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CHAPTER 13

Gases

The atmosphere is a homogeneous mixture (a solution) of gases.

Solids and liquids have essentially fixed volumes and are not able to be compressed easily. Gases
have volumes that depend on their conditions, and can be compressed or expanded by changes in
those conditions. Although the particles of matter in solids are essentially fixed in position (the so
lid is rigid), the particles in liquids and gases are free to move.

A small amount of water is added to a metal can and then the can is heated so as to boil the water
and fill the can with steam (gaseous water). The heat is then removed and the can is sealed off. As
the steam in the can cools, it condenses back to a liquid. Since the gas in the can has condensed, t
he pressure of the atmosphere is much larger than the pressure of gas in the can, and the atmosphe
ric pressure causes the can to collapse.

Figure 13.2 in the text shows a simple mercury barometer: a tube filled with mercury is inverted o
ver a reservoir (containing mercury) that is open to the atmosphere. When the tube is inverted, the
mercury falls to a level at which the pressure of the atmosphere is sufficient to support the colum
n of mercury. One standard atmosphere of pressure is taken to be the pressure capable of supporti
ng a column of mercury to a height of 760.0 mm above the reservoir level.

mix

Pressure units include mm Hg, torr, pascals, and psi. The unit “mm Hg” is derived from the baro
meter, since in a traditional mercury barometer, we measure the height of the mercury column (in
millimeters) above the reservoir of mercury.

1.00 atm = 760 torr = 760 mm Hg = 101.325 kPa = 14.70 psi

45.2 kPa 

1 atm

101.325 kPa

= 0.446 atm

755 mm Hg 

1 atm

760 mm Hg

= 0.993 atm

802 torr 

1 atm

101.325 kPa

760 torr

1 atm



= 107 kPa

1.04 atm 

760 mm Hg

1 atm

= 790. mm Hg

1.00 atm = 760 torr = 760 mm Hg = 101.325 kPa = 14.70 psi

14.9 psi ×

1 atm

14.70 psi

= 1.01 atm

272

Chapter 13: Gases

795 torr ×

1 atm

760 torr

= 1.05 atm

743 mm Hg ×

101.325 kPa

760 mm Hg

= 99.1 kPa

99,436 Pa ×

1 kPa

1000 Pa

= 99.436 kPa

1.00 atm = 760 torr = 760 mm Hg = 101.325 kPa = 14.70 psi

699 mm Hg 

1 atm

760 mm Hg

= 0.920 atm

18.2 psi 

760 mm Hg

14.70 psi

= 941 mm Hg

862 mm Hg = 862 torr

795 mm Hg 

14.70 psi

760 mm Hg

= 15.4 psi

10.

1.00 atm = 760 torr = 760 mm Hg = 101.325 kPa = 14.70 psi

17.3 psi ×

101.325 kPa

14.70 psi

= 119 kPa

1.15 atm ×

14.70 psi

1 atm

= 16.9 psi

4.25 atm ×

760 mm Hg

1 atm

= 3.23 × 103 mm Hg

224 psi ×

1 atm

14.70 psi

= 15.2 atm

11.

1.00 atm = 760 torr = 760 mm Hg = 101.325 kPa = 14.70 psi

1.54 × 105 Pa ×

1 atm

101,325 Pa

= 1.52 atm

1.21 atm ×

101,325 Pa

1 atm

= 1.23 × 105 Pa

97,325 Pa ×

760 mm Hg

101,325 Pa

= 730.14 mm Hg

1.32 kPa ×

1000 Pa

1 kPa

= 1.32 × 103 Pa

273

Chapter 13: Gases

12.

1.00 atm = 760 torr = 760 mm Hg = 101,325 Pa

774 torr 



= 1.03 × 105 Pa

0.965 atm 

= 9.78 × 104 Pa

112.5 kPa 

= 1.125 × 105 Pa

801 mm Hg 



= 1.07 × 105 Pa

13.

The volume of a sample of an ideal gas at constant temperature will decrease if the pressure on th
e gas is increased.

14.

Additional mercury increases the pressure on the gas sample, causing the volume of the gas upon
which the pressure is exerted to decrease (Boyle’s Law)

15.

pressure

16.

PV = k; P1V1 = P2V2

17.

P1 = 755 mm Hg

P2 = 780 mm Hg

V1 = 125 mL

V2 = ?

1 1

(755 mm Hg)(125 mL)

(780 mm Hg)



= 121 mL

P1 = 1.08 atm

P2 = 0.951 atm

V1 = 223 mL

V2 = ?

1 1

(1.08 atm)(223 mL)

(0.951 atm)



= 253 mL

P1 = 103 kPa

P2 = 121 kPa

V1 = 3.02 L

V2 = ?

1 1

(103 kPa)(3.02 L)

(121 kPa)



= 2.57 L

18.

P1 = 1.15 atm

P2 = 775 mm Hg = 1.020 atm

V1 = 375 mL

V2 = ?

1 1

(1.15 atm)(375 mL)

(1.020 atm)



= 423 mL

274

Chapter 13: Gases

P1 = 1.08 atm

P2 = 135 kPa = 1.33 atm

V1 = 195 mL

V2 = ?

1 1

(1.08 atm)(195 mL)

(1.33 atm)



= 158 mL

P1 = 131 kPa = 982.6 mm Hg

P2 = 765 mm Hg

V1 = 6.75 L

V2 = ?

1 1

(982.6 mm Hg)(6.75 L)

(765 mm Hg)



= 8.67 L

19.

P1 = 102.1 kPa

P2 = ? kPa

V1 = 19.3 L

V2 = 10.0 L

1 1

(102.1 kPa)(19.3 L)

(10.0 L)



= 197 kPa

P1 = 755 torr = 755 mm Hg

P2 = 761 mm Hg

V1 = 25.7 mL

V2 = ? mL

V2 =

1 1

(755 mm Hg)(25.7 mL)

(761 mm Hg)



= 25.5 mL

P1 = 1.05 atm

P2 = 112.2 kPa = 1.107 atm

V1 = 51.2 L

V2 = ?

V2 =

1 1

(1.05 atm)(51.2 L)

(1.107 atm)



= 48.6 L

20.

P1 = 785 mm Hg

P2 = 700. mm Hg

V1 = 53.2 mL

V2 = ?

= 59.7 mL

P1 = 1.67 atm

P2 = ?

V1 = 2.25 L

V2 = 2.00 L

= 1.88 atm

P1 = 695 mm Hg

P2 = 1.51 atm

V1 = 5.62 L

V2 = ?

= 3.40 L

275

Chapter 13: Gases

21.

P1 = 1.02 atm

P2 = 2.99 atm

V1 = 225 mL

V2 = ?

V2 =

1 1

(1.02 atm)(225 mL)

(2.99 atm)



= 76.8 mL

22.

P1 = P1

P2 = 2 × P1

V1 = 1.04 L

V2 = ? L

V2 =

1 1

( )(1.04 L)

(2 )





1.04 L

= 0.520 L

23.

P1 = 785 mm Hg

P2 = ?

V1 = 29.2 mL

V2 = 15.1 mL

P2 =

1 1

(785 mm Hg)(29.2 mL)

(15.1 mL)



= 1.52 × 103 mm Hg

24.

