UNIT 2 REVIEW
Part 1
(Pages 181–182)
1. A
2. C
3. B
4. 310
5. 121
6. 3.1
7. D
8. B
9. 1, 3, 4, 2
10. C
11. B
12. 1.2
13. 2, 2, 4, 1
14. A
15. 12.4
16. D
17. B
Solutions
4. T = (37 + 273) K = 310 K
5. P1V1 = P2V2
PV
V2 = 1 1
P2
102 kPa u 250 mL
210 kPa
= 121 mL
102 kPa
or
V2 = 250 mL u
= 121 mL
210 kPa
According to Boyle’s law, the final volume of air will be 121 mL.
=
6. T1 = (22 + 273) K = 295 K
T2 = (–15 + 273) K = 258 K
V1 V2
=
T1 T2
TV
V2 = 2 1
T1
258 K u 3.5 L
=
295 K
= 3.1 L
or
120
V2 = 3.5 L u
Unit 2 Solutions Manual
258 K
= 3.1 L
295 K
Copyright © 2007 Thomson Nelson
According to Charles’ law, the final volume of the balloon will be 3.1 L.
12. T1 = (–23 + 273) K = 250 K
T2 = (12 + 273) K = 285 K
PV
PV
1 1
= 2 2
T1
T2
PV T
V2 = 1 1 2
T1 P2
102 kPa u 1.00 kL u 285 K
250 K u 96 kPa
= 1.2 kL
285 K
102 kPa
V2 = 1.00 kL u
= 1.2 kL
u
250 K
96 kPa
=
or
According to the combined gas law, the final volume of air will be 1.2 kL.
15. 2 NH4NO3(s) Æ 2 N2(g) +
VO2 = 49.6 L u
4 H2O(g) +
49.6 L V
O2(g)
1
= 12.4 L
4
or
VO2 = 49.6 L H 2 O u
1 L O2
= 12.4 L
4 L H 2O
According to the law of combining volumes, the volume of oxygen produced will be
12.4 L.
Part 2
(Pages 182–185)
18. (a) T = (0 + 273) K = 273 K
(b) T = (21 + 273) K = 294 K
(c) T = (–273 + 273) K = 0 K
101.325 kPa
19. (a) 4.00 atm u
= 4.05 u 102 kPa = 0.405 MPa
1 atm
101.325 kPa
= 102 kPa
(b) 763 mm Hg u
760 mm Hg
101.325 kPa
= 4.56 u 104 kPa = 45.6 MPa
1 atm
1 mol
5.1 L u
0.21 mol
24.8 L
1 mol
20.7 m L u
0.924 mol
22.4 L
(c) 450 atm u
20. (a) nCO
(b) nF
2
[Note: These problems can also be solved using the ideal gas law.]
24.8 L
21. (a) VH
500 mol u
12.4 kL
2
1 mol
Copyright © 2007 Thomson Nelson
Unit 2 Solutions Manual
121
(b) VH S
2
56 k mol u
24.8 L
1 mol
1.4 ML
22. (a) The volume of a gas sample decreases proportionally with its pressure, assuming the
amount and temperature remain constant.
(b) The volume of a gas sample increases proportionally with its absolute temperature,
assuming the amount and pressure remain constant.
(c) The volume of a gas sample is directly proportional to the product of its amount and its
absolute temperature, and is inversely proportional to its pressure.
23. A law is empirical. For example, the volume–temperature relationship is a law. A theory is
based on non-observable ideas such as the random, colliding motion of molecules.
24. Avogadro’s idea is theoretical because it is based on a non-observable concept. Molecules in
a gas cannot be seen or counted directly.
25. (a) At low temperatures and high pressures, the behaviours of ideal and real gases differ the
most. The volume of an ideal gas will approach zero; the volume of a real gas will reach
a constant, non-zero value as the gas condenses to a liquid. According to the kinetic
molecular theory, the particles of an ideal gas have negligible size and no forces exist
between the molecules. However, a real gas has molecules with a definite size and
attractive intermolecular forces. As the molecules slow down (at low temperatures) and
become closer together (at high pressure), the intermolecular forces become significant
and the molecules bond together to form a liquid.
