CHAPTER 14 THE IDEAL GAS LAW
AND KINETIC THEORY
ANSWERS TO FOCUS ON CONCEPTS QUESTIONS
1. (d) This statement is not true, because the number of molecules that have a mass of ten grams
depends on the molecular mass of the substance, which is different for different substances.
2. 10.2 g
nRT
. Increasing n, T, and V together may or may not
V
cause P to increase. Increasing n and T certainly causes P to increase. However, increasing
V causes P to decrease. Therefore, the result of increasing all three variables together
depends on the relative amounts by which they are increased.
3. (b) For an ideal gas, we know that P =
4. 2.25 mol
5. 1.25 kg/m3
6. (a) If the speed of every atom in a monatomic ideal gas were doubled, the root-mean-square
speed of the atoms would be doubled. According to the kinetic theory of gases, the Kelvin
temperature is proportional to the square of the root-mean-square speed (see Equation 14.6).
Therefore, the Kelvin temperature would be multiplied by a factor of 22 = 4.
7. (d) According to the kinetic theory of gases, the average translational kinetic energy per
molecule is proportional to the Kelvin temperature (see Equation 14.6). Since each gas has
the same temperature, each has the same average translational kinetic energy. However, this
2
, and depends on the mass m. Since each type of molecule has the
kinetic energy is 12 mvrms
same kinetic energy, the molecules with the larger masses have the smaller translational rms
speeds vrms.
8. (c) According to the kinetic theory of gases, the internal energy U of a monatomic ideal gas
is U = 32 nRT (Equation 14.7). However, the ideal gas law indicates that PV = nRT, so that
U = 32 PV . Since P is doubled and V is reduced by one-half, the product PV and the internal
energy are unchanged.
Chapter 14 Answers to Focus on Concepts Questions
721
9. (e) Increasing the cross-sectional area of the diffusion channel and increasing the difference
in solute concentrations between its ends certainly would increase the diffusion rate.
However, increasing its length would decrease the rate (see Equation 14.8). Therefore,
increasing all three factors simultaneously could lead to a decrease in the diffusion rate,
depending on the relative amounts of the change in the factors.
10. 6.0 × 10−11 kg/s
722 THE IDEAL GAS LAW AND KINETIC THEORY
CHAPTER 14 THE IDEAL GAS LAW
AND KINETIC THEORY
PROBLEMS
______________________________________________________________________________
1.
SSM REASONING AND SOLUTION Since hemoglobin has a molecular mass of
64 500 u, the mass per mole of hemoglobin is 64 500 g/mol. The number of hemoglobin
molecules per mol is Avogadro's number, or 6.022 × 1023 mol–1. Therefore, one molecule of
hemoglobin has a mass (in kg) of
⎛ 64 500 g/mol ⎞ ⎛ 1 kg ⎞
= 1.07×10−22 kg
⎜
23
−1 ⎟ ⎜ 1000 g ⎟
⎠
⎝ 6.022×10 mol ⎠ ⎝
______________________________________________________________________________
2.
REASONING The number of molecules in a known mass of material is the number n of
moles of the material times the number NA of molecules per mole (Avogadro's number).
We can find the number of moles by dividing the known mass m by the mass per mole.
SOLUTION Using the periodic table on the inside of the text’s back cover, we find that the
molecular mass of Tylenol (C8H9NO2) is
Molecular mass
= 8 (12.011 u) + 9 (1.00794 u)
of Tylenol
Mass of 8
carbon atoms
Mass of 9
hydrogen atoms
+ 14.0067 u + 2 (15.9994) = 151.165 u
Mass of
nitrogen atom
Mass of 2
oxygen atoms
The molecular mass of Advil (C13H18O2) is
Molecular mass
= 13 (12.011 u) + 18 (1.00794 u) + 2 (15.9994) = 206.285 u
of Advil
Mass of 13
carbon atoms
Mass of 18
hydrogen atoms
Mass of 2
oxygen atoms
a. Therefore, the number of molecules of pain reliever in the standard dose of Tylenol is
Chapter 14 Problems
723
m
⎛
⎞
Number of molecules = n N A = ⎜
⎟ NA
⎝ Mass per mole ⎠
⎛ 325 × 10−3 g ⎞
23
21
−1
=⎜
⎟ ( 6.022 × 10 mol ) = 1.29 × 10
⎝ 151.165 g/mol ⎠
b. Similarly, the number of molecules of pain reliever in the standard dose of Advil is
m
⎛
⎞
Number of molecules = n N A = ⎜
⎟ NA
⎝ Mass per mole ⎠
⎛ 2.00 × 10−1 g ⎞
−1
23
20
=⎜
⎟ ( 6.022 × 10 mol ) = 5.84 ×10
⎝ 206.285 g/mol ⎠
______________________________________________________________________________
3.
REASONING AND SOLUTION
a. The molecular mass of a molecule is the sum of the atomic masses of its atoms. Thus, the
molecular mass of aspartame (C14H18N2O5) is (see the periodic table on the inside of the
text’s back cover)
Molecular mass = 14 (12.011 u) + 18 (1.00794 u)
Mass of 14
carbon atoms
Mass of 18
hydrogen atoms
+ 2 (14.0067 u) + 5 (15.9994) = 294.307 u
Mass of 2
nitrogen atoms
Mass of 5
oxygen atoms
b. The mass per mole of aspartame is 294.307 g/mol. The number of aspartame molecules
per mole is Avogadro’s number, or 6.022 × 1023 mol–1. Therefore, the mass of one aspartame
molecule (in kg) is
⎛ 294.307 g/mol ⎞ ⎛ 1 kg ⎞
–25
⎜
⎟ = 4.887 × 10 kg
−1 ⎟ ⎜
23
1000
g
⎝ 6.022 × 10 mol ⎠ ⎝
⎠
______________________________________________________________________________
4.
REASONING The mass of one of its atoms (in atomic mass units) has the same numerical
value as the element’s mass per mole (in units of g/mol). Atomic mass units can be
converted into kilograms using the fact that 1 u = 1.6605 × 10–27 kg. Dividing the mass of
the sample by the mass per mole gives the number of moles of atoms in the sample.
724 THE IDEAL GAS LAW AND KINETIC THEORY
SOLUTION
a. The mass per mole is 196.967 g/mol. Since the mass of one of its atoms (in atomic mass
units) has the same numerical value as the mass per mole, the mass of a single atom is
m = 196.967 u .
b. To convert the mass from atomic mass units to kilograms, we use the conversion factor of
1 u = 1.6605 × 10–27 kg:
⎛ 1.6605 × 10−27 kg ⎞
−25
m = (196.967 u ) ⎜
⎟ = 3.2706 × 10 kg
1
u
⎝
⎠
c. The number n of moles of atoms is equal to the mass m divided by the mass per mole:
285 g
m
=
= 1.45 mol
Mass per mole 196.967 g /mol
______________________________________________________________________________
n=
5.
SSM REASONING The mass (in grams) of the active ingredient in the standard dosage
is the number of molecules in the dosage times the mass per molecule (in grams per
molecule). The mass per molecule can be obtained by dividing the molecular mass (in
grams per mole) by Avogadro’s number. The molecular mass is the sum of the atomic
masses of the molecule’s atomic constituents.
SOLUTION Using N to denote the number of molecules in the standard dosage and
mmolecule to denote the mass of one molecule, the mass (in grams) of the active ingredient in
the standard dosage can be written as follows:
m = Nmmolecule
Using M to denote the molecular mass (in grams per mole) and recognizing that
M
mmolecule =
, where NA is Avogadro’s number and is the number of molecules per mole,
NA
we have
⎛ M ⎞
m = Nmmolecule = N ⎜⎜
⎟⎟
⎝ NA ⎠
M (in grams per mole) is equal to the molecular mass in atomic mass units, and we can
obtain this quantity by referring to the periodic table on the inside of the back cover of the
text to find the molecular masses of the constituent atoms in the active ingredient. Thus, we
have
Chapter 14 Problems
725
Molecular mass = 22 (12.011 u ) + 23 (1.00794 u ) + 1( 35.453 u ) + 2 (14.0067 u ) + 2 (15.9994 u )
Carbon
Hydrogen
Chlorine
Nitrogen
Oxygen
= 382.89 u
The mass of the active ingredient in the standard dosage is
⎛ M
m = N ⎜⎜
⎝ NA
6.
⎞
382.89 g/mol
⎛
⎞
19
−2
⎟⎟ = 1.572 × 10 molecules ⎜
⎟ = 1.00 × 10 g
23
×
6.022
10
molecules/mol
⎝
⎠
⎠
(
)
REASONING AND SOLUTION The number n of moles contained in a sample is equal to
the number N of atoms in the sample divided by the number NA of atoms per mole
(Avogadro’s number):
23
N
30.1×10
n=
=
= 5.00 mol
N A 6.022×1023 mol−1
Since the sample has a mass of 135 g, the mass per mole is
135 g
= 27.0 g/mol
5.00 mol
The mass per mole (in g/mol) of a substance has the same numerical value as the atomic
mass of the substance. Therefore, the atomic mass is 27.0 u. The periodic table of the
elements reveals that the unknown element is aluminum .
______________________________________________________________________________
7.
REASONING
a. The number n of moles of water is equal to the mass m of water divided by the mass per
mole: n = m/(Mass per mole). The mass per mole (in g/mol) of water has the same numerical
value as its molecular mass (which we can determine). According to Equation 4.5, the total
mass of the runner is equal to her weight W divided by the magnitude g of the acceleration
due to gravity, or W/g. Since 71% of the runner’s total mass is water, the mass of water is
m = (0.71)W/g.
b. The number N of water molecules is the product of the number n of moles and
Avogadro’s number NA (which is the number of molecules per mole), or N = n NA .
SOLUTION
a. Starting with n = m/(Mass per mole) and substituting in the relation m = (0.71)W/g, we
have
726 THE IDEAL GAS LAW AND KINETIC THEORY
(0.71)W
m
g
n=
=
Mass per mole Mass per mole
(1)
The molecular mass of water (H2O) is 2 (1.00794 u) + 15.9994 u = 18.0153 u. The mass per
mole is then 18.0153 g/mol. However, we need to convert this value into kilograms per mole
for use in Equation (1). This is because the value for the weight W in Equation (1) is given
in newtons (N), which is a SI unit. The SI unit for mass is the kilogram (kg), not the gram
(g). Converting the mass per mole value to kilograms per mole gives
Mass per mole = 18.0153
⎛
g ⎞ ⎛ 1 kg ⎞
g
= ⎜18.0153
⎟
mol ⎠ ⎜⎝ 103 g ⎟⎠
mol ⎝
Substituting this relation into Equation 1 gives
(0.71)W
(0.71) ( 580 N )
g
9.80 m/s 2
= 2.3 × 103 mol
n=
=
Mass per mole
⎛
g ⎞ ⎛ 1 kg ⎞
⎜18.0153
⎟
mol ⎠ ⎜⎝ 103 g ⎟⎠
⎝
b. The number of water molecules in the runner’s body is
N = n NA = ( 2.3 ×103 mol )( 6.022 × 1023 mol−1 ) = 1.4 ×1027
____________________________________________________________________________________________
8.
