Chapter 11
Gases and Their Properties
Chapter 11
Gases and Their Properties
INSTRUCTOR’S NOTES
It is important to emphasize that this chapter can be used almost anywhere in the sequence of topics, as long as the
basic ideas of stoichiometry have been covered. However, it works well to place it after the first 10 chapters since
we can then make a very nice connection between the phases of matter by covering the chapters on gases, liquids,
solids, and solutions in that order.
At the beginning of this series of topics we take the opportunity to emphasize two reasons for studying gases and
their behavior:

There are gases in the atmosphere - in every breathe we take! - and many common gases are commercially
important. Therefore we must understand how to deal with them conceptually and mathematically.

The behavior of gases is well understood and can be modeled. This has two benefits:
a) This has led to a better understanding of other aspects of the physical world (such as the energy
distribution in liquids).
b) It is important to recognize that nature and some aspects of social behavior can be subjected to
mathematical models, an approach that is important in business and industry today.
Another very useful point to make is that one can clearly see the difference between a law (the ideal gas law and the
laws on which it is based) and a theory (the kinetic molecular theory).
When discussing nonideal gases, we make the point that intermolecular forces play an important role, so it is a
natural transition to move from there to the discussion of liquids and solids in the next chapters.
Three to four lectures are normally given on this material.
SUGGESTED DEMONSTRATIONS
1.
General Gas Demonstrations

deGrys, H. “Thirty Feet and Rising: Constructing and Using a Water Barometer to Explore Chemical
Principles,” Journal of Chemical Education 2003, 80, 1156.

Campbell, D. J. “An Alcohol Rocket Car—A Variation on the ‘Whoosh Bottle’ Theme,” Journal of
Chemical Education 2001, 78, 910.

Bare, W. D.; Andrews, L. “A Demonstration of Ideal Gas Principles Using a Football,” Journal of
Chemical Education 1999, 76, 622.

Corkern, W. H.; Hughes, E., Jr. “The Methane Balloon,” Journal of Chemical Education 1999, 76, 794.

For a list of 26 demonstrations on “The Physical Behavior of Gases,” see Shakhashiri, B. Z. Chemical
Demonstrations: A Handbook for Teachers of Chemistry; University of Wisconsin Press, 1985; Vol. 2.
216
Chapter 11
2.
Gases and Their Properties
Boyle’s Law

To demonstrate Boyle’s law, bring a bicycle pump to lecture.

Vitz, E. “Ammonia Can Crush,” Journal of Chemical Education 1999, 76, 932.

Meyers, R. D.; Yee, G. T. “A More Dramatic Container to Crush by Atmospheric Pressure,” Journal of
Chemical Education 1999, 76, 933.

Shakhashiri, B. Z. “Collapsing Can,” Chemical Demonstrations: A Handbook for Teachers of Chemistry;
University of Wisconsin Press, 1985; Vol. 2, pp. 6-8.

Boyle's law has been demonstrated in Figure 11.4 using a large syringe. Instead of lead shot in a beaker, one
can also use a pile of textbooks to provide the pressure as described in Shakhashiri, B. Z. “Boyle’s Law and
the Mass of a Textbook,” Chemical Demonstrations: A Handbook for Teachers of Chemistry; University of
Wisconsin Press, 1985; Vol. 2, pp. 20-23. (For more on gas law demonstrations using syringes, see
Davenport, D. Journal of Chemical Education 1962, 39, 252.)
3.
Charles’s Law

A favorite demonstration with the students in shown in Figure 11.5. Perhaps 15–20 balloons are passed out
to students at the beginning of the lecture, and they are asked to blow them up and tie them off. When the
demonstration begins, each of these students comes to the front of the room and places the inflated balloon
in a large beaker of liquid nitrogen. After all the balloons have been placed in the beaker, they are poured
out again and reinflate to their original volume when warmed back to room temperature. (However, a few
may break, as they become twisted when frozen.)

Krnel, D.; Glazar, S. A. “’Experiment with a Candle’ Without a Candle,” Journal of Chemical Education
2001, 78, 914.

Shakhashiri, B. Z. “Charles Law,” Chemical Demonstrations: A Handbook for Teachers of Chemistry;
University of Wisconsin Press, 1985; Vol. 2, pp. 28-32.
4.
Avogadro’s Law

Shakhashiri, B. Z. “Avogadro’s Hypothesis,” Chemical Demonstrations: A Handbook for Teachers of
Chemistry; University of Wisconsin Press, 1985; Vol. 2, pp. 44-47.
5.
Density

Figure 11.8a illustrates the relative densities of two gases and is an easy demonstration. This can also be
used to illustrate diffusion, since He atoms diffuse from a balloon much faster than Ar or air (O 2 and N2).
6.
Molar Mass Determination

Shakhashiri, B. Z. “Determination of the Molecular Mass of the Gas from a Butane Lighter,” Chemical
Demonstrations: A Handbook for Teachers of Chemistry; University of Wisconsin Press, 1985; Vol. 2, pp.
48-50.
217
Chapter 11
7.
Gases and Their Properties
Diffusion and Effusion

Release a small quantity of a concentrated scent (rose, lavender, or lemon oil) at the beginning of the
lecture. The scent will fill the room.

Keller, P. C. “A Simple Apparatus to Demonstrate Differing Gas Diffusion Rates (Graham’s Law),”
Journal of Chemical Education 1990, 67, 160.

Shakhashiri, B. Z. “Graham’s Law of Diffusion,” Chemical Demonstrations: A Handbook for Teachers of
Chemistry; University of Wisconsin Press, 1985; Vol. 2, pp. 69-71.

Shakhashiri, B. Z. “Graham’s Law of Effusion,” Chemical Demonstrations: A Handbook for Teachers of
Chemistry; University of Wisconsin Press, 1985; Vol. 2, pp. 72-74.
218
Chapter 11
Gases and Their Properties
SOLUTIONS TO STUDY QUESTIONS
11.1
(a) 440 mm Hg ·
(b) 0.58 atm ·
1.013 bar
= 0.59 bar
1 atm
(c) 440 mm Hg ·
11.2
210 mm Hg ·
210 mm Hg ·
11.4
101.325 kPa
= 59 kPa
760 mm Hg
1 atm
= 0.28 atm
760 mm Hg
1.013 bar
= 0.28 bar
1 atm
0.28 atm ·
11.3
1 atm
= 0.58 atm
760 mm Hg
101.325 kPa
= 28 kPa
760 mm Hg
(a) 534 mm Hg ·
1.013 bar
= 0.712 bar
760 mm Hg
0.754 is the higher pressure
(b) 534 mm Hg ·
101.325 kPa
= 71.2 kPa
760 mm Hg
650 kPa is the higher pressure
(c) 1.34 bar ·
1  102 kPa
= 134 kPa
1 bar
363 mm Hg ·
1 atm
= 0.478 atm
760 mm Hg
363 kPa ·
934 kPa is the higher pressure
1atm
= 3.58 atm
101.325 kPa
0.523 bar ·
1 atm
= 0.516 atm
1.013 bar
0.256 atm < 363 mm Hg < 0.523 bar < 363 kPa
11.5
P2 =
PV
(67.5 mm Hg)(500. mL)
1 1
=
= 270. mm Hg
V2
125 mL
11.6
V2 =
PV
(56.5 mm Hg)(125 mL)
1 1
=
= 113 mL
P2
62.3 mm Hg
11.7
V 
 3.5 L 
V2 = T2  1  = (310. K) 
 = 3.7 L
T
 295.2 K 
 1
11.8
V 
 5.0 mL 
V2 = T2  1  = (273 K) 
 = 4.6 mL
 295 K 
 T1 
219
Chapter 11
Gases and Their Properties
11.9
 V  T 
 3.6 L  273.2 K 
P2 = P1  1  2  = (380 mm Hg) 