= 20.0 atm

25.

Absolute zero is the lowest temperature that can exist. Absolute zero is the temperature at which t
he volume of an ideal gas sample would be predicted to become zero. Absolute zero is the zero-p
oint on the Kelvin temperature scale (and corresponds to –273°C).

26.

Charles’s Law indicates that an ideal gas decreases by 1/273 of its volume for every degree Celsi
us its temperature is lowered. This means an ideal gas would approach a volume of zero at –273
C.

27.

directly

28.

V = kT; V1/T1 = V2/T2

29.

V1 = 1340 L

V2 = ? mL

T1 = 18°C = 291 K

T2 = 87°C = 360. K

30.

V1 = 375 mL

V2 = ? mL

T1 = 78C = 351 K

T2 = 22C = 295 K

V2 =

1 2

(375 mL)(295 K)

(351 K)

V T



= 315 mL

276

Chapter 13: Gases

31.

V1 = 2.03 L

V2 = 3.01 L

T1 = 24°C = 297 K

T2 = ?

T2 =

2 1

(3.01 L)(297 K)

(2.03 L)

V T



= 440 K = 167°C

V1 = 127 mL

V2 = ?

T1 = 273 K

T2 = 373 K

V2 =

1 2

(127 mL)(373 K)

(273 K)

V T



= 174 mL

V1 = 49.7 mL

V2 = ?

T1 = 34C= 307 K

T2 = 350 K

V2 =

1 2

(49.7 mL)(350 K)

(307 K)

V T



= 56.7 mL

32.

V1 = 25.0 L

V2 = 50.0 L

T1 = 0C = 273 K

T2 = ? C

T2 =

= 546 K = 273C

V1 = 247 mL

V2 = 255 mL

T1 = 25C = 298 K

T2 = ?C

T2 =

= 308 K = 35C

V1 = 1.00 mL

V2 = ? mL

T1 = 2272C = 2545 K

T2 = 25C = 298 K

V2 =

= 0.117 mL

33.

V1 = 9.14 L

V2 = ?

T1 = 24C = 297 K

T2 = 48C = 321 K

V2 =

1 2

(9.14 L)(321 K)

(297 K)

V T



= 9.88 L

V1 = 24.9 mL

V2 = 49.9 mL

T1 = –12C = 261 K

T2 = ?

T2 =

2 1

(49.9 mL)(261 K)

(24.9 mL)

V T



= 523 K = 250.°C

277

Chapter 13: Gases

V1 = 925 mL

V2 = ?

T1 = 25 K

T2 = 273 K

V2 =

1 2

(925 mL)(273 K)

(25 K)

V T



= 1.01 × 104 mL

34.

V1 = 2.01  102 L

V2 = 5.00 L

T1 = 1150C = 1423 K

T2 = ?C

T2 =

2 1

(5.00 L)(1423 K)

(201 L)

V T



= 35.4 K = –238C

V1 = 44.2 mL

V2 = ? mL

T1 = 298 K

T2 = 0

V2 =

1 2

(44.2 mL)(0 K)

(298 K)

V T



= 0 mL (0 K is absolute zero)

V1 = 44.2 mL

V2 = ? mL

T1 = 298 K

T2 = 0C = 273 K

V2 =

1 2

(44.2 mL)(273 K)

(298 K)

V T



= 40.5 mL

35.

V1 = 1.25L

V2 = ? mL

T1 = 291 K

T2 = 78 K

V2 =

1 2

(1.25 L)(78 K)

(291 K)

V T



= 0.335 L = 0.34 L

36.

V2 =

1 2

(125 mL)(250 K)

(450 K)

V T



= 69.4 mL = 69 mL to two significant figures

37.

24C + 273 = 297 K

72C + 273 = 345 K

V2 =

1 2

(375 mL)(345 K)

(297 K)

V T



= 436 mL

38.

V2 =

1 2

V T

Temp, C

Volume, mL

124

120.

117

113

110

107

103

100

39.

directly

40.

V = an; V1/n1 = V2/n2

278

Chapter 13: Gases

41.

V1/n1 = V2/n2

V1 = 242 mL

V2 = ? L

n1 = 0.00901 mol

n2 = 0.00703 mol

242 mL 

0.00703 mol
0.00901 mol

= 189 mL

42.

V = an; V1/n1 = V2/n2

Since 2.08 g of chlorine contains twice the number of moles of gas contained in the 1.04 g
sample, the volume of the 2.08 g sample will be twice as large = 1744 (1.74 × 103) mL

43.

V1 = 100. L

V2 = ? L

n1 = 3.25 mol

n2 = 14.15 mol

100. L 

14.15 mol

3.25 mol

= 435 L

44.

molar mass of Ar = 39.95 g

2.71 g Ar ×

1 mol

39.95 g

= 0.0678 mol Ar

4.21 L ×

1.29 mol

0.0678 mol

= 80.1 L

45.

Although the definition may seem a little strange, an ideal gas is one which obeys the ideal gas la
w, PV = nRT, exactly. That is, if knowledge of three of the properties of a gas (pressure, volume,
temperature, and amount) leads to the correct value for the fourth property when using this equati
on, then the gas under study is an ideal gas.

46.

Real gases most closely approach ideal gas behavior under conditions of relatively high temperat
ures (0°C or higher) and relatively low pressures (1 atm or lower).

47.

For an ideal gas, PV = nRT is true under any conditions. Consider a particular sample of gas (so t
hat n remains constant) at a particular fixed temperature (so that T remains constant also). Suppos
e that at pressure P1 the volume of the gas sample is V1. Then for this set of conditions, the ideal g
as equation would be given by

P1V1 = nRT.

If we then change the pressure of the gas sample to a new pressure P2, the volume of the gas
sample changes to a new volume V2. For this new set of conditions, the ideal gas equation would
be given by

P2V2 = nRT.

As the right-hand sides of these equations are equal to the same quantity (because we defined n
and T to be constant), then the left-hand sides of the equations must also be equal, and we obtain
the usual form of Boyle’s law.

P1V1 = P2V2

279

Chapter 13: Gases

48.

For an ideal gas, PV = nRT is true under any conditions. Consider a particular sample of gas (so t
hat n remains constant) at a particular fixed pressure (so that P remains constant also). Suppose th
at at temperature T1 the volume of the gas sample is V1. Then for this set of conditions, the ideal g
as equation would be given by

PV1 = nRT1.

If we then change the temperature of the gas sample to a new temperature T2, the volume of the
gas sample changes to a new volume V2. For this new set of conditions, the ideal gas equation
would be given by

PV2 = nRT2.

If we make a ratio of these two expressions for the ideal gas equation for this gas sample, and
cancel out terms that are constant for this situation (P, n, and R) we get

nRT



This can be rearranged to the familiar form of Charles’s law



49.