(b) At high temperatures and low pressures, real gases behave very much like ideal gases.
According to the kinetic molecular theory, the molecules are moving at high speeds (at
high temperature) and are relatively far apart (at low pressure). Any effect of actual
molecular size and intermolecular forces will be minimal. Under these conditions the
assumptions of the kinetic molecular theory for ideal gases will be valid.
26. P1V1 = P2V2
PV
V2 = 1 1
P2
100 kPa u 28.8 L
350 kPa
= 8.23 L
100 kPa
or V2 = 28.8 L u
350 kPa
V2 = 8.23 L
=
27.
According to Boyle’s law, the final volume of hydrogen gas will be 8.23 L, assuming the
chemical amount and temperature are constant.
T1
(23 273) K
296 K
V1
V2
T1
T2
T2
V2T1
3.5 L u 296 K
V1
4.0 L
t2 = (259 – 273) °C = -14 °C
259 K
According to Charles’s law, the outside temperature is -14 °C.
122
Unit 2 Solutions Manual
Copyright © 2007 Thomson Nelson
28. T = (140 + 273) K = 423 K
PV = nRT
PV
nBr2 =
RT
60 kPa u 18.8 L
=
8.314 kPa • L
u 413 K
mol • K
= 0.33 mol
1 mol • K
60 kPa
or nBr2
18.8 L u
u
413 K
8.314 kPa • L
0.33 mol
According to the ideal gas law, the chemical amount of bromine is 0.33 mol.
29. T1 = (25 + 273) K = 298 K
T2 = (–15 + 273) K = 258 K
PV
PV
1 1
= 2 2
T1
T2
T PV
V2 = 2 1 1
T1 P2
258 K u 100 kPa u 5.00 L
298 K u 91.5 kPa
= 4.73 L
100 kPa
258 K
V2 = 5.00 L u
u
298 K
91.5 kPa
V2 = 4.73 L
=
or
According to the combined gas law, the final volume of helium in the balloon will be 4.73 L.
30. Boyle’s Experiment
Problem
What is the effect of pressure on the volume of a gas?
Design
A volume of air is placed inside a syringe. The pressure on the air inside the syringe is
changed by adding different weights onto the end of the syringe plunger. The weight added is
the manipulated variable and the volume measured is the responding variable. Two important
controlled variables are temperature and chemical amount of air.
Charles’ Experiment
Problem
What is the effect of temperature on the volume of a gas?
Design
A volume of air is placed inside a syringe that is then immersed in a water bath. The
temperature of the water bath is manipulated and the volume of air is measured as the
responding variable. Two important controlled variables are pressure and chemical amount of
air.
1 atm
1.78 atm
31. (a) 180 kPa u
101.326 kPa
Copyright © 2007 Thomson Nelson
Unit 2 Solutions Manual
123
(b)
T1
(15 273) K
288 K
T2
(40 273) K
313 K
P1 V
P2 V
1
P2
2
T2
T1
PT
1 2
180 kPa u 313 K
T1
288 K
196 kPa
According to the combined gas law, the tire pressure becomes 196 kPa.
(c) According to the kinetic molecular theory, an increase in temperature means an increase
in the average speed of the molecules. Because the volume is constant, the molecules will
collide with each other and the walls of the tire more often and with more force,
therefore, the pressure increases.
(d) If the tire pressure is set when the tires are very hot, the pressure may become very low
when the tires cool to ambient temperatures. It should be noted that significant underinflation may damage tires.
T1 = (19.5 + 273) K = 292.5 K
P1 V1 P2 V2
32.
