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11
Gases
11.1 a. Gaseous particles have greater kinetic energies at higher temperatures. Because kinetic energy
is a measure of the energy of motion, the gaseous particles must be moving faster at higher
temperatures than at lower values.
b. Because particles in a gas are very far apart, gases can be easily compressed without the
particles bumping into neighboring gas particles. Neighboring particles are much closer
together in solids and liquids, and they will “bump” into each other and repel each other if the
sample is compressed.
c. Because the particles of a gas are very far apart, only a small amount of mass (due to the gas
particles themselves) is found in a given volume of space.
11.2 a. At the higher temperature in the fire, the number of collisions against the container’s walls
increases because the gaseous particles have greater velocities. This increases the pressure in
the can, and if that pressure exceeds what the container can endure, then the container will
explode.
b. The particles of a gas move faster at higher temperatures. This causes the particles to spread
farther apart, reducing the gas sample’s density. The density of air in the balloon is less, which
causes it to rise until its density becomes equal to the surrounding atmosphere.
c. Particles of a gas move rapidly and in straight lines until they reach a person who notices the
odor.
11.3 a. temperature
b. volume
c. amount of gas
d. pressure
11.4 a. temperature
b. pressure
c. volume
d. amount of gas
11.5 Some units used to describe the pressure of a gas are pounds per square inch (lb/in.2, which is
also abbreviated as psi), atmospheres (abbreviated atm), torr, mm Hg, in. Hg, pascals (Pa), and
kilopascals (kPa).
11.6 Statements a, d, and e describe the pressure of a gas.
760 torr
1 atm
14.7 lb/in.2
b. 2.00 atm !
1 atm
760 mm Hg
c. 2.00 atm !
1 atm
101.3 kPa
d. 2.00 atm !
1 atm
11.7 a. 2.00 atm !
" 1520 torr
" 29.4 lb/in.2
" 1520 mm Hg
" 203 kPa
1 atm
760 mm Hg
1 torr
b. 467 mm Hg !
1 mm Hg
1 cm
1 in. Hg
c. 467 mm Hg !
!
10 mm
2.54 cm
5
1.013 ! 10 Pa
d. 0.614 atm !
1 atm
11.8 a. 467 mm Hg !
" 0.614 atm
" 467 torr
" 18.4 in. Hg
" 62 200 Pa
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Chapter 11
11.9 The gases in the diver’s lungs (and dissolved in the blood) will expand because pressure decreases
as the diver ascends. Unless the diver exhales, the expanding gases could rupture the membranes
in the lung tissues. In addition, the formation of gas bubbles in the bloodstream could cause “the
bends.”
11.10 According to Boyle’s law, gases expand as the pressure is decreased. Because atmospheric
pressure decreases as altitude increases, the volume of gas sealed in the bag of chips will increase
at higher altitudes.
11.11 During expiration, the volume (capacity) of the lungs decreases as pressure increases.
11.12 A respirator inflates the lungs with air that is rich in oxygen (and has a low concentration of
carbon dioxide). This allows the blood to pick up more O2 to be delivered to the body’s cells as the
blood circulates.
11.13 a. According to Boyle’s law, for the pressure to increase while temperature and quantity of gas
remain constant, the gas volume must decrease. Thus, cylinder A would represent the final
volume.
1 atm
b. P1 " 650 mm Hg !
" 0.86 atm
P2 " 1.2 atm
760 mm Hg
V1 " 220 mL
V2 " 160 mL
Because P1V2 " P2V2, then V2 " P1V1/P2.
0.86 atm
V2 " 220 mL !
" 160 mL
1.2 atm
11.14 a. According to Boyle’s law, the volume must increase when the pressure is decreased, so
diagram C should represent the balloon’s final volume at an increased altitude.
b. When there is no change in the pressure, there is no change in the volume. Thus, diagram B
would represent the final volume of the balloon.
c. With an increase in pressure, there will be a decrease in volume. Thus, diagram A would
describe the balloon’s final volume.
11.15 a. The pressure doubles when the volume is halved.
b. The pressure falls to one-third the initial pressure when the volume expands to three times its
initial volume.
c. The pressure increases to ten times the original pressure when the volume decreases to 101 its
initial volume.
11.16 a. The volume decreases to one-third the initial volume when pressure increases to three times
its initial pressure.
b. The volume doubles when pressure falls to one-half its initial pressure.
c. The volume is five times greater when pressure is decreased to one-fifth its initial pressure.
11.17 From Boyle’s law, we know that pressure is inversely related to volume. (For example, the pressure
increases when the volume decreases.)
a. Volume increases; pressure must decrease.
10.0 L
" 328 mm Hg
655 mm Hg !
20.0 L
b. Volume decreases; pressure must increase.
10.0 L
655 mm Hg !
" 2620 mm Hg
2.50 L
c. The mL units must be converted to L for unit cancellation in the calculation, and because the
volume decreases, pressure must increase.
10.0 L
1000 mL
655 mm Hg !
!