P = 782.4 mm Hg = 1.029 atm; T = 26.2C = 299 K

V =

(0.1021 mol)(0.08206 L atm mol K )(299 K)

(1.029 atm)

nRT





= 2.44 L

V = 27.5 mL = 0.0275 L; T = 16.6C = 289.6 K (290. K)

P =

(0.007812 mol)(0.08206 L atm mol K )(290 K)

(0.0275 L)

nRT





= 6.75 atm

6.75 atm ×

760 mm Hg

1 atm

= 5.13 × 103 mm Hg

V = 45.2 mL = 0.0452 L

T =

(1.045 atm)(0.0452 L)

(0.002241 mol)(0.08206 L atm mol K )





= 257 K

T = 257 K – 273 K = –16C

50.

P = 782 mm Hg = 1.03 atm;

T = 27C = 300 K

V =

(0.210 mol)(0.08206 L atm mol K )(300 K)

(1.03 atm)

nRT





= 5.02 L

280

Chapter 13: Gases

V = 644 mL = 0.644 L

P =

(0.0921 mol)(0.08206 L atm mol K )(303 K)

(0.644 L)

nRT





= 3.56 atm

= 2.70  103 mm Hg

P = 745 mm = 0.980 atm

T =

(0.980 atm)(11.2 L)

(0.401 mol)(0.08206 L atm mol K )





= 334 K

51.

molar mass Ne = 20.18 g; 25°C = 298 K



(1.02 atm)(5.00 L)

(0.08206 L atm mol K )(298 K)





0.2086 mol Ne

0.2086 mol Ne ×

20.18 g Ne

1 mol Ne

= 4.21 g Ne

52.

Molar mass of O2 = 32.00 g; 56.2 kg = 5.62 × 104 g

T = 21°C = 294 K

53.

molar mass of He = 4.003 g; 100C = 373 K; 785 mm Hg = 1.033 atm

2.04 g He ×

1 mol He

4.003 g He

= 0.5096 mol He

V =

(0.5096 mol)(0.08206 L atm mol K )(373 K)

(1.033 atm)

nRT





= 15.1 L

54.

molar mass Ne = 20.18 g; 5.00 g = 0.248 mol

T =

= 379 K = 106°C

55.

T = 25C + 273 = 298 K; molar masses: He, 4.003 g; O2, 32.00 g



(255 atm)(100.0 L)

(0.08206 L atm mol K )(298 K)





1043 mol = 1.04  103 mol

1.04  103 mol of either He or O2 would be needed.

1.04  103 mol He 

4.003 g He

1 mol He

= 4.16  103 g He

281

Chapter 13: Gases

1.04  103 mol O2 

32.00 g O

1 mol O

= 3.33  104 g O2

56.

molar mass Ne = 20.18 g; 25C = 298 K; 50C = 323 K

1.25 g Ne ×

1 mol

20.18 g

= 0.06194 mol

P =

(0.06194 mol)(0.08206 L atm mol K )(298 K)

(10.1 L)

nRT





= 0.150 atm

P =

(0.06194 mol)(0.08206 L atm mol K )(323 K)

(10.1 L)

nRT





= 0.163 atm

57.

molar mass Ne = 20.18 g; P = 500. torr = 0.6579 atm

1.0 g Ne 

1 mol

20.18 g

= 0.0496 mol Ne

T =

= 809 K (810 K)

58.

molar mass O2 = 32.00 g; 784 mm Hg = 1.032 atm

4.25 g O2 ×

1 mol O

32.00 g O

= 0.1328 mol

T =

(1.032 atm)(2.51 L)

(0.1328 mol)(0.08206 L atm mol K )





= 238 K = –35C

59.

5.0 kg = 5.0  103 g; molar mass Ne = 20.18 g

5.0 × 103 g Ne ×

1 mol Ne

20.18 g Ne

= 247.8 mol Ne

60.

Molar masses: He, 4.003 g; Ar, 39.95 g

4.15 g He ×

1 mol He

4.003 g He

= 1.037 mol He

56.2 g Ar ×

1 mol Ar

39.95 g Ar

= 1.407 mol Ar

For He, P =

(1.037 mol)(0.08206 L atm mol K )(298 K)

(5.00 L)

nRT





= 5.07 atm

282

Chapter 13: Gases

For Ar, P =

(1.407 mol)(0.08206 L atm mol K )(303 K)

(10.00 L)

nRT





= 3.50 atm

The helium is at a higher pressure than the argon.

61.

P1 = 1.01 atm

P2 = ? atm

V1 = 24.3 mL

V2 = 15.2 mL

T1 = 25C = 298 K

T2 = 50C = 323 K

P2 =

2 1 1

1 2

(323 K)(1.01 atm)(24.3 mL)

(298 K)(15.2 mL)

T PV

= 1.75 atm

62.

molar mass Ar = 39.95 g; 29C = 302 K; 42C = 315 K

1.29 g Ar ×

1 mol Ar

39.95 g Ar

= 0.03229 mol Ar

63.

P1 = 1.05 atm

P2 = 0.997 atm

V1 = 459 mL

V2 = ? mL

T1 = 27C = 300. K

T2 = 15C = 288 K

V2 =

2 1 1

1 2

(288 K)(1.05 atm)(459 mL)

(300 K)(0.997 atm)

T PV

T P

= 464 mL

64.

Molar mass of H2O = 18.02 g; 2.0 mL = 0.0020 L; 225°C = 498 K

0.250 g H2O ×

1 mol H O

18.02 g H O

= 0.01387 mol H2O

P =

(0.01387 mol)(0.08206 L atm mol K )(498 K)

(0.0020 L)

nRT





= 283 atm = 2.8 × 102 atm

65.

In deriving the ideal gas law, we assume that the molecules of gas occupy no volume, and that the
molecules do not interact with each other. Under these conditions, there is no difference between
gas molecules of different substances (other than their masses) as far as the bulk behavior of the g
as is concerned. Each gas behaves independently of other gases present, and the overall properties
of the sample are determined by the overall quantity of gas present.

Ptotal = P1 + P2 + ... Pn where n is the number of individual gases present in the mixture.

66.

As a gas is bubbled through water, the bubbles of gas become saturated with water vapor, thus for
ming a gaseous mixture. The total pressure in a sample of gas that has been collected by bubbling
through water is made up of two components: the pressure of the gas of interest and the pressure

283

Chapter 13: Gases

of water vapor. The partial pressure of the gas of interest is then the total pressure of the sample
minus the vapor pressure of water.

67.

molar masses: He, 4.003 g; Ne, 20.18 g; 25C = 298 K

2.41 g He 

1 mol He

4.003 g He

= 0.602 mol He

2.79 g Ne 

1 mol Ne

20.18 g Ne

= 0.138 mol Ne

Phelium =

(0.602 mol)(0.08206 L atm mol K )(298 K)

(1.04 L)

helium





= 14.2 atm

Pneon =

(0.138 mol)(0.08206 L atm mol K )(298 K)

(1.04 L)

neon





= 3.25 atm

Ptotal = 14.2 atm + 3.25 atm = 17.5 atm

68.

molar masses: Ne, 20.18 g; Ar, 39.95 g; 27C = 300 K

1.28 g Ne ×

1 mol Ne

20.18 g Ne

= 0.06343 mol Ne

2.49 g Ar ×

1 mol Ar

39.95 g Ar

= 0.06233 mol Ar

Pneon =

(0.06343 mol)(0.08206 L atm mol K )(300 K)

(9.87 L)

neon





= 0.1582 atm

Pargon =

argon

(0.06233 mol)(0.08206 L atm mol K )(300 K)

(9.87 L)





= 0.1555 atm

Ptotal = 0.1582 atm + 0.1555 atm = 0.314 atm

69.

