97

CHAPTER FIVE

GASES

Questions

16. a. Heating the can will increase the pressure of the gas inside the can, P % T, V and n constant. Asthe pressure increases, it may be enough to rupture the can.

b. As you draw a vacuum in your mouth, atmospheric pressure pushing on the surface of the liquidforces the liquid up the straw.

c. The external atmospheric pressure pushes on the can. Since there is no opposing pressure fromthe air inside, the can collapses.

d. How "hard" the tennis ball is depends on the difference between the pressure of the air inside thetennis ball and atmospheric pressure. A "sea level" ball will be much "harder" at high altitudesince the external pressure is lower at high altitude. A “high altitude” ball will be "soft" at sealevel.

1 1 2 217. PV = nRT = constant at constant n and T. At two sets of conditions, P V = constant = P V .

1 1 2 2P V = P V (Boyle's law).

= constant at constant n and P. At two sets of conditions, = constant = .

(Charles's law)

18. Boyle's law: P % 1/V at constant n and T

In the kinetic molecular theory (kmt), P is proportional to the collision frequency which is proportionalto 1/V. As the volume increases there will be fewer collisions per unit area with the walls of thecontainer and pressure will decrease (Boyle's law).

Charles's law: V % T at constant n and P

When a gas is heated to a higher temperature, the speeds of the gas molecules increase and thus hit thewalls of the container more often and with more force. In order to keep the pressure constant, thevolume of the container must increase (this increases surface area which decreases the number ofcollisions per unit area which decreases the pressure). Therefore, volume and temperature are directlyrelated at constant n and P (Charles’s law).

19. The kinetic molecular theory assumes that gas particles do not exert forces on each other and that gas

CHAPTER 5 GASES98

particles are volumeless. Real gas particles do exert attractive forces for each other, and real gasparticles do have volumes. A gas behaves most ideally at low pressures and high temperatures. Theeffect of attractive forces is minimized at high temperatures since the gas particles are moving veryrapidly. At low pressure, the container volume is relatively large (P and V are inversely related) sothe volume of the container taken up by the gas particles is negligible.

20. Molecules in the condensed phases (liquids and solids) are very close together. Molecules in thegaseous phase are very far apart. A sample of gas is mostly empty space. Therefore, one would

2expect 1 mol of H O(g) to occupy a huge volume as compared to 1 mol of .

21. Method 1: molar mass =

Determine the density of a gas at a measurable temperature and pressure, then use the above equationto determine the molar mass.

Method 2:

Determine the effusion rate of the unknown gas relative to some known gas; then use Graham’s lawof effusion (the above equation) to determine the molar mass.

22. a. At constant temperature, the average kinetic energy of the He gas sample will equal the average

2kinetic energy of the Cl gas sample. In order for the average kinetic energy to be the same, the

2smaller He atoms must move at a faster average velocity as compared to Cl . Therefore, plot A,

2with the slower average velocity, would be for the Cl sample, and plot B would be for the Hesample. Note the average velocity in each plot is a little past the top peak.

2b. As temperature increases, the average velocity of a gas will increase. Plot A would be for O (g)

2at 273 K and plot B, with the faster average velocity, would be for O (g) at 1273 K.

2Since a gas behaves more ideally at higher temperatures, O (g) at 1273 K would behave mostideally.

23. Rigid container (constant volume): As reactants are converted to products, the mol of gas particlespresent decrease by one-half. As n decreases, the pressure will decrease (by one-half). Density is themass per unit volume. Mass is conserved in a chemical reaction, so the density of the gas will notchange since mass and volume do not change.

Flexible container (constant pressure): Pressure is constant since the container changes volume in orderto keep a constant pressure. As the mol of gas particles decrease by a factor of 2, the volume of thecontainer will decrease (by one-half). We have the same mass of gas in a smaller volume, so the gasdensity will increase (is doubled).

24. a. Containers ii, iv, vi, and viii have volumes twice that of containers i, iii, v, and vii. Containers iii,iv, vii, and viii have twice the number of molecules (mol) present as compared to containers i, ii,

CHAPTER 5 GASES 99

v, and vi. The container with the lowest pressure will be the one which has the fewest mol of gaspresent in the largest volume (containers ii and vi both have the lowest P). The smallest containerwith the most mol of gas present will have the highest pressure (containers iii and vii both havethe highest P). All the other containers (i, iv, v and viii) will have the same pressure between thetwo extremes. The order is: ii = vi < i = iv = v = viii < iii = vii.

b. All have the same average kinetic energy since the temperature is the same in each container.Only the temperature determines the average kinetic energy.

c. The least dense gas will be container ii since it has the fewest of the lighter Ne atoms present inthe largest volume. Container vii has the most dense gas since the largest number of the heavierAr atoms are present in the smallest volume. To figure out the ordering for the other containers,

1we will calculate the relative density of each. In the table below, m equals the mass of Ne in

1 1container i, V equals the volume of container i, and d equals the density of the gas in containeri.

Container

i ii iii iv v vi vii viii

mass,volume

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1m , V m , 2V 2m , V 2m , 2V 2m , V 2m , 2V 4m , V 4m , 2V

density

1 1 1 1 = d = 2d = d = 2d

1= d

1 1= 4d = 2d

From the table, the order of gas density is: ii < i = iv = vi < iii = v = viii < vii

rmsd. µ = (3 RT/M) ; the root mean square velocity only depends on the temperature and the molar1/2

mass. Since T is constant, the heavier argon molecules will have the slower root mean squarevelocity as compared to the neon molecules. The order is: v = vi = vii = viii < i = ii = iii = iv.