" 4400 mm Hg
1500 mL
1L
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Gases
d. The mL units must be converted to L for unit cancellation in the calculation, and because the
volume decreases, pressure must increase.
1L
" 0.12 L
120 mL !
1000 mL
10.0 L
655 mm Hg !
! " 55000 mm Hg
0.12 L
11.18 From Boyle’s law, we know that pressure is inversely related to volume.
5.00 L
a. 1.20 atm !
" 6.00 atm
1.00 L
b. The mL units must be converted to L for unit cancellation in the calculation.
1000 mL
5.00 L
!
" 2.4 atm
1.20 atm !
2500 mL
1L
1000 mL
5.00 L
c. 1.20 atm !
!
" 8.0 atm
750 mL
1L
5.00 L
d. 1.20 atm !
" 0.750 atm
8.00 L
11.19 760 mm Hg !
11.20 15.0 atm !
4.5 L
" 1700 mm Hg
2.0 L
20.0 L
" 1.00 atm
300.0 L
11.21 From Boyle’s law, we know that pressure is inversely related to volume. (For example, the pressure
increases when the volume decreases.)
a. Pressure increases; volume must decrease.
760 mm Hg
50.0 L !
" 25 L
1500 mm Hg
b. The mm Hg units must be converted to atm for unit cancellation in the calculation, and
because the pressure increases, volume must decrease.
1 atm
" 1.00 atm
760 mm Hg !
760 mm Hg
1.00 atm
50.0 L !
" 25 L
2.0 atm
c. The mm Hg units must be converted to atm for unit cancellation in the calculation, and
because the pressure decreases, volume must increase.
1 atm
" 1.00 atm
760 mm Hg !
760 mm Hg
1.00 atm
50.0 L !
" 100 L
0.500 atm
d. The mm Hg units must be converted to torr for unit cancellation in the calculation, and
because the pressure increases, volume must decrease.
760 mm Hg
1 torr
50.0 L !
!
" 45 L
850 torr
1 mm Hg
11.22 From Boyle’s law, we know that pressure is inversely related to volume.
0.80 atm
a. 25 mL !
" 50. mL
0.40 atm
0.80 atm
b. 25 mL !
" 10. mL
2.00 atm
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Chapter 11
760 mm Hg
0.80 atm
!
" 6.1 mL
2500 mm Hg
1 atm
760 torr
0.80 atm
d. 25 mL !
!
" 190 mL
80.0 torr
1 atm
c. 25 mL !
11.23 According to Charles’ law, there is a direct relationship between temperature and volume. For
example, volume increases when temperature increases, while the pressure and amount of gas
remain constant.
a. Diagram C describes an increased volume corresponding to an increased temperature.
b. Diagram A describes a decreased volume corresponding to a decreased temperature.
c. Diagram B shows no change in volume, which corresponds to no change in temperature.
11.24 According to Charles’ law, an increase in temperature will give an increase in volume.
a. Because the gas warms, its volume must also increase.
b. Because the warm gas will cool, its volume must decrease.
c. Because the gas warms, its volume must increase.
11.25 Heating a gas in a hot air balloon increases the volume of gas, which reduces its density and
allows the balloon to rise above the ground.
11.26 As the temperature decreases, the volume of the gas in the tire decreases, which makes the tire
appear lower or flat in the morning.
11.27 According to Charles’ law, gas volume is directly proportional to Kelvin temperature when P and
n are constant. In all gas law computations, temperatures must be in kelvin units. (Temperatures in
$C are converted to K by the addition of 273.)
a. When temperature decreases, volume must also decrease.
55$C % 273 " 328 K
75$C % 273 " 348 K
328 K
" 2400 mL
2500 mL !
348 K
b. When temperature increases, volume must also increase.
680 K
" 4900 mL
2500 mL !
348 K
c. #25$C % 273 " 248 K
248 K
" 1800 mL
2500 mL !
348 K
240 K
d. 2500 mL !
" 1700 mL
348 K
11.28 According to Charles’ law, a change in a gas’s volume is directly proportional to the change in its
Kelvin temperature. In all gas law computations, temperatures must be in Kelvin units. (Temperatures in $C are converted to K by the addition of 273.)
a. 0$C % 273 " 273 K
10.0 L
" 683 K
273 K !
683 K # 273 " 410$C
4.00 L
1.2 L
b. 273 K !
" 82 K
82 K # 273 " #191$C
4.00 L
2.50 L
c. 273 K !
" 171 K
171 K # 273 " #102$C
4.00 L
0.0500 L
d. 273 K !
3.4 K # 273 " #270$C
" 3.4 K
4.00 L
11.29 Because gas pressure increases with an increase in temperature, the gas pressure in an aerosol can
may exceed the tolerance of the can when it is heated and cause it to explode.
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Gases
11.30 The pressure in a tire is lower at the typical temperatures encountered on a winter morning, but it
increases as the temperature increases. If the pressure increases beyond the tire’s pressure rating,
then the tire may suffer a “blowout.”
