chapter 7 gases, liquids, and solids -...

Chapter 7–1

Chapter 7 Gases, Liquids, and Solids Solutions to In-Chapter Problems 7.1 Use Table 7.1 to compare the features of different states of methanol.

a. Density b. Intermolecular Spacing c. Intermolecular Attraction Gas Lowest Greatest Lowest

Liquid Higher Smaller Higher Solid Highest Smallest Highest

7.2 Use conversion factors to convert mm Hg to atm and psi.

mm Hg cancels.

630 mm Hg1 atm

760. mm Hg= 0.83 atmAnswer

a. b.

mm Hg cancels.

630 mm Hg = 12 psiAnswer760. mm Hg

14.7 psix x

7.3 Use conversion factors to solve the problems.

Atm cancels.

3.0 atm = 2,300 mm HgAnswer

a.

b. = 14 psiAnswer760. mm Hg

14.7 psi

x

x

1 atm

mm Hg cancels.

720 mm Hg

c. = 0.558 atmAnswer760. mm Hg

1 atmx424 mm Hg760. mm Hg

mm Hg cancels.


120 mm Hg × (1 atm/760 mm Hg) = 0.16 atm 80 mm Hg × (1 atm/760 mm Hg) = 0.1 atm 7.5 Use Boyle’s law to solve the problems as in Example 7.2.

P1V1

P2 =

(4.0 atm)(2.0 L) =

Atm cancels.5.0 atm

=Answer

V2 1.6 La.

P1V1

P2 =

(4.0 atm)(2.0 L) =

Atm cancels.2.5 atm

=Answer

V2 3.2 Lb.

P1V1

P2 =

(4.0 atm)(2.0 L) =

Atm cancels.10.0 atm

=Answer

V2 0.80 Lc.

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Gases, Liquids, and Solids 7–2

P1V1

P2 =

(3.0 x 103 mm Hg)(2.0 L) =

mm Hg cancels.380 mm Hg

=Answer

V2 16 Ld. =760. mm Hg1 atm

x4.0 atm 3.0 x 103 mm Hg

7.6 Use Boyle’s law to solve the problems as in Example 7.2.

a. P1V1P2 =(0.50 atm)(15.0 mL)

=30.0 mL

=AnswerV2

mL cancels.

0.25 atm

b. P1V1P2 =(0.50 atm)(15.0 mL)

=5.0 mL

=AnswerV2

mL cancels.

1.5 atm

c. P1V1P2 =(0.50 atm)(15.0 mL)

=100. mL

=AnswerV2

mL cancels.

0.075 atm

d. P1V1P2 =(0.50 atm)(15.0 mL)

=1.0 x 103 mL

=AnswerV2

mL cancels.

0.0075 atm1.0 L = 1.0 x 103 mL

7.7 Use Charles’s law to solve the problem as in Example 7.3. Remember to convert temperature to K.

K = °C + 273T1 = 37 °C + 273 = 310. K T2 = 0.0 °C + 273 = 273 K

V1T2

T1V2

310. K=

Kelvins cancel.

=(0.50 L)(273 K)

= 0.44 LAnswer

7.8 Use Charles’s law to solve the problems as in Example 7.3.

V1T2

T1V2

45 K=

Kelvins cancel.

=(25.0 L)(450 K)

= 250 LAnswer

a.

K = °C + 273T1 = 400. °C + 273 = 673 KT2 = 50. °C + 273 = 323 K

V1T2

T1V2

673 K=

Kelvins cancel.

=(50.0 mL)(323 K)

= 24 mLAnswer

b.


Chapter 7–3

7.9 Use Gay–Lussac’s law to solve the problem.

P1T1 T2

P2=

P2T1P1

= T2

1.00 atmAtm cancels.

=Answer

=(1.05 atm)(373 K)

392 K or 119 °CP2T1

P1=T2

K = °C + 273T1 = 100. °C + 273 = 373 K

7.10 Use Gay–Lussac’s law to solve the problem as in Sample Problem 7.4. T1 = 25 °C + 273 = 298 K.

P1T2

T1

P2 298 K

Kelvins cancel.

=Answer

= =(1.00 atm)(310. K)

1.04 atma.

