Page 453, Practice Problems 1. 480 kPa2. 0.408 atm3. 3.00 × 103 psi

Page 454, 8.1 Review Questions 1. A hyperbaric chamber will increase the height ofthe mercury column and hence the pressurereading. Increased pressure displaces the mercuryup.2.

mm Hg atm kPa 7.10 × 103 9.34 946 4170 5.49 555.696

3. mm Hg atm kPa 4850 6.375 646 711 0.936 94.8

4. a. 1126.8 mm Hgb. 524 mm Hg

5. 6860 torr6. 128 kPa7. 707 torr

Page 459, Practice Problems 1. 122 mL2. 90 kPa

Page 462, Practice Problems 1. 128 mL2. 2.4 °C

Page 465, Practice Problems 1. 500.0mmHg2. 21 °C

Page 467, Practice Problems 1. 907 L2. 5.41 L

Page 469, Quick Check Standard P Substitution Work

Shown R Value

1 atm 1 atm × 22.4 L 1 mol × 273 K 0.0821 L× atm

mol×K

760 mm Hg 760 mm Hg × 22.4L

1 mol × 273 K 62.4 L× mmHg

mol×K

14.7 psi 14.7 psi × 22.4 L 1 mol × 273 K

1.21 L× psimol

×K

101.3 kPa 101.3 kPa × 22.4 L 1 mol × 273 K

8.31 L× kPamol

×K

Page 471, Practice Problems 1. 1.70 L2. 27.9 g/mol3. 32.0 g/mol [O2]

Page 472, Quick Check Simple

Law Constant Variables

Derivation Shown

Final Form of Law

Charles's Law

Pressure and moles

P1V1n1 T1

=P2 V2n2 T2

V1T1= V2

T2

Gay-Lussac's

Law

Volume and moles

P1 V1n1 T1

=P2 V2n2 T2

P1T1= P2

T2

Combined Gas Law

Moles P1V1n1 T1

= P2V2n2 T2

P1V1T1

= P2V2T2

Page 472, Quick Check 2. 27.9 g/mol3. 32.0 g/mol

Page 473, Review Questions 1. a. Container One: i. Pressure will decrease due to

the temperature decrease (Gay-Lussac’s Law).P1V1/T1, hence, P1 = T1/V1 Volume cannot change. ii. Density remains the same (samemass/volume).Container Two: i. Volume will decrease due tothe temperature decrease (Charles’ Law). Sinceboth volume and temperature decrease,pressure will remain constant (Combined GasLaw). Volume able to decrease due totemperature decrease.ii. Density will increase (same mass/smallervolume).b. Container One: i. Pressure will increase(double) due to a doubling of the number ofmoles of gas. P = nRT/V. Assuming noconfounding temperature change.ii. Density will double (double the number ofparticles/same volume). Container Two: Pressure will remain constant as doubling n leads to doubling V (assuming T remains constant). ii. Density will not change (though the moles (mass) is doubled, so is the volume).

2. Boyle's Law, n and T are held constant, when V isdecreased, P is increased.

3. Charles' Law, P and n are held constant, when Tis increased, V is increased.

1. 900 mm Hg2. 612 mm Hg

Page 451, Practice Problems Chapter 8

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4. As n is doubled and T is doubled, and V isunchanged, P must be quadrupled. (It isincreased due to the increase in BOTH n AND T)

5. P is increased by 1/10 (4.8 atm) = 0.48 atm. Newpressure = 4.8 + 0.48 = 5.3 atm

6. 1.48 atm7. -10 °C (Check your significant figures!)8. 39.6 L9. 3780 kPa10. 44.0 psi11. 21.2 L12. 5.03 L13. 64.1 g/mol14. 0.46 L15. 3570 kPa16. 3.00 L17. a. 20.2 g/mol

b. 9.65 × 1022 atomsc. Neon

Page 481, Quick Check 1. a. The height of the water in the beaker will

decrease over time because some of the waterwill evaporate due to the vapor pressure itexerts. The vapor pressure causes the water toevaporate. Once enough water evaporates,some will begin to condense. Eventually the rateof evaporation will equal the rate ofcondensation and level will remain constant.b. If the temperature increased, the height ofthe water would decrease more and become constant faster than it did in part a.. The increased temperature provides more kinetic energy for more molecules to escape into the gas phase.c. Atmospheric pressure at sea level is 760mmHg. Even though the atmospheric pressurehas decreased to 500 mm Hg, there will be no change in the height of the water. Changes in atmospheric pressure only affects boiling points and not vapour pressures. Vapour pressure depends on temperature and the structure of the molecules.