Exercises

Pressure

25. a. 4.8 atm × = 3.6 × 10 mm Hg; b. 3.6 × 10 mm Hg × = 3.6 × 10 torr3 3 3

c. 4.8 atm × = 4.9 × 10 Pa; d. 4.8 atm × = 71 psi5

26. a. 2200 psi × = 150 atm; b. 150 atm × = 15 MPa

c. 150 atm × = 1.1 × 10 torr5

27. 6.5 cm × = 65 mm Hg or 65 torr; 65 torr × = 8.6 × 10 atm-2

CHAPTER 5 GASES100

8.6 × 10 atm = = 8.7 × 10 Pa-2 3

28. 20.0 in Hg × = 508 mm Hg = 508 torr; 508 torr × = 0.668 atm

29. If the levels of Hg in each arm of the manometer are equal, the pressure in the flask is equal toatmospheric pressure. When they are unequal, the difference in height in mm will be equal to thedifference in pressure in mm Hg between the flask and the atmosphere. Which level is higher will tellus whether the pressure in the flask is less than or greater than atmospheric.

flask atm flaska. P < P ; P = 760. - 118 = 642 mm Hg = 642 torr; 642 torr × = 0.845 atm

0.845 atm × = 8.56 × 10 Pa4

flask atm flaskb. P > P ; P = 760. torr + 215 torr = 975 torr; 975 torr × = 1.28 atm

1.28 atm × = 1.30 × 10 Pa5

flask flaskc. P = 635 - 118 = 517 torr; P = 635 + 215 = 850. torr

30. a. The pressure is proportional to the mass of the fluid. The mass is proportional to the volume ofthe column of fluid (or to the height of the column assuming the area of the column of fluid isconstant).

; In this case, the volume of silicon oil will be the same as the volume of Hg inExercise 5.29.

Hg oil oilV = ; V = V , , m =

Since P is proportional to the mass of liquid:

oil Hg Hg HgP = P = P = 0.0956 P

This conversion applies only to the column of liquid.

flaskP = 760. torr - (118 × 0.0956) torr = 760. - 11.3 = 749 torr

749 torr × = 0.986 atm; 0.986 atm × = 9.99 × 10 Pa4

CHAPTER 5 GASES 101

flaskP = 760. torr + (215 × 0.0956) torr = 760. + 20.6 = 781 torr

781 torr × = 1.03 atm; 1.03 atm × = 1.04 × 10 Pa5

b. If we are measuring the same pressure, the height of the silicon oil column would be 13.6 ÷ 1.30= 10.5 times the height of a mercury column. The advantage of using a less dense fluid thanmercury is in measuring small pressures. The quantity measured (length) will be larger for the lessdense fluid. Thus, the measurement will be more precise.

Gas Laws

1 1 2 231. From Boyle’s law, P V = P V at constant n and T.

2P = = 0.972 atm

As expected, as the volume increased, the pressure decreased.

32. The pressure exerted on the balloon is constant and the moles of gas present is constant. From

1 1 2 2Charles’s law, V /T = V /T at constant P and n.

2V = = 239 mL

As expected, as the temperature decreased, the volume decreased.

1 1 2 233. From Avogadro’s law, V /n = V /n at constant T and P.

2V = = 44.8 L

As expected, as the mol of gas present increases, volume increases.

2 2 434. As NO is converted completely into N O , the mol of gas present will decrease by one-half (from the2:1 mol ratio in the balanced equation). Using Avogadro’s law,

2 4 2N O (g) will occupy one-half the original volume of NO (g). This is expected since the mol of gas

2 2 4present decrease by one-half when NO is converted into N O .

35. a. PV = nRT, V = = 14.0 L

CHAPTER 5 GASES102

b. PV = nRT, n = = 4.72 × 10 mol-2

c. PV = nRT, T = = 678 K = 405°C

d. PV = nRT, P = = 133 atm

36. a. P = 7.74 × 10 Pa × = 0.0764 atm; T = 25 + 273 = 298 K3

PV = nRT, n = = 3.81 × 10 mol-5

b. PV = nRT, P = = 179 atm

c. V = = 3.6 L

d. T = = = 334 K = 61°C

37. n = = 1.11 × 10 mol3

For He: 1.11 × 10 mol × = 4.44 × 10 g He3 3

2 2For H : 1.11 × 10 mol × = 2.24 × 10 g H3 3

38. P = = 0.091 atm

CHAPTER 5 GASES 103

39. a. PV = nRT; 175 g Ar × = 4.38 mol Ar

T = = 69.6 K

b. PV = nRT, P = = 32.3 atm

240. 0.050 mL × × = 1.8 × 10 mol O-3

V = = 4.6 10 L = 46 mL-2

1 1 2 241. At constant n and T, PV = nRT = constant, P V = P V ; At sea level, P = 1.00 atm = 760. mm Hg.

2V = = 3.0 L

The balloon will burst at this pressure since the volume must expand beyond the 2.5 L limit of theballoon.

Note: To solve this problem, we did not have to convert the pressure units into atm; the units of mmHg canceled each other. In general, only convert units if you have to. Whenever the gas constant Ris not used to solve a problem, pressure and volume units must only be consistent, and not necessarilyin units of atm and L. The exception is temperature as T must always be converted to the Kelvinscale.