 = 250 mm Hg
 5.0 L  298 K 
 V2  T1 
11.10
 436.5 mm Hg   297.7 K 
 P  T 
V2 = V1  1  2  = (25.0 mL) 

 = 117 mL
 P2  T1 
 94.3 mm Hg   293.7 K 
11.11
T 
 268.2 K 
P2 = P1  2  = (360 mm Hg) 
 = 320 mm Hg
T
 298.7 K 
 1
11.12
 V  T 
 135 mL  273.2 K 
P2 = P1  1  2  = (165 mm Hg) 

 = 81.7 mm Hg
 252 mL  295.7 K 
 V2  T1 
11.13
 400. cm3   350. K 
 V  T 
P2 = P1  1  2  = (1.00 atm) 
 = 9.72 atm
3 
 V2  T1 
 50.0 cm   288 K 
11.14
 737 mm Hg   240. K 
 P  T 
7
V2 = V1  1  2  = (1.2  107 L) 

 = 1.2  10 L
P
T
600.
mm
Hg
289.2
K


 2  1 


The volume of gas is nearly the same (to 2 significant figures) at the higher altitude.
11.15
11.16
11.17
(a) 150 mL NO 
1 L O2
= 75 mL O2
2 L NO
(b) 150 mL NO 
2 L NO2
= 150 mL NO2
2 L NO
5.2 L C2 H6 
7 L O2
= 18 L O2
2 L C2 H6
5.2 L C2 H6 
6 L H2O
= 16 L H2O
2 L C2 H6
1.25 g 
1 mol CO2
= 0.0284 mol CO2
44.01 g
750. mL 
P=
11.18
220
nRT
(0.0284 mol)(0.082057 L  atm/K  mol)(295.7 K)
=
= 0.919 atm
V
0.750 L
30.0 kg 
V =
1L
= 0.750 L
103 mL
1000 g 1 mol He

= 7490 mol He
1 kg
4.003 g
nRT
(7490 mol)(0.082057 L  atm/K  mol)(295 K)
=
= 1.51  105 L
P
1.20 atm
Chapter 11
11.19
Gases and Their Properties
2.2 g 
1 mol CO2
= 0.050 mol CO2
44.0 g
313 mm Hg 
V =
11.20
nRT
(0.050 mol)(0.082057 L  atm/K  mol)(295 K)
=
= 2.9 L
P
0.418 atm
1.50 g 
1 mol C2 H5OH
= 0.0326 mol C2 H5OH
46.07 g
251 cm3 
P=
11.21
1 mL
1L

= 0.251 L
1 cm3
103 mL
nRT
(0.0326 mol C2 H5OH)(0.082057 L  atm/K  mol)(523 K)
=
= 5.6 atm
V
0.251 L
737 mm Hg 
n=
1 atm
= 0.418 atm
760 mm Hg
1 atm
= 0.970 atm
760 mm Hg
PV
(0.970 atm)(1.2  107 L)
=
= 4.8  105 mol He
RT
(0.082057 L  atm/K  mol)(298 K)
4.8  105 mol He 
11.22
11.23
n=
4.00 g
= 0.88 g He
1 mol He
0.20 mm Hg 
1 atm
= 2.6  10–4 atm
760 mm Hg
PM
(2.6  10–4 atm)(28.96 g/mol)
=
= 3.7  10–4 g/L
RT
(0.082057 L  atm/K  mol)(250 K)
233 mm Hg 
d=
11.25
PV
(1.1 atm)(5.0 L)
=
= 0.22 mol He
RT
(0.082057 L  atm/K  mol)(298 K)
0.22 mol He 
d=
11.24
4.00 g
= 1.9  106 g He
1 mol He
PM
(0.307 atm)(74.12 g/mol)
=
= 0.931 g/L
RT
(0.082057 L  atm/K  mol)(298 K)
189 mm Hg 
M =
1 atm
= 0.307 atm
760 mm Hg
1 atm
= 0.249 atm
760 mm Hg
dRT
(0.355 g/L)(0.082057 L  atm/K  mol)(290. K)
=
= 34.0 g/mol
P
0.249 atm
221
Chapter 11
11.26
Gases and Their Properties
195 mm Hg 
M =
11.27
dRT
(1.25 g/L)(0.082057 L  atm/K  mol)(298.2 K)
=
= 119 g/mol
P
0.257 atm
1.007 g
= 2.23 g/L
0.452 L
d=
715 mm Hg 
M =
11.28
1 atm
= 0.257 atm
760 mm Hg
1 atm
= 0.941 atm
760 mm Hg
dRT
(2.23 g/L)(0.082057 L  atm/K  mol)(296 K)
=
= 57.5 g/mol
P
0.941 atm
0.0125 g
= 0.0758 g/L
0.165 L
d=
13.7 mm Hg 
M =
1 atm
= 0.0180 atm
760 mm Hg
dRT
(0.0758 g/L)(0.082057 L  atm/K  mol)(295.7 K)
=
= 102 g/mol
P
0.0180 atm
102 g/mol
=2
51 g/mol
11.29
d=
The molecular formula is (CHF2)2 or C2H2F4
12.5  10–3 g
= 0.100 g/L
0.125 L
24.8 mm Hg 
M =
1 atm
= 0.0326 atm
760 mm Hg
dRT
(0.100 g/L)(0.082057 L  atm/K  mol)(298 K)
=
= 74.9 g/mol
P
0.0326 atm
(d) B6H10
11.30
d=
0.107 g
= 0.856 g/L
0.125 L
331 mm Hg 
M =
11.31
222
dRT
(0.856 g/L)(0.082057 L  atm/K  mol)(273.2 K)
=
= 44.1 g/mol
P
0.436 atm
2.2 g 
P=
1 atm
= 0.436 atm
760 mm Hg
1 mol Fe 1 mol H2