2. Acetone is volatile and thus has high vaporpressure in contrast to water which will have arelatively lower vapor pressure. Thesedifferences in vapor pressure are due to theIMFs present within each particle. Water hasstrong hydrogen bonds present, resulting infewer molecules with sufficient energy toovercome the attraction to enter the gas phasethan acetone.

Page 483, Practice Problems 1. a. 1.5 atm, 0.5 atm

b. 2.0 atm

2. 223 kPa

Page 484, Practice Problems 1. 0.24 L2. 3.70 L

Page 486, Practice Problems 1. 20 kPa2. CO2 = 44.1 kPa

CH4 = 19.9 kPaNH3 = 15.9 kPa

Page 487, 8.3 Review Questions 1. CH4 = 211.1 torr

O2 = 548.9 torr2. a.The vapor pressure of water at 25 ˚C at an

altitude of 3000 meters in the mountains is equal to the vapor pressure of water at sea level at 25˚ because the change in altitude (thus different atmospheric pressures) does not affect the vapor pressure. Vapor pressure is only dependent on temperature and the structure of the molecules/particles.b. Liquid Y has the stronger intermolecular forces because it requires a higher temperature, thus more kinetic energy, to overcome the attractions allowing some liquid particles to escape into the gas phase.c. As water evaporates due to vapour pressure, some will gradually re-condensate (because the container is sealed and limits the volume for the water vapour to expand) in order to create an equilibrium between the evaporation and condensation rates.

3. In an airtight pressure cooker, the water vapour/steam caused by vaporizing water is contained within the pot and does not escape. Vaporized water creates an artificial“atmosphere” at a very high pressure. Since the boiling point of water is the temperature at which the vapor pressure reaches atmospheric pressure, the boiling point is greatly elevated so the food cooks much faster at this high temperature.

4. He = 1.0×102 kPaO2 = 57 kPaN2 = 18 kPa

5. 337 kPa6. 285 kPa7. H2 = 0.40 atm

CO2 = 0.60 atm8. 13.33 psi9. P2N2 = 0.111 atm; P2NH3= 0.444 atm; PTotal = 0.555 atm

10. 3.6%

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11.a. 8.65%b. C10H12NO c. C20H24N2O2 12. 13 000 kPa

Page 492, Quick Check

a. Boyle’s Law: the relationship between pressureand volume If the volume of gas particles can beassumed to be negligible relative to the spacebetween them (1), a gas is manly empty space.This means an increase in the pressure on a gassample will cause the particles to move closer together resulting in a corresponding decreasein the sample’s volume.

b. Charles’s Law: the relationship between volumeand temperature. If the average kinetic energyof a gas sample is directly proportional to thetemperature of the gas in Kelvin degrees (4),then increased temperature must cause gasparticles to move faster and further apartcausing a proportional increase in the volume ofthe gas sample.

c. Gay-Lussac’s Law: the relationship betweenpressure and temperature If the average kineticenergy of a gas sample is directly proportionalto the temperature of the gas in Kelvin degrees(4), and the gas particles colliding with the wallsof their container produce pressure (2), then theKE directly proportional to the temperature ofthe gas is also directly proportional to thepressure produced by the gas particles.Increasing temperature will increase the KEpresent in gas molecules, resulting in morefrequent collisions with the container thus morepressure.

d. Dalton’s Law of Partial Pressures: allowscalculating the total pressure of a mixture ofdifferent gases. If the gas particles are inconstant motion and collide with the walls oftheir container to produce pressure (2), and gasparticles neither attract nor repel one another(3), then the sum of the partial pressures of eachcomponent gas must add up to the totalpressure of a mixture of gases.

Page 494, Practice Problems 1. 432 m/s2. 314 m/s

Page 497, Practice Problems 1. 132. a. The white ring of ammonium chloride will be

closer to the HCl end because HCl (g) is heavierand thus moves slower than NH3 (g).b. 1.47c. i. Both gases must be applied at the ends ofthe glass tube at the same time

ii. The tube must not be disturbedbefore a white ring deposits.

3. 1300 m/s

Page 501, Quick Check 1. a. O2 deviates the most from ideal behavior

because O2 is the largest gas present. Thedeviation from ideal behavior is most noticeablefor larger gases as they have greater volume.Larger gases also exert larger LDF on each other.b. 1000 K provides the most ideal behaviorbecause gases behave most ideally at hightemperatures. At higher temperatures, thevolume of the actual gas particles becomealmost zero compared to the total volume thatcontains the gas. Also the IMFs become muchless noticeable.