42. PV = nRT, n is constant. = nR = constant,

2 1V = 1.040 V , so

2P = = 100. psi × = 109 psi

43. PV = nRT, V and n constant, so = constant and .

2P = = 13.7 MPa × = 33.5 MPa

CHAPTER 5 GASES104

44. a. At constant n and V, = 40.0 atm × = 46.6 atm

b. = 273 K × = 1.02 × 10 K3

2c. T = = 273 K × = 171 K

45. PV = nRT, n constant; = nR = constant,

2P = = 710. torr × = 5.1 × 10 torr4

46. PV = nRT, V constant; = constant, ; mol × molar mass = mass

2mass = = 309 g Ar remains

2 47. PV = nRT, n is constant. = nR = constant, , V =

2V = 1.00 L × = 2.82 L; )V = 2.82 - 1.00 = 1.82 L

48. PV = nRT, P is constant.

= 0.921

Gas Density, Molar Mass, and Reaction Stoichiometry

49. STP: T = 273 K and P = 1.00 atm; n = = 6.7 × 10 mol He-2

CHAPTER 5 GASES 105

Or we can use the fact that at STP, 1 mol of an ideal gas occupies 22.42 L.

1.5 L × = 6.7 × 10 mol He; 6.7 × 10 mol He × = 0.27 g He-2 -2

2 2 2 250. CO (s) ÷ CO (g); 4.00 g CO × = 9.09 × 10 mol CO-2

2At STP, the molar volume of a gas is 22.42 L. 9.09 × 10 mol CO × = 2.04 L-2

6 12 6 2 2 251. C H O (s) + 6 O (g) ÷ 6 CO (g) + 6 H O(g)

6 12 6 25.00 g C H O × = 0.167 mol O

2V = = = 4.23 L O

Since T and P are constant, the volume of each gas will be directly proportional to the mol of gas

2 2 2present. The balanced equation says that equal mol of CO and H O will be produced as mol of O

2 2 2reacted. So the volumes of CO and H O produced will equal the volume of O reacted.

= 4.23 L

2 2 2 252. Since the solution is 50.0% H O by mass, the mass of H O decomposed is 125/2 = 62.5 g.

2 2 262.5 g H O × = 0.919 mol O

2V = = = 23.0 L O

53. = 2.1 × 10 mol5

2 22.1 × 10 mol H are in the balloon. This is 80.% of the total amount of H that had to be generated:5

CHAPTER 5 GASES106

2 2 20.80 (total mol H ) = 2.1 × 10 , total mol H = 2.6 × 10 mol H5 5

22.6 × 10 mol H × = 1.5 × 10 g Fe5 7

22.6 × 10 mol H × = 2.6 × 10 g of 98% sulfuric acid5 7

3 254. 2 NaN (s) ÷ 2 Na(s) + 3 N (g)

2 = 3.12 mol N needed to fill air bag.

3 2 3mass NaN reacted = 3.12 mol N × = 135 g NaN

3 2 2 2 3 2 2 255. CH OH + 3/2 O ÷ CO + 2 H O or 2 CH OH(l) + 3 O (g) ÷ 2 CO (g) + 4 H O(g)

350.0 mL × = 1.33 mol CH OH(l) available

2 = 1.85 mol O available

3 21.33 mol CH OH × = 2.00 mol O

2 32.00 mol O are required to react completely with all of the CH OH available. We only have 1.85 mol

2 2O , so O is limiting.

2 21.85 mol O × = 2.47 mol H O

56. For ammonia (in one minute):

3 = 1.1 × 10 mol NH3

3NH flows into the reactor at a rate of 1.1 × 10 mol/min.3

2For CO (in one minute):

2 = 6.6 × 10 mol CO2

CHAPTER 5 GASES 107

2CO flows into the reactor at 6.6 × 10 mol/min.2

3To react completely with 1.1 × 10 mol NH /min, we need:3

2 = 5.5 × 10 mol CO /min2

2Since 660 mol CO /min are present, ammonia is the limiting reagent.

= 3.3 × 10 g urea/min4

4 3 2 257. a. CH (g) + NH (g) + O (g) ÷ HCN(g) + H O(g); Balancing H first, then O, gives:

4 3 2 2 4 3 2 2CH + NH + O ÷ HCN + 3 H O or 2 CH (g) + 2 NH (g) + 3 O (g) ÷ 2 HCN(g) + 6 H O(g)

b. PV = nRT, T and P constant;

Since the volumes are all measured at constant T and P, the volumes of gas present are directlyproportional to the mol of gas present (Avogadro’s law). Because Avogadro’s law applies, the

4balanced reaction gives mol relationships as well as volume relationships. Therefore, 2 L of CH ,

3 22 L of NH and 3 L of O are required by the balanced equation for the production of 2 L of HCN.

4 3 2 2The actual volume ratio is 20.0 L CH :20.0 L NH :20.0 L O (or 1:1:1). The volume of O

4 3required to react with all of the CH and NH present is 20.0 L ×(3/2) = 30.0 L. Since only 20.0

2 2L of O are present, O is the limiting reagent. The volume of HCN produced is:

220.0 L O × = 13.3 L HCN

58. Since P and T are constant, V and n are directly proportional. The balanced equation requires 2 L of

2H to react with 1 L of CO (2:1 volume ratio due to 2:1 mol ratio in balanced equation). The actualvolume ratio present in one minute is 16.0 L/25.0 L = 0.640 (0.640:1). Since the actual volume ratio

2 3present is smaller than the required volume ratio, H is the limiting reactant. The volume of CH OH

2produced at STP will be one-half the volume of H reacted due to the 1:2 mol ratio in the balanced

3equation. In one minute, 16.0 L/2 = 8.00 L CH OH are produced (theoretical yield).