= 0.039 mol H2
55.9 g
1 mol Fe
nRT
(0.039 mol)(0.082057 L  atm/K  mol)(298 K)
=
= 0.096 atm
V
10.0 L
Chapter 11
11.32
Gases and Their Properties
356 mm Hg 
1 atm
= 0.468 atm
760 mm Hg
425 mm Hg 
1 atm
= 0.559 atm
760 mm Hg
n=
PV
(0.468 atm)(5.20 L)
=
= 0.0996 mol SiH4
RT
(0.082057 L  atm/K  mol)(298 K)
2 mol O2
= 0.199 mol O2
1 mol SiH4
nRT
(0.199 mol O2 )(0.082057 L  atm/K  mol)(298 K)
V =
=
= 8.71 L
P
0.559 atm
0.0996 mol SiH4 
11.33
n=
PV
(1.3 atm)(75.0 L)
=
= 4.0 mol N2
RT
(0.082057 L  atm/K  mol)(298 K)
4.0 mol N2 
11.34
0.048 g 
2 mol NaN3
65.0 g

= 170 g NaN3
3 mol N2
1 mol NaN3
1 mol C8 H18
= 4.2  10–4 mol C8 H18
114 g
18 mol H2 O
= 0.0038 mol H2 O
2 mol C8 H18
nH O RT
(0.0038 mol H2 O)(0.082057 L  atm/K  mol)(303.2 K)
= 2
=
= 0.020 atm
V
4.75 L
4.2  10–4 mol C8 H18 
PH2 O
25 mol O2
= 0.0053 mol O2
2 mol C6 H6
nO RT
(0.0053 mol O2 )(0.082057 L  atm/K  mol)(295 K)
= 2
=
= 0.027 atm
V
4.75 L
4.2  10–4 mol C8 H18 
PO2
11.35
1.00  103 g 
P=
11.36
nRT
(31.2 mol)(0.082057 L  atm/K  mol)(296 K)
=
= 1.7 atm
V
450 L
767 mm Hg 
n=
1 mol N2 H4
1 mol O2

= 31.2 mol O2
32.05 g
1 mol N2 H4
1 atm
= 1.01 atm
760 mm Hg
PV
(1.01 atm)(8.90 L)
=
= 0.371 mol CO2
RT
(0.082057 L  atm/K  mol)(295.2 K)
0.371 mol CO2 
4 mol KO2
71.10 g

= 52.7 g KO2
2 mol CO2
1 mol KO2
223
Chapter 11
11.37
Gases and Their Properties
1.0 g 
1 mol H2
= 0.50 mol H2
2.02 g
8.0 g 
1 mol Ar
= 0.20 mol Ar
39.9 g
PH2 =
nH2 RT
(0.50 mol)(0.082057 L  atm/K  mol)(300. K)
=
= 4.1 atm
V
3.0 L
PAr =
nAr RT
(0.20 mol)(0.082057 L  atm/K  mol)(300. K)
=
= 1.6 atm
V
3.0 L
Ptotal = 4.1 atm + 1.6 atm = 5.7 atm
11.38
%N = 100.0 – (4.5% H2S + 3.0% CO2) = 92.5% N
The partial pressure of each gas is proportional to its percentage:
PN2 = (46 atm)(0.925) = 43 atm
PH2S = (46 atm)(0.045) = 2.1 atm
PCO2 = (46 atm)(0.030) = 1.4 atm
11.39
(a)
mol halothane 170 mm Hg
=
= 0.30
mol O2
570 mm Hg
(b) 160 g 
11.40
(a) n =
1 mol O2
0.30 mol halothane
197 g


= 3.0  102 g halothane
32.0 g
1 mol O2
1 mol halothane
PV
(1.00 atm)(12.5 L)
=
= 0.517 mol He
RT
(0.082057 L  atm/K  mol)(294.7 K)
0.517 mol He 
(b) P =
4.003 g
= 2.07 g He
1 mol He
nRT
(0.517 mol He)(0.082057 L  atm/K  mol)(294.7 K)
=
= 0.48 atm
V
26 L
(c) PO2 = Ptotal – PHe = 1.00 atm – 0.48 atm = 0.52 atm
(d) XHe =
11.41
0.48 atm
= 0.48
1.00 atm
X O2 =
0.52 atm
= 0.52
1.00 atm
(a) Flask B is at a higher temperature (25 ºC) than flask A (0 ºC), so the average kinetic energy per
molecule is greater in flask B than in flask A.
(b) The molar mass of H2 (2.02 g/mol) is much less than the molar mass of CO2 (44.0 g/mol). Flask A has
molecules with higher average velocity. Note that even though the gases have different temperatures,
the difference in molar mass is much more significant and thus has a greater effect on the relative
average velocity.
224
Chapter 11
Gases and Their Properties
(c)
nH2
PH2 V/RTH2
(1 atm)(V)/(R)(273 K)
0.5 mol H2
=
=
=
nCO2
PCO2 V/RTCO2
(2 atm)(V)/(R)(298 K)
1mol CO2
Flask B contains a greater number of moles of gas and therefore contains more molecules of gas.
(d)
nH2
(1 atm)(V)/(R)(273 K)
0.5 mol H2
=
=
nCO2
(2 atm)(V)/(R)(298 K)
1mol CO2
1 mol CO2 (44 g) has a greater mass than 0.5 mol H2 (1.0 g). Flask B has a greater mass.
11.42
The molar mass of Ar (40 g/mol) is greater than the molar mass of N 2 (28 g/mol). Therefore, for samples
with equal mass there are more moles of N2 present than moles of Ar.
(a)
True. There are more moles of N2 present, so there are more molecules of N2 present.
(b)
False. Pressure is directly related to the number of moles of gas present. The pressure in the nitrogen
flask is greater because there are more moles of N2 present.
(c)
False. The gas with the smaller molar mass (N2) will have a greater velocity than the gas with the
greater molar mass (Ar).
(d)
True. The nitrogen molecules have a greater velocity than the argon molecules and there are more
molecules of nitrogen present, so they will collide more frequently with the walls of the flask.
11.43
rms speed O2
4.28  104 cm/s
=
=
rms speed CO2
u 2 (CO2 )
44.01 g/mol
32.00 g/mol
u 2 (CO2) = 3.65  104 cm/s
11.44
u2 =
3RT
=
M
3(8.3145 J/K  mol)(298 K)
= 515 m/s
28.01  10–3 kg/mol
rms speed CO
=
rms speed Ar
11.45
39.95 g/mol
= 1.194
28.01 g/mol
Increasing average molecular speed: CH2F2 < Ar < N2 < CH4
molar mass (g/mol)
11.46
40
28
16
Increasing average molecular speed: OSCl2 < Cl2O < Cl2 < SO2
molar mass (g/mol):
11.47
51
119
87
71
64
(a) F2 (38 g/mol) effuses faster than CO2 (44 g/mol).
(b) N2 (28 g/mol) effuses faster than O2 (32 g/mol).
(c) C2H4 (28.1 g/mol) effuses faster than C2H6 (30.1 g/mol).
(d) CFCl3 (137 g/mol) effuses faster than C2Cl2F4 (171 g/mol).
11.48
He will effuse faster
Rate of He effusion
=
Rate of Ar effusion
39.9 g/mol
= 3.16 times faster
4.00 g/mol
225
Chapter 11
11.49
Gases and Their Properties
Rate of He effusion
Rate of He effusion
=
=
Rate of unknown gas effusion 1/3(Rate of He effusion)
unknown gas molar mass
4.00 g/mol
unknown gas molar mass = 36 g/mol
11.50
Rate of I2
=
Rate of uranium fluoride
M uranium fluoride
M I2
 0.0150 g 1 mol I2 