52.5 g O2 = 1.641 mol O2; 65.1 g CO2 = 1.479 mol CO2; total moles = 3.120 mol

Poxygen = 9.21 atm 

1.641 mol O

3.120 mol total

= 4.84 atm O2

Pcarbon dioxide = 9.21 atm 

1.479 mol CO

3.120 mol total

= 4.37 atm CO2

Once the partial pressure of O2 had been calculated, we also could have calculated the partial
pressure of CO2 as the difference between the total pressure (9.21 atm) and the partial pressure of
O2 (4.84 atm).

70.

925 mm Hg = 1.217 atm; 26C = 299 K; molar masses: Ne, 20.18 g; Ar, 39.95 g



(1.217 atm)(3.00 L)

(0.08206 L atm mol K )(299 K)





0.1488 mol

284

Chapter 13: Gases

The number of moles of an ideal gas required to fill a given-sized container to a particular
pressure at a particular temperature does not depend on the specific identity of the gas. So 0.1488
mol of Ne gas or 0.1488 mol of Ar gas would give the same pressure in the same flask at the
same temperature.

mass Ne = 0.1488 mol Ne ×

20.18 g Ne

1 mol Ne

= 3.00 g Ne

mass Ar = 0.1488 mol Ar ×

39.95 g Ar

1 mol Ar

= 5.94 g Ar

71.

Poxygen = Ptotal – Pwater vapor = 772 – 26.7 = 745 torr

72.

The total pressure of the gases inside the container is 1.00 atm. The total number of gas particles i
nside the container is 10. Let x equal the pressure of each gas particle.

1.00 atm = 10x
x = 0.100 atm

There are 2 argon particles, 3 neon particles, and 5 helium particles present in the container.
Therefore, PTotal = PAr + PNe + PHe = 2x + 3x + 5x = 10x.
PAr = 2(0.100 atm) = 0.20 atm (taking into account significant figures when these pressures are added)
PNe = 3(0.100 atm) = 0.30 atm (taking into account significant figures when these pressures are added)
PHe = 5(0.100 atm) = 0.50 atm (taking into account significant figures when these pressures are added)

73.

Poxygen = Ptotal – Pwater vapor = (755 – 23) mm Hg = 732 mm Hg = 0.9632 atm

T = 24C + 273 = 297 K; V = 500. mL = 0.500 L



(0.9632 atm)(0.500 L)

(0.08206 L atm mol K )(297 K)





= 1.98  10–2 mol O2

74.

1.032 atm = 784.3 mm Hg; molar mass of Zn = 65.38 g

Phydrogen = 784.3 mm Hg – 32 mm Hg = 752.3 mm Hg = 0.990 atm

V = 240 mL = 0.240 L; T = 30C + 273 = 303 K

hydrogen



(0.990 atm)(0.240 L)

(0.08206 L atm mol K )(303 K)





0.00956 mol hydrogen

0.00956 mol H2 

1 mol Zn

1 mol H

= 0.00956 mol of Zn must have reacted

0.00956 mol Zn 

65.38 g Zn

1 mol Zn

= 0.625 g Zn must have reacted

75.

A law is a statement that precisely expresses generally observed behavior. A theory consists of a s
et of assumptions/hypotheses that is put forth to explain the observed behavior of matter. Theorie
s attempt to explain natural laws.

76.

A theory is successful if it explains known experimental observations. Theories that have been su
ccessful in the past may not be successful in the future (for example, as technology evolves, more
sophisticated experiments may be possible in the future).

285

Chapter 13: Gases

77.

assume that the volume of the molecules themselves in a gas sample is negligible compared to the
bulk volume of the gas sample: this helps us to explain why gases are so compressible.

78.

pressure

79.

kinetic energy

80.

81.

The temperature of a gas reflects, on average, how rapidly the molecules in the gas are moving. A
t high temperatures, the particles are moving very fast and collide with the walls of the container f
requently, whereas at low temperatures, the molecules are moving more slowly and collide with t
he walls of the container infrequently. The Kelvin temperature is directly proportional to the aver
age kinetic energy of the particles in a gas.

82.

If the temperature of a sample of gas is increased, the average kinetic energy of the particles of ga
s increases. This means that the speeds of the particles increase. If the particles have a higher spee
d, they will hit the walls of the container more frequently and with greater force, thereby increasin
g the pressure.

83.

The molar volume of a gas is the volume occupied by one mole of the gas under a particular set o
f temperature and pressure conditions (usually STP: 0C, 1 atm). When measured under the same
conditions, all ideal gases have the same molar volume (22.4 L at STP).

84.

Standard Temperature and Pressure, STP = 0C, 1 atm pressure. These conditions were chosen be
cause they are easy to attain and reproduce experimentally. The barometric pressure within a labo
ratory is likely to be near 1 atm most days, and 0C can be attained with a simple ice bath.

85.

molar masses: CaO, 56.08 g; CO2; 44.01 g

1.25 g CaO ×

1 mol CaO

56.08 g CaO

= 0.02229 mol CaO

From the balanced chemical equation, 0.02229 mol CaO would absorb 0.02229 mol CO2

0.02229 mol CO2 ×

44.01 g CO

1 mol CO

= 0.981 g CO2

Since one mole of an idea gas occupies 22.4 L at STP, 0.02229 mol of CO2 would occupy

0.02229 mol CO2 ×

22.4 L

1 mol

= 0.499 L at STP

86.

Molar mass of C = 12.01 g; 25°C = 298 K

1.25 g C 

1 mol

12.01 g

= 0.1041 mol C

Since the balanced chemical equation shows a 1:1 stoichiometric relationship between C and O2,
then 0.1041 mol of O2 will be needed

286

Chapter 13: Gases

V =

(0.1041 mol)(0.08206 L atm mol K )(298 K)

(1.02 atm)

nRT





= 2.50 L O2

87.

2C8H18(l) + 25O2(g)  16CO2(g) + 18H20(l)

molar mass C8H18 = 114.2 g

10.0 g C8H18 

1 mol C H

114.2 g C H

= 0.08757 mol C8H18

0.08757 mol C8H18 

25 mol O

2 mol C H

= 1.095 mol O2

At STP, one mole of an ideal gas occupies 22.4 L of volume.

1.095 mol O2 ×

22.4 L

mol

= 24.5 L O2 at STP

88.

Molar mass of Mg = 24.31 g; STP: 1.00 atm, 273 K

1.02 g Mg 

1 mol

24.31 g

= 0.0420 mol Mg

As the coefficients for Mg and Cl2 in the balanced equation are the same, for 0.0420 mol of Mg
reacting we will need 0.0420 mol of Cl2.

V = 0.0420 mol Cl2 ×

22.4 L

1 mol

= 0.940 L Cl2 at STP.

89.

27C = 300 K; 26 °C = 299 K; molar mass NH4Cl = 53.49 g

mol NH3 present =



(1.02 atm)(4.21 L)

(0.08206 L atm mol K )(300 K)





= 0.174 mol NH3

mol HCl present =



(0.998 atm)(5.35 L)

(0.08206 L atm mol K )(299 K)





= 0.218 mol HCl

NH3 and HCl react on a 1:1 basis: NH3 is the limiting reactant.