3= = 0.357 mol CH OH in one minute

3 30.357 mol CH OH × = 11.4 g CH OH (theoretical yield per minute)

% yield = × 100 = 46.5% yield

CHAPTER 5 GASES108

59. One of the equations developed in the text to determine molar mass is:

molar mass = where d = density in units of g/L

molar mass = = 42.1 g/mol

2The empirical formula mass of CH = 12.01 + 2(1.008) = 14.03 g/mol.

3 6 = 3.00; Molecular formula = C H

60. P × (molar mass) = dRT, d = , P × (molar mass) = × RT

Molar mass = M = = 96.9 g/mol

2 2 2Mass of CHCl . 12.0 + 1.0 + 35.5 = 48.5; = 2.00; Molecular formula is C H Cl .

61. P × (molar mass) = dRT, d = density =

4For SiCl , molar mass = M = 28.09 + 4(35.45) = 169.89 g/mol

4d = = 5.77 g/L for SiCl

3For SiHCl , molar mass = M = 28.09 + 1.008 + 3(35.45) = 135.45 g/mol

3d = = 4.60 g/L for SiHCl

62. = 12.6 g/L

CHAPTER 5 GASES 109

Partial Pressure

63. = 1.1 atm

2With air present, the partial pressure of CO will still be 1.1 atm. The total pressure will be the sum

total airof the partial pressures, P = + P .

totalP = 1.1 atm + = 1.1 + 0.97 = 2.1 atm

2 2 He64. = 1.00 g H × = 0.496 mol H ; n = 1.00 g He × = 0.250 mol He

= 12.2 atm

HeP = = 6.15 atm; = 12.2 atm + 6.15 atm = 18.4 atm

1 1 2 265. Use the relationship P V = P V for each gas, since T and n for each gas is constant.

2 2For H : P = = 475 torr × = 317 torr

2 2For N : P = 0.200 atm × = 0.0667 atm; 0.0667 atm × = 50.7 torr

totalP = = 317 + 50.7 = 368 torr

2 266. For H : P = = 360. torr × = 240. torr

TOTP = = 320. torr - 240. torr = 80. torr

2 1For N : P = = 80. torr × = 240 torr

467. a. mol fraction CH =

CHAPTER 5 GASES110

totalb. PV = nRT, n = = 0.161 mol

4c. = 0.412 × 0.161 mol = 6.63 × 10 mol CH-2

4 46.63 × 10 mol CH × = 1.06 g CH-2

2 2 2 = 0.588 × 0.161 mol = 9.47 × 10 mol O ; 9.47 × 10 mol O × = 3.03 g O-2 -2

68. If we had 100.0 g of the gas, we would have 50.0 g He and 50.0 g Xe.

He He total XeP = P P = 0.970 × 600. torr = 582 torr; P = 600. - 582 = 18 torr

69. , 1.032 atm = = 1.032 - 0.042 = 0.990 atm

2 = = 9.56 × 10 mol H-3

29.56 × 10 mol H × = 0.625 g Zn-3

total total tot tot He70. To calculate the volume of gas, we can use P and n (V = n RT/P ) or we can use P

He He He He Heand n (V = n RT/P ). Since is unknown, we will use P and n .

He He HeP + = 1.00 atm = 760. torr = P + 23.8 torr, P = 736 torr

Hen = 0.586 g × = 0.146 mol He

= 3.69 L

3 271. 2 NaClO (s) ÷ 2 NaCl(s) + 3 O (g)

CHAPTER 5 GASES 111

2 = 2.22 × 10 mol O-3

3 2Mass NaClO decomposed = 2.22 × 10 mol O × = 0.158 g-3

3 NaClO

3Mass % NaClO = × 100 = 18.0%

272. 10.10 atm - 7.62 atm = 2.48 atm is the pressure of the amount of F reacted.

PV = nRT, V and T are constant. = constant,

2 4 = 2.00; So Xe + 2 F ÷ XeF

Kinetic Molecular Theory and Real Gases

avg avg73. (KE) = (3/2) RT; At 273 K: (KE) = × 273 K = 3.40 × 10 J/mol 3

avgAt 546 K: (KE) = × 546 K = 6.81 × 10 J/mol3

avg 474. (KE) = (3/2) RT. Since the kinetic energy depends only on temperature, CH (Exercise 5.73) and

2 2N at the same temperature will have the same average kinetic energy. So, for N , the average kineticenergy is 3.40 × 10 J/mol (at 273 K) and 6.81 × 10 J/mol (at 546 K).3 3

rms 475. u = , where R = and M = molar mass in kg = 1.604 × 10 kg/mol for CH-2

4 rmsFor CH at 273 K: u = = 652 m/s

rms 4Similarly u for CH at 546 K is 921 m/s.

2 rmsFor N at 273 K: u = = 493 m/s

CHAPTER 5 GASES112

2 rmsSimilarly for N at 546 K, u = 697 m/s.

rms76. u = ; = =

We want the root mean square velocities to be equal, and this occurs when =. The ratio of the temperatures is:

= = = 87.93

6The heavier UF molecules would need a temperature 87.93 times that of the He atoms in order for theroot mean square velocities to be equal.

ave77. KE = (3/2) RT and KE = (1/2) mv ; As the temperature increases, the average kinetic energy of the2

gas sample will increase. The average kinetic energy increases because the increased temperatureresults in an increase in the average velocity of the gas molecules.

78. a b c d

avg. KE inc dec same (KE % T) same

avg. velocity inc dec same ( same

coll. freq wall inc dec inc inc

Average kinetic energy and average velocity depend on T. As T increases, both average kinetic energyand average velocity increase. At constant T, both average kinetic energy and average velocity areconstant. The collision frequency is proportional to the average velocity (as velocity increases it takesless time to move to the next collision) and to the quantity n/V (as molecules per volume increase,collision frequency increases).