REASONING The number n of moles of a species can be calculated from the mass m (in
grams) of the species and its molecular mass, or mass per mole M (in grams per mole),
m
according to n =
. To calculate the percentage, we divide the number n of moles of that
M
species by the total number nTotal of moles in the mixture and multiply that fraction by
100%. The total number of moles is the sum of the numbers of moles for each component.
The component with the greatest number of moles has the greatest percentage. For the three
components described, this would be helium, because it has the greatest mass and the
smallest mass per mole. The component with the smallest number of moles has the smallest
percentage. For the three components described, this would be argon, because it has the
smallest mass and the greatest mass per mole.
SOLUTION The percentage pArgon of argon is
Chapter 14 Problems
727
mArgon
pArgon =
nArgon
nArgon + nNeon + nHelium
× 100 =
M Argon
mArgon
M Argon
m
m
+ Neon + Helium
M Neon M Helium
× 100
1.20 g
39.948 g/mol
=
× 100 = 3.1 %
1.20 g
2.60 g
3.20 g
+
+
39.948 g/mol 20.180 g/mol 4.0026 g/mol
The percentage of neon is
pNeon =
nNeon
nArgon + nNeon + nHelium
×100 =
mArgon
M Argon
mNeon
M Neon
m
m
+ Neon + Helium
M Neon M Helium
× 100
2.60 g
20.180 g/mol
=
×100 = 13.4 %
1.20 g
2.60 g
3.20 g
+
+
39.948 g/mol 20.180 g/mol 4.0026 g/mol
The percentage of helium is
pHelium =
nHelium
nArgon + nNeon + nHelium
×100 =
mArgon
M Argon
mHelium
M Helium
m
m
+ Neon + Helium
M Neon M Helium
× 100
3.20 g
4.0026 g/mol
=
× 100 = 83.4 %
1.20 g
2.60 g
3.20 g
+
+
39.948 g/mol 20.180 g/mol 4.0026 g/mol
9.
SSM REASONING The number n of moles of water molecules in the glass is equal to
the mass m of water divided by the mass per mole. According to Equation 11.1, the mass of
water is equal to its density ρ times its volume V. Thus, we have
728 THE IDEAL GAS LAW AND KINETIC THEORY
n=
m
ρV
=
Mass per mole Mass per mole
The volume of the cylindrical glass is V = π r2h, where r is the radius of the cylinder and h is
its height. The number of moles of water can be written as
n=
ρV
Mass per mole
=
(
ρ π r 2h
)
Mass per mole
SOLUTION The molecular mass of water (H2O) is 2(1.00794 u) + (15.9994 u) = 18.0 u.
The mass per mole of H2O is 18.0 g/mol. The density of water (see Table 11.1) is
1.00 × 103 kg/m3 or 1.00 g/cm3.
n=
(
ρ π r 2h
)
(1.00 g /cm ) (π )( 4.50 cm )
=
3
2
(12.0 cm )
= 42.4 mol
Mass per mole
18.0 g /mol
______________________________________________________________________________
10. REASONING The initial solution contains N0 molecules of arsenic trioxide, and this
number is reduced by a factor of 100 with each dilution. After one dilution, in other words,
the number of arsenic trioxide molecules in the solution is N1 = N0/100. After 2 dilutions,
the number remaining is N2 = N1/100 = N0/1002. Therefore, after d dilutions, the number N
of arsenic trioxide molecules that remain in the solution is given by
N=
N0
(1)
100d
We seek the maximum value of d for which at least one molecule of arsenic trioxide
remains. Setting N = 1 in Equation (1), we obtain 100d = N0. Taking the common logarithm
of both sides yields log(100d ) = log N 0 , which we solve for d:
d log100 = log N0
or
2d = log N0
or
d = 12 log N 0
(2)
The initial number N0 of molecules of arsenic trioxide in the undiluted solution depends
upon the number n0 of moles of the substance present via N 0 = n0 N A , where NA is
Avogadro’s number. Because we know the original mass m of the arsenic trioxide, we can
m
find the number n0 of moles from the relation n0 =
. We will use the
Mass per mole
chemical formula As2O3 for arsenic trioxide, and the periodic table found on the inside of
the back cover of the textbook to determine the mass per mole of arsenic trioxide.
Chapter 14 Problems
729
SOLUTION The mass per mole of arsenic trioxide (As2O3) is the sum of the atomic masses
of its constituent atoms. There are two atoms of arsenic and three atoms of oxygen per
molecule, so we have
Molecular mass = 2 ( 74.9216 u ) + 3 (15.9994 u ) = 197.8414 u
Mass of 2
arsenic atoms
Mass of 3
oxygen atoms
Therefore, the mass per mole of arsenic trioxide is 197.8414 g/mol. From the relations
m
N 0 = n0 N A and n0 =
, we see that the initial number of arsenic trioxide
Mass per mole
molecules in the sample is
mN A
⎛
⎞
m
N 0 = n0 N A = ⎜
⎟ NA =
Mass per mole
⎝ Mass per mole ⎠
(3)
Substituting Equation (3) into Equation (2), we obtain
d=
1 log N
0
2
=
mN A
⎞
1 log ⎛
⎜
⎟
2
Mass per mole
⎝
⎠
=
(
)
⎡ (18.0 g ) 6.022 × 1023 mol−1 ⎤
⎥ = 11.4
⎥
197.8414 g/mol
⎣
⎦
1 log ⎢
2
⎢
It is only possible to conduct integral numbers of dilutions, so we conclude from this result
that d = 11 dilutions result in a solution with at least one molecule of arsenic trioxide
remaining, while d = 12 dilutions yield a solution in which there may or may not be any
molecules of arsenic trioxide present. Thus, the original solution may undergo at most
11 dilutions .
11. SSM REASONING Both gases fill the balloon to the same pressure P, volume V, and
temperature T. Assuming that both gases are ideal, we can apply the ideal gas law
PV = nRT to each and conclude that the same number of moles n of each gas is needed to
fill the balloon. Furthermore, the number of moles can be calculated from the mass m (in
m
. Using this
grams) and the mass per mole M (in grams per mole), according to n =
M
expression in the equation nHelium = nNitrogen will allow us to obtain the desired mass of
nitrogen.
SOLUTION Since the number of moles of helium equals the number of moles of nitrogen,
we have
730 THE IDEAL GAS LAW AND KINETIC THEORY
mHelium
=
M Helium
Number of moles
of helium
mNitrogen
M Nitrogen
Number of moles
of nitrogen
Solving for mNitrogen and taking the values of mass per mole for helium (He) and nitrogen
(N2) from the periodic table on the inside of the back cover of the text, we find
mNitrogen =
M Nitrogen mHelium
M Helium
=
( 28.0 g/mol )( 0.16 g ) = 1.1 g
4.00 g/mol
12. REASONING The number n of moles of an ideal gas that occupy a volume V is determined
by the Kelvin temperature T and pressure P of the gas, as we see from the ideal gas law
PV = nRT (Equation 14.1), where R = 8.31 J/ ( mol ⋅ K ) is the universal gas constant. We
will use Equation 14.1 to determine both the number n1 of moles of air present in the oven at
the initial temperature T1 and pressure P1, and also the final number n2 of moles of air at the
final temperature T2 and pressure P2. Because air is free to move in or out of the oven, the
air pressure inside the oven is the same as the air pressure of the environment outside. The
number Δn of moles that leave the oven is then the difference between n1 and n2:
Δn = n1 − n2
(1)
We note that the volume V of the oven is the same for both the initial and the final
conditions.
SOLUTION Solving PV = nRT (Equation 14.1) for the number n of moles of air in the
oven at a given temperature T and pressure P, we obtain
n=
PV
RT
(2)
Substituting Equation (2) into Equation (1), we obtain an expression for the number Δn of
moles that leave the oven as it heats up:
Δn = n1 − n2 =
⎛
⎞
PV
1 − P2V = V P1 − P2
⎜⎜
⎟
RT1 RT2 R ⎝ T1 T2 ⎟⎠
Using the given values of temperature and pressure in Equation (3), we find that
(3)
Chapter 14 Problems
731
⎡ 0.150 m3 ⎤ ⎛ 1.00 × 105 Pa 9.50 × 104 Pa ⎞
Δn = ⎢
−
⎟⎟ = 2.31 mol
⎥ ⎜⎜
8.31
J/
mol
K
296
K
453
K
⋅
(
)
⎠
⎣
⎦⎝
13. REASONING Since the absolute pressure, volume, and temperature are known, we may
use the ideal gas law in the form of Equation 14.1 to find the number of moles of gas.
When the volume and temperature are raised, the new pressure can also be determined by
using the ideal gas law.
SOLUTION
a. The number of moles of gas is
(
5
)(
3
)
1.72 × 10 Pa 2.81 m
PV
n=
=
= 201 mol
RT ⎡⎣8.31 J/ ( mol ⋅ K ) ⎤⎦ ⎡⎣( 273.15 + 15.5 ) K ⎤⎦
(14.1)
b. When the volume is raised to 4.16 m3 and the temperature raised to 28.2 °C, the pressure
of the gas is
P=
nRT ( 201 mol ) ⎡⎣8.31 J/ ( mol ⋅ K ) ⎤⎦ ⎡⎣( 273.15 + 28.2 ) K ⎤⎦
=
= 1.21 × 105 Pa
3
V
4.16 m
(
)
(14.1)
______________________________________________________________________________
14. REASONING According to the ideal gas law, PV = nRT , the temperature T is directly
proportional to the product PV, for a fixed number n of moles. Therefore, tanks with equal
values of PV have the same temperature. Using the data in the table given with the problem
statement, we see that the values of PV for each tank are (starting with tank A):
100 Pa ⋅ m3 , 150 Pa ⋅ m3 , 100 Pa ⋅ m3 , and 150 Pa ⋅ m3 . Tanks A and C have the same
temperature, while B and D have the same temperature.