0.174 mol NH3

1 mol NH Cl 53.49 g NH Cl

1 mol NH

1 mol NH Cl



= 9.31 g NH4Cl produced

90.

molar mass CaC2 = 64.10 g; 25C = 298 K

2.49 g CaC2 ×

1 mol

64.10 g

= 0.03885 mol CaC2

From the balanced chemical equation for the reaction, 0.03885 mol of CaC2 reacting completely
would generate 0.03885 mol of acetylene, C2H2

V =

(0.03885 mol)(0.08206 L atm mol K )(298 K)

(1.01 atm)

nRT





= 0.941 L

287

Chapter 13: Gases

V =

(0.03885 mol)(0.08206 L atm mol K )(273 K)

(1.00 atm)

nRT





= 0.870 L at STP

91.

CuSO45H2O(s)  CuSO4(s) + 5H2O(g)

350C = 623 K; molar mass CuSO45H2O = 249.7 g

5.00 g CuSO4=5H2O 

1 mol CuSO 5H O

249.7 g CuSO 5H O

= 0.02002 mol CuSO45H2O

0.02002 mol CuSO45H2O 

5 mol H O

1 mol CuSO 5H O

= 0.1001 mol H2O

V =

(0.1001 mol)(0.08206 L atm mol K )(623 K)

(1.04 atm)

nRT





= 4.92 L H2O

92.

Molar mass of Mg3N2 = 100.95 g; T = 24C = 297 K; P = 752 mm Hg = 0.989 atm

10.3 g Mg3N2 

1 mol

100.95 g

= 0.102 mol Mg3N2

From the balanced chemical equation, the amount of NH3 produced will be

0.102 mol Mg3N2 

2 mol NH

1 mol Mg N

= 0.204 mol NH3

V =

(0.204 mol)(0.08206 L atm mol K )(297 K)

(0.989 atm)

nRT





= 5.03 L

This assumes that the ammonia was collected dry.

93.

Molar masses: He, 4.003 g; H2, 2.016 g; 28C = 301 K

14.2 g He 

1 mol He

4.003 g He

= 3.55 mol He

21.6 g H2 

1 mol H

2.016 g H

= 10.7 mol H2

total moles = 3.55 mol + 10.7 mol = 14.25 mol

V =

(14.25 mol)(0.08206 L atm mol K )(301 K)

(0.985 atm)

nRT





= 357 L

94.

6Na + N2 → 2Na3N

T = 28°C = 301 K

Determine the number of moles of each reactant present.

288

Chapter 13: Gases

Determine how many moles of Na3N would be produced assuming each reactant is
completely consumed.

Once 0.145 mol Na3N is produced, all of the Na is used up (making Na the limiting
reactant). Thus,

0.145 mol Na

3N 

82.98 g Na

1 mol Na

= 12.0 g Na

95.

P1 = 892 mm Hg

P2 = 1.00 atm = 760 mm Hg

V1 = 25.2 mL

V2 = ?

T1 = 95C + 273 = 368 K

T2 = 273 K

V2 =

2 1 1

1 2

(273 K)(892 mm Hg)(25.2 mL)

(368 K)(760 mm Hg)

T PV

T P



= 21.9 mL

96.

P1V1/T1 = P2V2/T2

V1 = 50.0 L

V2 = ? L

T1 = 20.°C = 293 K

T2 = 273 K

P1 = 742 torr

P2 = 760. torr (1 atm)

97.

molar masses: O2, 32.00 g; N2, 28.02 g; CO2, 44.01 g; Ne, 20.18 g

5.00 g O2 

1 mol O

32.00 g O

= 0.1563 mol O2

5.00 g N2 

1 mol N

28.02 g N

= 0.1784 mol N2

5.00 g CO2 

1 mol CO

44.01 g CO

= 0.1136 mol CO2

5.00 g Ne 

1 mol Ne

20.18 g Ne

= 0.2478 mol Ne

Total moles of gas = 0.1563 + 0.1784 + 0.1136 + 0.2478 = 0.6961 mol

289

Chapter 13: Gases

22.4 L is the volume occupied by one mole of any ideal gas at STP. This would apply even if the
gas sample is a mixture of individual gases.

0.6961 mol 

22.4 L

1 mol

= 15.59 L = 15.6 L

The partial pressure of each individual gas in the mixture will be related to what fraction on a
mole basis each gas represents in the mixture.

Poxygen = 1.00 atm 

0.1563 mol O

0.6961 mol total

= 0.225 atm O2

Pnitrogen = 1.00 atm 

0.1784 mol N

0.6961 mol total

= 0.256 atm N2

Pcarbon dioxide = 1.00 atm 

0.1136 mol CO

0.6961 mol total

= 0.163 atm CO2

Pneon = 1.00 atm 

0.2478 mol Ne

0.6961 mol total

= 0.356 atm Ne

98.

Molar masses: He, 4.003 g; Ne, 20.18 g

6.25 g He 

1 mol He

4.003 g He

= 1.561 mol He

4.97 g Ne 

1 mol Ne

20.18 g Ne

= 0.2463 mol Ne

ntotal = 1.561 mol + 0.2463 mol = 1.807 mol

As 1 mol of an ideal gas occupies 22.4 L at STP, the volume is given by

1.807 mol 

22.4 L

1 mol

= 40.48 L = 40.5 L.

The partial pressure of a given gas in a mixture will be proportional to what fraction of the total
number of moles of gas the given gas represents

PHe =

1.561 mol He

1.807 mol total

 1.00 atm = 0.8639 atm = 0.864 atm

PNe =

0.2463 mol Ne

1.807 mol total

 1.00 atm = 0.1363 atm = 0.136 atm

99.

2Na(s) + Cl2(g)  2NaCl(s)

molar mass Na = 22.99 g

4.81 

1 mol Na

22.99 g Na

= 0.2092 mol Na

0.2092 mol Na ×

1 mol Cl

2 mol Na

= 0.1046 mol Cl2

290

Chapter 13: Gases

0.1046 mol Cl2 

22.4 L

1 mol

= 2.34 L Cl2 at STP

100.

2C2H2(g) + 5O2(g)  2H2O(g) + 4CO2(g)

molar mass C2H2 = 26.04 g

1.00 g C2H2 ×

1 mol

26.04 g



0.0384 mol C2H2

From the balanced chemical equation, 2 × 0.0384 = 0.0768 mol of CO2 will be produced.

0.0768 mol CO2 ×

22.4 L

1 mol

= 1.72 L at STP

101.

FeO(s) + CO(g)  Fe(s) + CO2(g)

molar mass FeO = 71.85 g; 1.45 kg = 1.45  103 g

1.45  103 g FeO 

1 mol FeO

71.85 g FeO

= 20.18 mol FeO

Since the coefficients of the balanced equation are all one, if 20.18 mol FeO reacts, then 20.18
mol CO(g) is required and 20.18 mol of CO2(g) is produced.

