Chapter 2
Answer Key
Section 2.1, Page
Section 2.2, Page 68
1. Compressibility and Expansion are the properties
of gas that are used to operate the ballast tanks
of the submarine. When water is pumped in, the
air in the tank is compressed. When water is
pumped out, the compressed air expands to fill
the empty tanks.
2. Compressed gases must be kept away from heat
sources and the containers must be handled with
care. Increased temperature would cause the gas
particles to move faster, and that would increase
the pressure inside the container. Damaging a gas
container could cause it to rupture or explode.
Page 68, continued
3. The compressibility of gases allows the gas
spring to absorb the shock of opening or closing
the hatchback.
4. At the same temperature, two gases will have
the same average kinetic energy. Their velocity
will vary in proportion to their respective
masses.
5. As the balloon rises, the pressure outside will
decrease, but the number of particles inside
stays the same. The particles will continue to hit
the side of the balloon pushing outwards. As
the balloon rises, the gas inside will expand,
filling all the space of the balloon.
Page 68, concluded
6. In diffusion, a gas mixes gradually with another
gas in the container, until the particles are
evenly distributed. In effusion, a gas escapes
from a container though one or more small
openings in the container, or through a porous
barrier.
7. Fastest  NH3, HF, CO, NO2, HI slowest
8. Answer: the molar mass of the unknown gas is
24.4 g/mol*
9. The molar mass of the unknown gas is 146 g/l
10.Curve 1 represents the gas with the greatest
mass. (the gas with the lowest rate should be
the one with the greatest mass.)
*full answer for this question on the next slide
Question 8, complete answer
The effusion rate of an unknown gas is 43.0 mL/min. The effusion rate of
CO2 is 32.0 mL/min. What is the molar mass of the unknown gas?
Data:
vx
= 43.0 mL/min
vCO
= 32.0 mL/min
2
To find:
MCO = 44.0 g/mol
Mx
= 24.4 g/mol
2
Answer: The molar
mass of the unknown
gas is about 24.4 g/mol
Step 1: find molar mass of CO2
MCO = 12.0 g/mol + 2(16.0 g/mol)
= 44.0 g/mol
2
Step 2: Use Graham’s Law
vx

vCO2
43

32
M CO2
Mx
44
Mx
43 6.6332

32
Mx
32  6.6332
Mx 
43
M x  (4.9363)2  24.367
Question 9, complete answer
It takes 192 s for an unknown gas to effuse. Nitrogen gas (N2) diffuses
across the same barrier in 84s. What is the molar mass of the unknown
gas.
Data:
tx
= 192 s
tN
= 84 s
2
To find:
MN
= 28.0 g/mol
Mx
= 146 g/mol
2
Answer: The molar
mass of the unknown
gas is about 146 g/mol
Step 1: find molar mass of N2
MN = 2(14.0 g/mol) = 28.0 g/mol
2
Step 2: Use Graham’s Law variation
Mx
tx