79. a. They will all have the same average kinetic energy since they are all at the same temperature.

2b. Flask C; H has the smallest molar mass. At constant T, the lightest molecules are the fastest (onthe average). This must be true in order for the average kinetic energies to be constant.

80. a. All the gases have the same average kinetic energy since they are all at the same temperature.

b. At constant T, the lighter the gas molecule, the faster the average velocity.

2 2 2Xe (131.3 g/mol) < Cl (70.90 g/mol) < O (32.00 g/mol) < H (2.016 g/mol)slowest fastest

CHAPTER 5 GASES 113

2 2c. At constant T, the lighter H molecules have a faster average velocity than the heavier Omolecules. As temperature increases, the average velocity of the gas molecules increases.

2 2Separate samples of H and O can only have the same average velocities if the temperature of the

2 2O sample is greater than the temperature of the H sample.

81. Graham’s law of effusion:

where M = molar mass; = 1.033

= 1.067, so M = 29.99 g/mol; Of the choices, the gas would be NO, nitrogen monoxide.

82.

1 1 = 0.502, 16.04 = (0.502) × M , M = 2

83. = 1.02; = 1.04

The relative rates of effusion of C O: C O: C O are 1.04: 1.02: 1.00.12 16 12 17 12 18

2Advantage: CO isn't as toxic as CO.

2Major disadvantages of using CO instead of CO:

21. Can get a mixture of oxygen isotopes in CO .

22. Some species, e.g., C O O and C O , would effuse (gaseously diffuse) at about the same12 16 18 12 17

rate since the masses are about equal. Thus, some species cannot be separated from eachother.

284. where M = molar mass; Let Gas (1) = He, Gas (2) = Cl

= 4.209, t = 19 min

CHAPTER 5 GASES114

85. a. P = = 12.24 atm

2b. × (V - nb) = nRT; For N : a = 1.39 atm L /mol and b = 0.0391 L/mol2 2

× (1.0000 L - 0.5000 × 0.0391 L) = 12.24 L atm

(P + 0.348 atm) × (0.9805 L) = 12.24 L atm

P = - 0.348 atm = 12.48 - 0.348 = 12.13 atm

c. The ideal gas law is high by 0.11 atm or × 100 = 0.91%.

86. a. P = = 1.224 atm

2b. × (V - nb) = nRT; For N : a = 1.39 atm L /mol and b = 0.0391 L/mol2 2

× (10.000 L - 0.5000 × 0.0391 L) = 12.24 L atm

(P + 0.00348 atm) × (10.000 L - 0.0196 L) = 12.24 L atm

P + 0.00348 atm = = 1.226 atm, P = 1.226 - 0.00348 = 1.223 atm

c. The results agree to ± 0.001 atm (0.08%).

d. In Exercise 5.85, the pressure is relatively high and there is a significant disagreement. In 5.86,the pressure is around 1 atm, and both gas law equations show better agreement. The ideal gaslaw is valid at relatively low pressures.

Atmospheric Chemistry

NO NO NO total87. P = 5 × 10 from Table 5.4. P = P × P = 5 × 10 × 1.0 atm = 5 × 10 atm-7 -7 -7

CHAPTER 5 GASES 115

PV = nRT, = 2 × 10 mol NO/L-8

= 1 × 10 molecules NO/cm13 3

He He He total88. P = 5.24 × 10 from Table 5.4. P = P × P = 5.24 × 10 × 1.0 atm = 5.2 × 10 atm-6 -6 -6

= 2.1 × 10 mol He/L-7

= 1.2 × 10 atoms He/cm14 3

89. At 100. km, T . - 75°C and P . 10 . 3 × 10 atm.-4.5 -5

PV = nRT, = nR = constant,

2V = = 4 × 10 L = 0.4 mL-4

90. At 15 km, T . -50°C and P = 0.1 atm. Use since n is constant.

2V = = 7 L

2 291. N (g) + O (g) ÷ 2 NO(g), automobile combustion or formed by lightning

2 2 22 NO(g) + O (g) ÷ 2 NO (g), reaction with atmospheric O

2 2 3 2 22 NO (g) + H O(l) ÷ HNO (aq) + HNO (aq), reaction with atmospheric H O

2 2S(s) + O (g) ÷ SO (g), combustion of coal

2 2 3 22 SO (g) + O (g) ÷ 2SO (g), reaction with atmospheric O

2 3 2 4 2H O(l) + SO (g) ÷ H SO (aq), reaction with atmospheric H O

3 3 3 2 2 292. 2 HNO (aq) + CaCO (s) ÷ Ca(NO ) (aq) + H O(l) + CO (g)

2 4 3 4 2 2H SO (aq) + CaCO (s) ÷ CaSO (aq) + H O(l) + CO (g)

CHAPTER 5 GASES116

Additional Exercises

93. a. PV = nRT b. PV = nRT c. PV = nRT

PV = Constant P = × T = Const × T T = × V = Const × V

d. PV = nRT e. P = f. PV = nRT

PV = Constant P = Constant × = nR = Constant

94. At constant T and P, Avogadro’s law applies; that is, equal volumes contain equal moles of molecules.In terms of balanced equations, we can say that mol ratios and volume ratios between the various

2 2reactants and products will be equal to each other. Br + 3 F ÷ 2 X; Two moles of X must contain

3two moles of Br and 6 moles of F; X must have the formula BrF .

x 295. Mn(s) + x HCl(g) ÷ MnCl (s) + H (g)

=

2mol Cl in compound = mol HCl = 0.100 mol H × = 0.200 mol Cl

= = = 4.00

CHAPTER 5 GASES 117

4The formula of compound is MnCl .

