Chapter 7 Gases, Liquids, and Solids - Riverside City...

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 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.

7.3 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.

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

7.4 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

Gases, Liquids, and Solids 7–2

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

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

TK = TC + 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

7.6 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.

TK = TC + 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.

7.7 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

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

Chapter 7–3

7.8 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.

7.9 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

7.10 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.


7.11 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 mol

22.4 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

7.13 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 TC to TK. TK = TC + 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

7.14 Use the ideal gas law to solve the problems as in Example 7.6 and Answer 7.13.

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

TK = TC + 273 = 25 °C + 273 = 298 Ka.

Chapter 7–5

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.

=

TK = TC + 273 = 20. °C + 273 = 293 Kb.

7.15 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

7.16 a. The percent of N2 and O2 is determined by how many molecules of each are present compared to the total number of molecules (eight).

N2:5 N2 molecules

8 total molecules= 0.625 = 62.5% 3 O2 molecules

8 total molecules= 0.375 = 37.5%O2:

b. Use the fractions from part (a) and the total pressure (3,500 psi) to determine the partial

pressures.

N2: 0.625 x 3500 psi = 2,200 psi O2: 0.375 x 3500 psi = 1,300 psi 7.17 Since the atmosphere contains 21% O2 regardless of location, use the partial pressure of O2 (PO2)

to determine the total pressure (Ptotal). Convert 21% to a decimal by moving the decimal point two places to the left (0.21).

7.18 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).


7.19 With dipole–dipole interactions, molecules align so that partial positive and partial negative charges are close together.

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

force is present, whereas a “–” means the force is absent.

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 –

7.21 London dispersion forces are present in all covalent compounds and atoms.

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. Ne atoms have stronger London dispersion forces because they are larger than He atoms.

7.22 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 can hydrogen bond, 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. 7.23 Water has stronger intermolecular forces because it can hydrogen bond, whereas CO2 has only

London dispersion forces. As a result, water is a liquid at room temperature, whereas CO2 is a gas.

7.24 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 can form hydrogen bonds.

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 can form hydrogen bonds.

Chapter 7–7

7.25 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.

7.26 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

7.27 • The larger the specific heat, the less the temperature of a substance will change when it absorbs

heat energy. • The larger the specific heat, the more heat that must be added to increase the temperature of a

substance a given number of degrees. a. Sand has the higher temperature because it has the lower specific heat. b. Ethanol requires the larger amount of heat because it has the larger specific heat.

7.28 Water has a high specific heat, so it can absorb or release a great deal of energy with only a small temperature change.

7.29 Use specific heat as a conversion factor to calculate the amount of heat absorbed using the known

mass and temperature change. 150. °C – 19 °C = 131 °C = ΔT

28.0 g x 131 °C x 0.107 cal1 g • 1 °C

= 392 cal

392 cal x 4.184 J1 cal

= 1,640 J

cal = mass x ΔT SHx

=

=

(heat absorbed in cal)

(heat absorbed in J) 7.30

55 °C – 12 °C = 43 °C = ΔT

200. g x 43 °C x 1.00 cal1 g • 1 °C

= 8,600 cal

8.6 kcal

cal = mass x ΔT SHx

=

=

7.31 Use specific heat as a conversion factor to determine the temperature change.

Cu: ΔT = calg • SH

= 20. cal10.0 g x 0.0900 cal/(g • °C)

= 22 °C

T2 = 15 °C + 22 °C = 37 °C

Hg: ΔT = calg • SH

= 20. cal10. g x 0.0335 cal/(g • °C)

= 60. °C

T2 = 15 °C + 60. °C = 75 °C


7.32 Use specific heat as a conversion factor to determine the temperature change. ΔT =

calg • SH

=950. cal

120. g x 0.583 cal/(g • °C)= 13.6 °C

T2 = 20. °C + 13.6 °C = 34 °C

7.33 Use the heat of fusion as a conversion factor to determine the amount of energy, as in Sample Problem 7.17.

79.7 calx =50.0 g

1 g3,985 cal rounded to 3,990 cal

Answera.

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

1,440 calAnswer

d. x1 g

79.7 cal

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

7.18.

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

7.35 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, which is exothermic.

7.36 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.

Chapter 7–9

7.37 Follow the steps in Sample Problem 7.20. Two calculations are needed—the energy released when the water is cooled and the energy released when the water freezes to form ice. [1] Calculate the heat released on cooling water (mass = 50.0 g) from 25 °C to 0.0 °C (25 °C –

0.0 °C = 25 °C = ΔT) using the specific heat.

1.00 cal1 g • 1 °C

= 1,250 cal rounded to 1,300 cal

heat = mass x ΔT x specific heat

= 50.0 g x 25 °C x

[2] Calculate the heat released when 50.0 g of water freezes to ice at 0.0 °C using the heat of

fusion. cal = 50.0 g x 79.7 cal

1 g= 3,985 cal rounded to 3,990 cal

[3] Add the answer in [1] to the answer in [2].