11.31 According to Gay-Lussac’s law, temperature is directly related to pressure. For example, the temperature increases when the pressure increases. In all gas law computations, temperatures must be
in kelvin units. (Temperatures in $C are converted to K by the addition of 273.)
a. When temperature decreases, pressure must also decrease.
P1 " 1200 torr
P2 " ?
T1 " 155$C % 273 " 428 K
T2 " 0$C % 273 " 273 K
273 K
" 770 torr
P2 " 1200 torr !
428 K
b. When pressure decreases, temperature must also increase.
12$C % 273 " 285 K
35$C % 273 " 308 K
308 K
" 1.51 atm
1.40 atm !
285 K
11.32 According to Gay-Lussac’s law, pressure is directly related to temperature. For example, pressure
increases when the temperature increases. In all gas law computations, temperatures must be in
Kelvin units. (Temperatures in $C are converted to K by the addition of 273.)
a. At constant volume,
P1 " 250 torr
P2 " 1500 torr
T1 " 0$C % 273 " 273 K
T2 " ?
1500 torr
" 1600 K # 273 " 1300$C
273 K !
250 torr
b. At constant volume,
P1 " 740 mm Hg
P2 " 680 mm Hg
T1 " 40.$C % 273 " 313 K
T2 " ?
680 torr
" 288 K # 273 " 15$C
313 K !
740 torr
11.33 a. boiling point
c. atmospheric pressure
b. vapor pressure
d. boiling point
11.34 Boiling would occur in b and c because the vapor pressure equals the atmospheric (external)
pressure in both.
11.35 a. Water boils at temperatures less than 100°C because the atmospheric pressure is less than
1 atm on top of a mountain. The boiling point is the temperature at which the vapor pressure of
a liquid becomes equal to the external (in this case, atmospheric) pressure.
b. The pressure inside a pressure cooker is greater than 1 atm; therefore, water boils above 100$C.
Foods cook faster at higher temperatures.
11.36 a. At a higher altitude, the external pressure is less, which means that water boils at a lower
temperature.
b. At an external pressure of 2.0 atm, water will boil at 120$C.
11.37 When Boyle’s, Charles’, and Gay-Lussac’s laws are brought together, they form the combined
gas law.
P1V1
PV
" 2 2
T1
T2
11.38 a. V2 "
P1V1T2
P2T1
b. P2 "
P1V1T2
V2T1
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Chapter 11
11.39 T1 " 298 K; V1 " 6.50 L; P1 " 845 mm Hg (1.11 atm)
a. T2 " 325 K; V2 " 1.85 L
6.50 L
325 K
1.11 atm !
!
" 4.25 atm
1.85 L
298 K
b. T2 " 12$C % 273 " 285 K; V2 " 2.25 L
285 K
6.50 L
!
" 3.07 atm
1.11 atm !
2.25 L
298 K
c. T2 " 47$C % 273 " 320 K; V2 " 12.8 L
6.50 L
320 K
1.11 atm !
!
" 0.605 atm
12.8 L
298 K
11.40 T1 " 112$C % 273 " 385 K; P1 " 1.20 atm; V1 " 735 mL
a. T2 " 281 K; P2 " 658 mm Hg (0.866 atm); V2 " ? mL
1.20 atm
281 K
!
" 743 mL
V2 " 735 mL !
0.866 atm
385 K
b. T2 " 75$C % 273 " 348 K; P2 " 0.55 atm; V2 " ? mL
1.20 atm
348 K
!
" 1400 mL
V2 " 735 mL !
0.55 atm
385 K
c. T2 " #15$C % 273 " 258 K; P2 " 15.4 atm; V2 " ? mL
1.20 atm
258 K
V2 " 735 mL !
!
" 3.84 mL
15.4 atm
385 K
11.41 T1 " 225$C % 273 " 498 K; P1 " 1.80 atm; V1 " 100.0 mL
T2 " #25$C % 273 " 248 K; P2 " 0.100 atm; V2 " ?
1.80 atm
248 K
V2 " 100.0 mL !
!
" 110 mL
0.80 atm
498 K
11.42 T1 " 8$C % 273 " 281 K; V1 " 50.0 mL; P1 " 3.00 atm
T2 " 37$C % 273 " 310. K; V2 " 150.0 mL; P2 " ?
50.0 mL
310 K
P2 " 3.00 atm !
!
" 1.10 atm
150.0 mL
281 K
11.43 Addition of more air molecules to a tire or basketball will increase its volume because more moles
of gas are added.
11.44 Air molecules are escaping from the balloon (it is deflating) as it flies around the room.
11.45 According to Avogadro’s law, the change in the gas volume is directly proportional to the change
in the number of moles of gas.
a. When moles decrease, volume must also decrease. (One-half of 4.00 mol " 2.00 mol.)
2.00 mol
" 4.00 L
8.00 L !
4.00 mol
b. When moles increase, volume must also increase.
1 mol Ne
" 1.24 mol Ne added
25.0 g Ne !