t N2
M N2
192

84
x
28
192
x

84 5.2915
192  5.2915
x
84
x  12.09485
x  146 .2855
Section 2.3, Page 74
• 1. Molecules striking the sides of a container
cause pressure.
• 2. the pressure is 0.905 atm or 91.7 kPa.
• 3. a) 450 mmHg or 60.0 kPa
b) 99.9 kPa or 749 mmHg
c) 160 kPa or 1199 mmHg
d) 159.9 kPa or
4. The pressure is 18mmHg or 2.4 kPa*
*I think the book author intended a different answer, but I have answered what the
question asked, not what I think it should have asked.
Section 2.4.1, Page 97
1.
2.
3.
4.
5.
6.
7.
8.
9.
V2= 240 mL
V2= 97.9 mL
P2= 140 kPa (or 1.4x102 kPa)
P2= 66.5 kPa
V2= 2.3 L
P2= 1.2 x 102 kPa (or 120 kPa)
V= 290 L (or 2.9 x 102 L
P= 1.5 x 102 kPa (or 150 kPa)
(a) P1= 108 kPa
(b) The pressure was very high (good weather)
but it is now falling (bad weather coming)
10. V= 8.32 L
Section 2.4.2, Page 97, 98
11. a) 298 K
b) 310 K
c) 423 K
12. a) 100°C
b) -175°C
c) 152°C
13. The volume occupied by the gas will be 33L.
14. The volume of the balloon will be 2.1 L
15. The final temperature will be 606°C
16. The variation will be 660°C
17. The volume of CO2 in the dough is 0.12L
18. The volume of the chlorine is 33 mL
19. The gas must be heated to 55°C
20. The nitrogen will occupy 386 mL
21. The temperature is 369 K or 96°C
Section 2.4.3, Page 98
22.Answer: The temperature will be 97°C.
23.The thermometer will show -3.7°C.
24.The pressure will be 759 mmHg.
25.The manometer will show a pressure of 109 kPa.
26.The temperature in the freezer is -16°C.
27.The gas must be heated to 313°C.
28.The valve will open when the butane is 240°C.
29.The gas will exert a pressure of 271 kPa
30.The pressure exerted by the helium is 18.9 atm
Section 2.4.4, Page 98
31.The volume of the container is 100 L (1x102 L)
32. The volume is
33. The balloon originally contained ≈4 mol of He.
You add 0.5 mol of oxygen, making the new
n≈4.5 mol. V  81L So V1≈72 L
1
4mol
4.5mol
34.Answer: 0.57 mol of air escaped from the tire.
35. Answer: 0.0098 mol of CO2 must be added to
the balloon.
36. There are 6.34x10-3 g of oxygen in the swim
bladder of the fish.
37. The total volume of argon is 77.3 L
38. a) 0.21 mol, b) 9.24x10-4 mol c) 3.7x10-3 mol
39. The volume occupied by the hydrogen is about
180 L
40. The amount of sulphur dioxide is 2.0x10-3 mol.
41. Neon will occupy 50.4 L
42. Hydrogen volume is 12 300 L
43. The volume of hydrogen sulphide is 1400 L
44. The volume of hydrogen is 24.5 L
45. There are 3.01x1023 molecules of helium
46. Seven or Eight envelopes (seven is the closest answer,
but if you needed all the air, you would need eight
envelopes)
47. SATP is a higher temperature than STP (25°C vs. 0°C).
According to kinetic theory at a higher temperature
the molecules would move faster, and try to expand
to a greater volume.
48. The water is a liquid. It does not have a standard
molar volume (only gases do), and liquid is a
condensed form of matter. It will have less volume.
49.The air pressure in the tire is 190 kPa
50.It would take 24 mol of gas
51.Since the rubber ball is elastic, the volume can
increase. If the volume increases the pressure
may not. In the propane tank the volume is
constant, so the pressure must increase.
52.Fifty two
P1
n1
P2
n2
c) 71.4 kPa
e) 175 mol
a) 150 kPa
b) 3.7 mol
d) 158 kPa
Page 104
1. P=104 kPa, V=2L, T=295K so n=PV / RT so
n=0.085mol (for all three containers)
a. The containers hold the same moles (0.085mol), and
therefore molecules of gas
b. The mass of the gases are: 0.34g, 2.7g, and 3.7g*
2. The pressure exerted by the methane is 971 kPa.
3. sample contains 4.10x10-2 mol of methane.
4. The temperature will be 497 K or 224°C
*It could be argued that the answers should be rounded to just one significant digit,
since the volume was given as 2L, rather than 2.0L, but that would render the
answers (0.3g, 3g and 4g) somewhat meaningless.
Page 104 continued
5. What quantity of methane gas is found in a sample with a
volume of 500mL at 35C and 210 kPa
P= 210 kPa
PV=nRT
V= 0.500 L
n = PV / RT
n= ?
n= (210 x 0.5) / (8.31 x 308)
T= 35+273= 308K
n= 0.0410 mol
The amount of methane is 0.041 moles (or 0.66g)
6. Determine the pressure exerted in a 50L cylinder if it
contains 30 mol of air and is heated to 40C?
P=?
PV=nRT
V= 50L
P= nRT / V
n= 30 mol
P= (30 x 8.31 x 313) / 50
T= 40+273=313K
P = 1560.6 kPa
The pressure is 1.6 x 103 kPa
Page 104 continued
9. The molar mass of the gas is 242 g/mol
(2.4x102 g/mol)
10. The mass of sulphur dioxide is 99g
11. The molar mass of the gas is 1.3 x 102 g/mol
12. The molar mass of the gas is 44.2 g/mol. The
gas could be Carbon dioxide or propane