SOLUTION The temperature of each gas can be found from the ideal gas law,
Equation 14.1:
( 25.0 Pa ) 4.0 m3
PAVA
=
= 120 K
TA =
nR
( 0.10 mol ) ⎡⎣8.31 J /( mol ⋅ K )⎤⎦
(
(
)
)
( 30.0 Pa ) 5.0 m
PV
TB = B B =
= 180 K
nR
( 0.10 mol ) ⎡⎣8.31 J /( mol ⋅ K )⎤⎦
3
732 THE IDEAL GAS LAW AND KINETIC THEORY
TC =
PCVC
nR
( 20.0 Pa ) ( 5.0 m3 )
=
=
( 0.10 mol ) ⎡⎣8.31 J /( mol ⋅ K )⎤⎦
(
120 K
)
( 2.0 Pa ) 75 m
PV
TD = D D =
= 180 K
nR
( 0.10 mol ) ⎡⎣8.31 J /( mol ⋅ K )⎤⎦
3
______________________________________________________________________________
15. REASONING AND SOLUTION The ideal gas law gives
⎛ P ⎞⎛ T ⎞
⎛ 65.0 atm ⎞ ⎛ 297 K ⎞ (
3
3
V2 = ⎜ 1 ⎟ ⎜ 2 ⎟ V1 = ⎜
⎟⎜
⎟ 1.00 m ) = 67.0 m
⎝ 1.00 atm ⎠ ⎝ 288 K ⎠
⎝ P2 ⎠ ⎝ T1 ⎠
______________________________________________________________________________
16. REASONING We can use the ideal gas law, Equation 14.1 (PV = nRT) to find the number
of moles of helium in the Goodyear blimp, since the pressure, volume, and temperature are
known. Once the number of moles is known, we can find the mass of helium in the blimp.
SOLUTION The number n of moles of helium in the blimp is, according to Equation 14.1,
n=
PV (1.1×105 Pa)(5400 m3 )
=
= 2.55 ×105 mol
RT [8.31 J/(mol ⋅ K)](280 K)
According to the periodic table on the inside of the text’s back cover, the atomic mass of
helium is 4.002 60 u. Therefore, the mass per mole is 4.002 60 g/mol. The mass m of
helium in the blimp is, then,
⎛ 1 kg ⎞
3
m = 2.55 ×105 mol ( 4.002 60 g/mol ) ⎜
⎟ = 1.0 ×10 kg
⎝ 1000 g ⎠
______________________________________________________________________________
(
)
17. REASONING The maximum number of balloons that can be filled is the volume of helium
available at the pressure in the balloons divided by the volume per balloon. The volume of
helium available at the pressure in the balloons can be determined using the ideal gas law.
Since the temperature remains constant, the ideal gas law indicates that
PV = nRT = constant, and we can apply it in the form of Boyle’s law, PiVi = PfVf. In this
expression Vf is the final volume at the pressure in the balloons, Vi is the volume of the
cylinder, Pi is the initial pressure in the cylinder, and Pf is the pressure in the balloons.
However, we need to remember that a volume of helium equal to the volume of the cylinder
will remain in the cylinder when its pressure is reduced to atmospheric pressure at the point
when balloons can no longer be filled.
Chapter 14 Problems
733
SOLUTION Using Boyle’s law we find that
Vf =
PV
i i
Pf
The volume of helium available for filling balloons is
Vf − 0.0031 m3 =
PV
i i
Pf
− 0.0031 m3
The maximum number of balloons that can be filled is
(
N Balloons
)(
)
PV
1.6 × 107 Pa 0.0031 m3
i i − 0.0031 m3
− 0.0031 m3
5
Pf
×
1.2
10
Pa
=
=
= 12
VBalloon
0.034 m3
18. REASONING According to the ideal gas law, PV = nRT , the absolute pressure P is
directly proportional to the temperature T when the volume is held constant, provided that
the temperature is measured on the Kelvin scale. The pressure is not proportional to the
Celsius temperature.
SOLUTION
a. According to Equation 14.1, the pressures at the two temperatures are
P1 =
nRT1
V
and P2 =
nRT2
V
Taking the ratio P2/P1 of the final pressure to the initial pressure gives
nRT2
P2
T
70.0 K
= V = 2 =
= 2.00
nRT1
P1
T1
35.0 K
V
b. Before determining the ratio of the pressures, we must convert the Celsius temperature
scale to the Kelvin temperature scale. Thus, the ratio of the pressures at the temperatures of
35.0 °C and 70.0 °C is
P2
T
( 273.15 + 70.0 ) K = 1.11
= 2 =
P1
T1 ( 273.15 + 35.0 ) K
______________________________________________________________________________
734 THE IDEAL GAS LAW AND KINETIC THEORY
19. SSM REASONING According to the ideal gas law (Equation 14.1), PV = nRT . Since n,
the number of moles, is constant, n1R = n2 R . Thus, according to Equation 14.1, we have
PV
1 1
T1
=
P2V2
T2
SOLUTION Solving for T2 , we have
⎛ P ⎞ ⎛V ⎞
⎛ 48.5 P1 ⎞ ⎡ V1 /16 ⎤
T2 = ⎜⎜ 2 ⎟⎟ ⎜⎜ 2 ⎟⎟ T1 = ⎜⎜
⎟⎟ ⎢
⎥ (305 K)= 925 K
⎝ P1 ⎠ ⎝ V1 ⎠
⎝ P1 ⎠ ⎣ V1 ⎦
______________________________________________________________________________
20. REASONING The pressure P of the water vapor in the container can be found from the
ideal gas law, Equation 14.1, as P = nRT / V , where n is the number of moles of water, R is
the universal gas constant, T is the Kelvin temperature, and V is the volume. The variables
R, T, and V are known, and the number of moles can be obtained by noting that it is equal to
the mass m of the water divided by its mass per mole.
SOLUTION Substituting n = m/(Mass per mole) into the ideal gas law, we have
m
⎛
⎞
RT
⎜
nRT ⎝ Mass per mole ⎟⎠
=
P=
V
V
The mass per mole (in g/mol) of water (H2O) has the same numerical value as its molecular
mass. The molecular mass of water is 2 (1.00794 u) + 15.9994 u = 18.0153 u. The mass per
mole of water is 18.0153 g/mol. Thus, the pressure of the water vapor is
⎛
⎞
4.0 g
m
⎛
⎞
[8.31 J/ ( mol ⋅ K )] ( 388 K )
⎜
RT
⎜ 18.0153 g /mol ⎟⎟
⎜ Mass per mole ⎟
⎠
⎠
P=⎝
= 2.4 × 104 Pa
=⎝
3
V
0.030 m
____________________________________________________________________________________________
21. REASONING According to Equation 14.2, PV = NkT , where P is the pressure, V is the
volume, N is the number of molecules in the sample, k is Boltzmann's constant, and T is the
Kelvin temperature. The number of gas molecules per unit volume in the atmosphere is
N / V = P /(kT) . This can be used to find the desired ratio for the two planets.
Chapter 14 Problems
735
SOLUTION We have
( N / V ) Venus
( N / V ) Earth
=
PVenus / (kTVenus ) ⎛ PVenus ⎞ ⎛ TEarth ⎞ ⎛ 9.0 × 10 6 Pa ⎞ ⎛ 320 K ⎞
⎟ = 39
=⎜
⎟⎜
⎟=⎜
⎟⎜
PEarth / (kTEarth ) ⎝ PEarth ⎠ ⎝ T Venus ⎠ ⎝ 1.0 × 10 5 Pa ⎠ ⎝ 740 K ⎠
Thus, we can conclude that the atmosphere of Venus is 39 times "thicker" than that of Earth.
______________________________________________________________________________
22. REASONING Pushing down on the pump handle lowers the piston from its initial height
hi = 0.55 m above the bottom of the pump cylinder to a final height hf. The distance d the
biker must push the handle down is the difference between these two heights: d = hi − hf .
The initial and final heights of the piston determine the initial and final volumes of air inside
the pump cylinder, both of which are the product of the piston height h and its crosssectional area A:
Vi = Ahi
and
Vf = Ahf
(1)
We will assume that the air in the cylinder may be treated as an ideal gas, and that
compressing it produces no change in temperature. Since none of the air escapes from the
cylinder up to the point where the pressure inside equals the pressure in the inner tube, we
can apply Boyle’s law:
PV
i i = Pf Vf
(14.3)
We will use Equations (1) and Equation 14.3 to determine the distance d = hi − hf through
which the pump handle moves before air begins to flow. We note that the final pressure Pf is
equal to the pressure of air in the inner tube.
SOLUTION Since the cross-sectional area A of the piston does not change, substituting
Equations (1) into Equation 14.3 yields
Pi A hi = Pf A hf
Pi hi = Pf hf
or
(2)
Solving Equation (2) for the unknown final height hf, we obtain
hf =
Pi hi
Pf
(3)
With Equation (3), we can now calculate the distance d = hi − hf that the biker pushes down
on the handle before air begins to flow from the pump to the inner tube:
736 THE IDEAL GAS LAW AND KINETIC THEORY
d = hi − hf = hi −
⎛
P
= hi ⎜⎜ 1 − i
Pf
⎝ Pf
Ph
i i
⎞
⎟⎟
⎠
⎛ 1.0 × 105 Pa ⎞
= ( 0.55 m ) ⎜⎜ 1 −
⎟⎟ = 0.32 m
5
⎝ 2.4 × 10 Pa ⎠
23. SSM WWW REASONING AND SOLUTION
a. Since the heat gained by the gas in one tank is equal to the heat lost by the gas in the
other tank, Q1 = Q2, or (letting the subscript 1 correspond to the neon in the left tank, and
letting 2 correspond to the neon in the right tank) cm1ΔT1 = cm2 ΔT2 ,
cm1 (T − T1 ) = cm2 (T2 − T )
m1 (T − T1 ) = m2 (T2 − T )
Solving for T gives
T=
m2T2 + m1T1
(1)
m2 + m1
The masses m1 and m2 can be found by first finding the number of moles n1 and n2. From
the ideal gas law, PV = nRT, so
n1 =
( 5.0 ×105 Pa )( 2.0 m3 )
PV
1 1 =
= 5.5 × 102 mol
RT1 [8.31 J/(mol ⋅ K) ] ( 220 K )
This corresponds to a mass m1 = (5.5×102 mol) ( 20.179 g/mol ) = 1.1×104 g = 1.1×101 kg .
