392 Kð119 CÞ
P 1 V1
ð65 cmÞð1:75 LÞ
63. To double the volume of a gas, at constant pressure, the temperature (K) must be doubled.
V1 V2
¼
T1 T2
V1 2 V1
¼
T1
T2
V2 ¼ 2 V1
T2 ¼
2 V1 T 1
V1
T2 ¼ 2 T1
T2 ¼ 2ð300: KÞ ¼ 600: K ¼ 327 C
64. V ¼ volume at 22 C and 740 torr
2 V ¼ volume after change in temperatureðP constantÞ
V ¼ volume after change in pressureðT constantÞ
P 1 V1
Since temperature is constant, P1 V1 ¼ P2 V2 or P2 ¼
V2
2V
P2 ¼ ð740 torrÞ
¼ 1:5 103 torr ðpressure to change 2 V to VÞ
V
- 139 -
- Chapter 12 -
65. Volume is constant, so
T2 ¼
P1 P2
¼
T1 T2
or
T2 ¼
T1 P2
;
P1
ð500: torrÞð295 KÞ
¼ 211 K ¼ 62 C
700: torr
66. The volume of the tires remains constant (until they burst), so
P1 P2
¼
T1 T2
or T2 ¼
T1 P2
;
P1
71:0 F ¼ 21:7 C ¼ 295 K
T2 ¼
ð44 psiÞð295 KÞ
¼ 433 K ¼ 160 C ¼ 320 F
30: psi
67. Use the combined gas laws.
P 1 V1 P 2 V2
¼
T1
T2
P2 ¼
or P2 ¼
P 1 V1 T 2
V2 T1
P1 and T1 are at STP
ð1:00 atmÞð800: mLÞð303 KÞ
¼ 3:55 atm
ð250: mLÞð273 KÞ
68. Use the combined gas law
P1 V1 P2 V2
¼
T1
T2
or V2 ¼
P1 V1 T2
P2 T1
First calculate the volume at STP.
V2 ¼
ð400: torrÞð600: mLÞð273 KÞ
¼ 275 mL ¼ 0:275 L
ð760: torrÞð313 KÞ
At STP, a mole of any gas has a volume of 22.4 L
1 mol
6:022 1023 molecules
ð0:275 LÞ
¼ 7:39 1021 molecules
22:4 L
1 mol
Each molecule of N2O contains 3 atoms, so:
3 atoms
21
7:39 10 molecules
¼ 2:22 1022 atoms
1 molecule
69. Pressure varies directly with absolute temperature.
P1 P2
¼
T1 T2
P2 ¼
P1 T2
T1
T1 ¼ 25 C þ 273 ¼ 298 K
T2 ¼ 212 F ¼ 100 C ¼ 373 K
ð32 lb=in:2 Þð373 KÞ
¼ 40: lb=in:2
298 K
At 212 F the tire pressure is 40. lb/in.2
The tire will not burst.
P2 ¼
- 140 -
- Chapter 12 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.
ð760 mmÞð13:6Þ ¼ 1:03 104 mmð33:8 ftÞ
71. Use the ideal gas equation
PV ¼ nRT
n¼
RT
PV
Change 2:20 103 lb=in:2 to atmosphere
1 atm
3
2
¼ 150: atm
2:20 10 lb=in:
14:7 lb=in:2
n¼
ð150: atmÞð55 LÞ
¼ 3:3 102 mol O2
0:0821 L atm
ð300: KÞ
mol K
72. The conversion is: m3 ! cm3 ! mL ! L ! mol
100 cm 3 1 mL
1L
1 mol
ð1:00 m3 Þ
¼ 44:6 mol Cl2
1m
1 cm3
1000 mL 22:4 L
73. First calculate the moles of gas and then convert moles to molar mass.
1 mol
ð0:560 LÞ
¼ 0:0250 mol
22:4 L
1:08 g
¼ 43:2 g=molðmolar massÞ
0:0250 mol
74. The conversion is: g=L ! g=mol
1:78 g 22:4 L
¼ 39:9 g=molðmolar massÞ
L
mol
75. PV ¼ nRT
nRT ð0:510 molÞð0:0821 L atm=mol KÞð320: KÞ
(a) V ¼
¼
¼ 8:4 L H2
P
1:6 atm
(b)
(c)
n¼
RV
ð0:789 atmÞð16:0 LÞ
¼
¼ 0:513 mol CH4
RT ð0:0821 L atm=mol KÞð300: KÞ
The molar mass for CH4 is 16.04 g=mol
ð16:04 g=molÞð0:513 molÞ ¼ 8:23 g CH4
g
PV ¼ nRT, but n ¼ where M is the molar mass and g is the grams of the gas.
M
gRT
. To determine density, d ¼ g=V.
Thus, PV ¼
M
gRT
g
g PM
Solving PV ¼
for produces ¼
.
M
V
V RT
g ð4:00 atmÞð44:01 g=molÞ
d¼ ¼
¼ 8:48 g=L CO2
V
ð0:0821 L atm=mol KÞ
- 141 -
- Chapter 12 -
(d)
Since d ¼
M¼
g PM
¼
from part (c), solve for M (molar mass)
V RT
dRT ð2:58 g=LÞð0:0821 L atm=mol KÞð300: KÞ
¼
¼ 63:5 g=mol ðmolar massÞ
P
1:00 atm
! C2 H4 F2 ðgÞ
76. C2 H2 ðgÞ þ 2HFðgÞ ! 1:0 mol C2 H4 F2
1:0 mol C2 H2 1 mol C2 H4 F2
¼ 2:5 mol C2 H4 F2
ð5:0 mol HFÞ
2 mol HF
C2H2 is the limiting reactant. 1.0 mol C2H4F2 forms, no moles C2H2 remain.