P1T2

T1

P2 298 K

Kelvins cancel.

=Answer

= =(1.00 atm)(150. K)

0.503 atmb.

P1T2

T1

P2 298 K

Kelvins cancel.

=Answer

= =(1.00 atm)(323 K)

1.08 atmc.

P1T2

T1

P2 298 K

Kelvins cancel.

=Answer

= =(1.00 atm)(473 K)

1.59 atmd.

7.11 As a sealed container is heated in the microwave, the gases inside expand and increase the container’s internal pressure, eventually popping the lid off.

7.12 Use the combined gas law to solve the problem as in Example 7.4.

T1 T2

P1V1 P2V2 =

P2T1V2

P1V1T2 =

P2 T1V2

P1V1T2 =

(298 K)(2.0 L)(750 mm Hg)(1.0 L)(233 K)

= = 290 mm HgAnswer



7.13 Use Avogadro’s law to solve the problems as in Example 7.5.

V1

n1 n2

V2= Solve for V2 by multiplying both sides by n2.V2=

n1

n2V1

V2 =n1

n2V1 =Answer

=(2.5 mol)(3.5 L)

(5.0 mol)Mol cancel.

1.8 La.

V2 =n1

n2V1 =Answer

=(3.65 mol)(3.5 L)


2.6 Lb.

V2 =n1

n2V1 =Answer

=(21.5 mol)(3.5 L)


15 Lc.

7.14 Convert moles to volume at STP as in Sample Problem 7.7.

4.5 mol O2 x =Answer1 mol

22.4 L 1.0 x 102 La.

0.35 mol O2 x =

Answer1 mol22.4 L 7.8 Lb.

18.0 g O2 x =Answer1 mol

22.4 L 12.6 Lc. x32.00 g1 mol


1.5 L x =Answer

1 mol22.4 L

0.067 mola.

8.5 L x =Answer

1 mol22.4 L

0.38 molb.

25 mL x =Answer

1 mol22.4 L

1.1 x 10–3 molc. x1 L

1000 mL


Chapter 7–5

7.16 Use the ideal gas law to solve the problem as in Example 7.6.

P = 175 atmT = 20. oC

known quantities desired quantityV = 5.0 L n = ? mol

Convert °C to K. K = °C + 273 = 20. °C + 273 = 293 K

Use the value of R with atm since the pressure is given in atm; that is, R = 0.0821 L•atm/(mol•K).

PV = nRT Solve for n by dividing both sides by RT.

Answer=(5.0 L)

0.0821 L • atmmol • K

(175 atm)= 36 mol

(293 K)

n =RTPV

7.17 Use the ideal gas law to solve the problems as in Example 7.6 and Answer 7.16.

R = 0.0821 L•atm/(mol•K)PV = nRT Solve for P by dividing both sides by V.

Answer=

(10.0 L)

0.0821L • atmmol • K

0.45 mol

= 1.1 atm(298 K)

P =V

nRT

K = °C + 273 = 25 °C + 273 = 298 Ka.

R = 0.0821 L•atm/(mol•K)PV = nRT Solve for P by dividing both sides by V.

P =V

Answer=

(5.0 L)

0.0821 L • atmmol • K

0.357 mol

= 1.7 atm(293 K)

nRT

P =V

nRT

10.0 g N21 mol N2

28.02 g N2x

molar massconversion factor

0.357 mol N2

Grams cancel.

=

K = °C + 273 = 20. °C + 273 = 293 Kb.

7.18 Use Dalton’s law to solve the problem as in Example 7.7.

Ptotal = PO2 + PCO2 4.0 atm = 2.5 atm + PCO2 CO2: 1.5 atm O2: 2.5 atm




Methane: 0.85 × 750 mm Hg = 637.5 rounded to 640 mm Hg Ethane: 0.10 × 750 mm Hg = 75 mm Hg Propane: 0.050 × 750 mm Hg = 37.5 rounded to 38 mm Hg

7.20 Use Dalton’s law to solve the problem.

Chamber A: pure O2 = 2.5 atm = higher pressure of O2. Chamber B: 40.% O2 = 0.40 × 5.5 atm = 2.2 atm for O2

7.21 London dispersion forces are very weak interactions due to the momentary changes in electron

density in a molecule. All molecules exhibit these forces (a–d). 7.22 With dipole–dipole interactions, molecules align so that partial positive and partial negative

charges are close together.