Similarly, n2 = 2.4 × 102 mol and m2 = 4.9 × 103 g = 4.9 kg. Substituting these mass values
into Equation (1) yields
T=
( 4.9 kg ) ( 580 K ) + (1.1×101 kg ) ( 220 K )
=
( 4.9 kg ) + (1.1× 101 kg )
3.3 × 102 K
b. From the ideal gas law,
nRT ⎡⎣( 5.5 ×102 mol ) + ( 2.4 × 102 mol ) ⎤⎦ [8.31 J/ ( mol ⋅ K )] ( 3.3 × 102 K )
=
= 2.8 × 105 Pa
3
3
V
2.0 m + 5.8 m
______________________________________________________________________________
P=
Chapter 14 Problems
737
24. REASONING Since the temperature of the confined air is constant, Boyle's law applies,
and PsurfaceVsurface = PhVh , where Psurface and Vsurface are the pressure and volume of the
air in the tank when the tank is at the surface of the water, and Ph and Vh are the pressure
and volume of the trapped air after the tank has been lowered a distance h below the surface
of the water. Since the tank is completely filled with air at the surface, Vsurface is equal to
the volume Vtank of the tank. Therefore, the fraction of the tank's volume that is filled with
air when the tank is a distance h below the water's surface is
Vh
Vtank
=
Vh
Vsurface
=
Psurface
Ph
We can find the absolute pressure at a depth h using Equation 11.4. Once the absolute
pressure is known at a depth h, we can determine the ratio of the pressure at the surface to
the pressure at the depth h.
SOLUTION According to Equation 11.4, the trapped air pressure at a depth h = 40.0 m is
Ph = Psurface + ρ gh = (1.01× 105 Pa) + ⎡⎣(1.00 × 103 kg/m3 )(9.80 m/s 2 )(40.0 m) ⎤⎦
= 4.93 ×105 Pa
where we have used a value of ρ = 1.00 ×103 kg/m3 for the density of water. The desired
volume fraction is
1.01 × 10 5 Pa
= 0.205
Vtank
Ph
4.93 × 10 5 Pa
______________________________________________________________________________
Vh
=
Psurface
=
25. REASONING The mass (in grams) of the air in the room is the mass of the nitrogen plus
the mass of the oxygen. The mass of the nitrogen is the number of moles of nitrogen times
the molecular mass (in grams/mol) of nitrogen. The mass of the oxygen can be obtained in a
similar way. The number of moles of each species can be found by using the given
percentages and the total number of moles. To obtain the total number of moles, we will
apply the ideal gas law. If we substitute the Kelvin temperature T, the pressure P, and the
volume V (length × width × height) of the room in the ideal gas law, we can obtain the total
PV
number nTotal of moles of gas as nTotal =
, because the ideal gas law does not distinguish
RT
between types of ideal gases.
SOLUTION Using m to denote the mass (in grams), n to denote the number of moles, and
M to denote the molecular mass (in grams/mol), we can write the mass of the air in the room
as follows:
738 THE IDEAL GAS LAW AND KINETIC THEORY
m = mNitrogen + mOxygen = nNitrogen M Nitrogen + nOxygen M Oxygen
We can now express this result using f to denote the fraction of a species that is present and
nTotal to denote the total number of moles:
m = nNitrogen M Nitrogen + nOxygen M Oxygen = f Nitrogen nTotal M Nitrogen + f Oxygen nTotal M Oxygen
According to the ideal gas law, we have nTotal =
PV
. With this substitution, the mass of
RT
the air becomes
(
)
(
m = f Nitrogen M Nitrogen + f Oxygen M Oxygen nTotal = f Nitrogen M Nitrogen + fOxygen M Oxygen
) PV
RT
The mass per mole for nitrogen (N2) and for oxygen (O2) can be obtained from the periodic
table on the inside of the back cover of the text. They are, respectively, 28.0 and
32.0 g/mol. The temperature of 22 ºC must be expressed on the Kelvin scale as 295 K (see
Equation 12.1). The mass of the air is, then,
(
m = f Nitrogen M Nitrogen + f Oxygen M Oxygen
) PV
RT
1.01× 105 Pa ) ⎡⎣( 2.5 m )( 4.0 m )( 5.0 m ) ⎤⎦
(
= ⎡0.79 ( 28.0 g/mol ) + 0.21( 32.0 g/mol ) ⎤
⎣
⎦
⎡⎣8.31 J/ ( mol ⋅ K ) ⎤⎦ ( 295 K )
= 5.9 × 104 g
26. REASONING The ideal gas law is PV = nRT. Since the number of moles is constant, this
PV
PV
is the same
= nR = constant . Thus, the value of
equation can be written as
T
T
initially and finally, and we can write
P0V0
T0
=
Pf Vf
Tf
This expression can be solved for the final temperature Tf.
(1)
Chapter 14 Problems
739
We have no direct data for the initial and final pressures. However, we can deal with this by
realizing that pressure is the magnitude of the force applied perpendicularly to a surface
divided by the area of the surface. Thus, the magnitudes of the forces that the initial and
final pressures apply to the piston (and, therefore, to the spring) are given by Equation 11.3
as
(2)
F0 = P0 A
and
Ff = Pf A
Force applied by
final pressure
Force applied by
initial pressure
The forces in Equations (2) are applied to the spring, and the force FxApplied that must be
applied to stretch an ideal spring by an amount x with respect to its unstrained length is
given by Equation 10.1 as
FxApplied = kx
(3)
where k is the spring constant.
Lastly, we note that the final volume is the initial volume plus the amount by which the
volume increases as the spring stretches. The increased volume due to the additional
stretching is A ( xf − x0 ) . Therefore, we have
Vf = V0 + A ( xf − x0 )
(4)
SOLUTION The final temperature can be obtained by rearranging Equation (1) to show
that
⎛ PV ⎞
(5)
Tf = ⎜ f f ⎟ T0
⎜ PV ⎟
⎝ 0 0⎠
Into this result we can now substitute expressions for P0 and Pf. These expressions can be
obtained by using Equations (2) in Equation (3) as follows (and recognizing that, for the
initial and final forces, P0A = FxApplied and Pf A = FxApplied):
= kx0
P0 A
and
Pf A
= kxf
(6)
Force applied to
spring by final
pressure
Force applied to
spring by initial
pressure
Thus, we find that
P0 =
kx0
A
and
Pf =
kxf
A
(7)
Finally, we substitute Equations (7) for the initial and final pressure and Equation (4) for the
final volume into Equation (5). With these substitutions Equation (5) becomes
740 THE IDEAL GAS LAW AND KINETIC THEORY
⎛ kxf
⎜ A
⎛ PV ⎞
Tf = ⎜ f f ⎟ T0 = ⎝
⎜ PV ⎟
⎝ 0 0⎠
=
⎞
⎟ ⎡⎣V0 + A ( xf − x0 ) ⎤⎦ T0 x ⎡V + A ( x − x ) ⎤ T
f
0
f
0 ⎦ 0
⎠
= ⎣
x0V0
⎛ kx0 ⎞
⎜ A ⎟ V0
⎝
⎠
( 0.1000 m ) ⎡⎣6.00 ×10−4 m3 + ( 2.50 ×10−3 m 2 ) ( 0.1000 m − 0.0800 m ) ⎤⎦ ( 273 K )
( 0.0800 m ) ( 6.00 ×10−4 m3 )
= 3.70 ×102 K
27. SSM REASONING The graph that accompanies Problem 75 in Chapter 12 can be used
to determine the equilibrium vapor pressure of water in the air when the temperature is
30.0 °C (303 K). Equation 12.6 can then be used to find the partial pressure of water in the
air at this temperature. Using this pressure, the ideal gas law can then be used to find the
number of moles of water vapor per cubic meter.
SOLUTION According to the graph that accompanies Problem 75 in Chapter 12, the
equilibrium vapor pressure of water vapor at 30.0 °C is approximately 4250 Pa. According
to Equation 12.6,
(
)(
)
Partial
1
pressure of = Percent relative Equilibrium vapor pressure of ×
water at the existing temperature 100
humidity
water vapor
=
( 55)( 4250 Pa ) = 2.34 × 103 Pa
100
The ideal gas law then gives the number of moles of water vapor per cubic meter of air as
P
(2.34 × 10 3 Pa)
n
=
=
= 0.93 mol/m 3
(14.1)
V RT [8.31 J/(mol ⋅ K)](303 K)
______________________________________________________________________________
28. REASONING AND SOLUTION If the pressure at the surface is P1 and the pressure at a
depth h is P2, we have that P2 = P1 + ρ gh. We also know that P1V1 = P2V2. Then,
V1
V2
=
P2
P1
=
P1 + ρ gh
P1
= 1+
ρ gh
P1
Chapter 14 Problems
741
Therefore,
(1.000 ×103 kg/m3 )(9.80 m/s 2 )(0.200 m)
= 1.02
V2
1.01×105 Pa
______________________________________________________________________________
V1
= 1+
29. REASONING The desired percentage is the volume the atoms themselves occupy divided
by the total volume that the gas occupies, multiplied by the usual factor of 100. The volume
VAtoms that the atoms themselves occupy is the volume of an atomic sphere ( 43 π r 3 , where r
is the atomic radius), times the number of atoms present, which is the number n of moles
times Avogadro’s number NA. The total volume VGas that the gas occupies can be taken to
be that calculated from the ideal gas law, because the atoms themselves occupy such a small
volume.
SOLUTION The total volume VGas that the gas occupies is given by the ideal gas law as
VGas = nRT/P, where the temperature and pressure at STP conditions are 273 K and
1.01 × 105 Pa. Thus, we can write the desired percentage as
Percentage =
=
VAtoms
VGas
4π
3
4 π r 3 nN
4 π r3 N P
(
)
(
) A ×100
A
3
3
× 100 =
× 100 =
nRT
P
RT
( 2.0 ×10−10 m ) ( 6.022 ×1023 mol−1 )(1.01×105 Pa ) ×100 = 0.090 %
3
⎡⎣8.31 J/ ( mol ⋅ K ) ⎤⎦ ( 273 K )
30. REASONING AND SOLUTION We need to determine the amount of He inside the
balloon. We begin by using Archimedes’ principle; the balloon is being buoyed up by a
force equal to the weight of the air displaced. The buoyant force, Fb, therefore, is equal to
3
Fb = mg = ρ Vg = (1.19 kg/m3 ) ⎡⎣ 43 π (1.50 m ) ⎤⎦ ( 9.80 m/s 2 ) = 164.9 N
Since the material from which the balloon is made has a mass of 3.00 kg (weight = 29.4 N),
the He inside the balloon weighs 164.9 N – 29.4 N = 135.5 N. Hence, the mass of the
helium present in the balloon is m = (135.5 N ) / 9.80 m/s 2 = 13.8 kg . Now we can
(
)
determine the number of moles of He present in the balloon:
n=
m
13.8 kg
=
= 3450 mol
M 4.0026 ×10−3 kg/mol
742 THE IDEAL GAS LAW AND KINETIC THEORY
Using the ideal gas law to find the pressure, we have
P=
nRT ( 3450 mol ) [8.31 J/(mol ⋅ K) ] ( 305 K )
=
= 6.19 × 105 Pa
4 π (1.50 m )3
V
3
______________________________________________________________________________
31. SSM REASONING According to the ideal gas law (Equation 14.1), PV = nRT . Since
n, the number of moles of the gas, is constant, n1R = n2 R . Therefore, PV
1 1 / T1 = P2V2 / T2 ,
where T1 = 273 K and T2 is the temperature we seek. Since the beaker is cylindrical, the
volume V of the gas is equal to Ad, where A is the cross-sectional area of the cylindrical
volume and d is the height of the region occupied by the gas, as measured from the bottom
of the beaker. With this substitution for the volume, the expression obtained from the ideal
gas law becomes
P1d1 P2 d 2
=
(1)
T1
T2
where the pressures P1 and P2 are equal to the sum of the atmospheric pressure and the
pressure caused by the mercury in each case. These pressures can be determined using
Equation 11.4. Once the pressures are known, Equation (1) can be solved for T2 .