253.8 g 
M uranium fluoride
 1 hr
=
253.8 g/mol
1
 0.0177 mg




1 hr
M uranium fluoride 

M
(0.00334)2 (M uranium fluoride )2 = uranium fluoride
253.8 g/mol
M uranium fluoride = 353 g/mol
11.51
Pideal =
nRT
(4.00 mol)(0.082057 L  atm/K  mol)(373.2 K)
=
= 30.6 atm
V
4.00 L

P + a


n
V 
 
2

 (V – bn) = nRT


2

2
2  4.00 mol 
P
+
(6.49
atm

L
/
mol
)

 4.00 L    4.00 L – [0.0562 L/mol][4.00 mol]


 

 (4.00 mol)(0.082057 L  atm/K  mol)(373.2 K)
P = 26.0 atm
11.52
165 g 
(a) P =
1 mol CO2
= 3.75 mol CO2
44.01 g
nRT
(3.75 mol)(0.082057 L  atm/K  mol)(298 K)
=
= 7.33 atm
V
12.5 L
2

n 
 P + a    (V – bn) = nRT

V  

2
(b)  P + (3.59 atm  L2 / mol2 )  3.75 mol   12.5 L – [0.0427 L/mol][3.75 mol]

 12.5 L   


 

 (3.75 mol)(0.08206 L  atm/K  mol)(298 K)
P = 7.11 atm
11.53
226
atm
mm Hg
kPa
bar
Standard atmosphere
1
760
101.325
1.013
Partial pressure of N2 in the atmosphere
0.780
593
79.1
0.791
Tank of compressed H2
131
99800
13300
133
Atmospheric pressure at the top of Mt. Everest
0.333
253
33.7
0.337
Chapter 11
11.54
Gases and Their Properties
At STP, 1 mol of a gas occupies 22.414 L
1 mol CO2
1 mol C

= 0.089 mol C
22.414 L
1 mol CO2
1 mol N2
2 mol N
0.5 L N2 

= 0.045 mol N
22.414 L 1 mol N2
2.0 L CO2 
0.31 mol H
7 mol H
=
0.045 mol N
1 mol N
11.55
u2 =
3RT
=
M
0.089 mol C
2 mol C
=
0.045 mol N 1 mol N
The empirical formula is C2 H7 N
3(8.314 J/K  mol)(240 K)
= 1220 m/s
0.00400 kg/mol
New speed = (1220 m/s)(1.10) = 1350 m/s
1350 m/s =
11.56
3(8.314 J/K  mol)T
0.00400 kg/mol
T = 290. K = 17 C
mass (g)
T 
n2 = n1  1  and n =
32.00 g/mol
 T2 
T 
 300. K 
mass2 = mass1  1  = (12.0 g) 
 = 12.9 g
 278.2 K 
 T2 
11.57
2 C4H9SH(g) + 15 O2(g)  2 SO2(g) + 8 CO2(g) + 10 H2O(g)
95.0  10–3 g ·
Ptotal =
1 mol C4 H9SH
20 mol gas
·
= 0.0105 mol gas
90.19 g
2 mol C4 H9SH
ntotal RT
(0.0105 mol)(0.082057 L·atm/K·mol)(298 K)
=
= 0.0491 atm
V
5.25 L
PSO2 = 0.0491 atm ·
2 mol SO2
= 0.00491 atm
20 mol gas
PCO2 = 0.0491 atm ·
8 mol CO2
= 0.0196 atm
20 mol gas
PH2 O = 0.0491 atm ·
10 mol H2 O
= 0.0246 atm
20 mol gas
PV
(7.25 atm)(1.52 L)
=
= 331 K = 58 º C
nR
(0.406 mol)(0.082057 L  atm/K  mol)
11.58
T=
11.59
Assume the Mars atmosphere behaves as an ideal gas.


1 atm
103 L 
3

 8 mm Hg 
 10. m 
760 mm Hg  
1 m3 
PV
n=
= 
= 4 mol
RT
(0.082057 L  atm/K  mol)(300. K)
227
Chapter 11
11.60
Gases and Their Properties
2.25 g 
1 mol Si
= 0.0801 mol Si
28.09 g


1 atm
 585 mm Hg 
 (6.56 L)
760 mm Hg 

nCH3Cl =
= 0.206 mol CH3Cl
(0.082057 L  atm/K  mol)(298 K)
0.206 mol CH3Cl
2.58 mol CH3Cl
2 mol CH3Cl
=
>
Si is the limiting reactant
0.0801 mol Si
1 mol Si
1 mol Si
0.0801 mol Si 
P(CH3 )2SiCl2 =
11.61
0.450 g 
1 mol (CH3 )2SiCl2
129.1 g

= 10.3 g (CH3 )2SiCl2
1 mol Si
1 mol (CH3 )2SiCl2
(0.0801 mol)(0.082057 L  atm/K  mol)(368 K)
= 0.369 atm
6.56 L
1 mol Ni
= 0.00767 mol Ni
58.69 g


1 atm
 418 mm Hg 
 (1.50 L)
760
mm
Hg


nCO =
= 0.0337 mol CO
(0.082057 L  atm/K  mol)(298.2 K)
0.0337 mol CO
4.40 mol CO
4 mol CO
=
>
Ni is the limiting reactant
0.00767 mol Ni
1 mol Ni
1 mol Ni
0.00767 mol Ni 
11.62
1 mol Ni(CO)4
170.7 g

= 1.31 g Ni(CO)4
1mol Ni
1 mol Ni(CO)4
(a) molar mass (g/mol): H2O, 18; CO2, 44; O2, 32; C2H6, 30
Rms speed: CO2<O2<C2H6<H2O
(b) n = PV/rT = (0.337 atm)(3.26 L)/(0.08206 L·atm/mol·K)(298 K) = 0.0449 mol C 2H6
0.0449 mol C2H6 · 7 mol O2/2 mol C2H6 = 0.157 mol O2
Total moles = 0.0449 mol C2H6 + 0.157 mol O2 = 0.202 mol
P = nRT/V = (0.202 mol)(0.08206 L·atm/mol·K)(298 K)/(3.26 L) = 1.52 atm total pressure
PO2 = (0.157 mol/0.202 mol)(1.52 atm) = 1.18 atm
11.63
228
1.
1.0 L H2 
1 mol H2
= 0.045 mol H 2
22.414 L
2.
1.0 L Ar 
1 mol Ar
= 0.045 mol Ar
22.414 L
3.
n=
PV
(1.0 atm)(1.0 L)
=
= 0.041 mol H2
RT
(0.082057 L  atm/K  mol)(300. K)
Chapter 11
4.
Gases and Their Properties