13. The sample contains 0.500 moles
14. The volume of the balloon is 2.5x102 L
15. The pressure of the argon in the bulb is 78 kPa
16. The temperature of the chlorine gas is 23°C
Page 104 continued
7. What volume does 50kg of oxygen gas occupy at a
pressure of 150 kPa and a temperature of 125°C
P=150 kPa
V=?
n= 1562 mol
T=125+273=398K
n= 50000g / 32g/mol= 1562 mol
PV=nRT
V= (nRT) / P
V = (1562 x8.31x398) / 150
V= 34462L
The volume is 3.4 x104 L
8. At what temperature does 10.5 g of ammoina (NH3) gas
exert a pressure of 85 kPa in a 30 L container?
P=85kPa
n= 10.5g / 17 g/mol = 0.618 mol
V=30L
PV = nRT
m=10.5g
T = PV / nR
n=0.618 mol
T= (85x30) / (0.618x8.31)
T=?
T=496.5 k
The temperature would be about 497 K or 224° C
Page 107
1. The gas must be brought to a temperature of
99.55 K or -173.45°C (1.0x102 K or -173°C)
2. The volume of the gas would be 67.34 L (67 L)
3. The volume of the gas would be 5.8L
4. The final temperature will be 231°C (or 504K)
5. The new volume of the balloon is 1.0 L
6. The missing quantities are:
a. 88C b. 15.9 C c. 3.8x102 kPa d. 5 L* e. 1.03x103L
* 4.605 L rounded to 1 SF. (a nasty trick to play on you!)
7. This is a nasty question, since both the volume and
pressure can change, there are too many variables, so
it cannot be solved using the general gas law
a. No answer.
b. No answer.
8. The volume of methane is 127 mL (or 0.127 L)
9. The new volume of the gas is 163 mL
10. The new temperature of the gas is 1.5x102K or
1.2x102°C
11. The temperature of the bulb is 310°C
12. The final volume of the gas is 1.5 times the original
volume.
13. A pressure of 642 kPa must be applied.
Page 113
1. The resulting pressure is 300.65 kPa
2. The partial pressure of Ar is 50.2 kPa
3. Find the percentages of the total pressure.
1. Partial pressure of Ne is 13.92 kPa(14kPa)
2. Partial pressure of He is 26.68 kPa(27kPa)
3. Partial pressure of Rd is 75.4 kPa (75 kPa)
4. The total pressure of mixture is 110 kPa
(1.1x102 kPa)
5. The total number of moles is 2.92 mol. The
partial pressures are:
1.
2.
3.
4.
Methane (CH4) is 748.99 kPa ( 7.5x102 kPa)
Oxygen (O2) is 484.64 kPa ( 4.8x102 kPa)
Nitrogen (N2) is 1101.45 kPa (1.1x103 kPa)
Propane (C3H8) is 237.9 kPa (2.4x102 kPa)
6. The partial pressure of CO2 is 0.7 atm.
7. The partial pressures of the gases are:
1. Carbon dioxide = 3.591 kPa (3.6 kPa)
2. Oxygen = 4.104 kPa ( 4.1 kPa)
3. Water vapour= 94.905 kPa (95 kPa)
8. The container has 64% chlorine inside.
Chapter 2. Physical Properties of Gases
(lines are drawn through questions which may have misprints in the textbook)
1.
2.
3.
4.
Answer is Pressure of the gas is 69.6 kPa
Pressure of CO2 is 103 kPa
It moves towards the open end
Since the end is closed, atmospheric pressure
does not exert force on the mercury, so the
difference in height is equal to the gas
pressure.
5. (explanation pending)
6. The pressure in the cylinder is 57 atmospheres
7. The hydrogen will occupy 375 dm3 (or 375 L)
8. The final temperature of the gas is 49.3°C
9. About 1.7 L of water vapour will form in the cake
10.The volume of chlorine is 170 L
11.The final temperature will be 74°C
12.The final pressure will be 3.0x102 kPa
13. a)318 K
b) 340 K c) 623 K
14. a) 200°C
b) -165°C c) -48°C
15.The volume of CO2 in the dough is 0.56 L
16.The volume of argon is 2.6 L
17.The new volume of gas is 8.2 L
18.The heat increases the kinetic energy inside the
balloon, this can increase the pressure until the
balloon bursts.
19.The nitrogen will occupy 0.50 L
20.The volume of dry air is 21 L
21.The volume of the balloon is 2.1 L
22.The final temperature is 606°C
23.The final temperature of the oxygen is -233°C
24.The pressure in the tire is 252 kPa
25.A pressure of 6.4x102 kPa must be exerted.
26.The nitrogen will occupy 2.8 L
27.The volume of the balloon will be 2.2 L
28.The container can be safely heated to 374°C
29.The final temperature of argon is 273°C
30.The final temperature of neon is -67°C
31.The pressure of oxygen is 9.5x102 kPa
32.The pressure in the bulb will be 24 atm.
33.The volume of the bubble will be 21 mL
34.The molar mass of the gas is 48.2 g/mol
35. The molar mass of halothane is 197 g/mol
36. The molar mass of the gas is71.2 g/mol
37. The oxygen occupies a volume of L
38. The nitrogen occupies a volume of 0.33 L
39. a) 2.41g b) 1.75 g
40. The temperature is 12°C
41. The gas will occupy 5.0 L at 0°C
42. The increase in pressure is 127%
43. There are 0.18 mol of oxygen
44. The effusion rate of nitrogen is higher than the
unknown gas, therefore, by Graham’s law, the molar
mass of the unknown gas must be higher than
nitrogen.