SOLUTION Using Equation 11.4, we obtain the following values for the pressures P1 and
P2 . Note that the initial height of the mercury is h1 = 12 (1.520 m) = 0.760 m , while the final
height of the mercury is h2 =
1 (1.520
4
m) = 0.380 m .
P1 = P0 + ρ gh1 = (1.01× 105 Pa ) + ⎡⎣(1.36 × 104 kg/m3 )(9.80 m/s2 )(0.760 m) ⎤⎦ = 2.02 × 105 Pa
P2 = P0 + ρ gh2 = (1.01× 105 Pa ) + ⎡(1.36 × 104 kg/m3 ) ( 9.80 m/s 2 ) (0.380 m) ⎤ = 1.52 × 105 Pa
⎣
⎦
In these pressure calculations, the density of mercury is ρ = 1.36 × 104 kg/m3 . In
Equation (1) we note that d1 = 0.760 m and d 2 = 1.14 m . Solving Equation (1) for T2 and
substituting values, we obtain
⎡ (1.52 × 105 Pa ) (1.14 m) ⎤
⎛Pd ⎞
T2 = ⎜⎜ 2 2 ⎟⎟ T1 = ⎢
⎥ (273 K) = 308 K
5
⎝ P1d1 ⎠
⎣⎢ ( 2.02 × 10 Pa ) ( 0.760 m ) ⎦⎥
______________________________________________________________________________
Chapter 14 Problems
743
32. REASONING AND SOLUTION The volume of the cylinder is V = AL where A is the
cross-sectional area of the piston and L is the length. We know P1V1 = P2V2 so that the new
pressure P2 can be found. We have
⎛V ⎞
⎛ AL ⎞
⎛L ⎞
P2 = P1 ⎜ 1 ⎟ = P1 ⎜ 1 1 ⎟ = P1 ⎜ 1 ⎟
⎝ V2 ⎠
⎝ A2 L2 ⎠
⎝ L2 ⎠
(since A1 = A2 )
⎛ L ⎞
4
= (1.01× 105 Pa ) ⎜
⎟ = 5.05 × 10 Pa
⎝ 2L ⎠
The force on the piston and spring is, therefore,
F = P2A = (5.05 × 104 Pa)π (0.0500 m)2 = 397 N
The spring constant is k = F/x (Equation 10.1), so
F
397 N
=
= 1.98 × 103 N/m
x 0.200 m
______________________________________________________________________________
k=
33. REASONING The average kinetic energy per molecule is proportional to the Kelvin
temperature of the carbon dioxide gas. This relation is expressed by Equation 14.6 as
2
3
1
2 mvrms = 2 k T , where m is the mass of a carbon dioxide molecule. The mass m is equal to
the molecular mass of carbon dioxide (44.0 u), expressed in kilograms.
SOLUTION Solving Equation 14.6 for the temperature of the gas, we have
⎛ 1.66 × 10−27 kg ⎞
2
44.0 u ⎜
⎟ ( 650 m /s )
2
1u
mv
⎝
⎠
= 750 K
T = rms =
−23
3k
3 1.38 × 10 J/K
(
)
(
)
______________________________________________________________________________
34. REASONING According to the kinetic theory of gases, the average kinetic energy of an
2
atom is related to the temperature of the gas by 12 mvrms
= 23 kT (Equation 14.6). We see that
the temperature is proportional to the product of the mass and the square of the rms-speed.
2
Therefore, the tank with the greatest value of mvrms
has the greatest temperature. Using the
information from the table given with the problem statement, we see that the values of
2
mvrms for each tank are:
744 THE IDEAL GAS LAW AND KINETIC THEORY
Product of the mass and the
square of the rms-speed
2
mvrms
Tank
A
B
2
m ( 2vrms ) = 4mvrms
C
2
2
= 2mvrms
( 2m ) vrms
D
2
2m ( 2vrms ) = 8mvrms
2
2
Thus, tank D has the greatest temperature, followed by tanks B, C, and A.
SOLUTION The temperature of the gas in each tank can be determined from Equation 14.6:
(
)
2
3.32 × 10 kg (1223 m /s )
mvrms
=
= 1200 K
TA =
−23
3k
3 1.38 × 10 J /K
TB =
TC =
TD =
(
m 2vrms
3k
−26
(
(
2
2m 2vrms
3k
)
) = (3.32 × 10 kg ) ( 2 × 1223 m /s )
3 (1.38 × 10 J /K )
2 ( 3.32 × 10 kg ) (1223 m /s )
=
=
3 (1.38 × 10 J /K )
2
−26
−23
−23
)
2
=
(
2 3.32 × 10
(
= 4800 K
2
−26
2
( 2m ) vrms
3k
2
−26
)
2400 K
kg ( 2 × 1223 m /s )
3 1.38 × 10
−23
J /K
)
2
= 9600 K
These results confirm the conclusion reached in the REASONING.
______________________________________________________________________________
35. REASONING Assuming that neon (a monatomic gas) behaves as an ideal gas, its internal
energy U can be found from U = 32 nRT (Equation 14.7), where n is the number of moles of
neon, R = 8.31 J/ ( mol ⋅ K ) is the universal gas constant, and T is the Kelvin temperature.
We seek the increase ΔU = Uf − Ui in the internal energy of the neon as its temperature
increases from Ti to Tf. Because the gas is in a closed, rigid container, neither the number n
of moles nor the volume V of the neon change when its temperature is raised. Consequently,
we will employ the ideal gas law, PV = nRT (Equation 14.1), to determine the value of the
quantity nR from the volume V, the initial pressure Pi, and the initial Kelvin temperature Ti
of the neon.
Chapter 14 Problems
745
SOLUTION From U = 32 nRT (Equation 14.7), the increase in the internal energy of the
gas is given by
Δ U = U f − U i = 32 nRTf − 32 nRTi = 32 nR (Tf − Ti )
Solving PV
i = nRTi (Equation 14.1) for the quantity nR yields nR =
(1)
PV
i
Ti
. Substituting this
result into Equation (1), we find that
⎛ PV ⎞
ΔU = 32 ⎜ i ⎟ (Tf − Ti ) =
⎝ Ti ⎠
3
2
(1.01×105 Pa )( 680 m3 )
293.2 K
( 294.3 K − 293.2 K ) = 3.9 × 105 J
36. REASONING AND SOLUTION To find the rms-speed of the CO2 we need to first find the
temperature. We can do this since we know the rms-speed of the H2O molecules. Using
1 m v2
rms
2
=
3 kT,
2
2
we can solve for the temperature to get T = m vrms
/(3k), where the mass of
an H2O molecule is (18.015 g/mol)/(6.022 × 1023 mol–1) = 2.99 × 10–23 g. Thus,
T=
( 2.99 ×10−26 kg ) ( 648 m/s )2 = 303 K
3 (1.38 ×10−23 J/K )
The molecular mass of CO2 is 44.01 u, hence the mass of a CO2 molecule is 7.31 × 10–26 kg.
The rms-speed for CO2 is
3kT
3(1.38 ×10−23 J/K)(303 K)
=
= 414 m/s
vrms =
m
7.31 × 10−26 kg
______________________________________________________________________________
37. SSM
REASONING
The behavior of the molecules is described by Equation 14.5:
2
PV = 23 N ( 12 mvrms
) . Since the pressure and volume of the gas are kept constant, while the
number of molecules is doubled, we can write PV
1 2 = P2V2 , where the subscript 1 refers to
the initial condition, and the subscript 2 refers to the conditions after the number of
molecules is doubled. Thus,
2
3
1
2
1
N1 ⎡ m (vrms )12 ⎤ = N 2 ⎡ m (vrms )22 ⎤
⎣2
⎦ 3
⎣2
⎦
or
N1 (vrms )12 = N 2 (vrms )22
The last expression can be solved for (vrms )2 , the final translational rms speed.
746 THE IDEAL GAS LAW AND KINETIC THEORY
SOLUTION
Since the number of molecules is doubled, N 2 = 2 N1 . Solving the last
expression above for (vrms )2 , we find
N1
N1
463 m/s
= 327 m/s
N2
2 N1
2
______________________________________________________________________________
(vrms ) 2 = (vrms )1
= (463 m/s)
=
38. REASONING According to the kinetic theory of gases, the average kinetic energy of an
atom is KE = 32 kT (Equation 14.6), where k is Boltzmann’s constant and T is the Kelvin
temperature. Therefore, the ratio of the average kinetic energies is equal to the ratio of the
Kelvin temperatures of the gases. We are given no direct information about the
temperatures. However, we do know that the temperature of an ideal gas is related to the
pressure P, the volume V, and the number n of moles of the gas via the ideal gas law,
PV = nRT. Thus, the ideal gas law can be solved for the temperature, and the ratio of the
temperatures can be related to the other properties of the gas. In this way, we will obtain the
desired kinetic-energy ratio.
SOLUTION Using Equation 14.6 and the ideal gas law in the form T =
PV
, we find that
nR
the desired ratio is
⎛
3
KE Krypton 2 kTKrypton
= 3
=
kT
KE Argon
Argon
2
⎞
PV
3k⎜
⎟
2 ⎜n
⎟
R
⎝ Krypton ⎠
⎛
⎞
3 k ⎜ PV
⎟
2 ⎜n
R⎟
⎝ Argon ⎠
=
nArgon
nKrypton
Here, we have taken advantage of the fact that the pressure and volume of each gas are the
same. While we are not given direct information about the number of moles of each gas, we
do know that their masses are the same. Furthermore, the number of moles can be
calculated from the mass m (in grams) and the mass per mole M (in grams per mole),
m
. Substituting this expression into our result for the kinetic-energy ratio
according to n =
M
gives
m
nArgon
M Argon
M Krypton
KE Krypton
=
=
=
m
nKrypton
M Argon
KE Argon
M Krypton
Taking the masses per mole from the periodic table on the inside of the back cover of the
text, we find
Chapter 14 Problems
747
KE Krypton M Krypton
83.80 g/mol
=
=
= 2.098
M Argon
39.948 g/mol
KE Argon
______________________________________________________________________________
39. REASONING The smoke particles have the same average translational kinetic energy as
2
the air molecules, namely, 12 mvrms
= 32 kT , according to Equation 14.6. In this expression m
is the mass of a smoke particle, vrms is the rms speed of a particle, k is Boltzmann’s
constant, and T is the Kelvin temperature. We can obtain the mass directly from this
equation.