20.18 g Ne
(1.50 mol % 1.24 mol " 2.74 mol)
2.74 mol
" 14.6 L
8.00 L !
1.50 mol
c. 1.50 mol % 3.50 mol " 5.00 mol
5.00 mol
8.00 L !
" 26.7 L
1.50 mol
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Gases
1 mol O2
" 0.150 mol O2
32.00 g O2
New moles " 0.650 mol O2
" 0.65 mol
n2 " 0.150 mol % 0.50 mol
0.65 mol O2
" 65 L
15.0 L !
0.150 mol O2
10.0 L
" 0.100 mol O2 remain
b. 0.150 mol !
15.0 L
1 mol He
" 1.00 mol He
c. 4.00 g He !
4.00 g He
New moles " 1.00 mol % 0.15 mol " 1.15 mol gas
1.15 mol gas
15.0 L !
" 115 L
0.15 mol O2
11.46 a. 4.80 g O2 !
11.47 At STP, the molar volume of any gas is 22.4 L/mol.
1 mol O2
a. 44.8 L O2 !
" 2.00 mol O2
22.4 L
1 mol CO2
b. 4.00 L CO2 !
" 0.179 mol CO2
22.4 L
1 mol O2
22.4 L O2
c. 6.40 g O2 !
!
" 4.48 L O2
32.00 g O2
1 mol O2
1 mol Ne
22.4 L Ne
1000 mL
d. 50.0 g Ne !
!
!
" 55 500 mL Ne
20.18 g Ne
1 mol Ne
1 L Ne
11.48 At STP, the molar volume of any gas is 22.4 L/mol.
22.4 L
a. 2.5 mol N2 !
" 56 L N2
1 mol N2
22.4 L
1000 mL
b. 0.420 mol He !
!
" 9410 mL
1 mol He
1L
1 mol Ne
20.18 g Ne
c. 11.2 L !
!
" 10.1 g Ne
22.4 L
1 mol Ne
1L
1 mol H2
d. 1600 mL !
!
" 0.071 mol H2
1000 mL
22.4 L
11.49 At STP, the volume of any gas is 22.4 L/mol.
a. For F2, the molar mass is 38.00 g/mol.
1 mol F2
38.00 g F2
!
" 1.70 g F2/L
1 mol F2
22.4 L
b. For CH4, the molar mass is 16.04 g/mol.
1 mol CH4
16.04 g CH4
!
" 0.716 g CH4/L
1 mol CH4
22.4 L
c. For Ne, the molar mass is 20.18 g Ne/mol.
1 mol Ne
20.18 g Ne
!
" 0.901 g Ne/L
1 mol Ne
22.4 L
d. For SO2, the molar mass is 64.06 g SO2/mol.
64.06 g SO2
1 mol SO2
!
" 2.86 g SO2/L
1 mol SO2
22.4 L
11.50 At STP, the volume of any gas is 22.4 L/mol.
a. For C3H8, the molar mass is 44.09 g/mol.
44.09 g C3H8
1 mol C3H8
!
" 1.97 g C3H8/L
1 mol C3H8
22.4 L
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Chapter 11
b. For NH3, the molar mass is 17.03 g/mol.
1 mol NH3
17.03 g NH3
!
" 0.760 g NH3/L
1 mol NH3
22.4 L
c. For Cl2, the molar mass is 20.18 g Cl2/mol.
1 mol Cl2
79.90 g Cl2
" 3.17 g Cl2/L
!
1 mol Cl2
22.4 L
d. For Ar, the molar mass is 64.06 g Ar/mol.
39.95 g Ar
1 mol Ar
!
" 1.78 g Ar/L
1 mol Ar
22.4 L
11.51 P "
nRT
(2.00 mol)(0.0821 L&atm)(300 K)
"
" 4.93 atm
V
(10.0 L)(mol&K)
11.52 PV " nRT
nRT
(4.0 mol)(0.0821 L&atm)(291 K)
V"
"
" 68 L
P
(1.40 atm)(mol&K)
11.53 n "
PV
(845 mm Hg)(20.0 L)
32.00 g O2
"
!
" 29.4 g O2
RT
(62.4 L&mm Hg)(295 K)
1 mol O2
mol&K
1 mol Kr
" 0.119 mol Kr
83.80 g Kr
nRT
(0.119 mol)(62.4 L&mm Hg)(298 K)
V"
"
" 3.85 L, or 3850 mL
P
(575 mm Hg)(mol&K)
11.54 10.0 g Kr !
1 mol N2
" 0.892 mol
28.02 g
PV
(630. mm Hg)(50.0 L)
T"
"
" 566 K # 273 " 293$C
nR
(0.892 mol)(62.4 L&mm Hg)
mol&K
11.55 25.0 g N2 !
1 mol CO2
" 0.00514 mol
44.01 g CO2
PV
(455 mm Hg)(0.525 L)
T"
"
" 745 K # 273 " 472$C
nR
(0.00514 mol)(62.4 L&mm Hg)
mol&K
11.56 n " 0.226 g CO2 !
11.57 a. 0.450 L !
