Instructor’s Manual for Cracolice/Peters Introductory Chemistry, 4th Edition
CHAPTER 15
GASES, LIQUIDS, AND SOLIDS
Authors’ Comments on Chapter 15
Chapter 15 covers a number of topics related to the transitions among the gas, liquid, and solid states.
Students should be familiar with most of the topics covered through Chapter 13 before studying this
chapter. Chapter 14 is not a prerequisite.
Answers Only for Black-Numbered Unanswered Questions, Exercises, and Problems
denotes problems assignable in OWL
2.
0.560 atm
4.
658 torr
6.
728 torr; 0.320 mol
8. Gas particles are very widely spaced; liquid particles are “touchingly close.” Density is mass per unit
volume. The same number of liquid particles will occupy a smaller volume than they occupy as a gas.
10. There is no space between water molecules; they cannot be forced closer together. Molecules in air are
widely separated; they can be pushed closer together.
12. The stronger the intermolecular attractions, the slower the evaporation rate, other things being equal.
Condensation rate therefore equals evaporation rate at a lower vapor concentration, which means lower
partial pressure at equilibrium.
14. Liquids with strong intermolecular attractions have high viscosity, an internal resistance to flow.
16. Motor oil is more viscous than water. This predicts that intermolecular attractions are stronger in motor
oil, as strong attractions lead to internal resistance to flow, which is viscosity.
18. An isolated drop is normally spherical because of surface tension. When sitting on a plate, a drop tends
to be flattened by gravity. The honey drop remains closer to spherical than the water drop, indicating
stronger surface tension and stronger intermolecular attractions.
20. Soap reduces intermolecular attractions, thus lowering the surface tension of water. This makes the
soapy water able to penetrate fabrics and clean them throughout.
15.1
Instructor’s Manual for Cracolice/Peters Introductory Chemistry, 4th Edition
22. NO, N2O, NO2
24. Of the three compounds, only N2O is a liquid at –90°C. Therefore, only N2O possesses the property of
viscosity. If a solid is considered more “viscous” than a liquid, NO2 is the most viscous.
26. A compound with a large nonpolar section and an —OH or an —NH2 group at the end would exhibit
hydrogen bonding, but the induced dipole forces resulting from the large nonpolar section would dominate
the intermolecular attractive forces. An example is CH3CH2CH2CH2CH2CH2CH2OH.
28.
NOCl: dipole forces; NH2Cl: hydrogen bonds; SiCl4: induced dipole forces.
30. Ionic compounds have ionic bonds as the interparticle forces. Polar molecular compounds have dipole
forces acting between the particles. Ionic bonds are much stronger than any intermolecular forces,
including dipole forces. The ionic compound would have the higher melting point because more energy is
needed to overcome the stronger forces holding the solid together.
32. CCl4, because it is larger, as suggested by its higher molecular mass.
34. H2S, because it is slightly more polar.
36. For hydrogen bonding to occur, a hydrogen atom must be bonded to another element with a high
enough electronegativity to shift the bonding electron pair away from the hydrogen atom. Fluorine, oxygen,
and nitrogen are the only elements that satisfy this requirement.
38. (a) Induced dipole forces: (b) Dipole forces and induced dipole forces.
40. a, c, and d: dipole forces and induced dipole forces. b and e: induced dipole forces.
42.
d
44. CO2 molecules are smaller than SO2 molecules, and CO2 molecules are linear and nonpolar while SO2
molecules are bent and polar. Both differences predict weaker intermolecular attractions and therefore
higher vapor pressure for CO2.
46. The term dynamic refers to the “active” character of the equilibrium. The rate of change from liquid to
vapor equals the rate of change from vapor to liquid. Although the concentrations remain constant, there is
continual change between the two states at the particulate level.
48.
Since C7H16 must be raised to a higher temperature than CH3COOCH3 to have a vapor pressure of
400 torr, the intermolecular forces are stronger in C7H16. The vapor pressure of CH3COOCH3 is 400 torr at
40°C. Since vapor pressure increases as temperature increases, its vapor pressure will be higher at 78°C.
15.2
Instructor’s Manual for Cracolice/Peters Introductory Chemistry, 4th Edition
50. Boiling point is the temperature at which the vapor pressure of a liquid is equal to the pressure above its
surface. Your vapor-pressure-versus-temperature curve should be similar to one from Figure 15.16. The
normal boiling point is found at a vapor pressure of 760 torr.
52. A gas can be condensed by increasing the pressure. Higher pressure forces the particles closer to one
another, where intermolecular attractive forces become significant, thus condensing a gas into a liquid.
54.
Since C8H18 must be raised to a higher temperature than CS2 to have a vapor pressure of 400 torr,
the intermolecular forces are stronger in C8H18. The normal boiling point of a liquid is the temperature at
which the vapor pressure of the liquid is 1 atm or 760 torr. Since C8H18 must be raised to a higher
temperature than CS2 to have a vapor pressure of 760 torr, C8H18 will have the higher normal boiling point.
56. More energy is required to vaporize X, so it would have the higher boiling point and lower vapor
pressure.