SOLUTION Solving Equation 14.6 for the mass m, we find
(
)
−23
J/K ( 301 K )
3kT 3 1.38 ×10
m= 2 =
= 1.6 × 10−15 kg
2
vrms
2.8 ×10−3 m/s
(
)
______________________________________________________________________________
40. REASONING The average kinetic energy KE of each molecule in the gas is directly
proportional to the Kelvin temperature T, according to KE = 32 kT (Equation 14.6), where k
is the Boltzmann constant. Solving for the temperature, we obtain
T=
( )
2 KE
3k
(1)
The average kinetic energy KE of one molecule is the total average kinetic energy
of all the molecules divided by the total number N of molecules:
KE
( )total
KE =
( KE )total
N
(2)
We will determine the number N of molecules in the gas by multiplying the number n of
moles by Avogadro’s number NA:
N = nN A
(3)
( )total of all the molecules is equal to the kinetic energy
The total average kinetic energy KE
KEbullet of the bullet, which has a mass m and a speed v. Thus, according to Equation 6.2,
748 THE IDEAL GAS LAW AND KINETIC THEORY
( KE )total = KE bullet = 12 mv2
(6.2)
SOLUTION Substituting Equation (2) into Equation (1) yields
( )
⎡ KE
⎤
total
⎢
⎥
2
⎢
⎥ 2 KE
N
2 KE
⎦=
total
T=
= ⎣
3k
3k
3kN
( )
( )
(4)
Substituting Equation (3) and Equation 6.2 into Equation (4), we obtain
T=
( )total = 2 ( 12 mv2 ) =
2 KE
3kN
3k ( nN A )
mv 2
3knN A
(5)
The Boltzmann constant k is equal to the ratio of the universal gas constant R to Avogadro’s
number: k = R N A . Making this substitution into Equation (5) yields the Kelvin
temperature of the gas:
T=
41.
(
⎛
3⎜
⎜
⎝
SSM WWW
)
8.0 ×10−3 kg ( 770 m/s )
mv 2
mv 2
=
=
= 95 K
3
Rn
3
8.31
J/
mol
⋅
K
2.0
mol
⎡
⎤
⎞
(
)
(
)
⎣
⎦
R
⎟ n NA
⎟
NA
⎠
2
REASONING The rms-speed vrms of the sulfur dioxide molecules is
2
related to the Kelvin temperature T by 12 mvrms
= 32 kT (Equation 14.6), where m is the mass
of a SO2 molecule and k is Boltzmann’s constant. Solving this equation for the rms-speed
gives
vrms =
3kT
m
(1)
The temperature can be found from the ideal gas law, Equation 14.1, as T = PV / ( nR ) ,
where P is the pressure, V is the volume, n is the number of moles, and R is the universal gas
constant. All the variables in this relation are known. Substituting this expression for T into
Equation (1) yields
Chapter 14 Problems
vrms
749
⎛ PV ⎞
3k ⎜
⎟
⎝ nR ⎠ = 3 k PV
=
m
nmR
The mass m of a single SO2 molecule will be calculated in the Solutions section.
SOLUTION Using the periodic table on the inside of the text’s back cover, we find the
molecular mass of a sulfur dioxide molecule (SO2) to be
32.07 u
+ 2 (15.9994 u) = 64.07 u
Mass of a single
sulfur atom
Mass of two
oxygen atoms
Since 1 u = 1.6605 × 10−27 kg (see Section 14.1), the mass of a sulfur dioxide molecule is
⎛ 1.6605 × 10−27 kg ⎞
−25
m = ( 64.07 u ) ⎜
kg
⎟ = 1.064 × 10
1
u
⎝
⎠
The translational rms-speed of the sulfur dioxide molecules is
vrms =
3 k PV
nmR
3 (1.38 × 10−23 J/K )( 2.12 × 104 Pa )( 50.0 m3 )
=
= 343 m/s
( 421 mol ) (1.064 × 10−25 kg ) ⎡⎣8.31 J/ ( mol ⋅ K )⎤⎦
______________________________________________________________________________
Power is the rate at which energy is produced, according to
42. REASONING
Energy
(Equation 6.10b), so the time required for the engine to produce a certain
Power =
t
amount of energy is given by
t=
Energy
Power
(1)
Assuming that helium (a monatomic gas) behaves as an ideal gas, its internal energy U can
be found from U = 32 nRT (Equation 14.7), where n is the number of moles of helium in the
container, R = 8.31 J/ ( mol ⋅ K ) is the universal gas constant, and T is the Kelvin
temperature. We will determine the quantity nRT in Equation 14.7 from the pressure P and
volume V of the helium via the ideal gas law PV = nRT (Equation 14.1).
750 THE IDEAL GAS LAW AND KINETIC THEORY
SOLUTION The engine must produce an amount of energy equal to the internal energy
U = 32 nRT (Equation 14.7) of the helium, so from Equation (1) we have that
3 nRT
U
Energy
t=
=
= 2
Power Power Power
(2)
Substituting PV = nRT (Equation 14.1) into Equation (2) yields
t=
3 nRT
2
Power
=
3PV
2 ( Power )
(3)
Using the equivalence 1 hp = 746 W in Equation (3), we obtain
t=
3PV
3 ( 6.2 ×105 Pa )( 0.010 m3 )
= 5.0 ×101 s
=
(
)
2 Power
⎛ 746 W ⎞
2 0.25 hp ⎜
⎟
⎝ 1 hp ⎠
(
)
43. SSM WWW REASONING AND SOLUTION
a. Assuming that the direction of travel of the bullets is positive, the average change in
momentum per second is
Δp/Δt = mΔv/Δt = (200)(0.0050 kg)[(0 m/s) – 1200 m/s)]/(10.0 s) = –120 N
b. The average force exerted on the bullets is F = Δp/Δt. According to Newton’s third law,
the average force exerted on the wall is − F = 120 N .
c. The pressure P is the magnitude of the force on the wall per unit area, so
P = (120 N)/(3.0 × 10–4 m2) = 4.0 × 10 5 Pa
______________________________________________________________________________
44. REASONING Because the container holds 1.000 mol of neon, we know that the number of
neon atoms inside is equal to Avogadro’s number: N = NA. These NA atoms are continually
bouncing back and forth between the walls of the cubical container, with an rms speed vrms
2
found from 12 mvrms
= 23 kT (Equation 14.6), where m is the mass of a single neon atom, k is
the Boltzmann constant, and T is the Kelvin temperature. Following the development of
kinetic theory given in Section 14.3 of the text, the time t that elapses between successive
collisions of one atom with one wall of the container is
Chapter 14 Problems
t=
2L
vrms
751
(1)
This is because we assume that, between successive collisions with a given wall, the atom
travels to the opposite wall and back, which is a total distance of 2L. We will use
1 mv 2 = 3 kT (Equation 14.6) to determine the rms speed of the neon atoms for use in
rms
2
2
Equation (1).
There are three identical pairs of walls, so on average 1/3 of the NA atoms in the container
collide with a given wall in the time t. Therefore, the rate at which atoms collide with a
given wall is
N
(2)
Collision rate = A
3t
SOLUTION Substituting Equation (1) into Equation (2) yields
Collision rate =
Solving
1 mv 2
rms
2
1
2
NA
3t
=
NA
⎛ 2L ⎞
3 ⎜⎜
⎟⎟
⎝ vrms ⎠
=
N A vrms
(3)
6L
= 32 kT (Equation 14.6) for the rms speed of the atoms, we obtain
2
mvrms
=
3
2
kT
or
2
vrms
=
3kT
m
or
vrms =
3kT
m
(4)
Substituting Equation (4) into Equation (3) yields
Collision rate =
N A vrms
6L
=
NA
6L
3kT
m
Therefore, the rate at which neon atoms collide with each wall of the container is
6.022 × 1023 atoms ) 3 (1.38 ×10−23 J/K ) ( 293 K )
(
Collision rate =
= 2.01×1026 atoms/s
6 ( 0.300 m )
3.35 × 10−26 kg
(5)
752 THE IDEAL GAS LAW AND KINETIC THEORY
45. REASONING The mass m of oxygen that diffuses in a time t through a trachea is given by
Equation 14.8 as m = ( DA ΔC ) t / L , where ΔC is the concentration difference between the
two ends of the trachea, A and L are, respectively, its cross-sectional area and length, and D
is the diffusion constant. The concentration difference ΔC is the higher concentration C2
minus the lower concentration C1, or ΔC = C2 − C1 .
SOLUTION Substituting the relation ΔC = C2 − C1 into Equation 14.8 and solving for C1
gives
⎛ L ⎞⎛ m ⎞
C1 = C2 − ⎜
⎟⎜ ⎟
⎝ DA ⎠ ⎝ t ⎠
We recognize that m/t is the mass per second of oxygen diffusing through the trachea. Thus,
the oxygen concentration at the interior end of the trachea is
⎡
⎤
1.9 × 10−3 m
3
−12
C1 = 0.28 kg/m3 − ⎢
⎥ (1.7 × 10 kg/s ) = 0.14 kg/m
−5 2
−9
2)
(
⎢⎣ (1.1 × 10 m /s ) 2.1 × 10 m ⎥⎦
______________________________________________________________________________
46. REASONING The mass m of methane that diffuses out of the tank in a time t is given by
( DA ΔC ) t (Equation 14.8), where D is the diffusion constant for methane, A is the
m=
L
cross-sectional area of the pipe, L is the length of the pipe, and ΔC = C1 − C2 is the
difference in the concentration of methane at the two ends of the pipe. The higher
concentration C1 at the tank-end of the pipe is given, and the concentration C2 at the end of
the pipe that is open to the atmosphere is zero.