1 mol
" 0.0201 mol
22.4 L
0.84 g
" 42 g/mol
0.0201 mol
22.4 L
1.28 g
!
" 28.7 g/mol
b.
1L
1 mol
c. T " 295 K V " 1.00 L P " 685 mm Hg
(685 mm Hg)(1.00 L)
PV
"
" 0.0372 mol
n"
RT
(62.4 L&mm Hg)(295 K)
mol&K
1.48 g
Molar mass "
" 39.8 g/mol
0.372 mol
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Gases
(0.95 atm)(2.30 L)
PV
"
" 0.090 mol
RT
(0.0821 L&atm)(297 K)
mol&K
2.96 g
Molar mass "
" 33 g/mol
0.090 mol
1 mol
" 0.0893 mol
2.00 L !
22.4 L
11.6 g
" 130. g/mol
0.0893 mol
22.4 L
0.715 g
!
" 16.0 g/mol
1L
1 mol
(1.20 atm)(1.00 L)
PV
"
" 0.0429 mol
n"
RT
(0.0821 L&atm)(295 K)
mol&K
0.726 g
" 16.9 g/mol
Molar mass "
0.0429 mol
(685 mm Hg)(1.23 L)
PV
"
" 0.0453 mol
n"
RT
(62.4 L&mm Hg)(298 K)
mol&K
2.32 g
Molar mass "
" 51.2 g/mol
0.0453 mol
1 mol Mg
" 0.339 mol Mg
8.25 g Mg !
24.31 g Mg
1 mol H2
22.4 L
1 mol Mg !
!
" 7.60 L H2 released at STP
1 mol Mg
1 mol H2
(735 mm Hg)(5.00 L)
PV
"
" 0.202 mol H2
n"
RT
(62.4 L&mm Hg)(291 K)
mol&K
1 mol Mg
24.31 g Mg
0.202 mol H2 !
!
" 4.92 g Mg
1 mol H2
1 mol Mg
d. n "
11.58 a.
b.
c.
d.
11.59 a.
b.
1 mol NH4NO3
" 0.322 mol NH4NO3
80.05 g NH4NO3
4 mol H2O
" 0.644 mol H2O
0.322 mol NH4NO3 !
2 mol NH4NO3
(0.644 mol)(0.0821 L&atm)(623 K)
nRT
"
" 34.7 L
V"
P
(0.950 atm)(mol&K)
PV
2 mol NH4NO3
(0.950 atm)(10.0 L)
b. n "
" 0.186 mol O2 !
" 0.372 mol
"
RT
(0.0821 L&atm)(623 K)
1 mol O2
mol&K
80.05 g NH4NO3
0.372 mol NH4NO3 !
" 29.8 g NH4NO3
1 mol NH4NO3
11.60 a. 25.8 g NH4NO3 !
1 mol C4H10
13 mol O2
!
" 6.17 mol O2
58.12 g C4H10
2 mol C4H10
nRT
(6.17 mol)(0.0821 L&atm)(298 K)
V"
"
" 178 L
P
(0.850 atm)(mol&K)
11.61 55.2 g C4H10 !
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Chapter 11
1 mol KNO3
1 mol O2
!
" 0.247 mol O2
101.1 g KNO3
2 mol KNO3
nRT
(0.247 mol O2)(0.0821 L&atm)(308 K)
V"
"
" 5.25 L O2
P
(1.19 atm)(mol&K)
11.62 50.0 g KNO3 !
1 mol Al
3 mol O2
!
" 0.15 mol O2
26.98 g Al
4 mol Al
22.4 L
0.15 mol O2 !
" 3.4 L O2 (STP)
1 mol O2
11.63 5.4 g Al !
(725 mm Hg)(4.00 L)
PV
"
" 0.0676 mol NO2
RT
(62.4 L&mm Hg)(688 K)
mol&K
4 mol NH3
17.03 g NH3
0.0676 mol NO2 !
!
" 1.15 g NH3
4 mol NO2
1 mol NH3
11.64 n "
11.65 a. PO2 " 765 mm Hg # 22 mm Hg " 743 mm Hg
PV
(743 mm Hg)(0.256 L)
"
" 0.0103 mol O2
b. n "
RT
(62.4 L&mm Hg)(297 K)
mol&K
11.66 a. PO2 " 765 # 14 mm Hg " 751 mm Hg
PV
(751 mm Hg)(0.425 L)
"
" 0.0177 mol NO2
b. n "
RT
(62.4 L&mm Hg)(289 K)
mol&K
11.67 Each gas particle in a gas mixture exerts a pressure as it strikes the walls of the container. The
total gas pressure for any gaseous sample is thus a sum of all the individual pressures. When the
portion of the pressure due to a particular type of gaseous particle is discussed, it is only part of
the total. Accordingly, these “portions” are referred to as “partial” pressures.
11.68 Because the helium and oxygen “particles” in the sample must have the same average kinetic
energies, each particle will exert the same average pressure against the container’s walls. For the
partial pressures for each gas to be the same, there must be equal numbers of these two types of
gaseous particles.