H Cl H Cl!" !"!+ !+

7.23 Use Table 7.3 to decide which types of forces are present in each molecule. “+” means the force

is present.

London Dispersion Dipole–Dipole Hydrogen Bonding a. Cl2 + no net dipole – b. HCN + + – c. HF + + + (the only molecule listed with an

O–H, N–H, or H–F bond) d. CH3Cl + + – e. H2 + no net dipole –

7.24 London dispersion forces are present in all covalent compounds.

Dipole–dipole interactions are present only in polar compounds with a permanent dipole. Hydrogen bonding occurs only in compounds that contain an O–H, N–H, or H–F bond.

a. H2O has hydrogen bonding, which is stronger than the London forces in CO2. b. HBr has dipole–dipole interactions, which are stronger than the London forces in CO2. c. H2O has hydrogen bonding, which is stronger than the dipole–dipole interactions in HBr. d. C2H6 has stronger London dispersion forces because it is larger than CH4.

7.25 The stronger the intermolecular forces, the higher the boiling point and melting point.

a. C2H6 has stronger London dispersion forces, so it has a higher boiling point and melting point than CH4.

b. CH3OH has hydrogen bonding, so it has a higher boiling point and melting point than C2H6. c. Br is larger than Cl, so HBr has a higher boiling point and melting point than HCl. d. CH3Br has dipole–dipole interactions, so it has a higher boiling point and melting point than

C2H6, which has only London dispersion forces.


Chapter 7–7

7.26 Water has stronger intermolecular forces since it can hydrogen bond, whereas CO2 has only London dispersion forces. This explains why water is a liquid at room temperature, whereas CO2 is a gas.

7.27 The stronger the intermolecular forces, the lower the vapor pressure at a given temperature.

a. CH4 has a higher vapor pressure because it has only London dispersion forces, whereas NH3 has hydrogen bonding.

b. CH4 has a higher vapor pressure because it has only London dispersion forces and is a smaller molecule than C2H6.

c. C2H6 has a higher vapor pressure because it has only London dispersion forces, whereas CH3OH has hydrogen bonding.

7.28 When you get out of a pool, the water on your body evaporates and this cools your skin. When

you re-enter the water, the water feels warmer because your skin is cooler. 7.29 Benzene cannot hydrogen bond, whereas water can, so benzene has much weaker intermolecular

forces than water and is thus less viscous. Ethylene glycol, on the other hand, has two OH groups capable of hydrogen bonding, so it has stronger intermolecular forces than water and is thus more viscous.

7.30 The stronger the intermolecular forces, the stronger surface molecules are pulled down toward

the interior of a liquid and the higher the surface tension.

The surface tension of gasoline should be lower than that of water because gasoline cannot hydrogen bond, whereas water can, giving it higher surface tension.

7.31 An ionic solid is composed of oppositely charged ions.

A molecular solid is composed of individual molecules arranged regularly. A network solid is composed of a vast number of atoms covalently bonded together forming sheets or three-dimensional arrays. A metallic solid, such as copper or silver, can be treated as a lattice of metal cations surrounded by a cloud of electrons that move freely.

a. CaCl2 = ionic c. sugar (C12H22O11) = molecular b. Fe = metallic d. NH3(s) = molecular

7.32 Use the heat of fusion as a conversion factor to determine the amount of energy, as in Sample

Problem 7.14.

79.7 calx =50.0 g1 g

3,985 cal rounded to 3,990 calAnswer

a.

79.7 calx =35.0 g1 g

2,790 calAnswer

b.

79.7 calx =35.0 g1 g

2.79 kcalAnswer

c. 1 kcalx1000 cal



18.02 gx =1.00 mol1 mol

1,430 calAnswer

d. x1 g

79.7 cal

7.33 Use the heat of vaporization to convert grams to an energy unit, calories, as in Sample Problem

7.15.

42 g 540 cal1 g

x = 23,000 calAnswer

a.

42 g 540 cal1 g

x = 23,000 calAnswer

b.