58. Amorphous solids have no long-range ordering on the particulate level. Crystalline solids have particles
arranged in a repeating pattern. Polycrystalline solids have small orderly crystals arranged in a random
fashion. Because of the disorderly arrangement of particles in an amorphous solid, there is a range of
strengths of intermolecular attractive forces throughout the solid, and thus no definite physical properties.
Crystalline solids have distinct physical properties.
60. C, molecular; D, metallic
62.
2.20 kJ/g
64.
11.2 kJ
66.
28.5 g
68.
– 1.22 kJ
70. 2.2 kJ
72. 151 J/g
74. 108 g Ag
76.
0.15 J/g • °C
78.
181 J
80.
– 35 J
82.
35°C
15.3
Instructor’s Manual for Cracolice/Peters Introductory Chemistry, 4th Edition
84. Vertically, temperature; horizontally, energy
86. E definitely; maybe D and F
88. G
90. The solid substance melts completely at constant temperature K.
92. O – N
94.
5.213 kJ + 9.333 kJ + 1.253 kJ = 15.799 kJ
96.
0.34 kJ + 10.39 kJ + 7.30 kJ = 18.03 kJ
98.
(– 0.9709 kJ) + (– 1.199 kJ) + (– 0.6981 kJ) = – 2.868 kJ
Solutions for Black-Numbered Unanswered Questions, Exercises, and Problems
denotes problems assignable in OWL
2.
0.326 atm + 0.234 atm = 0.560 atm
4.
540 torr ×
1.00 L
= 270 torr
(1.00 + 1.00) L
776 torr ×
1.00 L
= 388 torr
(1.00 + 1.00) L
270 torr + 388 torr = 658 torr
6.
n=
745 torr – 17.5 torr = 728 torr
PV
mol ⋅ K
1
= 728 torr × 8.04 L ×
×
= 0.320 mol
RT
62.4 L ⋅ torr
(20 + 273) K
8. Gas particles are very widely spaced; liquid particles are “touchingly close.” Density is mass per unit
volume. The same number of liquid particles will occupy a smaller volume than they occupy as a gas.
10. There is no space between water molecules; they cannot be forced closer together. Molecules in air are
widely separated; they can be pushed closer together.
12. The stronger the intermolecular attractions, the slower the evaporation rate, other things being equal.
Condensation rate therefore equals evaporation rate at a lower vapor concentration, which means lower
partial pressure at equilibrium.
14. Liquids with strong intermolecular attractions have high viscosity, an internal resistance to flow.
16. Motor oil is more viscous than water. This predicts that intermolecular attractions are stronger in motor
oil, as strong attractions lead to internal resistance to flow, which is viscosity.
15.4
Instructor’s Manual for Cracolice/Peters Introductory Chemistry, 4th Edition
18. An isolated drop is normally spherical because of surface tension. When sitting on a plate, a drop tends
to be flattened by gravity. The honey drop remains closer to spherical than the water drop, indicating
stronger surface tension and stronger intermolecular attractions.
20. Soap reduces intermolecular attractions, thus lowering the surface tension of water. This makes the
soapy water able to penetrate fabrics and clean them throughout.
22. NO, N2O, NO2
24. Of the three compounds, only N2O is a liquid at –90°C. Therefore, only N2O possesses the property of
viscosity. If a solid is considered more “viscous” than a liquid, NO2 is the most viscous.
26. A compound with a large nonpolar section and an —OH or an —NH2 group at the end would exhibit
hydrogen bonding, but the induced dipole forces resulting from the large nonpolar section would dominate
the intermolecular attractive forces. An example is CH3CH2CH2CH2CH2CH2CH2OH.
28.
NOCl: dipole forces; NH2Cl: hydrogen bonds; SiCl4: induced dipole forces.
30. Ionic compounds have ionic bonds as the interparticle forces. Polar molecular compounds have dipole
forces acting between the particles. Ionic bonds are much stronger than any intermolecular forces,
including dipole forces. The ionic compound would have the higher melting point because more energy is
needed to overcome the stronger forces holding the solid together.
32. CCl4, because it is larger, as suggested by its higher molecular mass.
34. H2S, because it is slightly more polar.
36. For hydrogen bonding to occur, a hydrogen atom must be bonded to another element with a high
enough electronegativity to shift the bonding electron pair away from the hydrogen atom. Fluorine, oxygen,
and nitrogen are the only elements that satisfy this requirement.
38. (a) Induced dipole forces: (b) Dipole forces and induced dipole forces.
40. a, c, and d: dipole forces and induced dipole forces. b and e: induced dipole forces.
42.
d
44. CO2 molecules are smaller than SO2 molecules, and CO2 molecules are linear and nonpolar while SO2
molecules are bent and polar. Both differences predict weaker intermolecular attractions and therefore
higher vapor pressure for CO2.
15.5
Instructor’s Manual for Cracolice/Peters Introductory Chemistry, 4th Edition
46. The term dynamic refers to the “active” character of the equilibrium. The rate of change from liquid to
vapor equals the rate of change from vapor to liquid. Although the concentrations remain constant, there is
continual change between the two states at the particulate level.
48.