11.69 To obtain the total pressure in a gaseous mixture, add the partial pressures (provided each carries
the same pressure unit).
Ptotal " PNitrogen % POxygen % PHelium
" 425 torr % 115 torr % 225 torr " 765 torr
11.70 To obtain the total pressure in a gaseous mixture, add the partial pressures (provided each carries
the same pressure unit).
Ptotal " PArgon % PNeon % PNitrogen
" 415 mm Hg % 75 mm Hg % 125 mm Hg " 615 mm Hg
1 atm
615 mm Hg !
" 0.809 atm
760 mm Hg
11.71 Because the total pressure in a gaseous mixture is the sum of the partial pressures (same pressure
unit), addition and subtraction are used to obtain the “missing” partial pressure.
PNitrogen " Ptotal # (POxygen % PHelium)
" 925 torr # (425 torr % 75 torr) " 425 torr
122
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Gases
11.72 Because the total pressure in a gaseous mixture is the sum of the partial pressures, addition and
subtraction are used to obtain the “missing” partial pressure.
PHelium " Ptotal # PMixture of Oxygen, Nitrogen, and Neon
" 1.50 atm # 1.20 atm " 0.30 atm
11.73 a. 2; fewest number of gas particles exerts the lowest pressure.
b. 1; greatest number of gas particles exerts the highest pressure.
11.74 a.
b.
c.
d.
e.
3; volume increases.
1; volume decreases.
3; volume decreases.
3; volume increases.
2; volume stays the same.
11.75 a.
b.
c.
d.
A; volume decreases when temperature decreases.
C; volume increases when pressure decreases.
A; volume decreases when the moles of gas decrease.
B; doubling temperature doubles the volume, but losing half the gas particles decreases the
volume by half. The two effects cancel and no change in volume occurs.
e. C; increasing the moles increases the volume to keep T and P constant.
11.76 a. Volume decreases; pressure increases.
b. Increase particles; pressure increases.
c. Decrease temperature; pressure decreases.
11.77 V1 " 425 mL T1 " 297 K P1 " 745 mm Hg !
1 atm
" 0.980 atm
760 mm Hg
T2 " 178 K P2 " 0.115 atm
V2 " ? L
(425 mL)(0.980 atm)(178 K)
V2 "
" 2170 mL
(297 K)(0.115 atm)
11.78 a. The volume of the chest and lungs will decrease when compressed during the Heimlich
maneuver.
b. A decrease in volume causes the pressure to increase. A piece of food would be dislodged with
a sufficiently high pressure.
11.79 T1 " 15$C % 273 " 288 K; P1 " 745 mm Hg; V1 " 4250 mL
760 mm Hg
1000 mL
" 912 mm Hg; V2 " 2.50 L !
" 2.50 ! 103 mL
P2 " 1.20 atm !
1 atm
1L
2500 mL
912 mm Hg
288 K !
!
" 207 K # 273 " #66$C
4250 mL
745 mm Hg
11.80 Remember to convert temperatures to Kelvin units and solve for new volume, V2.
380 torr
1 atm
228 K
750 L !
!
!
" 1500 L
0.20 atm
760 torr
281 K
PV
(12 atm)(35.0 L)
" 1.8 mol CO2
"
RT
(0.0821 L&atm)(278 K)
mol&K
6.022 ! 1023 molecules
1.8 mol CO2 !
" 1.1 ! 1024 molecules CO2
1 mol CO2
11.81 n "
1 mol N2
" 1.78 mol N2
28.02 g N2
nRT
(1.78 mol N2)(0.0821 L&atm)(298 K)
P"
"
" 2.92 atm
V
15.0 L(mol&K)
11.82 50.0 g N2 !
123
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Chapter 11
PV
(2500 mm Hg)(2.00 L)
"
" 0.28 mol CH4
RT
(62.4 L&mm Hg)(291 K)
mol&K
16.04 g CH4
0.28 mol CH4 !
" 4.5 g CH4
1 mol CH4
11.83 n "
1 mol O2
" 0.066 mol O2
6.02 ! 1023 molecules
nRT
(0.066 mol)(62.4 L&mm Hg)(278 K)
1000 mL
V"
"
" 1.4 L !
" 1400 mL
P
845 mm Hg(mol&K)
1L
11.84 4.0 ! 1022 molecules !
11.85 T " 297 K
V " 4.60 L
1 mol CO2
1.00 g CO2 !
" 0.0227 mol CO2
44.01 g CO2
760 mm Hg
(0.0227 mol)(0.0821 L&atm)(297 K)
nRT
"
" 0.120 atm !
P"
V
(4.60 L)(mol&K)
1 atm
" 91.5 mm Hg
11.86 a. Ptotal " PNitrogen % Pwater
PNitrogen " Ptotal # Pwater
" 745 mm Hg # 32 mm Hg " 713 mm Hg
PV
(713 mm Hg)(0.25 L)
"
" 0.0094 mol N2
b. n "
RT
(62.4 L&mm Hg)(303 K)
mol&K
28.02 g N2
0.0094 mol N2 !