1 kcal1000 cal

540 cal18.02 gx =1.00 mol1 mol

9.7 kcalAnswer

c. x1 g

x

1 kcal1000 cal

540 cal18.02 gx =3.5 mol1 mol

34 kcalAnswer

d. x1 g

x

7.34 The first graphic represents a gas (randomly placed spheres that are far apart) and the second

graphic represents a liquid (closely packed but randomly arranged spheres). The molecular art therefore represents condensation. This is exothermic.

7.35 a. Since heat is added, the graph is a heating curve.

b. The melting point of the substance is at the plateau B→C, which is 40 °C. c. The boiling point of the substance is at the plateau D→E, which is 77 °C. d. Solid and liquid are present at the plateau B→C. e. Only liquid is present along the C→D diagonal.

7.36 a. gas b. solid and liquid c. liquid d. solid e. liquid and gas

•

• •

• •75 °C

50 °C

Tem

pera

ture

85 °C = gas

65 °C = liquid

10 °C = solid

50 °Csolid and liquid

75 °Cliquid and gas

Heat added Solutions to End-of-Chapter Problems 7.37 Use a conversion factor to convert from mm Hg to atm.

814.3 mm Hg1 atm


x


Chapter 7–9

7.38 Use a conversion factor to convert from mm Hg to atm.

652.5 mm Hg1 atm


x


520 mm Hg1 atm


x

2.8 atm = 41 psiAnswer

a. x1 atm

14.7 psi

b. 100. mm Hg101,325 Pa760. mm Hg

= 13,300 PaAnswer

x

20.0 atm = 15,200 torrAnswer

c. x1 atm

760. torr

d.


x 1 atm14.7 psi

3200 psi = 220 atmAnswer

220 atm x1 atm

760. mm Hg = 170,000 mm HgAnswer

x 1 atm14.7 psi

825 psi = 56.1 atmAnswer

56.1 atm x1 atm

760. mm Hg = 42,700 mm HgAnswer

7.41 a. The volume decreases from X to (a) with the same number of gas particles; the pressure

increases.

b. The volume remains the same, but the number of particles decreases; the pressure decreases. c. The volume remains the same, but the number of particles increases; the pressure increases.

7.42 Use the gas laws to draw the balloons.

A b. volume halved

a. volume halved

c. volume increased d. volume doubled



7.43 Use the gas laws to draw the balloons.

given balloon

a. b. c.

increased temperatureVolume increases.

decreased pressureVolume increases.

decreased pressureVolume increases.

7.44 a. [1] Volume decreases when temperature is lowered. b. [1] Volume decreases when gas leaks out. c. [3] Increased altitude results in decreased atmospheric pressure and larger volume. 7.45 a. Volume increases as outside atmospheric pressure decreases as the balloon floats to a higher

altitude. b. Volume decreases at the lower temperature when the balloon is placed in liquid nitrogen. c. Volume decreases as external pressure increases in the hyperbaric chamber. d. Volume increases as temperature increases in the microwave.

7.46 a. As the pressure is doubled, the volume will be reduced by a factor of two. b. As the temperature is doubled, the volume will double. c. The volume will remain the same when both the pressure and temperature are halved. 7.47 Use Boyle’s law to fill in the table as in Example 7.2.

P1 V1 P2 V2 a. 2.0 atm 3.0 L 8.0 atm 0.75 L b. 55 mm Hg 0.35 L 18 mm Hg 1.1 L c. 705 mm Hg 215 mL 99.7 mm Hg 1.52 L

7.48 Use Boyle’s law to fill in the table as in Example 7.2.

P1 V1 P2 V2 a. 2.5 atm 1.5 L 3.8 atm 0.99 L b. 2.0 atm 350 mL 750 mm Hg 710 mL c. 75 mm Hg 9.1 mL 0.77 mm Hg 890 mL

7.49 Use Boyle’s law to solve the problem as in Example 7.2.

V21.0 atm

= =(3.5 atm)(10. mL)

= 35 mLAnswer

P1V1

P2

7.50 Use Boyle’s law to solve the problem as in Example 7.2.

P25.6 L

= =(756 mm Hg)(4.5 L)