T1
TP
T2 = 1 2
P1
T2
292.5 K u 195 kPa
96.7 kPa
= 590 K
t2 = (590 – 273) °C = 317 °C
195 kPa
or T2 = 292.5 K u
96.7 kPa
= 590 K
t2 = (590 – 273) °C = 317 °C
=
According to the combined gas law, the final gas temperature when the container breaks will
be 317 °C.
33. (a) T1 = (150 + 273) K = 423 K
T2 = (110 + 273) K = 383 K
PV
PV
1 1
= 2 2
T1
T2
T PV
P2 = 2 1 1
T1V2
383 K u 600 kPa u 10.0 kL
423 K u 18.0 kL
= 302 kPa
10.0 kL
383 K
P2 = 600 kPa u
u
18.0 kL
423 K
= 302 kPa
=
or
124
Unit 2 Solutions Manual
Copyright © 2007 Thomson Nelson
According to the combined gas law, the final pressure of the turbine steam will be
302 kPa.
(b) PV = nRT
nH 2 O
PV
RT
mH2O
600 kPa u 10.0 kL
8.314 kPa • L
u 423 K
mol • K
1.71 kmol
18.02 g
1.71 k mol u
30.7 kg
1 mol
or mH 2O
10.0 k L u
1 mol • K
8.314 kPa • L
u
600 kPa
423 K
u
18.02 g
1 mol
30.7 kg
According to the ideal gas law, the mass of steam is 30.7 kg.
34. According to the kinetic molecular theory, warm air has faster-moving molecules that occupy
a larger volume at a constant pressure compared with cooler air. The volume increase makes
such air less dense than surrounding air, so it rises, as denser, cooler air around it falls.
Æ
4 NO(g) + 6 H2O(g)
35. (a) 4 NH3(g) + 5 O2(g)
V
V
1.00 L
All gas volumes measured at the same temperature and pressure.
4
VNH3 = 1.00 L u
= 1.00 L
4
5
VO2 = 1.00 L u
= 1.25 L
4
4 L NH 3
VNH3 = 1.00 L NO u
= 1.00 L
or
4 L NO
5 L O2
VO2 = 1.00 L NO u
= 1.25 L
4 L NO
According to the law of combining volumes, the required volume of ammonia is 1.00 L
and of oxygen is 1.25 L.
(b) Avogadro’s theory states that equal volumes of gases at the same temperature and
pressure contain equal numbers of molecules. Therefore, the ratio of coefficients (number
of molecules) in the equation is the same as the ratio of volumes measured.
36. (a) When 50 mL of oxygen reacts with excess dissolved glucose, the balanced chemical
equation indicates that the same volume (chemical amount) of carbon dioxide gas will be
produced because they both have the same coefficients in the balanced chemical
equation. Therefore, 50 mL of carbon dioxide gas will be produced.
(b) The first reaction will produce greater leavening. The leavening depends on the chemical
amount of carbon dioxide produced. More carbon dioxide is produced per mole of
glucose consumed in the first reaction compared to the second reaction. In the first
reaction, three times more carbon dioxide is produced than in the second reaction (6 mol
versus 2 mol).
37. d CH
16.05 g
4
1 mol
u
1 mol
24.8 L
Copyright © 2007 Thomson Nelson
0.647 g/L
Unit 2 Solutions Manual
125
28.02 g
1 mol
u
1.13 g/L
24.8 L
1 mol
You should be near the floor because methane is less dense (0.647 g/L) than nitrogen
(1.13 g/L), which is the largest component of air.
38. (a) According to Boyle’s law, the pressure and volume are inversely related. The new
volume will be:
1
= 150 mL.
300 mL u
2
(b) The solubility of carbon dioxide gas in the beverage is also important.
(c) The pop is carbonated because there is a large pressure of carbon dioxide above the
liquid, which keeps most of the gas dissolved in the pop. When the can is opened, there is
a sudden decrease in pressure reducing the solubility and causing bubbles to form so
rapidly that some of the liquid is carried out of the container.
d N2
39. Purpose
The purpose of this investigation is to create a possible relationship between two variables.