20.18 mol 

22.4 L

1 mol

= 452 L

4.52  104 L CO(g) is required for reaction and 4.52  104 L CO2(g) are produced by the
reaction.

102.

125 mL = 0.125 L

0.125 L 

1 mol

22.4 L

= 0.00558 mol H2

From the balanced chemical equation, one mole of zinc is required for each mole of hydrogen
produced. Therefore, 0.00558 mol of Zn will be required.

0.00558 mol Zn 

65.38 g Zn

1 mol

= 0.365 g Zn

103.

kelvin (absolute)

104.

twice

105.

Gases consist of tiny particles, which are so small that the fraction of the bulk volume of the gas o
ccupied by the particles is negligible. The particles of a gas are in constant random motion and co
llide with the walls of the container (giving rise to the pressure of the gas). The particles of a gas
do not attract or repel each other. The average kinetic energy of the particles of a gas is reflected i
n the temperature of the gas sample.

291

Chapter 13: Gases

106.

In both cases, the gas particles will uniformly distribute throughout both flasks.

Case 1:

Case 2:

Case 1: The drawing will have four particles in the left and two particles in the right.

Case 2: The drawing will have three particles in the left and three particles in the right.

107.

sum

108.

First determine what volume the helium in the tank would have if it were at a pressure of 755 mm
Hg (corresponding to the pressure the gas will have in the balloons).

8.40 atm = 6384 mm Hg

V2 = (25.2 L) 

6384 mm Hg

755 mm Hg

= 213 L

Allowing for the fact that 25.2 L of He will have to remain in the tank, this leaves 213 – 25.2 =
187.8 L of He for filling the balloons.

187.8 L He 

1 balloon

1.50 L He

= 125 balloons

109.

A decrease in temperature would tend to make the volume of the weather balloon decrease. As th
e overall volume of a weather balloon increases when it rises to higher altitudes, the contribution
to the new volume of the gas from the decrease in pressure must be more important than the decre
ase in temperature (the temperature change in kelvins is not as dramatic as it seems in degrees Cel
sius).

110.

According to the balanced chemical equation, when 1 mol of (NH4)2CO3 reacts, a total of 4 moles
of gaseous substances is produced.

molar mass (NH4)2CO3 = 96.09 g; 453 °C = 726 K

52.0 g 

1 mol

96.09 g

= 0.541 mol

As 0.541 mol of (NH4)2CO3 reacts, 4(0.541) = 2.16 mol of gaseous products result.

V =

(2.16 mol)(0.08206 L atm mol K )(726 K)

(1.04 atm)

nRT





= 124 L

111.

CaCO3(s)  CaO(s) + CO2(g)

774 torr = 1.018 atm; 55°C + 273 = 328 K; molar mass CaCO3 = 100.1 g

292

Chapter 13: Gases

10.0 g CaCO3 

1 mol

100.1 g

= 0.0999 mol CaCO3

From the balanced equation, 0.0999 mol CO2 will be produced.

V =

(0.0999 mol)(0.08206 L atm mol K )(328 K)

(1.018 atm)

nRT





= 2.64 L CO2

112.

CaCO3(s) + 2H+(aq)  Ca2+(aq) + H2O(l) + CO2(g)

molar mass CaCO3 = 100.1 g; 60°C + 273 = 333 K

10.0 g CaCO3 

1 mol

100.1 g

= 0.0999 mol CaCO3 = 0.0999 mol CO2 also

Pcarbon dioxide = Ptotal – Pwater vapor

Pcarbon dioxide = 774 mm Hg – 149.4 mm Hg = 624.6 mm Hg = 0.822 atm

Vwet =

(0.0999 mol)(0.08206 L atm mol K )(333 K)

(0.822 atm)

nRT





= 3.32 L wet CO2

Vdry = 3.32 L 

624.6 mm Hg

774 mm Hg

= 2.68 L

113.

2S(s) + 3O2(g)  2SO3(g)

350.°C + 273 = 623 K; molar mass S = 32.07 g

5.00 g 

1 mol S

32.07 g S

= 0.1559 mol S

0.1559 mol S 

3 mol O

2 mol S

= 0.2339 mol O2

V =

(0.2339 mol)(0.08206 L atm mol K )(623 K)

(5.25 atm)

nRT





= 2.28 L O2

114.

(b)

115.

molar mass He = 4.003 g

10.0 g He 

1 mol He

4.003 g He

= 2.498 mol He

2.498 mol 

22.4 L

1 mol

= 56.0 L He

116.

n1 = 0.214 mol

n2 = 0.375 mol

V1 = 652 mL

V2 = ?

V2 =

= 1140 mL = 1.14 L

293

Chapter 13: Gases

117.

0.903 atm 

760 mm Hg

1 atm

= 686 mm Hg

2.1240  106 Pa 

760 mm Hg

101,325 Pa

= 1.5931  104 mm Hg

445 kPa 

760 mm

101.325 kPa

= 3.34  103 mm Hg

342 torr = 342 mm Hg

118.

645 mm Hg 

101,325 Pa

760 mm Hg

= 8.60  104 Pa

221 kPa = 221  103 Pa = 2.21  105 Pa

0.876 atm 

101,325 Pa

1 atm

= 8.88  104 Pa

32 torr 

101,325 Pa

760 torr

= 4.3  103 Pa

119.

1002 mm Hg = 1.318 atm

V = 123 L 

4.56 atm

1.318 atm

= 426 L

25.2 mm Hg = 0.0332 atm

P = 0.0332 atm 

634 mL

166 mL

= 0.127 atm

511 torr = 6.81  104 Pa = 68.1 kPa

V = 443 L 

68.1 kPa

1.05 kPa

= 2.87  104 L

120.

1.00 mm Hg = 1.00 torr

V = 255 mL 

1.00 torr

2.00 torr

= 128 mL

1.0 atm = 101.325 kPa

V = 1.3 L 

1.0 kPa

101.325 kPa

= 1.3  10–2 L

1.0 mm Hg = 0.133 kPa

V = 1.3 L 

1.0 kPa

0.133 kPa

= 9.8 L

121.

Assume the pressure at sea level to be 1 atm (760 mm Hg). Calculate the volume the balloon wou
ld have if it rose to the point where the pressure has dropped to 500 mm Hg. If this calculated vol
ume is greater than the balloon’s specified maximum volume (2.5 L), the balloon will burst.

294

Chapter 13: Gases

2.0 L 

760 mm Hg
500 mm Hg

= 3.0 L > 2.5 L. The balloon will burst.

122.

1.52 L = 1.52  103 mL

755 mm Hg 

1.52 10 mL

450 mL



= 2.55  103 mm Hg

123.

22C + 273 = 295 K; 100C + 273 = 373 K

729 mL 

373 K
295 K

= 922 mL

124.

74C + 273 = 347 K; –74C + 273 = 199 K

100. mL 

199 K

347 K

= 57.3 mL

100C + 273 = 373 K

373 K 

600 mL
500 mL

= 448 K (175C)

zero (the volume of any gas sample becomes zero at 0 K)

125.

0C + 273 = 273 K

273 K 

44.4 L
22.4 L

= 541 K (268°C)

–272C + 273 = 1 K; 25C + 273 = 298 K

1.0  10–3 mL 

298 K

1 K

= 0.30 mL

–40C + 273 = 233 K

233 K 

1000 L

32.3 L

= 7.21  103 K (6940°C)

126.