2. Moving down the noble gas family:“a” (accounting for the attraction between gasparticles) increases due to the increase in size,thus an increase in LDF“b” (accounting for the volume occupied by thegas) increases due to the increase in energylevels/layers of electrons, thus a larger volume

3. Moving from N2 to CO2 to CCl4:“a” (accounting for the attraction between gas particles) increases due to the increase in size,thus an increase in LDF“b” (accounting for the volume occupied by thegas) increases due to the increase in energylevels/layers of electrons, thus a larger volume

4. Moving from N2 to CO to HCl to H2O:“a” (accounting for the attraction between gasparticles) increases due to the increase in size,thus an increase in LDF as well as the presence of dipole-dipole in HCl and H-bonding in H2O.“b” (accounting for the volume occupied by thegas) is approximately the same for nitrogenand CO as they contain the same number ofelectrons and so the volume of their electronclouds are equal while the number of electronsand hence size of electron cloud is muchgreater for HCl and much smaller for water.

Page 502, 8.4 Review Questions 1. KE Ne = 3710 J/mol, KE N2 = 7430 J/mol2. Ne = 606 m/s

N2 = 728 m/sAs temperature increases, μrms increases(doubling T increased V by 2 ).

3. In order of deviation from ideal behavior: SO2 >O2 > CH4 > H2 because:SO2 has the most attractive forces present (LDF)as it is larger in size and exerts dipole-dipoleforces due to its bent shape. It also has thegreatest volume due to its layers of electrons.

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4. Van der Waal’s equation for real gases:

P + anV

⎛⎝⎜

⎞⎠⎟

2

× (v-nb) = nRT

a. H2O has the higher value for ‘a’ due to H- bonds (along with dipole-dipole and LDFs),

while H2S exerts only dipole-dipole forces (due to polarity caused by bent shape) and LDFs. H2S has the higher value for ‘b’ because it is greater in volume than H2O. b. The Van der Waals constant ‘a’ can becorrelated with the bpt of a substance because‘a’ accounts for the attractive intermolecularforces that real gases demonstrate. Theseattractive Intermolecular forces are broken as asubstance boils. As ‘a’ increases, bpt decreases.

5. a. The balloon containing CO2(g) contains thegreats mass of gas because of all the five gases,CO2 has the largest molar mass. Using PV=nRT,because all other factors are constant, the molesof gas are identical in each balloon. Thus theheaviest balloon depends on the molar massesof the gases.b. The average kinetic energies of the gasmolecules in the balloons are identical becausekinetic energy is only dependent on

temperature: KEavg Ne(g) =32

RT

c. The balloon containing CO2 contains the gasthat would be expected to deviate most fromthe behavior of an ideal gas because CO2possesses the greatest attraction due to its size(more LDF interactions), as well as the greatestvolume due to its layers of electrons (largerelectron cloud).d. The balloon containing He will be thesmallest because the effusion of He is thequickest due to its small size.

6. 5.0 L of He(g) at STP:i. Kelvin temperature is increased by100˚C:

a. KE increases

b. average velocity increasesc. frequency of collisions with thecontainer walls (pressure) increases

ii. The volume is halved:a. KE remains unchangedb. average velocity remains unchangedc. frequency of collisions with thecontainer walls (pressure) doubles

iii. The number of moles of gas isdoubled:

a. KE remains unchangedb. average velocity remains unchangedc. frequency of collisions with thecontainer walls (pressure) increases by2X)

7. Assuming STP conditions:Root mean square velocities of Br2 =206 m/s, Cl2 = 310 m/s, F2 = 423 m/sAverage Kinetic energy of Br2, Cl2, F2 =3.40 × 103 J

a. greatest velocity: F2b. greatest kinetic energy: all equalc. greatest effusion rate: F2

8. Helium9. 4, H2 (g)10. 25 min11. a. 12.2 atm

b. 11.9 atmc. The difference in values are due to thecorrection factor “b” subtracted from thevolume to account for the volume of gasparticles as well as a correction factor “a” subtracted from the ideal gas pressure toaccount for the attraction between gas particles

12. a. 0.122 atmb. 0.122 atmc. The answer with the ideal Gas Law is identicalto the answer with Van der Waal's equationwithin 3 significant figures!

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