Total energy = 1,300 + 3,990 = 5,290 cal rounded to 5,300 cal 7.38 Follow the steps in Sample Problem 7.20. Three calculations are needed—the energy needed to

melt ice to water, the energy needed to heat liquid water to its boiling point, and the energy needed to vaporize liquid water to steam. [1] Calculate the heat needed to melt ice to water (mass = 25.0 g) using the heat of fusion.

cal = 25.0 g x 79.7 cal1 g

= 1,993 cal rounded to 1,990 cal [2] Calculate the heat needed to heat liquid water (25.0 g) from 0.0 °C to 100. °C (100. °C –

0.0 °C = 100. °C = ΔT) using the specific heat.

1.00 cal1 g • 1 °C

= 2,500 cal = 2.50 x 103 cal

heat = mass x ΔT x specific heat

= 25.0 g x 100. °C x

[3] Calculate the heat needed to vaporize water to steam (25.0 g) using the heat of vaporization.

cal = 25.0 g x 540 cal1 g

= 13,500 cal rounded to 14,000 cal [4] Add up the answers in steps [1]–[3].

Total energy = 1,990 + (2.50 x 103) + 14,000

= 18,490 cal rounded to 18,000 cal


Solutions to End-of-Chapter Problems 7.39 a. From the pressure gauge: 2,600 psi.

b. Use a psi–mm Hg conversion factor.

2600 psi x 760. mm Hg14.7 psi

= 130,000 mm Hg = 1.3 x 105 mm Hg

7.40 a. From the pressure gauge: 1700 psi.

b. Use a psi–Pa conversion factor.

1700 psi x 101,325 Pa

14.7 psi= 1.2 x 107 Pa


520 mm Hg1 atm

760. mm Hg= 0.68 atmAnswer

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

7.43 a. Largest number of particles = highest pressure = [3]. b. Smallest number of particles = lowest pressure = [2]. . 7.44 a. The pressure will increase as the temperature is increased at constant volume. b. The pressure will double when the volume is halved at constant temperature. c. The pressure will increase by a factor of 4 when the volume is halved and the temperature is

doubled. 7.45 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.

Chapter 7–11

7.46 Use the gas laws to draw the balloons.

A b. volume halved

a. volume halved

c. volume increased d. volume doubled 7.47 Use the gas laws to draw the balloons.

given balloon

a. b. c.

increased temperatureVolume increases.

decreased pressureVolume increases.

decreased pressureVolume increases.

7.48 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. 7.49 a. The volume increases as the outside atmospheric pressure decreases when the balloon floats to

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

7.50

Pressure doubledTemperature constant

Volume halved

Pressure constantTemperature doubled

Volume doubled

a.

b.

c.Pressure halved

Temperature halved

No change


7.51 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

7.52 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

7.53 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 7.54 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

7.55 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

7.56 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

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

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

V1T2

T1V2

298 K= =

(2.2 L)(195 K)= 1.4 L

Answer

Chapter 7–13

7.58 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 7.59 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

7.60 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 atmTK = TC + 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

7.63 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

7.64 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

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

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 TC to TK. TK = TC + 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

Chapter 7–15


Convert °C to K. K = °C + 273 = 25 °C + 273 = 298. K

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


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



=(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





= (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.

6 red spheres11 spheres total

x total pressure = 480 mm Hg

total pressure = 480 mm Hg x 11 spheres total6 red spheres

= 880 mm Hg

Chapter 7–17

7.77 Use Dalton’s law to solve the problem as in Example 7.8.

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

7.78 a. The percent of N2 and O2 is determined by how many molecules of each are present compared

to the total number of molecules (20).

N2: 13 N2 molecules20 total molecules

= 0.65 = 65% O2: 7 O2 molecules20 total molecules

= 0.35 = 35%

b. Use the fractions from part (a) and the total pressure (3,200 psi) to determine the partial

pressures.

N2: 0.65 x 3,200 psi = 2,000 psi O2: 0.35 x 3,200 psi = 1,100 psi

7.79 If the overall pressure is three times as great, the partial pressure is three times higher.

593 mm Hg × 3.00 = 1,780 mm Hg 7.80 Use Dalton’s law to solve the problem.

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.

H C CH

HC

OH

H

HH C Cl

H

Brb.a.

London dispersion forcesdipole–dipole interactions

London dispersion forcesdipole–dipole interactions

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. 7.88 Compound A has a higher boiling point than B because the two N–H bonds in A make it possible

to hydrogen bond with other molecules of itself. Compound B has N–C bonds which cannot participate in hydrogen bonding.