Problem
What effect does the pressure of nitrogen gas have on its solubility in water at a fixed
temperature?
(a) Analysis
According to the evidence of the graph, the solubility of nitrogen in water increases with
increased pressure.
(b) From the graph, approximately 1.5 mmol/L of nitrogen will dissolve at 225 kPa.
Therefore, in 5.00 L of blood (mostly water), the chemical amount of nitrogen that will
dissolve is:
1.5 mmol
nN 2
5.00 L u
7.5 mmol
1L
(c) From the graph, at 100 kPa the solubility of nitrogen gas is 0.65 mmol/L.
The difference in the solubility of nitrogen from 300 kPa to 100 kPa is (2.00 – 0.65)
mmol/L = 1.35 mmol/L at 25 °C.
1.35 mmol
5.00 L u
6.75 mmol
nN 2
1L
126
Unit 2 Solutions Manual
Copyright © 2007 Thomson Nelson
The chemical amount of nitrogen that comes out of solution is 6.75 mmol.
PV = nRT
nRT
VN 2 =
P
6.75 m mol u
=
8.314 kPa • L
u 298 K
mol • K
100 kPa
= 167 mL
According to the evidence and the ideal gas law, the volume of nitrogen gas that will come
out of solution is 167 mL.
(d) Scuba diving involves using compressed gases (usually air) for breathing under water. The
process of breathing under water also requires that the pressure of the air be the same as the
external water pressure in order to properly and safely expand and contract the lungs. Finally,
the solubility of air (nitrogen) decreases as the pressure decreases, causing nitrogen to come
out of solution. If a diver ascends too quickly, nitrogen comes out of solution and forms
bubbles in the blood vessels, causing a painful condition known as “the bends.” Therefore,
knowledge of gas properties is important for scuba diving.
40. Purpose
The purpose of this investigation is to use gas concepts in a chemical analysis.
Problem
Is the HCFC sample tested CHF2Cl(g), C2H3FCl2(g), or C2H3F2Cl(g)?
Analysis
m = 457.64 g – 454.26 g = 3.38 g
T = (22.0 + 273) K = 295 K
§m·
PV = ¨ ¸ RT
©M ¹
mRT
M=
PV
kPa • L
3.38 g u 8.314
u 295 K
mol • K
=
100.1 kPa u 0.840 L
= 98.6 g/mol
kPa • L
295 K
1
or M
3.38 g u 8.314
u
u
mol • K
100.1 kPa
0.840 L
98.6 g/mol
According to the ideal gas law, the molar mass of the gas is 98.6 g/mol. The substance tested
must be C2H3F2Cl(g) (M = 100.50 g/mol), which has the closest molar mass to the
experimental value.
41. Problem
What is the relationship between the volume of a sealed balloon and the temperature to which
it is subjected?
Prediction
As the temperature of the water surrounding the balloon increases, the volume of the balloon
will also increase.
Copyright © 2007 Thomson Nelson
Unit 2 Solutions Manual
127
Design
An inflated balloon is tied (sealed) and then placed in a series of water baths of varying
temperatures. After the balloon has been submerged in each water bath for 3 min, the
circumference of the balloon (measured consistently along the same line of the balloon) is
measured. The results are then compiled, compared, and interpreted.
Materials
Ɣ balloon
Ɣ pail
Ɣ water of varying temperatures
Ɣ outdoor alcohol thermometer
Ɣ waterproof marker
Ɣ tape measure
Procedure
1. Inflate a balloon to about half of its maximum volume and tie it.
2. Draw a line around the centre of the balloon (circumference) about which you will
measure. (Use a waterproof marker).