1 atm
 900 mm Hg 
 (1.0 L)
760 mm Hg 
PV

n=
=
= 0.05 mol He
RT
(0.082057 L  atm/K  mol)(273 K)
(a) The number of molecules is proportional to number of moles, so sample 4 contains the greatest
number of molecules.
(b) Sample 3 contains the smallest number of gas molecules.
(c) Argon has the greatest molar mass, so sample 2 contains the largest mass of gas
(0.045 mol Ar · 39.9 g/mol = 1.8 g Ar).
11.64
1 mol C3 H8
· 288 mm Hg = 48.0 mm Hg
6 mol gas
PC3H8 =
PO2 =
5 mol O2
· 288 mm Hg = 240. mm Hg
6 mol gas
PH2 O = 240. mm Hg ·
11.65
3.52 g ·
nCO
4 mol H2 O
= 192 mm Hg
5 mol O2
1 mol Fe
= 0.0630 mol Fe
55.85 g


1 atm
 732 mm Hg 
 (5.50 L)
760
mm
Hg
PV

=
= 
= 0.218 mol CO
RT
(0.082057 L  atm/K  mol)(296 K)
0.218 mol CO
3.46 mol CO
5 mol CO
=
<
0.0630 mol Fe
1 mol Fe
1 mol Fe
0.218 mol CO ·
11.66
CO is the limiting reactant
1 mol Fe(CO)5
195.9 g
·
= 8.55 g Fe(CO)5
5 mol CO
1 mol Fe(CO)5
100.00% – (11.79% C + 69.57% Cl) = 18.64% F
Assume 100.00 g compound
11.79 g ·
1 mol C
= 0.9816 mol C
12.011 g
69.57 g ·
1 mol Cl
= 1.962 mol Cl
35.453 g
18.64 g ·
1 mol F
= 0.9812 mol F
18.998 g
0.9816 mol C 1 mol C
=
0.9812 mol F
1 mol F
1.962 mol Cl
2 mol Cl
=
0.9812 mol F
1 mol F
The empirical formula is CCl2F
229
Chapter 11
Gases and Their Properties
 0.107 g 

 (0.082057 L  atm/K  mol)(298 K)
dRT
0.458 L 
M =
= 
= 204 g/mol
1 atm
P
21.3 mm Hg 
760 mm Hg
204 g/mol
=2
102 g/mol
11.67
The molecular formula is (CCl2F)2 or C2Cl4F2
1 mol S
= 0.7868 mol S
32.066 g
1 mol F
74.77 g 
= 3.935 mol F
18.998 g
3.935 mol F
5 mol F
=
The empirical formula is SF5
0.7868 mol S 1 mol S
25.23 g 
 0.0955 g 

 (0.082057 L  atm/K  mol)(318 K)
dRT
0.089 L 
M =
= 
= 254 g/mol
1 atm
P
83.8 mm Hg 
760 mm Hg
254 g/mol
=2
127 g/mol
11.68
The molecular formula is (SF5)2 or S2F10
0.95 g ·
1 mol (NH4 )2 Cr2 O7
5 mol gas
·
= 0.019 mol gas produced
252 g
1 mol (NH4 )2 Cr2 O7
Ptotal =
ntotal RT
(0.018 mol)(0.082057 L  atm/K  mol)(296 K)
=
= 0.031 atm
V
15.0 L
PN2 = 0.031 atm ·
1 mol N2
= 0.0061 atm
5 mol gas
PH2 O = 0.031 atm ·
11.69
4 mol H2 O
= 0.024 atm
5 mol gas
 92 g 1 m3 
·

 (0.082057 L  atm/K  mol)(210. K)
1 m3 103 L 
dRT
= 
= 29 g/mol
(a) M =
1 atm
P
42 mm Hg ·
760 mm Hg
(b) X O2 + X N2 = 1
29 g/mol = X O2 ·
X O2 = 0.18
X N2 = 0.82
230
32.0 g
28.0 g
+ (1 – X O2 ) ·
1 mol O2
1 mol N2
Chapter 11
11.70
Gases and Their Properties
He: P2 =
PV
(145 mm Hg)(3.0 L)
1 1
=
= 87 mm Hg
V2
5.0 L
Hg: P2 =
PV
(355 mm Hg)(2.0 L)
1 1
=
= 140 mm Hg
V2
5.0 L
Ptotal = 87 mm Hg + 140 mm Hg = 230 mm Hg
11.71


1 atm
17.2 mm Hg ·
 (1.850 L)
760
mm
Hg
PV


n=
=
= 0.00174 mol gas
RT
(0.082057 L  atm/K  mol)(294 K)
M =
11.72
0.150 g
= 86.4 g/mol
0.00174 mol
The gas is probably ClO2F (86.4 g/mol)
PF2 (consumed) = Ptotal – PXe – PF2 (unreacted) = 0.72 atm – 0.12 atm – 0.36 atm = 0.24 atm
nF =
(0.24 atm)(0.25 L)
2 mol F

= 0.0054 mol F
(0.082057 L  atm/K  mol)(273.2 K) 1 mol F2
(0.12 atm)(0.25 L)
= 0.0013 mol Xe
(0.082057 L  atm/K  mol)(273.2 K)
0.0054 mol F
4 mol F
=
The empirical formula is XeF4
0.0013 mol Xe 1 mol Xe
nXe =
11.73
d is not correct. The rate of effusion is inversely proportional to the square of a gas’s molar mass.
11.74
(a) (12.0 g CO2)(1mol/44.01 g) = 0.273 mol
P = nRT/V = (0.273 mol)(0.08206 L·atm/mol·K)(298 K)/0.500 L = 13.4 atm
P = nRT/(V-bn) – a(n/V)2 =
(0.273 mol)(0.08206 L·atm/mol·K)(298 K)/(0.500 L – (0.0427 L/mol) (0.273 mol) )
– (3.59 atm·L2/mol2)(0.273 mol/0.500 L)2
P = 12.6 atm
(13.4 atm – 12.6 atm)/12.6 atm x 100 % = 6 %
(b) P = nRT/V = (0.273 mol)(0.08206 L·atm/mol·K)(203 K)/0.500 L = 9.10 atm
P = nRT/(V-bn) – a(n/V)2 =
(0.273 mol)(0.08206 L·atm/mol·K)(203 K)/(0.500 L – (0.0427 L/mol) (0.273 mol))
– (3.59 atm·L2/mol2)(0.273 mol/0.500 L)2
P = 8.25 atm
(9.10 atm – 8.25 atm)/8.25 atm x 100 % = 10 %
231
Chapter 11
11.75
Gases and Their Properties
104 s (1 min/60 s) = 1.73 min
(0.033 mol/min)/(0.033 mol/1.73 min) = (X/44.0 g/mol) 1/2
X = 1.3 x 102 g/mol
11.76
rateCH 4
rateCF4