2 2 296. 2 H (g) + O (g) ÷ 2 H O(g); Since P and T are constant, volume ratios will equal mol

f i f i 2 2 2 2ratios (V /V = n /n ). Let x = mol H = mol O present initially. H will be limiting since a 2:1 H

2to O mol ratio is required by the balanced equation, but only a 1:1 mol ratio is present. Therefore,

2 2no H will be present after the reaction goes to completion. However, excess O (g) will be present as

2well as the H O(g) produced.

2 2 2mol O reacted = x mol H × = x/2 mol O

2 2 2 2mol O remaining = x mol O initially - x/2 mol O reacted = x/2 mol O

2 2 2mol H O produced = x mol H × = x mol H O

2 2Total mol gas initially = x mol H + x mol O = 2 x

2 2Total mol gas after reaction = x/2 mol O + x mol H O = 1.5 x

f i = 0.75; V /V = 0.75:l or 3:4

1 1 2 297. We will apply Boyle’s law to solve. PV = nRT = contstant, P V = P V

Let condition (1) correspond to He from the tank that can be used to fill balloons. We must leave

1 11.0 atm of He in the tank, so P = 200. atm - 1.00 = 199 atm and V = 15.0 L. Condition (2) will

2 2correspond to the filled balloons with P = 1.00 atm and V = N(2.00 L) where N is the number offilled balloons, each at a volume of 2.00 L.

199 atm × 15.0 L = 1.00 atm × N(2.00 L), N = 1492.5; We can’t fill 0.5 of a balloon, so N =1492 balloons or to 3 significant figures, 1490 balloons.

98. mol of He removed = = 7.16 × 10 mol-5

In the original flask, 7.16 × 10 mol of He exerted a partial pressure of 1.960 - 1.710 = 0.250 atm.-5

V = = 7.00 × 10 L = 7.00 mL-3

2 1 1 2 299. For O , n and T are constant, so P V = P V .

1P = = 785 torr × = 761 torr =

totP = , = 785 - 761 = 24 torr

CHAPTER 5 GASES118

3 5 3 9 2 2 2 2100. 4 C H N O (s) 6 12 CO (g) + 6 N (g) + 10 H O(g) + O (g); For every 4 mol of nitroglycerin reacted,12 + 6 + 10 + 1 = 29 mol of gas are produced.

3 5 3 9mol gas produced = 25.0 g C H N O × = 0.798 mol

totP = = = 5.06 atm

101. 1.00 × 10 kg Mo × = 1.04 × 10 mol Mo3 4

21.04 × 10 mol Mo × = 3.64 × 10 mol O4 4

2 = 8.66 × 10 L of O5

28.66 × 10 L O × = 4.1 × 10 L air5 6

21.04 × 10 mol Mo × = 3.12 × 10 mol H4 4

2 = 7.42 × 10 L of H5

3 2102. For NH : P = = 0.500 atm × = 0.333 atm

2 2For O : P = = 1.50 atm × = 0.500 atm

After the stopcock is opened, V and T will be constant, so P % n. The balanced equation requires:

= = 1.25

The actual ratio present is: = = 1.50

CHAPTER 5 GASES 119

3The actual ratio is larger than the required ratio, so NH in the denominator is limiting. Since equal

3mol of NO will be produced as NH reacted, the partial pressure of NO produced is 0.333 atm (thesame as reacted).

2 5103. 750. mL juice × = 90. mL C H OH present

2 5 290. mL C H OH × × = 1.5 mol CO

2The CO will occupy (825 - 750. =) 75 mL not occupied by the liquid (headspace).

= 490 atm

2 2Actually, enough CO will dissolve in the wine to lower the pressure of CO to a much morereasonable value.

104. PV = nRT, V and T are constant.

When V and T are constant, pressure is directly proportional to moles of gas present, and pressureratios are identical to mol ratios.

2 2 2 2At 25°C: 2 H (g) + O (g) ÷ 2 H O(l); H O(l) is produced.

2 2The balanced equation requires 2 mol H for every mol O reacted. The same ratio (2:1) holds true

2 2for pressure units. The actual pressure ratio present is 2 atm H to 3 atm O , well below the required

22:1 ratio. Therefore, H is the limiting reactant. The only gas present at 25°C after the reaction goes

2to completion will be the excess O .

2 2(reacted) = 2.00 atm H × = 1.00 atm O

2 total(excess) = (initially) - (reacted) = 3.00 atm - 1.00 atm = 2.00 atm O = P

2 2 2 2At 125°C: 2 H (g) + O (g) ÷ 2 H O(g); H O(g) is produced.

The major difference in the problem is that gaseous water is now a product, which will increase the

CHAPTER 5 GASES120

total pressure.

2 2 (produced) = 2.00 atm H × = 2.00 atm H O

total 2 2P = (excess) + (produced) = 2.00 atm O + 2.00 atm H O = 4.00 atm

5 7 2 3105. If Be , the formula is Be(C H O ) and M . 13.5 + 15(12) + 21(1) + 6(16) = 311 g/mol.3+

5 7 2 2If Be , the formula is Be(C H O ) and M . 9.0 + 10(12) + 14(1) + 4(16) = 207 g/mol.2+

Data Set I (M = dRT/P and d = mass/V):

M = = 209 g/mol

Data Set II:

M = = 202 g/mol

5 7 2 2These results are close to the expected value of 207 g/mol for Be(C H O ) . Thus, we conclude fromthese data that beryllium is a divalent element with an atomic mass of 9.0 amu.