SOLUTION Solving m =
( DA ΔC ) t
(Equation 14.8) for the cross-sectional area A, we
L
obtain
A=
mL
( D ΔC ) t
The elapsed time t must be converted from hours to seconds:
⎛ 3600 s ⎞
4
t = 12 hr ⎜
⎟ = 4.32 ×10 s
⎝ 1 hr ⎠
(
)
Therefore, the cross-sectional area A of the pipe is
(1)
Chapter 14 Problems
753
9.00 × 10 −4 kg ) (1.50 m )
(
A=
= 2.29 × 10 −3 m 2
−5
2
3
3
4
( 2.10 ×10 m /s )( 0.650 kg/m − 0 kg/m )( 4.32 ×10 s )
47. SSM REASONING AND SOLUTION
According to Fick's law of diffusion
(Equation 14.8), the mass of ethanol that diffuses through the cylinder in one hour (3600 s)
is
(DA ΔC)t (12.4 × 10 –10 m 2 /s)(4.00 × 10 –4 m 2 )(1.50 kg/m 3 )(3600 s)
m=
=
= 1.34 × 10 –7 kg
L
(0.0200 m)
______________________________________________________________________________
48. REASONING According to the kinetic theory of gases, the average kinetic energy KE of
an atom in an ideal gas is related to the Kelvin temperature of the gas by KE = 32 k T
(Equation 14.6). Since the temperature is the same for both gases, the A and B atoms have
the same average kinetic energy. The average kinetic energy is related to the rms-speed by
2
. Since both gases have the same average kinetic energy and gas A has the
KE = 12 mvrms
smaller mass, it has the greater rms-speed. Therefore, gas A has the greater diffusion rate.
SOLUTION For a fixed temperature, the ratio RA/RB of the diffusion rates for the two
types of atoms is equal to the ratio vrms, A/vrms, B of the rms-speeds. According to
Equation 14.6, it follows that vrms = 2KE/m , so that
v
RA
= rms, A =
RB
vrms, B
2KE
mA
2KE
mB
=
mB
=
mA
2.0 u
= 1.4
1.0 u
______________________________________________________________________________
49. REASONING AND SOLUTION
a. As stated, the time required for the first solute molecule to traverse a channel of length L
is t = L2 /(2D) . Therefore, for water vapor in air at 293 K, where the diffusion constant is
D = 2.4 × 10 –5 m 2 /s , the time t required for the first water molecule to travel L = 0.010 m
is
( 0.010 m )
L2
t=
=
= 2.1 s
2 D 2 2.4 × 10−5 m 2 /s
2
(
)
754 THE IDEAL GAS LAW AND KINETIC THEORY
b. If a water molecule were traveling at the translational rms speed v rms for water, the time t
it would take to travel the distance L = 0.010 m would be given by t = L / vrms , where,
according to Equation 14.6 ( KE = 2 mv2rms ), v rms = 2(KE)/ m . Before we can use the last
expression for the translation rms speed v rms , we must determine the mass m of a water
molecule and the average translational kinetic energy KE .
1
Using the periodic table on the inside of the text’s back cover, we find that the molecular
mass of a water molecule is
2(1.00794 u) + 15.9994 u = 18.0153 u
Mass of one
oxygen atom
Mass of two
hydrogen atoms
The mass of a single molecule is
m=
18.0153 × 10 –3 kg/mol
6.022 × 10
23
mol
–1
= 2.99 × 10 –26 kg
The average translational kinetic energy of water molecules at 293 K is, according to
Equation 14.6,
3
2
3
2
KE = kT = (1.38 × 10 –23 J/K) (293 K ) = 6.07 × 10 –21 J
Therefore, the translational rms speed of water molecules is
v rms =
2(KE )
m
=
(
2 6.07 × 10 –21 J
2.99 × 10 –26 kg
)
= 637 m/s
Thus, the time t required for a water molecule to travel the distance L = 0.010 m at this
speed is
0.010 m
L
t=
=
= 1.6 × 10 –5 s
v rms 637 m/s
c. In part (a), when a water molecule diffuses through air, it makes millions of collisions
each second with air molecules. The speed and direction change abruptly as a result of each
collision. Between collisions, the water molecules move in a straight line at constant speed.
Although a water molecule does move very quickly between collisions, it wanders only very
slowly in a zigzag path from one end of the channel to the other. In contrast, a water
molecule traveling unobstructed at its translational rms speed [as in part (b)], will have a
larger displacement over a much shorter time. Therefore, the answer to part (a) is much
longer than the answer to part (b).
______________________________________________________________________________
Chapter 14 Problems
755
50. REASONING Since mass is conserved, the mass flow rate is the same at all points, as
described by the equation of continuity (Equation 11.8). Therefore, the mass flow rate at
which CCl4 enters the tube is the same as that at point A. The concentration difference of
CCl4 between point A and the left end of the tube, ΔC , can be calculated by using Fick's
law of diffusion (Equation 14.8). The concentration of CCl4 at point A can be found from
CA = Cleft end – ΔC.
SOLUTION
a. As discussed above in the reasoning, the mass flow rate of CCl4 as it passes point A is
the same as the mass flow rate at which CCl4 enters the left end of the tube; therefore, the
–13
kg/s .
mass flow rate of CCl4 at point A is 5.00 × 10
b. Solving Fick's law for ΔC , we obtain
ΔC =
mL (m / t ) L
=
DAt
DA
=
(5.00×10
(20.0×10
−13
−3
m)
–4
2
kg/s)(5.00×10
–10
2
m /s)(3.00×10
–2
kg/m ) − (4.2 ×10
m )
= 4.2×10
−3
kg/m
3
Then,
CA = Cleft end − ΔC = (1.00 × 10
3
–3
3
kg/m )= 5.8 ×10
–3
kg/m
3
______________________________________________________________________________
51. SSM WWW REASONING AND SOLUTION
a. The average concentration is Cav = (1/2) (C1 + C2) = (1/2)C2 = m/V = m/(AL), so that
C2 = 2m/(AL). Fick's law then becomes m = DAC2t/L = DA(2m/AL)t/L = 2Dmt/L2. Solving
for t yields
t = L2 / ( 2 D )
b. Substituting the given data into this expression yields
t = (2.5 × 10–2 m)2/[2(1.0 × 10–5 m2/s)] = 31 s
______________________________________________________________________________
52. REASONING AND SOLUTION Equation 14.8,
DA ΔC t
m=
. Solving for the time t gives
L
mL
t=
DA ΔC
gives
Fick's
law
of
diffusion:
(1)
756 THE IDEAL GAS LAW AND KINETIC THEORY
The time required for the water to evaporate is equal to the time it takes for 2.0 grams of
water vapor to traverse the tube and can be calculated from Equation (1) above. Since air in
the tube is completely dry at the right end, the concentration of water vapor is zero,
C1 = 0 kg/m3, and ΔC = C2 – C1 = C2. The concentration at the left end of the tube, C2, is
equal to the density of the water vapor above the water. This can be found from the ideal
gas law:
ρ RT
PV = nRT
P=
⇒
M
where M = 0.0180152 kg/mol is the mass per mole for water (H2O). From the figure that
accompanies Problem 75 in Chapter 12, the equilibrium vapor pressure of water at 20 °C is
2.4 × 103 Pa. Therefore,
C2 = ρ =
PM ( 2.4 × 103 Pa ) ( 0.0180 kg/mol )
=
= 1.8 ×10−2 kg/m3
RT
[8.31 J/(mol ⋅ K)] (293 K)
Substituting values into Equation (1) gives
t=
( 2.0×10−3 kg ) (0.15 m)
=
( 2.4 ×10−5 m 2 /s )( 3.0 ×10−4 m 2 ) (1.8 ×10−2 kg/m3 )
2.3 × 106 s
This is about 27 days!
______________________________________________________________________________
53. REASONING AND SOLUTION To find the temperature T2, use the ideal gas law with n
and V constant. Thus, P1/T1 = P2/T2. Then,
⎛ 3.01× 105 Pa ⎞
⎛P ⎞
T2 = T1 ⎜ 2 ⎟ = ( 284 K ) ⎜
⎟ = 304 K
5
P
×
2.81
10
Pa
⎝
⎠
⎝ 1⎠
______________________________________________________________________________
54. REASONING AND SOLUTION According to the ideal gas law (Equation 14.1), the total
number of moles n of fresh air in a normal breath is
n=
PV (1.0 × 105 Pa)(5.0 × 10 –4 m 3 )
=
= 1.94 × 10 –2 mol
[8.31 J/(mole ⋅ K)] (310 K)
RT
The total number of molecules in a normal breath is nNA , where NA is Avogadro's number.
Since fresh air contains approximately 21% oxygen, the total number of oxygen molecules
in a normal breath is (0.21) nNA or
(0.21)(1.94 ×10
−2
mol)(6.022 ×10
23
−1
mol )= 2.5 ×10
21
Chapter 14 Problems
757
______________________________________________________________________________
55.
SSM REASONING According to Equation 11.1, the mass density ρ of a substance is
defined as its mass m divided by its volume V: ρ = m / V . The mass of nitrogen is equal to
the number n of moles of nitrogen times its mass per mole: m = n (Mass per mole). The
number of moles can be obtained from the ideal gas law (see Equation 14.1) as
n = (PV)/(RT). The mass per mole (in g/mol) of nitrogen has the same numerical value as its
molecular mass (which we know).
SOLUTION Substituting m = n (Mass per mole) into ρ = m / V , we obtain
ρ=
m n (Mass per mole)
=
V
V
(1)
Substituting n = (PV)/(RT) from the ideal gas law into Equation 1 gives the following result:
⎛PV
n (Mass per mole) ⎜⎝ RT
ρ=
=
V
⎞
⎟(Mass per mole) P (Mass per mole)
⎠
=
RT
V
The pressure is 2.0 atmospheres, or P = 2 (1.013 × 105 Pa). The molecular mass of nitrogen
is given as 28 u, which means that its mass per mole is 28 g/mol. Expressed in terms of
kilograms per mol, the mass per mole is
⎛
g ⎞ ⎛ 1 kg ⎞
Mass per mole = ⎜ 28
⎟
⎝ mol ⎠ ⎜⎝ 103 g ⎟⎠
The density of the nitrogen gas is
⎛
g ⎞ ⎛ 1 kg ⎞
2 (1.013 × 105 Pa ) ⎜ 28
⎟
⎝ mol ⎠ ⎜⎝ 103 g ⎟⎠
P (Mass per mole)
= 2.2 kg/m3
ρ=
=
[8.31 J/ ( mol ⋅ K )] ( 310 K )
RT
_____________________________________________________________________________________________
56. REASONING The diffusion of the glycine is driven by the concentration difference
ΔC = C2 – C1 between the ends of the tube, where C2 is the higher concentration and
C1 = C2 − ΔC is the lower concentration. The concentration difference is related to the mass
m DA ΔC
rate of diffusion by Fick’s law:
(Equation 14.8), where D is the diffusion
=
t
L
constant of glycine in water, and A and L are, respectively, the cross-sectional area and
length of the tube.