" 0.26 g N2
1 mol N2
11.87 D "
Mass
32.00 g O2
1 mol
"
" 1.43 g/L
!
Volume
1 mol
22.4 L (STP)
1L
1 mol gas
!
" 0.0100 mol gas
1000 mL
22.4 L (STP)
g
1.15 g gas
Molar mass "
"
" 115 g/mol
mol
0.0100 mol gas
11.88 225 mL !
Using ideal gas law:
1
molar mass (MM)
g RT
g RT
MM "
PV " nRT PV "
MM
PV
(1.15 g)(0.0821 L&atm)(273 K)
MM "
" 115 g/mol
(1.0 atm)(0.225 L)(mol&K)
n"g!
11.89 1 torr " 1 mm Hg Molar mass " g of gas/mol of gas
(748 torr)(0.941 L)
PV
" 0.0385 mol
"
n"
RT
(62.4 L&mm Hg)(293 K)
mol&K
1.62 g
Molar mass "
" 42.1 g/mol
0.0385 mol
124
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Gases
1L
1 mol gas
!
" 0.0340 mol gas
1000 mL
22.4 L (STP)
Molar mass " g/mol " 1.02 g gas/0.0340 mol gas " 30.0 g/mol
11.90 762 mL !
Using ideal gas law:
PV " nRT PV " (g/MM)RT MM "
g RT
PV
(1.02 g)(0.0821 L&atm)(273 K)
" 30.0 g/mol
(1.0 atm)(0.762 L)(mol&K)
Because CH3 has a molar mass of 15.0, there must be two CH3 units in a molecular formula of C2H6.
30.0 g/mol
" 2.00 (CH3)2 " C2H6
15.0 g/EF
MM "
11.91 25.0 g Zn !
1 mol Zn
1 mol H2
22.4 L H2 (STP)
!
!
" 8.57 L H2 (STP)
65.38 g Zn
1 mol Zn
1 mol H2
1 mol Mg
1 mol H2
!
" 0.494 mol H2
24.31 g Mg
1 mol Mg
nRT
(0.494 mol)(62.4 L&mm Hg)(297 K)
"
" 11.0 L
V"
P
(835 mm Hg)(mol&K)
11.92 12.0 g Mg !
1 mol
" 0.415 mol NO2
6.022 ! 1023 molecules NO2
7 mol O2
22.4 L O2
!
" 16 L O2
0.415 mol NO2 !
4 mol NO2
1 mol O2
(725 mm Hg)(5.00 L)
PV
b. n "
"
" 0.0897 mol H2O
RT
(62.4 L&mm Hg)(648 K)
mol&K
4 mol NH3
17.03 g NH3
!
" 1.02 g NH3
0.0897 mol H2O !
6 mol H2O
1 mol NH3
11.93 a. 2.5 ! 1023 molecules NO2 !
(748 torr)(1.88 L)
PV
"
" 0.0769 mol
RT
(62.4 L&mm Hg)(293 K)
mol&K
3.24 g
" 42.1 g/mol
Molar mass "
0.0769 mol
1 mol C
b. 2.78 g C !
" 0.232 mol C ' 0.232 " 1 mol C
12.01 g C
1 mol H
" 0.46 mol H ' 0.232 " 2 mol H
0.46 g H !
1.008 g H
Empirical formula " CH2 " 14.0 g/EF
42.1 g/mol
n"
" 3 (CH2)3 " C3H6 molecular formula
14.0 g/EF
11.94 a. n "
(752 mm Hg) ! (0.782 L)
PV
" 0.0318 mol
"
RT
(62.4 L&mm Hg)(296 K)
mol&K
2.23 g
" 70.1 g/mol
Molar mass " g/mol "
0.0318 mol
Empirical formula mass of CH2 " 14.0 g/mol
70.1 g/mol ' 14.0 g/mol " 5.01
CH2 ! 5 " C5H10
11.95 n "
125
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Chapter 11
11.96 n(N2) "
PV
(840 mm Hg)(2.0 L)
2 mol NO2
46.0 g NO2
"
!
!
" 8.3 g NO2
RT
(62.4 L&mm Hg)(297 K)
1 mol N2
1 mol NO2
mol&K
11.97 The partial pressure of each gas is proportional to the number of particles of each type of gas
that is present. Thus, a ratio of partial pressure to total pressure is equal to the ratio of moles of
that gas to the total number of moles of gases that are present:
PHelium/Ptotal " nHelium/ntotal
Solving the equation for the partial pressure of helium yields:
2.0 mol
" 600 torr
PHelium " 2400 torr !
8.0 mol
6.0 mol
" 1800 torr
POxygen " 2400 torr !
8.0 mol
11.98 Because the total pressure is to be reported in mm Hg, the atm and torr units (for argon and
nitrogen, respectively) must be converted to mm Hg, as follows:
760 mm Hg
" 190 mm Hg
0.25 atm !
1 atm
1 mm Hg
360 torr !