106. Out of 100.00 g compounds, there are:

58.51 g C × = 4.872 mol C; = 2.001

7.37 g H × = 7.31 mol H; = 3.00

34.12 g N × = 2.435 mol N; = 1.000

2 3Empirical formula: C H N

2; Let Gas (1) = He; 3.20 = , M = 41.0 g/mol

2 3Empirical formula mass of C H N . 2(12.0) + 3(1.0) + 1(14.0) = 41.0. So, the molecular formula

2 3is also C H N.

107. , = 726 torr - 23.8 torr = 702 torr × = 0.924 atm

CHAPTER 5 GASES 121

2PV = nRT, = 1.20 × 10 mol N-3

Mass of N in compound = 1.20 × 10 mol × = 3.36 × 10 g-3 -2

% N = × 100 = 13.3% N

2108. 0.2766 g CO × = 7.548 × 10 g C; % C = × 100 = 73.78% C-2

20.0991 g H O × = 1.11× 10 g H; % H = × 100 = 10.9% H-2

2PV = nRT, = 1.23 × 10 mol N-3

21.23 × 10 mol N × = 3.45 × 10 g nitrogen-3 -2

% N = × 100 = 7.14% N

% O = 100.00 - (73.78 + 10.9 + 7.14) = 8.2% O

Out of 100.00 g of compound, there are:

73.78 g C × = 6.143 mol C; 7.14 g N × = 0.510 mol N

10.9 g H × = 10.8 mol H; 8.2 g O × = 0.51 mol O

12 21Dividing all values by 0.51 gives an empirical formula of C H NO.

= 392 g/mol

12 21Empirical formula mass of C H NO .195 g/mol and

24 42 2 2Thus, the molecular formula is C H N O .

109. At constant T, the lighter the gas molecules, the faster the average velocity. Therefore, the pressure

CHAPTER 5 GASES122

2will increase initially because the lighter H molecules will effuse into container A faster than air willescape. However, the pressures will eventually equalize once the gases have had time to mixthoroughly.

3 8110. The van der Waals' constant b is a measure of the size of the molecule. Thus, C H should have thelargest value of b since it has the largest molar mass (size).

2 2 2 4111. The values of a are: H , ; CO , 3.59; N , 1.39; CH , 2.25

2Since a is a measure of interparticle attractions, the attractions are greatest for CO .

Challenge Problems

112. PV = nRT, V and T are constant.

We will do this limiting reagent problem using an alternative method. Let's calculate the partial

3 3pressure of C H N that can be produced from each of the starting materials assuming each reactantis limiting. The reactant that produces the smallest amount of product will run out first and is thelimiting reagent.

3 6 3 6 = 0.500 MPa C H × = 0.500 MPa if C H is limiting.

3 3 = 0.800 MPa NH × = 0.800 MPa if NH is limiting.

2 2 = 1.500 MPa O × = 1.000 MPa if O is limiting.

3 6 3 2Thus, C H is limiting. Although more product could be produced from NH and O , there is only

3 6 3 3 3 3enough C H to produce 0.500 MPa of C H N. The partial pressure of C H N after the reaction is:

0.500 × 10 Pa × = 4.94 atm6

3 3n = = 30.3 mol C H N

3 330.3 mol × = 1.61 × 10 g C H N can be produced.3

2 3 2 3113. BaO(s) + CO (g) ÷ BaCO (s); CaO(s) + CO (g) ÷ CaCO (s)

i 2 2n = = initial moles of CO = = 0.0595 mol CO

CHAPTER 5 GASES 123

f 2 2n = = final moles of CO = = 0.0182 mol CO

20.0595 - 0.0182 = 0.0413 mol CO reacted.

2Since each metal reacts 1:1 with CO , the mixture contains 0.0413 mol of BaO and CaO. The molarmasses of BaO and CaO are 153.3 g/mol and 56.08 g/mol, respectively.

Let x = g BaO and y = g CaO, so:

x + y = 5.14 g and mol

Solving by simultaneous equations:

x + 2.734 y = 6.33-x -y = -5.14

1.734 y = 1.19

y = 0.686 g CaO and 5.14 - y = x = 4.45 g BaO

% BaO = × 100 = 86.6% BaO; % CaO = 100.0 - 86.6 = 13.4% CaO

3 2 2 2114. Cr(s) + 3 HCl(aq) ÷ CrCl (aq) + 3/2 H (g); Zn(s) + 2 HCl(aq) ÷ ZnCl (aq) + H (g)

2 2mol H produced = n = = 9.02 × 10 mol H-3

2 2 29.02 × 10 mol H = mol H from Cr reaction + mol H from Zn reaction-3

2From the balanced equation: 9.02 × 10 mol H = mol Cr × (3/2) + mol Zn × 1-3

Let x = mass of Cr and y = mass of Zn, then:

x + y = 0.362 g and 9.02 × 10 = -3

We have two equations and two unknowns. Solving by simultaneous equations:

9.02 × 10 = 0.02885 x + 0.01530 y-3

CHAPTER 5 GASES124

-0.01530 × 0.362 = -0.01530 x - 0.01530 y 3.48 × 10 = 0.01355 x x = mass Cr = = 0.257 g-3

y = mass Zn = 0.362 g - 0.257 g = 0.105 g Zn; mass % Zn = × 100 = 29.0% Zn

4 2 2 2115. a. The reaction is: CH (g) + 2 O (g) ÷ CO (g) + 2 H O(g)

PV = nRT, = RT = constant,

2 4The balanced equation requires 2 mol O for every mol of CH that reacts. For three times as

2 4much oxygen, we would need 6 mol O per mol of CH reacted . Air is 21% mol

2 air 2percent O , so = 0.21 n . Therefore, the mol of air we would need to delivery the excess Oare:

air air = 0.21 n = 6 , n = 29 = 29

In one minute:

= 200. L × 29 × = 8.7 × 10 L air/min3

4 2 2b. If x moles of CH were reacted, then 6 x mol O were added, producing 0.950 x mol CO and

20.050 x mol of CO. In addition, 2 x mol H O must be produced to balance the hydrogens.