= 610 mm HgAnswer

P1V1

V2


Chapter 7–11

7.51 Use Charles’s law to fill in the table as in Example 7.3.

V1 T1 V2 T2 a. 5.0 L 310 K 4.0 L 250 K b. 150 mL 45 K 1.1 L 45 oC c. 60.0 L 0.0 oC 180 L 820 K

7.52 Use Charles’s law to fill in the table as in Example 7.3.

V1 T1 V2 T2 a. 10.0 mL 210 K 21 mL 450 K b. 255 mL 55 °C 120 mL 150 K c. 13 L –150 °C 52 L 490 K

7.53 Use Charles’s law to solve the problem as in Example 7.3.

K = °C + 273T1 = 25 °C + 273 = 298 KT2 = –78 °C + 273 = 195 K

V1T2

T1V2

298 K= =

(2.2 L)(195 K)= 1.4 L

Answer

7.54 Use Charles’s law to solve the problem as in Example 7.3.

K = °C + 273T1 = 20 °C + 273 = 293 K

V2T1

V1T2

750. L= = (1000. L)(293 K) = 391 K

Answer 7.55 Use Gay–Lussac’s law to fill in the table as in Sample Problem 7.4.

P1 T1 P2 T2 a. 3.25 atm 298 K 4.34 atm 398 K b. 550 mm Hg 273 K 350 mm Hg –100. oC c. 0.50 atm 250 oC 955 mm Hg 1,300 K

7.56 Use Gay–Lussac’s law to solve the problem as in Sample Problem 7.4.

P1 T1 P2 T2 a. 1.74 atm 120 °C 1.3 atm 20. °C b. 220 mm Hg 150 °C 160 mm Hg 300.K c. 0.75 atm 198 °C 220 mm Hg 180 K


P1T2

T1P2 373 K

=Answer

= =(1.0 atm)(423 K)

1.1 atmK = °C + 273T1 = 100. °C + 273 = 373 KT2 = 150. °C + 273 = 423 K


P1T2

T1P2 274 K

=Answer

= =(750. mm Hg)(353 K)

966 mm HgK = °C + 273T1 = 1.0 °C + 273 = 274 KT2 = 80. °C + 273 = 353 K



7.59 Use the combined gas law to solve the problems as in Example 7.4.

P1 V1 T1 P2 V2 T2 a. 0.90 atm 4.0 L 265 K 1.4 atm 3.0 L 310 K b. 1.2 atm 75 L 5.0 oC 700. mm Hg 110 L 50 oC c. 200. mm Hg 125 mL 298 K 100. mm Hg 0.62 L 740 K

7.60 Use the combined gas law to solve the problems as in Example 7.4.

P1 V1 T1 P2 V2 T2 a. 0.55 atm 1.1 L 340 K 0.18 atm 3.0 L 298 K b. 735 mm Hg 1.2 L 298 K 1.1 atm 0.97 L 0.0 °C c. 7.5 atm 230 mL –120 °C 15 atm 0.45 L 600 K


Solve for V2 by multiplying both sides byT1 T2

P1V1 P2V2 = P2

T2

V2 T1P2

P1V1T2 =

(291 K)(1.0 atm)(200. atm)(6.0 L)(298 K)

= = 1,200 LAnswer

K = °C + 273T1 = 18 °C + 273 = 291 KT2 = 25 °C + 273 = 298 K

.


T1 T2

P1V1 P2V2 = 1/4 P1 =V2 = 2 V1T2 = 1/2 T1 T1

P1V1 =1/2 T1

P2 (2V1) P2a.

Answer

T1 T2

P1V1 P2V2 = P1 =V2 = 2 V1T2 = 2 T1 T1

P1V1 =2 T1

P2 (2V1) P2b.

Answer

T1 T2

P1V1 P2V2 = 4 P1 =V2 = 1/2 V1T2 = 2 T1 T1

P1V1 =2 T1

P2 (1/2V1) P2c.

Answer 7.63 STP is “standard temperature and pressure,” or 0 oC at 760 mm Hg. The standard molar volume is

the volume that one mole of a gas occupies at STP, or 22.4 liters. 7.64 The same number of moles of two gases at STP will have the same volume since one mole of a

gas occupies 22.4 liters at STP. The masses of the two gases will depend on their molar masses. 7.65 Convert volume to moles at STP as in Sample Problem 7.7.