758 THE IDEAL GAS LAW AND KINETIC THEORY
SOLUTION Solving
m DA ΔC
(Equation 14.8) for ΔC, we obtain
=
t
L
⎛m⎞
L⎜ ⎟
ΔC = ⎝ t ⎠
DA
(1)
Substituting Equation (1) into C1 = C2 − ΔC yields
⎛m⎞
L⎜ ⎟
C1 = C2 − ΔC = C2 − ⎝ t ⎠
DA
(2)
We note that the length of the tube is L = 2.0 cm = 0.020 m. Substituting this and the other
given values into Equation (2), we find that the lower concentration is
C1 = 8.3 × 10−3 kg/m3 −
( 0.020 m ) ( 4.2 × 10−14 kg/s )
(1.06 ×10
−9
m /s )(1.5 × 10
2
−4
m
2
)
= 3.0 × 10−3 kg/m3
2
57. SSM REASONING AND SOLUTION Using the expressions for v and ( v ) given in
the statement of the problem, we obtain:
2
1
3
a.
v 2 = (v12 + v22 + v32 ) =
b.
(v)
2
1⎡
(3.0
3 ⎣⎢
m/s) 2 + (7.0 m/s)2 + (9.0 m/s) 2 ⎤⎥ = 46.3 m 2 /s 2
⎦
2
2
= ⎡⎣ 13 (v1 + v2 + v3 ) ⎤⎦ = ⎡⎣ 13 (3.0 m/s + 7.0 m/s + 9.0 m/s) ⎤⎦ = 40.1 m 2 /s 2
v 2 and ( v ) are not equal, because they are two different physical quantities.
______________________________________________________________________________
2
58. REASONING The translational rms-speed vrms is related to the Kelvin temperature T by
1 mv 2
rms
2
= 23 kT (Equation 14.6), where m is the mass of the oxygen molecule and k is
Boltzmann’s constant. Solving this equation for the rms-speed gives vrms = 3kT / m . This
relation will be used to find the ratio of the speeds.
SOLUTION The rms-speeds in the ionosphere and near the earth’s surface are
( vrms )ionosphere =
3kTionosphere
m
and
( vrms )earth's surface =
3kTearth's surface
m
Chapter 14 Problems
759
Dividing the first equation by the second gives
( vrms )ionosphere
=
v
( rms )earth's surface
3kTionosphere
m
3kTearth's surface
m
=
Tionosphere
Tearth's surface
= 3 = 1.73
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59.
SSM REASONING The mass m of nitrogen that must be removed from the tank is equal
to the number of moles withdrawn times the mass per mole. The number of moles
withdrawn is the initial number ni minus the final number of moles nf, so we can write
m=
( ni − nf ) ( Mass per mole )
(1)
Number of
moles withdrawn
The final number of moles is related to the initial number by the ideal gas law.
SOLUTION From the ideal gas law (Equation 14.1), we have
nf =
Pf V
RT
and
ni =
PV
i
RT
Note that the volume V and temperature T do not change. Dividing the first by the second
equation gives
Pf V
nf
P
⎛P ⎞
RT
nf = ni ⎜ f ⎟
=
= f
or
PV
ni
Pi
i
⎝ Pi ⎠
RT
Substituting this expression for nf into Equation 1 gives
⎡
⎛P
m = ( ni − nf ) ( Mass per mole ) = ⎢ ni − ni ⎜ f
⎢⎣
⎝ Pi
⎞⎤
⎟ ⎥ ( Mass per mole )
⎠ ⎥⎦
The molecular mass of nitrogen (N2) is 2 (14.0067 u) = 28.0134 u. Therefore, the mass per
mole is 28.0134 g/mol. The mass of nitrogen that must be removed is
760 THE IDEAL GAS LAW AND KINETIC THEORY
⎡
⎛P
m = ⎢ ni − ni ⎜ f
⎢⎣
⎝ Pi
⎞⎤
⎟ ⎥ ( Mass per mole )
⎠ ⎥⎦
⎡
g ⎞
⎛ 25 atm ⎞ ⎤ ⎛
= ⎢ 0.85 mol − ( 0.85 mol ) ⎜
⎟ ⎥ ⎜ 28.0134
⎟ = 8.1 g
mol ⎠
⎝ 38 atm ⎠ ⎦ ⎝
⎣
______________________________________________________________________________
3
60. REASONING AND SOLUTION Since the density of aluminum is 2700 kg/m , the
number of atoms of aluminum per cubic meter is, using the given data,
⎛ 2700 × 103 g/m 3 ⎞
23
28
3
−1
N /V = ⎜
⎟ 6.022 × 10 mol = 6.0 × 10 atoms/m
⎝ 26.9815 g/mol ⎠
(
)
Assuming that the volume of the solid contains many small cubes, with one atom at the
center of each, then there are (6.0 × 10 28 )1/3 atoms/m = 3.9 × 10 9 atoms/m along each
edge of a 1.00-m3 cube. Therefore, the spacing between the centers of neighboring atoms is
1
–10
= 2.6 × 10
m
9
3.9 × 10 /m
______________________________________________________________________________
d=
61. REASONING Since the xenon atom does not interact with any other atoms or molecules
on its way up, we can apply the principle of conservation of mechanical energy (see Section
6.5) and set the final kinetic plus potential energy equal to the initial kinetic plus potential
energy. Thus, during the rise, the atom’s initial kinetic energy is converted entirely into
gravitational potential energy, because the atom comes to a momentary halt at the top of its
trajectory. The initial kinetic energy 12 mv02 is equal to the average translational kinetic
energy. Therefore,
1 mv 2
0
2
= KE = 32 kT , according to Equation 14.6, where k is Boltzmann’s
constant and T is the Kelvin temperature. The gravitational potential energy is mgh,
according to Equation 6.5.
SOLUTION Equation 6.9b gives the principle of conservation of mechanical energy:
1 mv 2
f
2
+ mghf
=
Final mechanical energy
In this expression, we know that
1 mv 2
0
2
1 mv 2
0
2
+ mgh0
Initial mechanical energy
= KE = 32 kT and that
1 mv 2
f
2
= 0 J (since the atom
comes to a halt at the top of its trajectory). Furthermore, we can take the height at the
earth’s surface to be h0 = 0 m. Taking this information into account, we can write the
energy-conservation equation as follows:
Chapter 14 Problems
mghf = 32 kT
or
hf =
761
3kT
2mg
Using M to denote the molecular mass (in kilograms per mole) and recognizing that
M
m=
, where NA is Avogadro’s number and is the number of xenon atoms per mole, we
NA
have
3kN AT
3kT
3kT
hf =
=
=
2mg
2 Mg
⎛ M ⎞
2⎜
⎟g
⎝ NA ⎠
Recognizing that kNA = R and that M = 131.29 g/mol = 131.29 × 10−3 kg/mol, we find
hf =
3 ⎡⎣8.31 J/ ( mol ⋅ K ) ⎤⎦ ( 291 K )
3kN AT 3RT
=
=
= 2820 m
2 Mg
2 Mg 2 131.29 ×10−3 kg/mol 9.80 m/s 2
(
)(
)
62. REASONING AND SOLUTION Since we are treating the air as a diatomic ideal gas
(PV = nRT), it follows that
U = 52 nRT = 52 PV =
5
2
( 7.7×106 Pa )(5.6 ×105 m3 ) = 1.1× 1013 J
The number of joules of energy consumed per day by one house is
J ⋅ h ⎞ ⎛ 3600 s ⎞
⎛
8
30.0 kW ⋅ h = ⎜ 30.0 × 103
⎟⎜
⎟ = 1.08 ×10 J
s
1
h
⎝
⎠⎝
⎠
The number of homes that could be served for one day by 1.1 × 1013 J of energy is
⎛
⎞
1 home
⎟=
(1.1×1013 J ) ⎜ 1.08
×108 J
1.0 ×105 homes
⎝
⎠
______________________________________________________________________________
63. REASONING When perspiration absorbs heat from the body, the perspiration vaporizes.
The amount Q of heat required to vaporize a mass mperspiration of perspiration is given by
Equation 12.5 as Q = mperspirationLv, where Lv is the latent heat of vaporization for water at
body temperature. The average energy E given to a single water molecule is equal to the
heat Q divided by the number N of water molecules.
762 THE IDEAL GAS LAW AND KINETIC THEORY
SOLUTION Since E = Q/N and Q = mperspirationLv, we have
E=
Q mperspiration Lv
=
N
N
But the mass of perspiration is equal to the mass mH2O molecule of a single water molecule
times the number N of water molecules. The mass of a single water molecule is equal to its
molecular mass (18.0 u), converted into kilograms. The average energy given to a single
water molecule is
E=
mperspiration Lv
N
=
mH O molecule N Lv
2
N
⎛ 1.66 × 10−27 kg ⎞
−20
6
E = mH O molecule Lv = (18.0 u ) ⎜
⎟ 2.42 × 10 J/kg = 7.23 × 10 J
2
1u
⎝
⎠
______________________________________________________________________________
(
)
64. REASONING AND SOLUTION At the instant just before the balloon lifts off, the
buoyant force from the outside air has a magnitude that equals the magnitude of the total
weight. According to Archimedes’ principle, the buoyant force is the weight of the
displaced outside air (density ρ0 = 1.29 × 103 kg/m3). The mass of the displaced outside air
is ρ0V, where V = 650 m3. The corresponding weight is the mass times the magnitude g of
the acceleration due to gravity. Thus, we have
(ρ V ) g
0
Buoyant force
=
mtotal g
Total weight
of balloon
(1)
The total mass of the balloon is mtotal = mload + mair, where mload = 320 kg and mair is the mass
of the hot air within the balloon. The mass of the hot air can be calculated from the ideal
gas law by using it to obtain the number of moles n of air and multiplying n by the mass per
mole of air, M = 29 × 10–3 kg/mol:
⎛ PV
mair = n M = ⎜
⎝ RT
⎞
⎟M
⎠
Thus, the total mass of the balloon is mtotal = mload + PVM/(RT) and Equation (1) becomes
⎛ PV
⎝ RT
ρ0V = mload + ⎜
⎞
⎟M
⎠
Chapter 14 Problems
763
Solving for T gives
T=
PVM
( ρ0V – mload ) R
(1.01 × 10 Pa )( 650 m )( 29 × 10 kg/mol)
=
=
⎡(1.29 × 10 kg/m )( 650 m ) – 320 kg ⎤ ⎡8.31 J/ ( mol ⋅ K ) ⎤
⎦
⎣
⎦⎣
5
3
3
3
3
–3
440 K
______________________________________________________________________________