12C + 273 = 285 K; 192C + 273 = 465 K

75.2 mL 

465 K
285 K

= 123 mL

127.

5.12 g O2 = 0.160 mol; 25.0 g O2 = 0.781 mol

6.21 L 

0.781 mol
0.160 mol

= 30.3 L

128.

Three changes you can make to double the volume are:

1) increase the temperature (double the temperature in the kelvin scale); If the temperature is
increased, the gas particles have more kinetic energy and will hit the piston with more force (and

295

Chapter 13: Gases

more pressure). Therefore, the piston will move up until the pressure inside the container is the
same as outside the container (causing the volume to increase).
2) add more moles of gas to the container (double the amount); By adding more moles of gas to
the container, gas particles will hit the walls of the container more frequently (and thus exert
more pressure). The piston will move up until the pressure inside the container is the same as
outside the container (causing the volume to increase).
3) decrease the pressure outside the container (by half); By decreasing the pressure outside the
container, the pressure inside becomes greater than the pressure outside. The gas particles inside
will push the piston up until the pressure inside the container is the same as outside the container
(causing the volume to increase).

129.

V = 142 mL = 0.142 L

T =

(21.2 atm)(0.142 L)

(0.432 mol)(0.08206 L atm mol K )





= 84.9 K

V = 1.23 mL = 0.00123 L

P =

(0.000115 mol)(0.08206 L atm mol K )(293 K)

(0.00123 L)

nRT





= 2.25 atm

P = 755 mm Hg = 0.993 atm; T = 131C + 273 = 404 K

V =

(0.473 mol)(0.08206 L atm mol K )(404 K)

(0.993 atm)

nRT





= 15.8 L = 1.58  104 mL

130.

V = 21.2 mL = 0.0212 L

T =

(1.034 atm)(0.0212 L)

(0.00432 mol)(0.08206 L atm mol K )





= 61.8 K

V = 1.73 mL = 0.00173 L

P =

(0.000115 mol)(0.08206 L atm mol K )(182 K)

(0.00173 L)

nRT





= 0.993 atm

P = 1.23 mm Hg = 0.00162 atm; T = 152C + 273 = 425 K

V =

(0.773 mol)(0.08206 L atm mol K )(425 K)

(0.00162 atm)

nRT





= 1.66  104 L

131.

molar mass N2 = 28.02 g; T = 26C + 273 = 299 K

n = 14.2 g N2 

1 mol N

28.02 g N

= 0.507 mol N2

P =

(0.507 mol)(0.08206 L atm mol K )(299 K)

(10.0 L)

nRT





= 1.24 atm

132.

27C + 273 = 300 K

The number of moles of gas it takes to fill the 100. L tanks to 120 atm at 27C is independent of
the identity of the gas.

296

Chapter 13: Gases



(120 atm)(100. L)

(0.08206 L atm mol K )(300 K)





= 487 mol

487 mol of any gas will fill the tanks to the required specifications.

molar masses: CH4, 16.0 g; N2, 28.0 g; CO2, 44.0 g

for CH4: (487 mol)(16.0 g/mol) = 7792 g = 7.79 kg CH4

for N2: (487 mol)(28.0 g/mol) = 13,636 g = 13.6 kg N2

for CO2: (487 mol)(44.0 g/mol) = 21,428 g = 21.4 kg CO2

133.

molar mass He = 4.003 g

n = 4.00 g He 

1 mol He

4.003 g He

= 0.999 mol He

T =

(1.00 atm)(22.4 L)

(0.999 mol)(0.08206 L atm mol K )





= 273 K = 0°C

134.

molar mass of O2 = 32.00 g; 55 mg = 0.055 g

n = 0.055 g 

1 mol O

32.00 g O

= 0.0017 mol

V = 100. mL = 0.100 L; T = 26C + 273 = 299 K

P =

(0.0017 mol)(0.08206 L atm mol K )(299 K)

(0.100 L)

nRT





= 0.42 atm

135.

P1 = 1.0 atm

P2 = 220 torr = 0.289 atm

V1 = 1.0 L

V2 = ?

T1 = 23C + 273 = 296 K

T2 = –31C = 242 K

V2 =

2 1 1

1 2

(242 K)(1.0 atm)(1.0 L)

(296 K)(0.289 atm)

T PV

T P



= 2.8 L

136.

molar mass N2 = 28.02 g; 3.20 g = 0.114 mol; 8.80 g = 0.314 mol

n1 = 0.114 mol

n2 = 0.114 mol + 0.314 mol = 0.428 mol

V1 = 1.71 L

V2 = ?

V2 =

= 6.41 L

137.

molar mass of O2 = 32.00 g; 25C + 273 = 298 K

50. g O2 

1 mol O

32.00 g O

= 1.56 mol O2

total number of moles of gas = 1.0 mol N2 + 1.56 mol O2 = 2.56 mol

297

Chapter 13: Gases

P =

(2.56 mol)(0.08206 L atm mol K )(298 K)

(5.0 L)

nRT





= 13 atm

138.

PTotal = 325 torr + 475 torr + 650. torr = 1450. torr

Since the volume and temperature are constant, there is a direct relationship between the
pressure and number of moles. The gas with the highest pressure (which is O2) must
contain the greatest number of moles and collide with the walls more frequently.

139.

The pressures must be expressed in the same units, either mm Hg or atm.

Phydrogen = Ptotal – Pwater vapor

1.023 atm = 777.5 mm Hg

Phydrogen = 777.5 mm Hg – 42.2 mm Hg = 735.3 mm Hg

42.2 mm Hg = 0.056 atm

Phydrogen = 1.023 atm – 0.056 atm = 0.967 atm

140.

N2(g) + 3H2(g)  2NH3(g)

molar mass of NH3 = 17.03 g; 11C + 273 = 284 K

5.00 g NH3 

1 mol NH

17.03 g NH

= 0.294 mol NH3 to be produced

0.294 mol NH3 

1 mol N

2 mol NH

= 0.147 mol N2 required

0.294 mol NH3 

3 mol H

2 mol NH

= 0.441 mol H2 required

Vnitrogen =

(0.147 mol)(0.08206 L atm mol K )(284 K)

(0.998 atm)

nRT





= 3.43 L N2

Vhydrogen =

(0.441 mol)(0.08206 L atm mol K )(284 K)

(0.998 atm)

nRT





= 10.3 L H2

141.

C6H12O6(s) + 6O2(g)  CO2(g) + 6H2O(g)

molar mass of C6H12O6 = 180. g; 28C + 273 = 301 K

5.00 g C6H12O6 

1 mol C H O

180. g C H O

= 0.02778 mol C6H12O6

0.02778 mol C6H12O6 

6 mol O

1 mol C H O

= 0.1667 mol O2

Voxygen =

(0.1667 mol)(0.08206 L atm mol K )(301 K)

(0.976 atm)

nRT





= 4.22 L

298

Chapter 13: Gases

Because the coefficients of CO2(g) and H2O(g) in the balanced chemical equation happen to be
the same as the coefficient of O2(g), the calculations for the volumes of these gases produced are
identical: 4.22 L of each gaseous product is produced.