5.0 L He x =Answer

1 mol22.4 L

0.22 mola.

11.2 L He x =Answer

0.500 molb.1 mol22.4 L


Chapter 7–13

50.0 mL He x =Answer

0.002 23 molc. 1 mol22.4 L

1 L1000 mL

x

7.66 Convert volume to moles at STP as in Sample Problem 7.7.

4.0 L Ar x =Answer

1 mol22.4 L

0.18 mola.

31.2 L Ar x =Answer1.39 molb.

1 mol22.4 L

120 mL Ar x =Answer

0.005 4 molc. 1 mol22.4 L

1 L1000 mL

x

7.67 Convert moles and grams to volume at STP as in Sample Problem 7.7.

4.2 mol Ar x =Answer

94 La.1 mol22.4 L

3.5 g CO2 x =Answer1.8 Lb. 22.4 L

1 mol1 mol

44.01 gx

2.1 g N2 x =Answer1.7 Lc.

1 mol22.4 L1 mol

28.02 gx

7.68 Use conversion factors to solve the problem.

3.01 x 1021 molecules N2 x =Answer1 mol

22.4 L1 mol6.02 x 1023 molecules

x 0.112 L or 112 mL


Convert °C to K. K = °C + 273 = 37 °C + 273 = 310. K

Use the value of R with mm Hg since the pressure is given in mm Hg; that is, R = 62.4 L•mm Hg/(mol•K).


Answer

=(0.45 L)

62.4 L • mm Hgmol • K

(747 mm Hg)= 0.017 mol

(310. K)

n =RTPV





Use the value of R with atm since the pressure is given in atm; that is, R = 0.0821 L•atm/mol•K.


Answer

=(7.0 L)

0.0821 L • atmmol • K

(210 atm)= 60. mol

(298. K)

n =RTPV





=(5.0 L)


(740 mm Hg)=

0.19 mol

(310. K)

n =RTPV

1.1 x 1023 moleculesx1 mol

6.02 x 1023 molecules

Answer=

Answer0.19 mol


PV = nRT Solve for P by dividing both sides by V.

= 0.0821 L • atmmol • K(0.227 mol) =(325 K)P =

VnRT

Answer0.606 atm

10.0 g CO2 x44.0 g

mol = 0.227 mol

10.0 L


Chapter 7–15




= (2.0 L)


= 0.059 mol

(273 K)

n =RTPV 1.9 g O2x

1 mol

32.00 g

O2 has more moles and more mass.

=

=(1.5 L)


(650 mm Hg)= 0.052 mol

(298 K)

n =RTPV 1.5 g N2x

1 mol

28.02 g=

(500. mm Hg)



Answer

=(30.0 L)

0.0821 L • atmmol • K

(2.1 atm)= 2.6 mol

(298. K)

n =RTPV

If the gas is helium: 2.6 mol He x 4.00 g1 mol Answer

= 10. g

If the gas is argon: 2.6 mol Ar x 40.0 g1 mol Answer

= 1.0 x 102 g

7.75 Use Dalton’s law.

A:3 red spheres

9 spheres totalx 630 mm Hg = 210 mm Hg

B:6 blue spheres

9 spheres totalx 630 mm Hg = 420 mm Hg

7.76 Use Dalton’s law.

5 blue spheres6 red spheres

x 630 mm Hg = 525 mm Hg = pressure of B

total pressure = 630 mm Hg + 525 mm Hg = 1155 mmg Hg




Oxygen: 0.21 × 460 mm Hg = 97 mm Hg for O2 Nitrogen: 0.78 × 460 mm Hg = 360 mm Hg for N2

7.78 Use Dalton’s law to solve the problem as in Example 7.8. Oxygen: 0.21 × 175 atm = 37 atm 7.79 If the overall pressure is three times as great, the partial pressure is three times higher.

593 mm Hg × 3 = 1,780 mm Hg 7.80 Use Dalton’s law to solve the problem as in Example 7.8.