M CF4 M CH 4
rateCH 4
0.14 mol/min
= (88.0 g/mol/16.0 g/mol)1/2
= 0.33 mol/min
0.66 mol(1/0.33 mol/min) = 2.0 min
11.77
Helium pressure = gauge pressure + barometric pressure = 22 mm Hg + 755 mm Hg = 777 mm Hg


1 atm
 777 mm Hg ·
 (0.305 L)
760
mm
Hg
PV


n=
=
= 0.0128 mol He
RT
(0.082057 L  atm/K  mol)(298 K)
11.78
1 atm
23.8 mm Hg 
n
P
6.022  1023 molecules
1L
760 mm Hg
=
=


3
V
RT
(0.082057 L  atm/K  mol)(298 K)
1 mol
10 cm3
= 7.71  1017 molecules/cm3
11.79
nO2


1 atm
 735 mm Hg ·
 (0.327 L)
760 mm Hg 
= 
= 0.0132 mol O2
(0.082057 L  atm/K  mol)(292 K)
0.0132 mol O2 
2 mol KClO3
122.5 g

= 1.08 g KClO3
3 mol O2
1 mol KClO3
1.08 g KClO3
 100% = 69.1%
1.56 g mixture
11.80
Ptotal = PH2O + PO2 + PCO2 + PN2
PN2 = 253 mm Hg – 47.1 mm Hg – 35 mm Hg – 7.5 mm Hg
PN2 = 163 mm Hg
11.81
232
(a) NO2 < O2 < NO
(b) 150 mm Hg ·
1 mol O2
= 75 mm Hg
2 mol NO
(c) 150 mm Hg ·
2 mol NO2
= 150 mm Hg
2 mol NO
Chapter 11
11.82
Gases and Their Properties
(a) 562 g ·
nRT
(49.5 mol)(0.082057 L  atm/K  mol)(329 K)
=
= 1360 L
1 atm
P
745 mm Hg ·
760 mm Hg
VH2 =
(b) 562 g ·
Vair =
11.83
PNH3 =
PF2 =
1 mol NH3
3 mol H2
·
= 49.5 mol H2
17.03 g
2 mol NH3
1 mol NH3
1 mol N2
100.0 mol air
·
·
= 21.1 mol air
17.03 g
78.1 mol N2
2 mol NH3
nRT
(21.1 mol)(0.082057 L  atm/K  mol)(302 K)
=
= 534 L
1 atm
P
745 mm Hg ·
760 mm Hg
4 mol NH3
· 120 mm Hg = 69 mm Hg
7 mol gas
3 mol F2
· 120 mm Hg = 51 mm Hg
7 mol gas
Ptotal = 120 mm Hg ·
11.84
(a) nClF3
1 mol NF3
= 17 mm Hg
7 mol gaseous reactants


1 atm
 250 mm Hg ·
 (2.5 L)
760
mm
Hg
PV


=
=
= 0.034 mol ClF3
RT
(0.082057 L  atm/K  mol)(293 K)
0.034 mol ClF3 
6 mol NiO
74.7 g

= 3.8 g NiO
4 mol ClF3 1 mol NiO
(b) Partial pressures:
3 mol O2
= 0.026 mol O2
4 mol ClF3
2 mol Cl2
0.034 mol ClF3 
= 0.017 mol Cl2
4 mol ClF3
0.034 mol ClF3 
nRT
(0.026 mol)(0.082057 L  atm/K  mol)(293 K)
=
= 0.25 atm = 190 mm Hg
V
2.5 L
nRT
(0.017 mol)(0.082057 L  atm/K  mol)(293 K)
PCl2 =
=
= 0.16 atm = 120 mm Hg
V
2.5 L
Ptotal = 190 mm Hg + 120 mm Hg = 310 mm Hg
PO2 =
11.85
 45 mm Hg 
(a) PH2 O = (relative humidity)(H2 O vapor pressure) = 
 (17.5 mm Hg) = 7.9 mm Hg at 20 ºC
 100 mm Hg 


1 atm
 7.9 mm Hg ·
 (18.0 g/mol)
760 mm Hg 
PM
d=
= 
= 0.0078 g/L
RT
(0.082057 L  atm/K  mol)(293 K)
233
Chapter 11
Gases and Their Properties
 95 mm Hg 
(b) PH2 O = (relative humidity)(H2 O vapor pressure) = 
 (4.6 mm Hg) = 4.4 mm Hg at 0 ºC
 100 mm Hg 


1 atm
 4.4 mm Hg ·
 (18.0 g/mol)
760
mm
Hg
PM


d=
=
= 0.0046 g/L
RT
(0.082057 L  atm/K  mol)(273 K)
11.86
 55 mm Hg 
PH2 O = (relative humidity)(H2 O vapor pressure) = 
 (21.1 mm Hg) = 12 mm Hg
 100 mm Hg 


1 atm
1L 
2
12 mm Hg ·
 4.5 m · 3.5 m · –3 3 
760 mm Hg  
PV
10 m 
n=
= 
= 10. mol CO2
RT
(0.082057 L  atm/K  mol)(296 K)
10. mol ·
11.87
44.0 g
= 4.4 x 10-2 g CO2
1 mol CO2
PCO2 = 1.56 atm – 1.34 atm = 0.22 atm
nCO2 =
PV
(0.22 atm)(0.55 L)
=
= 0.0050 mol CO2
RT
(0.082057 L  atm/K  mol)(297 K)
0.0050 mol CO2 ·
44.0 g
= 0.22 g CO2
1 mol CO2

1 mol O2 
 0.0870 g ·
 (0.082057 L  atm/K  mol)(297 K)
32.00 g 
nRT

PO2 =
=
= 0.12 atm
V
0.55 L
PCO = 1.56 atm – 0.22 atm – 0.12 atm = 1.22 atm
nCO =
PV
(1.22 atm)(0.55 L)
=
= 0.028 mol CO
RT
(0.082057 L  atm/K  mol)(297 K)
0.028 mol CO ·
11.88
28.0 g
= 0.77 g CO
1 mol CO
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g)
Assume a one-minute time period:

1 atm 
 773 mm Hg ·
 (5.0 L)
760 mm Hg 
nCH4 = 
= 0.21 mol CH4
(0.082057 L  atm/K  mol)(301 K)
2 mol O2
0.21 mol CH4 ·
= 0.41 mol O2
1 mol CH4
(0.41 mol O2 )(0.082057 L  atm/K  mol)(299 K)
VO2 =
= 10 L O2
1 atm
742 mm Hg ·
760 mm Hg
The oxygen must be supplied to the burner at a rate of 10 L/min.
234
Chapter 11
11.89
nCO2
Gases and Their Properties


1 atm
 44.9 mm Hg ·
 (1.50 L)
760 mm Hg 
PV

=
=
= 0.00362 mol CO2
RT
(0.082057 L  atm/K  mol)(298 K)
0.00362 mol CO2 ·
1 mol CO
28.01 g
·
= 0.102 g CO
1 mol CO2 1 mol CO
mass of Fe in sample = 0.142 g sample – 0.102 g CO = 0.040 g Fe
0.040 g Fe ·
1 mol Fe
= 7.3  10–4 mol Fe
55.8 g
0.00362 mol CO
5 mol CO
=
–4
1 mol Fe
7.3  10 mol Fe
11.90
nCO2 = nMCO3
The empirical formula is Fe(CO)5


1 atm
 69.8 mm Hg ·
 (0.285 L)
760 mm Hg 
= 
= 0.00107 mol MCO3
(0.082057 L  atm/K  mol)(298 K)
0.158 g MCO3
= 148 g/mol
0.00107 mol MCO3
148 g/mol = Mmetal + MCO3 = Mmetal + 60.0 g/mol
Mmetal = 88 g/mol The metal is probably Sr (87.6 g/mol)
11.91
(a) 0.136 g ·
P=
1 mol NaBH4
1 mol B2 H6
·
= 0.00180 mol B2 H6
37.83 g
2 mol NaBH4
 0.00180 mol  (0.082057 L  atm/K  mol)(298 K)
nRT
=
= 0.0160 atm
V
2.75 L
(b) PH2 = 0.0160 atm ·
2 mol H2
= 0.0320 atm
1 mol B2 H6
Ptotal = 0.0160 atm + 0.0320 atm = 0.0480 atm
11.92
nN2


1 atm
 713 mm Hg 
 (0.295 L)
760
mm
Hg
PV

=
= 
= 0.0115 mol N2
RT
(0.082057 L  atm/K  mol)(294.2 K)
0.0115 mol N2 ·
1 mol NaNO2
69.00 g
·
= 0.791 g NaNO2
1 mol N2
1 mol NaNO2
0.791 g NaNO2
· 100% = 64.2%
1.232 g sample
11.93
0.0120 L ·
1.50 mol HCl
= 0.0180 mol HCl
1L
Set up two equations with two unknowns.
1.249 g = x mol NaHCO3 ·
84.007 g
105.99 g
+ y mol Na2CO3 ·
1 mol NaHCO3
1 mol Na2 CO3
235
Chapter 11
Gases and Their Properties

1 mol HCl  
2 mol HCl 
0.0180 mol HCl =  x mol NaHCO3 ·
 +  y mol Na 2 CO3 ·

1 mol NaHCO3  
1 mol Na 2 CO3 

Substitute (0.0180 – 2y) for x and solve.
y = 0.00424 mol Na2CO3; x = 0.00952 mol NaHCO3
nCO2 = 0.00424 mol Na2 CO3 ·
V =
11.94
nCO2
1 mol CO2
1 mol CO2
+ 0.00952 mol Na 2CO3 ·
= 0.0138 mol CO2
1 mol Na2 CO3
1 mol NaHCO3
nRT
(0.0138 mol)(0.082057 L  atm/K  mol)(298 K)
=
= 0.343 L
1 atm
P
745 mm Hg ·
760 mm Hg


1 atm
 735 mm Hg ·
 (0.665 L)
760
mm
Hg
PV

=
= 
= 0.0263 mol CO2
RT
(0.082057 L  atm/K  mol)(298 K)
Set up two equations with two unknowns.
0.0263 mol CO2 = x mol Na2 CO3 ·
2.50 g = x mol NaHCO3 ·
1 mol CO2
1 mol CO2
+ y mol Na2 CO3 ·
1 mol Na2 CO3
1 mol NaHCO3
84.01 g
106.0 g
+ y mol Na2CO3 ·
1 mol NaHCO3
1 mol Na2 CO3
Substitute (0.0263 – y) for x and solve.
y = 0.0132 mol Na2CO3; x = 0.0131 mol NaHCO3
84.01 g
1 mol NaHCO3
· 100% = 44.0% NaHCO3
2.50 g mixture
0.0131 mol ·
106.0 g
1 mol Na 2 CO3
· 100% = 56.0% Na2CO3
2.50 g mixture
0.0132 mol ·
11.95
2 Cu(NO3)2(s)  4 NO2(g) + O2(g) + 2 CuO(s)


1 atm
 725 mm Hg ·
 (0.125 L)
760 mm Hg 
PV

n=
=
= 0.00472 mol
RT
(0.082057 L·atm/K·mol)(308 K)
0.195 g
= 41.3 g/mol
0.00472 mol
Set up two equations with two unknowns.
0.00472 mol = x mol NO2 + y mol O2
0.195 g = x mol NO2 ·
46.01 g
32.00 g
+ y mol O2 ·
1 mol NO2
1 mol O2
Substitute (0.00472 – y) for x and solve.
y = 0.00158 mol O2; x = 0.00314 mol NO2
236
Chapter 11
Gases and Their Properties
X NO2 =
0.00314 mol NO2
= 0.665
(0.00314 + 0.00158) mol
XO2 = 1 – X NO2 = 0.335
If some NO2 molecules combine to form N2O4, the apparent mole fraction of NO2 would be smaller than
expected (0.8). As this is the case, it is apparent that some N2O4 has been formed (as is observed in the
experiment).
11.96
0.09912 g H2O ·
1 mol H2 O
2 mol H
·
= 0.01100 mol H
1 mol H2 O
18.015 g
0.01100 mol H ·
1.0079 g
= 0.01109 g H
1 mol H
H2CO3(aq) + 2 NaOH(aq)  2 H2O() + Na2CO3(aq)
0.02881 L ·
0.3283 mol NaOH 1mol H2 CO3
1 mol C
·
·
= 0.004729 mol C
1L
2 mol NaOH 1 mol H2 CO3
0.004729 mol C ·
nN2
12.011 g
= 0.05680 g C
1 mol C