4 2 2 2 4 2 2CH (g) + 2 O (g) ÷ CO (g) + 2 H O(g); CH (g) + 3/2 O (g) ÷ CO(g) + 2 H O(g)

2Amount O reacted:

2 20.950 x mol CO × = 1.90 x mol O

20.050 x mol CO × = 0.075 x mol O

2 2Amount of O left in reaction mixture = 6.00 x - 1.90 x - 0.075 x = 4.03 x mol O

2 2 2Amount of N = 6.00 x mol O × = 22.6 x . 23 x mol N

The reaction mixture contains:

2 2 20.950 x mol CO + 0.050 x mol CO + 4.03 x mol O + 2.00 x mol H O

2 + 23 x mol N = 30. x total mol of gas

= 0.0017; = 0.032; = 0.13;

CHAPTER 5 GASES 125

= 0.067; = 0.77

116. The reactions are:

2 2 2C(s) + 1/2 O (g) ÷ CO(g) and C(s) + O (g) ÷ CO (g)

PV = nRT, P = n = n (constant)

Since the pressure has increased by 17.0%, the number of moles of gas has also increased by 17.0%.

final initialn = 1.170 n = 1.170 (5.00) = 5.85 mol gas =

= 5.00 (balancing moles of C)

= 0.85

2 2If all C was converted to CO , no O would be left. If all C was converted to CO, we would get 5 mol

2CO and 2.5 mol excess O in the reaction mixture. In the final mixture:

2 = 1.70 mol CO; 1.70 + = 5.00, = 3.30 mol CO

= 0.291; = 0.564; = 0.145 . 0.15

117. a. Volume of hot air: V = Br = B(2.50 m) = 65.4 m3 3 3

(Note: radius = diameter/2 = 5.00/2 = 2.50 m)

65.4 m × = 6.54 × 10 L3 4

n = = 2.31 × 10 mol air3

Mass of hot air = 2.31 × 10 mol × = 6.70 × 10 g3 4

Mass of air displaced:

CHAPTER 5 GASES126

n = = 2.66 × 10 mol air3

Mass = 2.66 × 10 mol × = 7.71 × 10 g of air displaced3 4

Lift = 7.71 × 10 g - 6.70 × 10 g = 1.01 × 10 g4 4 4

b. Mass of air displaced is the same, 7.71 × 10 g. Moles of He in balloon will be the same as moles4

of air displaced, 2.66 × 10 mol, since P, V and T are the same.3

Mass of He = 2.66 × 10 mol × = 1.06 × 10 g3 4

Lift = 7.71 × 10 g - 1.06 × 10 g = 6.65 × 10 g4 4 4

c. Mass of hot air:

n = = 1.95 × 10 mol air3

1.95 × 10 mol × = 5.66 × 10 g of hot air3 4

Mass of air displaced:

n = = 2.25 × 10 mol air3

2.25 × 10 mol × = 6.53 × 10 g of air displaced3 4

Lift = 6.53 × 10 g - 5.66 × 10 g = 8.7 × 10 g4 4 3

118.

At low P and high T, the molar volume of a gas will be relatively large. The an /V and an b/V terms2 3 2

become negligible because V is large. Since nb is the actual volume of the gas molecules themselves,then nb << V and the -nbP term is negligible compared to PV. Thus PV = nRT.

119. a. If we have 1.0 × 10 L of air, then there are 3.0 × 10 L of CO.6 2

CHAPTER 5 GASES 127

CO CO total CO COP = P × P ; P = since V % n; P = × 628 torr = 0.19 torr

COb. n = ; Assuming 1.0 cm of air = 1.0 mL = 1.0 × 10 L:3 -3

COn = = 1.1 × 10 mol CO-8

1.1 × 10 mol × = 6.6 × 10 molecules CO in the 1.0 cm of air-8 15 3

to t120. a. Initially, = 1.00 atm and the total pressure is 2.00 atm (P = ). The total

pressure after reaction will also be 2.00 atm since we have a constant pressure container. Since

2 2V and T are constant before the reaction takes place, there must be equal moles of N and H

2 2present initially. Let x = mol N = mol H that are present initially. From the balanced equation,

2 2 3 2 2N (g) + 3 H (g) 6 2 NH (g), H will be limiting since three times as many mol of H are required

2to react as compared to mol of N .

2After the reaction occurs, none of the H remains (it is the limiting reagent).

3 2mol NH produced = x mol H × = 2x/3

2 2mol N reacted = x mol H × = x/3

2 2 2 2mol N remaining = x mol N present initially - x/3 mol N reacted = 2x/3 mol N remaining

2 3After the reaction goes to completion, equal mol of N (g) and NH (g) are present (2x/3). Sinceequal mol are present, then the partial pressure of each gas must be equal

totP = 2.00 atm = ; Solving:

b. V % n since P and T are constant. The mol of gas present initially are:

= x + x = 2x mol

After reaction, the mol of gas present are:

= = 4x/3 mol

=

CHAPTER 5 GASES128

The volume of the container will be two-thirds the original volume so:

V = 2/3(15.0 L) = 10.0 L