Ptotal = PO2 + PCO2 + PN2 850 mm Hg = 450 mm Hg + 150 mm Hg + PN2 PN2 = 850 mm Hg – 450 mm Hg – 150 mm Hg = 250 mm Hg N2: 250 mm Hg O2: 450 mm Hg CO2: 150 mm Hg

7.81 Water is a liquid at room temperature because it is capable of hydrogen bonding and these strong

intermolecular attractive forces give it a higher boiling point than H2S, which can’t hydrogen bond.

7.82 Cl2, Br2, and I2 all have London forces only. The strength of the intermolecular forces increase as

molar mass increases, so the melting points and boiling points increase with increasing molar mass. As a result, Cl2 (the lightest of the three halogens) is a gas, Br2 is a liquid, and I2 (the heaviest of the three halogens) is a solid at room temperature.

7.83 Use Table 7.3 to decide which types of forces are present in each molecule.

a. chloroethane: London forces, dipole–dipole b. cyclopropane: London forces only

7.84 Use Table 7.3 to decide which types of forces are present in each molecule. a. acetaldehyde: London forces, dipole–dipole b. acetic acid: London forces, dipole–dipole, hydrogen bonding 7.85 Hydrogen bonding occurs only in compounds that contain an O–H, N–H, or H–F bond. Only

compound (d) can hydrogen bond. 7.86 Hydrogen bonding occurs only in compounds that contain an O–H, N–H, or H–F bond. Only

compound (d) can hydrogen bond. 7.87 No, H2C=O has no H on the O atom, so it cannot form hydrogen bonds with another molecule of

formaldehyde.


Chapter 7–17

7.88 NaCl is an ionic solid and is composed of sodium cations and chloride anions held together by the electrostatic attraction of opposite charges. Water is a covalent molecule that forms intermolecular hydrogen bonds with other water molecules. The attractive interactions between the ions in an ionic compound are much stronger than the attractive interactions in the hydrogen bonds of a covalent molecule. Therefore, NaCl has a much higher melting point than water.

7.89 a. Ethylene has London forces only, whereas methanol has London forces, dipole–dipole forces,

and hydrogen bonding. b. Methanol has a higher boiling point because it can hydrogen bond. c. Ethylene has a higher vapor pressure at any given temperature because it has weaker

intermolecular forces. 7.90 a. Ethanol has London forces, dipole–dipole forces, and hydrogen bonding. Dimethyl ether has

London forces and dipole–dipole forces. b. Ethanol has the higher boiling point because it can hydrogen bond.

c. Dimethyl ether has the higher vapor pressure at any given temperature because it has weaker intermolecular forces.

7.91 Vapor pressure and boiling point are inversely related: lower vapor pressure corresponds to a

higher boiling point.

Increasing boiling point: butane, acetaldehyde, Freon-113 7.92 Boiling point and vapor pressure are inversely related—that is, a lower boiling point corresponds

to a higher vapor pressure, whereas a higher boiling point corresponds to a lower vapor pressure. a. Ethanol will have a higher vapor pressure than 1-propanol. b. Hexane will have a higher vapor pressure than octane. 7.93 Glycerol is more viscous than water since it has three OH groups and many opportunities for

hydrogen bonding. Acetone cannot hydrogen bond, so its intermolecular forces are weaker and it has low viscosity.

7.94 A needle floats when it is carefully placed on the surface of the water because the water

molecules on the surface are affected only by those water molecules at or below the surface layer. The inward force of attraction on the surface molecules causes the liquid to behave as if it had a skin on which the needle floats. The needle sinks to the bottom when its tip is pushed below the surface because the “skin” has been disrupted and can no longer support the needle.

7.95 Use the definitions from Answer 7.31 to classify each solid.

a. KI: ionic d. diamond: network b. CO2: molecular e. the plastic polyethylene: amorphous c. bronze, an alloy of Cu and Sn: metallic

7.96 Use the definitions from Answer 7.31 to classify each solid. a. CaCO3: ionic d. graphite: network b. CH3COOH: molecular e. the plastic polypropylene: amorphous c. Ag: metallic



7.97 Evaporation is an endothermic process by which a liquid enters the gas phase. Condensation is an exothermic process that occurs when a gas enters the liquid phase.

7.98 Sublimation is an endothermic process by which a solid transforms directly to the gas phase.

Deposition is an exothermic process that occurs when a gas transforms directly to the solid phase. 7.99 The first graphic represents a liquid (closely packed but randomly arranged spheres) and the

second graphic represents a gas (randomly placed spheres that are far apart). The molecular art therefore represents vaporization. Energy is absorbed.