142.

2Cu2S(s) + 3O2(g)  2Cu2O(s) + 2SO2(g)

molar mass Cu2S = 159.2 g; 27.5C + 273 = 301 K

25 g Cu2S 

1 mol Cu S

159.2 g Cu S

= 0.1570 mol Cu2S

0.1570 mol Cu2S 

3 mol O

2 mol Cu S

= 0.2355 mol O2

Voxygen =

(0.2355 mol)(0.08206 L atm mol K )(301 K)

(0.998 atm)

nRT





= 5.8 L O2

0.1570 mol Cu2S 

2 mol SO

2 mol Cu S

= 0.1570 mol SO2

Vsulfur dioxide =

(0.1570 mol)(0.08206 L atm mol K )(301 K)

(0.998 atm)

nRT





= 3.9 L SO2

143.

2NaHCO3(s)  Na2CO3(s) + H2O(g) + CO2(g)

molar mass NaHCO3 = 84.01 g; 29C + 273 = 302 K; 769 torr = 1.012 atm

1.00 g NaHCO3 

1 mol NaHCO

84.01 g NaHCO

= 0.01190 mol NaHCO3

0.01190 mol NaHCO3 

1 mol H O

2 mol NaHCO

= 0.00595 mol H2O

Because H2O(g) and CO2(g) have the same coefficients in the balanced chemical equation for the
reaction, if 0.00595 mol H2O is produced, then 0.00595 mol CO2 must also be produced. The total
number of moles of gaseous substances produced is thus 0.00595 + 0.00595 = 0.0119 mol.

Vtotal =

(0.0119 mol)(0.08206 L atm mol K )(302 K)

(1.012 atm)

nRT





= 0.291 L

144.

One mole of any ideal gas occupies 22.4 L at STP.

35 mol N2 

22.4 L

1 mol

= 7.8  102 L

145.

P1 = 0.987 atm

P2 = 1.00 atm

V1 = 125 L

V2 = ?

T1 = 25C + 273 = 298 K

T2 = 273 K

V2 =

2 1 1

1 2

(273 K)(0.987 atm)(125 L)

(298 K)(1.00 atm)

T PV

T P



= 113 L

299

Chapter 13: Gases

146.

Assuming the temperature is the same on both sides, the initial pressure of helium is great
er because there are more gas particles colliding with the walls of its container (within the
same volume).

When the valve is opened, the gases disperse uniformly throughout the entire apparatus
(i.e., two Ne and three He in each vessel).

From the diagram, when the volume doubles the pressure is halved so

The original pressure of helium is 1.5 times the original pressure of neon.

Po(He) = 1.5 Po(Ne)

The final pressure is

147.

CaCO3(s)  CaO(s) + CO2(g)

molar mass of CaCO3 = 100.1 g

27.5 g CaCO3 

1 mol CaCO

100.1 g CaCO

= 0.275 mol CaCO3

From the balanced chemical equation, if 0.275 mol of CaCO3 reacts, then 0.275 mol of CaCO3
will be produced.

0.275 mol H2 

22.4 L

1 mol

= 6.16 L

148.

The solution is only 50% H2O2. Therefore 125 g solution = 62.5 g H2O2

molar mass of H2O2 = 34.02 g; T = 27C = 300 K; P = 764 mm Hg = 1.01 atm

62.5 g H2O2 

1 mol

34.02 g

= 1.84 mol H2O2

1.84 mol H2O2 

1 mol O

2 mol H O

= 0.920 mol O2

V =

(0.920 mol)(0.08206 L atm mol K )(300 K)

(1.01 atm)

nRT





= 22.4 L

149.

2NaN3(s)  2Na(s) + 3N2(g)

300

Chapter 13: Gases

molar mass of NaN3 = 65.02 g; T = 0C = 273 K; P = 1.00 atm

= 3.12 mol N2

3.12 mol N2 

= 2.08 mol NaN3

2.08 mol NaN3 

= 135 g NaN3

150.

V = .04  500 mL = 20 mL CO2 = 0.02 L

molar mass of CO2 = 44.01 g; T = 25C = 298 K; P = 1.00 atm

= 8  10–4 mol CO2

8  10–4 mol CO2 

= 0.04 g CO2

151.

PV = nRT

P(atm)

V(L)

n(mol)

6.74

10.4

2.00

155°C

0.300

1.74

0.0410

155 K

4.47

25.0

2.19

349°C

140.

2.25

10.5

93°C

152.

(b), (c), (d)

molar mass of N2 = 28.02 g; 28 g = 1.0 mol

molar mass of O2 = 32.00 g; 28 g = 0.88 mol; 32 g = 1.0 mol

Double the moles of gas would need to be present in order to double the pressure. Adding 28 g of
O2 is not double the moles, as in (a), but adding 32 g of O2 is double the moles, as in (d). (n1/P1 =
n2/P2)

T = –73C = 200. K; T = 127C = 400. K; T = 30C = 303 K; T = 60C = 333 K

Doubling the temperature (in Kelvin) will double the pressure, as in (b), but not in (e). (P1/T1 =
P2/T2)

Cutting the volume by half will double the pressure, as in (c). (P1V1 = P2V2)

153.

n1 = 150.0 mol

n2 = ?

P1 = 8.93 MPa

P2 = 2.00 MPa

T1 = 25C = 298 K

T2 = 19C = 292 K

n2 =

= 34.3 mol Ar

34.3 mol Ar 

= 1.37 × 103 g Ar

301

Chapter 13: Gases

154.

V1 = 855 L

V2 = ?

P1 = 730 torr

P2 = 605 torr

T1 = 25C = 298 K

T2 = 15C = 288 K

V2 =

= 997 L

ΔV = 997 L – 855 L = 142 L

155.

V = 936 mL = 0.936 L

P = 0.967 atm

T = 31C = 304 K

mass of gas = 135.87 g – 134.66 g = 1.21 g

= 0.0363 mol

molar mass of gas = 1.21 g/0.0363 mol = 33.3 g/mol

156.

Xe + 2F2  XeF4

molar mass of XeF4 = 207.3 g; T = 400C = 673 K; V = 20.0 L;

PXenon = 0.859 atm; PFluorine = 1.37 atm

= 0.311 mol Xe

= 0.496 mol F2

F2 is the limiting reactant (only 0.248 mol Xe needed).

0.496 mol F2 



= 51.4 g XeF4

157.

CaSiO3(s) + 6HF(g)  CaF2(aq) + SiF4(g) + 3H2O(l)

molar mass of CaSiO3 = 116.17 g; molar mass of SiF4 = 104.09 g; molar mass of H2O = 18.016 g

T = 27C = 300. K; V = 31.8 L; P = 1.00 atm

32.9 g CaSiO3 

= 0.283 mol CaSiO3

= 1.29 mol HF

HF is the limiting reactant (only 0.215 mol CaSiO3 needed).

1.29 mol HF 



= 22.4 g SiF4

1.29 mol HF 



= 11.6 g H2O

302

Chapter 13: Gases

158.

(a), (c), and (d); Only doubling the temperature in Kelvin will double the volume, thus (b) is not
true.

303