According to the equation, 2.0 mol HF yields 1.0 mol C2H4F2. Therefore,
5:0 mol HF 2:0 mol HF ¼ 3:0 mol HF unreacted
The flask contains 1.0 mol C2H4F2 and 3.0 mol HF when the reaction is complete.
The flask contains 4.0 mol of gas.
P¼
nRT ð4:0 molÞð0:0821 L atm=mol KÞð273 KÞ
¼
¼ 9:0 atm
V
10:0 L
3 mol H2
22:4 L
77. ð8:30 mol AlÞ
¼ 279 L H2 at STP
2 mol Al
mol
78. According to Graham’s Law of Effusion, the rates of effusion are inversely proportional
to the molar mass.
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffi
rate He
molar mass N2
28:02 pffiffiffiffiffiffiffiffiffiffiffi
¼
¼
¼ 7:000 ¼ 2:646
molar mass He
rate N2
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.
rffiffiffiffiffiffiffiffiffiffiffi
rate He
16:04 pffiffiffiffiffiffiffiffiffiffiffi
¼
¼ 4:007 ¼ 2:002
rate CH4
4:003
Helium effuses twice as fast as CH4.
(b)
x ¼ distance He travels
100 x ¼ distance CH4 travels
DHe ¼ 2 DCH4
D ¼ distance traveled
x ¼ 2ð100 xÞ
3x ¼ 200
x ¼ 66:7 cm
The gases meet 66.7 cm from the helium end.
- 142 -
- Chapter 12 80. Assume 100. g of material to start with. Calculate the empirical formula.
1 mol
ð85:7 gÞ
¼ 7:14 mol
12:01 g
1 mol
ð14:3 gÞ
¼ 14:2 mol
1:008 g
C
H
7:14
¼ 1:00 mol
7:14
14:2
¼ 1:99 mol
7:14
The empirical formula is CH2. To determine the molecular formula, the molar mass
must be known.
2:50 g 22:4 L
¼ 56:0 g=molðmolar massÞ
L
mol
56:0
The empirical formula mass is 14:0
¼4
14:0
Therefore, the molecular formula is ðCH2 Þ4 ¼ C4 H8
81. 2 COðgÞ þ O2 ðgÞ ! 2 CO2 ðgÞ
ð10:0 mol COÞ
2 mol CO2
2 mol CO
Determine the limiting reactant
¼ 10:0 mol CO2 ðfrom COÞ
2 mol CO2
ð8:0 mol O2 Þ
¼ 16 mol CO2 ðfrom O2 Þ
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)
P¼
nRT ð13 molÞð0:0821 L atm=mol KÞð273 KÞ
¼
¼ 29 atm
V
10: L
D
! 2 KC1ðsÞ þ 3 O2ðgÞ
82. 2 KClO3 ðsÞ First calculate the moles of O2 produced. Then calculate the grams of KClO3 required to produce
the O2. Then calculate the % KC1O3.
1 mol
ð0:25 L O2 Þ
¼ 0:011 mol CO2
22:4 L
2 mol KClO3
122:6 g
ð0:011 mol O2 Þ
¼ 0:90 g KClO2 in the sample
3 mol O2
mol
0:90 g
ð100Þ ¼ 75% KClO3 in the mixture
1:20 g
- 143 -
- Chapter 12 83. Assume 1.00 L of air. The mass of 1.00 L of air is 1.29 g.
P1 V1 P2 V2
¼
T1
T2
V2 ¼
d¼
P1 V1 T2 ð760 torrÞð1:00 LÞð2:90 KÞ
¼
¼ 1:8 L
P2 T1
ð450 torrÞð273 KÞ
m 1:29 g
¼
¼ 0:72 g=L
V
1:8 L
84. Each gas behaves as though it were alone in a 4.0 L system.
(a) After expansion: P1 V1 ¼ P2 V2
P1 V1 ð150: torrÞð3:0 LÞ
¼
¼ 1:1 102 torr
V2
4:0 L
P1 V1 ð50: torrÞð1:0 LÞ
P2 ¼
¼
¼ 13 torr
V2
4:0 L
P2 ¼
For CO2
For H2
(b)
85.
Ptotal ¼ PH2 þ PCO2 ¼ 110 torr þ 13 torr ¼ 120 torr ð2 sig: figuresÞ
P1 ¼ 40:0 atm
P2 ¼ P2
V1 ¼ 50:0 L
V2 ¼ 50:0 L
T1 ¼ 25 C ¼ 298 K
T2 ¼ 25 C þ 152 C ¼ 177 C ¼ 450: K
Gas cylinders have constant volume, so pressure varies directly with temperature.
P2 ¼
P1 T2 ð40:0 atmÞð450: KÞ
¼
¼ 60:4 atm
T1
298 K
86. You can identify the gas by determining its density.
mass of gas ¼ 1:700 g 0:500 g ¼ 1:200 g
volume of gas: Charles law problem. Correct volume to 273 K
V1 V2
¼
T1 T2
V2 ¼
V1 T2 ð0:4478 LÞð273 KÞ
¼
¼ 0:3785 L
T1
323 K
m
1:200 g
¼
¼ 3:170 g=L
V 0:3785 L
gas is chlorine (see Table 12.3)
d¼
- 144 -