" 360 mm Hg
1 torr
and Ptotal " PArgon % PHelium % PNitrogen
" 190 mm Hg % 350 mm Hg % 360 mm Hg
" 900 mm Hg (9.0 ! 102 mm Hg)
11.99 Because the partial pressure of nitrogen is to be reported in torr, the atm and mm Hg units (for
oxygen and argon, respectively) must be converted to torr, as follows:
760 torr
" 460 torr (O2)
0.60 atm !
1 atm
1 torr
" 425 torr (Ar)
425 mm Hg !
1 mm Hg
and PNitrogen " Ptotal # (POxygen % PArgon)
" 1250 torr # (460 torr % 425 torr) " 370 torr
11.100 Because we need the answer in mm Hg, we must include a pressure unit conversion factor for
oxygen.
760 mm Hg
" 342 mm Hg
POxygen " 0.450 atm !
1 atm
1 mm Hg
" 255 mm Hg
255 torr !
1 torr
Ptotal " 255 mm Hg % 342 mm Hg " 597 mm Hg
11.101 a. PHydrogen " Ptotal # Pwater vapor " 755 mm Hg # 21 mm Hg " 734 mm Hg
(734 mm Hg) ! (0.415 L)
PV
"
b. n "
" 0.0165 mol H2
RT
(62.4 L mm Hg)(296 K)
mol K
2 mol Al
27.0 g Al
c. 0.0165 mol H2 !
!
" 0.297 g Al
3 mol H2
1 mol Al
11.102 a. PO2 " 744 mm Hg # 25 mm Hg " 719 mm Hg
PV
(719 mm Hg) ! (0.226 L)
b. n "
"
" 0.00871 mol O2
RT
(62.4 L mm Hg)(299 K)
mol K
126
47374_11_p113-128.qxd 2/9/07 9:09 AM Page 127
Gases
2 mol KClO3
" 0.00581 mol KClO3
3 mol O2
122.55 g KClO3
d. 0.00581 mol KClO3 !
" 0.712 g KClO3 reacted
1 mol KClO3
c. 0.00871 mol O2 !
11.103 a. False. The flask containing helium has more moles and thus more atoms of helium.
b. False. There are different numbers of moles in the flasks, which means the pressures are
different.
c. True. There are more moles of helium, which makes the pressure of helium greater than that
of neon.
d. False. Density is molar mass divided by molar volume. Because the molar masses are
different, their densities are different.
11.104 a. V " 25.0 L T " 413 K P " 0.900 atm
(0.900 atm)(25.0 L)
PV
"
n"
" 0.664 mol
RT
(0.0821 L&atm)(413 K)
mol&K
92.9 g
" 140. g/mol
Molar mass "
0.664 mol
PV
(1.30 atm)(25.0 L)
b. T "
"
" 596 # 273 " 323$C
nR
(0.664 mol)(0.0821 L&atm)
mol&K
11.105 0.762 L !
1 mol
" 0.0340 mol
22.4 L
1.02 g
" 30.0 g/mol
0.0340 mol
1 mol C
" 0.799 mol C/0.799 mol " 1.00 mol C
9.60 g C !
12.01 g C
1 mol H
" 2.40 mol H/0.799 mol " 3.00 mol H
2.42 g H !
1.008 g H
Empirical formula " CH3 Empirical mass " 12.01 % 3(1.008) " 15.03 g/EF
30.0 g/mol
" 2 (CH3)2 " C2H6
15.03 g/EF
1 mol NaN3
3 mol N2
" 2.03 mol NaN3 !
" 3.04 mol N2
65.02 g NaN3
2 mol NaN3
22.4 L
" 68.2 L
At STP 3.04 mol N2 !
1 mol N2
11.106 132 g NaN3 !
PV
6 mol H2O
(1.00 atm)(7.50 L)
" 0.295 mol O2 !
" 0.295 mol H2O
"
RT
(0.0821 L&atm)(310 K)
6 mol O2
mol&K
1 mol C6H12O6
6 mol H2O
" 0.100 mol C6H12O6 !
18.0 g C6H12O6 !
180.2 g C6H12O6
1 mol C6H12O6
" 0.600 mol H2O
0.295 mol H2O is the smaller number of moles of product. Thus, O2 is the limiting reactant.
18.02 g H2O
" 5.32 g H2O
0.295 mol H2O !
1 mol H2O
11.107 n "
127
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Chapter 11
PV
(1.08 atm)(2.00 L)
"
" 0.0883 mol N2
RT
(0.0821 L&atm)(298 K)
mol&K
PV
(0.118 atm)(4.00 L)
" 0.0193 mol O2 (smaller amount of reactant)
"
n O2 "
RT
(0.0821 L&atm)(298 K)
mol&K
2 mol NO
30.01 g NO
!
" 1.16 g NO
0.0193 mol O2 !
1 mol O2
1 mol NO
If the initial volume is 1 V, the final volume is 1⁄2 V, or 0.500 V. (Temperature remains constant).
V1
P2 " 597 mm Hg !
" 1190 mm Hg
0.500 V2
11.108 nN2 "
128