Since C7H16 must be raised to a higher temperature than CH3COOCH3 to have a vapor pressure of
400 torr, the intermolecular forces are stronger in C7H16. The vapor pressure of CH3COOCH3 is 400 torr at
40°C. Since vapor pressure increases as temperature increases, its vapor pressure will be higher at 78°C.
50. Boiling point is the temperature at which the vapor pressure of a liquid is equal to the pressure above its
surface. Your vapor-pressure-versus-temperature curve should be similar to one from Figure 15.16. The
normal boiling point is found at a vapor pressure of 760 torr.
52. A gas can be condensed by increasing the pressure. Higher pressure forces the particles closer to one
another, where intermolecular attractive forces become significant, thus condensing a gas into a liquid.
54.
Since C8H18 must be raised to a higher temperature than CS2 to have a vapor pressure of 400 torr,
the intermolecular forces are stronger in C8H18. The normal boiling point of a liquid is the temperature at
which the vapor pressure of the liquid is 1 atm or 760 torr. Since C8H18 must be raised to a higher
temperature than CS2 to have a vapor pressure of 760 torr, C8H18 will have the higher normal boiling point.
56. More energy is required to vaporize X, so it would have the higher boiling point and lower vapor
pressure.
58. Amorphous solids have no long-range ordering on the particulate level. Crystalline solids have particles
arranged in a repeating pattern. Polycrystalline solids have small orderly crystals arranged in a random
fashion. Because of the disorderly arrangement of particles in an amorphous solid, there is a range of
strengths of intermolecular attractive forces throughout the solid, and thus no definite physical properties.
Crystalline solids have distinct physical properties.
60. C, molecular; D, metallic
62.
58.2 kJ
= 2.20 kJ/g
26.4 g
64.
28.6 g ×
66.
10.5 kJ ×
0.393 kJ
= 11.2 kJ
g
1g
1000 J
×
= 28.5 g
368 J
kJ
15.6
Instructor’s Manual for Cracolice/Peters Introductory Chemistry, 4th Edition
22.4 g × –
68.
70. 35.4 g Au ×
72.
54.4 J
1 kJ
×
= – 1.22 kJ
g
1000 J
63 J
= 2.2 × 103 J = 2.2 kJ
g Au
7.08 kJ
= 0.151 kJ/g = 151 J/g
46.9 g
74. 11.3 kJ ×
1000 J 1 g Ag
×
= 108 g Ag
105 J
kJ
q
1
1
= 27.6 J ×
×
= 0.15 J/g • °C
m × ΔT
10.7 g
(39.9 – 22.7)°C
76.
c=
78.
q = m × c × ∆T = 10.4 g ×
1.02 J
× (39.5 – 22.4)°C = 181 J
g ⋅ °C
80.
q = m × c × ∆T = 10.0 g ×
0.24 J
× (23.6 – 38.1)°C = – 35 J
g ⋅ °C
82.
∆T =
q
1
g ⋅ °C
= 105 J ×
×
= 11°C; 24.4°C + 11°C = 35°C
m × c
12.4 g
0.74 J
84. Vertically, temperature; horizontally, energy
86. E definitely; maybe D and F
88. G
90. The solid substance melts completely at constant temperature K.
92. O – N
94.
q (heat liquid) = 26.10 g ×
q (boil) = 26.10 g ×
2.280 J
1 kJ
× [36.20 – (– 51.40)]°C ×
= 5.213 kJ
g ⋅ °C
1000 J
357.6 J
1 kJ
×
= 9.333 kJ
g
1000 J
q (heat gas) = 26.10 g ×
1.650 J
1 kJ
× (65.30 – 36.20)°C ×
= 1.253 kJ
g ⋅ °C
1000 J
qtotal = 5.213 kJ + 9.333 kJ + 1.253 kJ = 15.799 kJ
96.
q (heat solid) = 36.90 g ×
q (melt) = 36.90 g ×
0.4600 J
1 kJ
× (1857 – 1837)°C ×
= 0.34 kJ
1000 J
g ⋅ °C
281.5 J
1 kJ
×
= 10.39 kJ
g
1000 J
15.7
Instructor’s Manual for Cracolice/Peters Introductory Chemistry, 4th Edition
q (heat liquid) = 36.90 g ×
0.9370 J
1 kJ
× (2068 – 1857)°C ×
= 7.30 kJ
g ⋅ °C
1000 J
qtotal = 0.34 kJ + 10.39 kJ + 7.30 kJ = 18.03 kJ
98.
q (cool liquid) = 22.80 g ×
q (freeze) = 22.80 g × –
0.1510 J
1 kJ
× (271.0 – 553.0)°C ×
= – 0.9709 kJ
g ⋅ °C
1000 J
52.60 J
1 kJ
×
= – 1.199 kJ
g
1000 J
q (cool solid) = 22.80 g ×
0.1260 J
1 kJ
× (28.0 – 271.0)°C ×
= – 0.6981 kJ
g ⋅ °C
1000 J
qtotal = (– 0.9709 kJ) + (– 1.199 kJ) + (– 0.6981 kJ) = – 2.868 kJ
15.8