7.89 Undecane has the higher melting point. Although both pentane and undecane have only London

dispersion forces, undecane is larger in size with greater surface area, so the London forces are stronger.

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 because 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 7.26 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 7.26 to classify each solid. a. CaCO3: ionic d. graphite: network b. CH3COOH: molecular e. the plastic polypropylene: amorphous c. Ag: metallic

Chapter 7–19

7.97 The larger the specific heat, the less the temperature will change when a substance absorbs a

particular amount of energy. Since the temperature of Y is less than the temperature of X, the specific heat of Y is higher than the specific heat of X.

7.98 a. The larger the specific heat, the less the temperature will change when a substance absorbs a

particular amount of energy. 2-Propanol has a larger specific heat than ethanol (see Table 7.4). If the same amount of heat is added to all three samples, sample A will have the highest temperature because ethanol has a smaller specific heat than 2-propanol and the sample size in A is smaller than in B. b. C will have the lowest temperature because the specific heat of 2-propanol is greater than the specific heat of ethanol..

7.99 Use specific heat as a conversion factor to calculate the energy change.

a. cal = g x ΔT x SH

= 50. g x 35 °C x 1.00 cal1 g • 1 °C

= 1.8 x 103 cal= 7.5 x 103 J

b. cal = g x ΔT x SH

= 250 g x 75 °C x 0.214 cal1 g • 1 °C

= 4.0 x 103 cal= 1.7 x 104 J

7.100 Use specific heat as a conversion factor to calculate the energy change.

cal = g x ΔT x SH ΔT = 48 °C − 18 °C = 30 °C

= 55 g x 30 °C x 0.583 cal1 g 1°C

= 960 cal 7.101

a. H2O ΔT = calg • SH

= 200. cal100. g • 1.00 cal/(g • °C) = 2.00 °C T2 = 16.0 °C + 2.00 °C = 18.0 °C

b. H2O ΔT = 350. J

50.0 g • 4.18 J/(g • °C)= 1.67 °C T2 = 16.0 °C + 1.67 °C = 17.7 °C

7.102

cal = g x ΔT x SH

37.0 cal = 12.0 g x 8.5 °C x SH

SH = 0.36 cal/g °C 7.103 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, so energy is absorbed.

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

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

7.105 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.

7.106 Use the heat of fusion and the heat of the vaporization as conversion factors to determine the amount of energy required.

79.7 calx =250 g

1 g2.0 x 104 cal

Answer

540 calx =50.0 g1 g

2.7 x 104 calmelting vaporizing

Thus, vaporizing 50.0 g of water takes more energy than melting 250 g of ice.

7.107

•

••

•

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

7.108 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).

7.109

•• •

• •

A

BC

DE

Heat added

–57

126130

–70

Tem

pera

ture

°C

Chapter 7–21

7.110

•• •

• •

Heat added

0

100120

–20

Tem

pera

ture

°C

A

B

C

7.111 Use Sample Problem 7.20 and Answer 7.37 to solve the problem.

a. [1] cal = 45 g x

= 3,600 cal

[2] heat = mass x ΔT x SH

79.7 cal1 g

= 45 g x 55 °C x 1.00 cal1 g • 1 °C

= 2,500 cal

[3] Total energy = 3,600 + 2,500 = 6,100 cal

b. 6,100 cal

c. [1] cal = 35 g x = 19,000 cal


540 cal1 g

= 35 g x 100. °C x 1.00 cal1 g • 1 °C

= 3,500 cal

[4] Total energy = 19,000 + 3,500 + 2,800 = 25,000 cal

[3] cal = 35 g x = 2,800 cal79.7 cal1 g

= 25 kcal 7.112 Use Sample Problem 7.20 and Answer 7.37 to solve the problem.

[1]a. heat = mass x ΔT x SH

= 150 g x 65 °C x 1.00 cal1 g °C

= 9,800 cal

[2] cal = 150 g x 540 cal1 g

= 81,000 cal

[3] Total energy = 9,800 cal + 81,000 cal = 91,000 cal

b. [1] cal = 42 g x 540 cal1 g

= 23,000 cal


= 42 g x 100. °C x 1.00 cal1 g °C

= 4200 cal

[3] cal = 42 g x 79.7 cal1 g

= 3300 cal

[4] Total energy = 23,000 cal + 4200 cal + 3300 cal = 30,500 cal


7.113 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

= =

7.114 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. 7.115 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. 7.116 You feel cool when you get out of a swimming pool, because the water at the surface of your skin

evaporates and evaporation removes heat from the skin causing cooling. The water feels warmer when you get back into the swimming pool because your skin has cooled, therefore making the water feel warmer.

7.117 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.

7.118

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.

7.119 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

(273 K)

n =RTPV

Molar mass 4.0 g/mol = Helium

7.120

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 - Riverside City...

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