7.100 The first graphic represents a solid (closely packed arranged spheres) and the second graphic

represents a gas (randomly placed spheres that are far apart). The molecular art therefore represents sublimation. Energy is absorbed.

7.101 a. Melting 100 g of ice is endothermic: energy is absorbed.

b. Freezing 25 g of water is exothermic: energy is released. c. Condensing 20 g of steam is exothermic: energy is released. d. Vaporizing 30 g of water is endothermic: energy is absorbed.

7.102 The heat of fusion is the amount of energy required to melt 1 gram of a substance. The heat of vaporization is the amount of energy required to vaporize 1 gram of a substance.

7.103 Use the heat of fusion and heat of the vaporization as conversion factors to determine the amount of energy required, as in Sample Problem 7.14.

79.7 calx =250 g

1 g2.0 x 104 cal

Answer

540 calx =50.0 g1 g

2.7 x 104 calmelting vaporizing

7.104 Use the heat of vaporization as a conversion factor to determine the amount of energy required.

Answer

540 calx =255 g1 g

140 kcal1 kcalx1000 cal

7.105

•

••

•

V

W X

Y

Heat removed

85 oC

10 oC

[1]

[2]

a.

b. melting point = 10 °C

c. boiling point = 85 °C

ZTem

pera

ture


Chapter 7–19

7.106 Line segment VW represents the physical state shown in A (vapor state). Line segment XY represents the physical state shown in B (liquid state). Z represents the physical state shown in C (solid state).

7.107

•• •

• •

A

BC

DE

Heat added

–57

126130

–70

Tem

pera

ture

°C

7.108

7.109 The gases inside the bag had a volume of 250 mL at 760 mm Hg and take up a greater volume at

the reduced pressure of 650 mm Hg.

(760 mm Hg)(250 mL)

650 mm Hg= 290 mLP1V1 = P2V2

P1V1V2 P2

= =

7.110 A bubble at the bottom of a glass of a soft drink get larger as it rises to the surface because the

pressure is less at the surface and the volume increases as the pressure decreases. 7.111 Water is one of the few substances that expands as it enters the solid phase. This causes the bottle

to crack because the water occupies a larger volume when it freezes.

•• •

• •

Heat added

0

100120

–20

Tem

pera

ture

°C

A

B

C



7.112 The density of a gas, which can be expressed as n/V = P/RT, will decrease if the temperature is increased while the pressure is held constant. Heated air expands (volume increases), so the density decreases since the number of molecules remain the same. This allows the hotter air to rise. Cooler air then rushes in to fill the space vacated by the hotter air, thus creating wind.

7.113 a. As a person breathes faster, he eliminates more CO2 from the lungs; therefore, the measured

value of CO2 is lower than the normal value of 40 mm Hg.

b. Many people with advanced lung disease have lost lung tissue over time and therefore cannot exchange adequate amounts of oxygen through the lungs and into the blood, leading to a lower-than-normal partial pressure of oxygen. In addition, they often breathe more slowly and with lower volumes than normal, so they cannot eliminate enough CO2 and therefore the partial pressure of CO2 climbs.

7.114

V21.0 atm

= =(4.0 atm)(0.50 L)

= 2.0 LAnswer

P1V1

P2

The volume of air will increase from 0.50 L to 2.0 L as the scuba diver ascends to the surface.

Therefore, it is necessary for the scuba diver to exhale as he rises to the surface of the water in order to eliminate the excess volume of gas. Otherwise, the diver may suffer lung damage.

7.115 Use the ideal gas law to solve the problem.

Use the value of R with atm since the pressure is given in atm; that is, R = 0.0821 L•atm/(mol•K).


=(11.2 L)

0.0821L • atm

mol • K

(2 atm)= 1 mol = molar mass 4.0 g/mol = Helium

(273 K)

n =RTPV

7.116

100. g NaN3 x =Answer51.7 L

1 mol22.4 L1 mol NaN3

65.0 g NaN3x

2 mol NaN3

3 mol N2x


chapter 7 gases, liquids, and solids -...

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