3. Measure the circumference of the balloon.
4. Record the room temperature.
5. Fill a pail with enough cold water to completely submerge the balloon.
6. Measure and record the temperature of the water in the pail.
7. Totally submerge the balloon in water for at least 3 min.
8. Remove the balloon and immediately measure its circumference.
9. Gradually increase the temperature of the water in the pail and repeat steps 5 to 8 for as
wide a range of temperatures as possible.
Analysis
Ɣ Use the circumference to calculate the volume of the balloon after each trial (assuming
that the balloon is spherical).
Ɣ Convert the temperatures in Celsius to Kelvin.
Ɣ Graph the volume of the balloon versus the absolute temperature.
42. Normally, the air temperature decreases with altitude near Earth’s surface. Combustion
products from automobiles and industrial processes usually rise as part of a convection cell.
As these combustion products rise, the polluted air is carried away from ground level, and
fresh air comes in to replace it. When a temperature inversion occurs, warmer air at higher
altitudes moves on top of the air near the ground. Convection can no longer occur. Under
these conditions, the polluted air is unable to rise through the warmer air aloft, so it remains
trapped under a layer of warm air.
43. Ɣ
Ɣ
Gases assume the volume and shape of a container, flow easily, and are very
compressible. These familiar observations can easily be explained by the concepts of the
kinetic molecular theory. The familiar observation that hot air rises can be explained by a
combination of kinetic molecular theory and the concepts of molar volume and molar
mass (related to atomic theory). The observation that many gases can be condensed to
liquids is related to the concept of real versus ideal gases and intermolecular forces.
Pressure, temperature, volume and chemical amount are all related by the ideal gas law,
PV = nRT.
Extension
44. The earliest gases used as anaesthetics were nitrous oxide, ether, and chloroform.
Unfortunately, some of these gases were unreliable and some were flammable. Cyclopropane
was later discovered to be an anaesthetic but (like ether) it was flammable. By the 1950s,
128
Unit 2 Solutions Manual
Copyright © 2007 Thomson Nelson
continued research and development led to other non-flammable and safer anaesthetics like
halothane mixed with oxygen or air. Halothane and related compounds are still used today.
Compressed air is the common gas for shallow dives. There are many different mixtures that
are used for deep dives. For example, heliox is a mixture of helium and oxygen. Trimix is a
mixture of oxygen, nitrogen, and helium that is used for very deep dives. The percentage
composition used depends on the depth of the dive. Other noble gases are also occasionally
used in mixtures for breathing or as a decompression gas.
45. Bernoulli proposed that a gas contained tiny particles (“corpuscles”) that moved very rapidly.
These particles collide with the walls of the container creating the gas pressure. This idea was
not accepted because scientists at the time believed that particles of a gas repelled each other
and stayed more or less in one place in a fluid called the ether. The accepted belief was
supported by other work from people like Isaac Newton.
The lack of acceptance of Bernoulli’s hypothesis illustrates several things about the
nature of science. It is difficult to get new ideas accepted unless there is some experimental
evidence to clearly contradict the existing belief and support the new idea. Scientists tend to
stick to accepted beliefs until they are forced to change. At the time, the new hypothesis did
not provide any better explanation than the accepted beliefs which fit with other ideas at the
time. Finally, this may also illustrate that well-known and established scientists have a greater
influence than relatively unknown scientists.
46. (a) A large amount of methane rising through the ocean would create a large area of bubbles.
This mixture would not be dense enough to float a ship. Vessels caught in such an area
would sink abruptly.
(b) Methane rising through the atmosphere would create an area of gas much less dense than
air. Airplanes caught in such an area would lose lift, drop in altitude abruptly, and crash if
the drop took them down to the surface of the water.
(c) One popular hypothesis for the supposed anomalies of the Bermuda triangle is that the
disappearances were due to aliens from outer space. The problem with this or any other
explanation is that evidence clearly shows that some disappearances were in fact due to
severe weather and this region is no more dangerous than any other part of the ocean.
Copyright © 2007 Thomson Nelson
Unit 2 Solutions Manual
129