1 atm
 65.12 mm Hg ·
 (0.2250 L)
760
mm
Hg
PV


=
=
= 7.88  10–4 mol
RT
(0.082057 L·atm/K·mol)(298 K)
7.88  10–4 mol N2 ·
2 mol N
= 0.00158 mol N
1 mol N2
0.00158 mol N ·
14.007 g
= 0.0221 g N
1 mol N
0.1152 g compound – (0.01109 g H) – (0.05680 g C) – (0.0221 g N) = 0.0252 g O
0.0252 g O ·
1 mol O
= 0.00158 mol O
16.00 g
0.01100 mol H
= 7 mol H/1 mol N
0.00158 mol N
0.00158 mol O
= 1 mol O/1 mol N
0.00158 mol N
150 g/mol
= 2.1
73.1 g/mol
11.97
0.004729 mol C
= 3 mol C/1 mol N
0.00158 mol N
The empirical formula is C3 H7 NO (73.1 g/mol)
The molecular formula is C6H14N2O2
(a) M = dRT/P
M = (5.6 g/L)(0.082057 L·atm/mol·K)(291.4 K)/(0.971 atm) = 138 g/mol
Molar mass: PF3, 88 g/mol; PF5, 126 g/mol; P2F4, 138 g/mol
Gas is P2F4
(b)
rate unknown
=
rate CO2
M CO2
M unknown
237
Chapter 11
Gases and Their Properties
(0.028 mol/min)/(0.050 mol.min) = [(44.01 g/mol)/X] 1/2
X = 1.4 x 102 g/mol; this is consistent with the molar mass of 138 g/mol
11.98
(a) C3H8(g) + 5 O2(g)  3 CO2(g) + 4 H2O()
(b) n = PV/RT = (0.10 atm)(1.50 L)/(0.08206 L·atm/mol·K)(293 K) = 0.0062 mol
(c) X =
PCO 2
PT
= (0.10 atm/(0.10 atm + 5.0 atm) = 0.020
(d) n = PV/RT = (5.0 atm)(1.50 L)/(0.08206 L·atm/mol·K)(293 K) = 0.31 mol
moles O2 remaining = 0.31 mol – 0.019 mol = 0.29 mol O2
(e) (3 mol CO2/1 mol C3H8)0.0062 mol = 0.019 mol CO2
PCO2 = (0.019 mol)(0.08206)(296.4 K)/(1.50 L) = 0.31 atm
(f)
11.99
PO2 = (0.29 mol)(0.08206)(296.4 K)/(1.50 L) = 4.7 atm
(a) There are more moles of O2 in the flask (oxygen has a smaller molar mass), so the oxygen has a
greater partial pressure.
(b) Oxygen has a smaller molar mass so its molecules have the greater average speed.
(c) The gases are at the same temperature so the average kinetic energy is the same for both gases.
11.100 (a) True
(b) False
Nitrogen (N2) has a smaller molar mass than O2, so an equal mass of N2 contains more
moles and more molecules than the O2 sample
11.101 (a) Acetylene has a smaller molar mass, so there are more moles of gas in the acetylene cylinder and thus
the pressure is greater.
(b) The acetylene cylinder has a greater number of molecules.
11.102 At a constant pressure, the number of moles of a gas is inversely proportional to the temperature of the gas.
Therefore flask B (at a lower temperature) contains more moles (and more molecules) of oxygen.
11.103 (a) probably not a gas (a gas would expand more)
(b) probably not a gas (density = 8.2 g/mL, too large for a gas)
(c) insufficient information (could also be a liquid or a solid)
(d) gas
11.104 (a) All four tires have the same pressure, temperature, and volume. They have the same number of gas
molecules.
(b) 160. g/16.0 g = 10 times heavier
(c) All the molecules have the same kinetic energy (they have the same temperature). Helium is the
lightest gas of the four, so its molecules have the greatest average speed.
238
Chapter 11
Gases and Their Properties
11.105
(a)
Helium balloon
Hydrogen balloon
Volume
V
2V
Pressure
2 atm
1 atm
Temperature
296 K
268 K
nHe
PV / RT
(2 atm)(V )/(R )(296 K)
0.9 mol He
=
=
=
nH2
PV / RT
(1 atm)(2  V )/(R )(268 K)
1 mol H2
There are more moles of H2 than He, so there are more molecules of H2 than He.
(b) Assume that the hydrogen balloon contains 1 mol (2.02 g) of H 2. This must mean that the helium
balloon contains 0.9 mol (3.6 g) of He. Therefore, there is a greater mass of helium present.
11.106 (a) 65.0 g 
1 mol Na
= 2.83 mol Na
23.00 g
(2.12 atm)(35.0 L)
= 3.05 mol N2 O
(0.082057 L  atm/K  mol)(296 K)
3.05 mol N2 O 1.08 mol N2 O
0.75 mol N2 O
=
>
2.83 mol Na
1 mol Na
1 mol Na
1 mol NaN3
65.02 g
2.83 mol Na 

= 45.9 g NaN3
4 mol Na
1 mol NaN3
nN2 O =
(b)
N
0
N
+1
N
–
–2
N
–1
N
+1
N
–1
–
Na is the limiting reactant
N
–2
N
N
+1
0
–
The formal charges (shown below the resonance structures) indicate that the center resonance structure
is most likely.
(c) The azide ion is linear.
11.107 Speed of gas molecules is related to the square root of the absolute temperature, so a doubling of the
temperature will lead to an increase of about (2)1/2 or 1.4.
11.108 (a) 7 + 2(6) = 19 valence electrons
(b)
(c)
–
O
sp3
Cl
O
The ion is bent
(d) Ozone has a larger bond angle. The central atom is sp2 hybridized. O
(e) 15.6 g ·
nCl2
O
O
1 mol NaClO2
= 0.172 mol NaClO2
90.44 g


1 atm
1050 mm Hg ·
 (1.45 L)
760
mm
Hg
PV


=
=
= 0.0828 mol Cl2
RT
(0.082057 L  atm/K  mol)(295 K)
239
Chapter 11
Gases and Their Properties
0.172 mol NaClO2
2.08 mol NaClO2
2 mol NaClO2
=
>
0.0828 mol Cl2
1 mol Cl2
1 mol Cl2
0.0828 mol Cl2 ·
Cl2 is the limiting reactant
2 mol ClO2
67.45 g
·
= 11.2 g ClO2
1 mol Cl2
1 mol ClO2
SOLUTIONS TO APPLYING CHEMICAL PRINCIPLES: THE GOODYEAR BLIMP
1.
d = MP/RT = (4.002 g/mol)(1.00 atm)/(0.082057 L·atm/mol·K)(298 K) = 0.164 g/L
2.
Average molar mass of dry air = 2.8960 g/mol
d = MP/RT = (28.960 g/mol)(1.00 atm)/(0.082057 L·atm/mol·K)(298 K) = 1.18 g/L
3.
340. m3 (10 dm/m)3 = 3.40 x 105 L
5740 m3 (10 dm/m)3 = 5.74 x 106 L
d = (gross weight + mass air + mass passengers and ballast)/volume blimp = density of dry air
d = 5.820 x 106 g + (1.18 g/L)(3.40 x 105 L) + mass passengers and ballast/5.74 x 106 L = 1.18 g/L
mass passengers and ballast = 552000 g = 5.52 x 105 g = 5.52 x 102 kg
240