CHAPTER 14 THE IDEAL GAS LAW
AND KINETIC THEORY
CONCEPTUAL QUESTIONS
____________________________________________________________________________________________
1.
REASONING AND SOLUTION
a. Avogadro's number NA is the number of particles per mole of substance. Therefore, one
mole of hydrogen gas (H2) and one mole of oxygen gas (O2) contain the same number
(Avogadro's number) of molecules.
b. One mole of a substance has a mass in grams that is equal to the atomic or molecular
mass of the substance. The molecular mass of oxygen is greater than the molecular mass of
hydrogen. Therefore, one mole of oxygen has more mass than one mole of hydrogen.
____________________________________________________________________________________________
2.
REASONING AND SOLUTION Substances A and B have the same mass densities.
Therefore, the mass per unit volume of substance A is equal to that of substance B.
a. The mass of one mole of a substance depends on the molecular mass of the substance. In
general, the molecular masses of substances A and B will differ, and one mole of each
substance will not occupy the same volume; therefore, even though substances A and B
have the same mass density, one mole of substance A will not have the same mass as
substance B.
b. Since the mass per unit volume of substance A is the same as the mass per unit volume of
substance B, 1 m3 of substance A has the same mass as 1 m3 of substance B.
____________________________________________________________________________________________
3.
REASONING AND SOLUTION A tightly sealed house has a large ceiling fan that blows
air out of the house and into the attic. The fan is turned on, and the owners forget to open
any windows or doors. As the fan transports air molecules from the house into the attic, the
number of air molecules in the house decreases. Since the house is tightly sealed, the
volume of the house remains constant. If the temperature of the air inside the house remains
constant, then from the ideal gas law, PV = nRT , the pressure in the house must decrease.
The air pressure in the attic, however, increases. The fan must now blow air from a lower
pressure region to a higher pressure region. Thus, it becomes harder for the fan to do its job.
____________________________________________________________________________________________
4.
SSM REASONING AND SOLUTION According to the ideal gas law (Equation 14.1),
PV = nRT . Therefore, the pressure of a gas confined to a fixed volume is directly
proportional to the Kelvin temperature. Therefore, when the temperature is increased, the
pressure increases proportionally.
The gas above the liquid in a can of hair spray is at a relatively high pressure. If the
temperature of the can is increased to a sufficiently high temperature, the pressure of the
confined gas could increase to a value larger than the walls of the can will sustain. The can
704 THE IDEAL GAS LAW AND KINETIC THEORY
would then explode. Therefore, the label on hair spray cans usually contains the warning
"Do not store at high temperatures."
____________________________________________________________________________________________
5.
REASONING AND SOLUTION When an electric furnace in a tightly sealed house is
turned on for a while, the temperature of the air in the house increases. Since the house is
tightly sealed, both the volume of the air and the number of air molecules remain constant.
If we assume that air behaves like an ideal gas, then, from the ideal gas law, PV = nRT , the
pressure in the house must increase.
____________________________________________________________________________________________
6.
REASONING AND SOLUTION At the sea coast, there is a cave that can be entered only
by swimming beneath the water through a submerged passage and emerging into a pocket of
air within the cave. Since the cave is not vented to the external atmosphere, the air in the
cave contains a fixed number of moles n. As the tide comes in, the water level in the cave
rises, thereby decreasing the volume of the air above the water. According to the ideal gas
law (Equation 14.1), PV = nRT . The pressure of the air in the cave is, therefore, inversely
proportional to the air volume. Since the volume of the air decreases, the pressure increases.
Thus, if you are in the cave when the tide comes in, your ears will "pop" inward in a manner
that is analogous to what happens when you climb down a mountain.
____________________________________________________________________________________________
7.
REASONING AND SOLUTION Atmospheric pressure decreases with increasing altitude.
Helium filled weather balloons are under-inflated when they are launched. As the balloon
rises, the pressure exerted on the outside of it decreases. The number of helium molecules
in the balloon is fixed. If we assume that the temperature of the atmosphere remains
constant, then from the ideal gas law, PV = nRT , we see that the volume of the helium in
the balloon will increase, thereby further inflating the balloon. If the balloon is fully
inflated when it is launched from earth, it would burst when it reaches an altitude where the
expanded volume of the helium is greater than the maximum volume of the balloon.
____________________________________________________________________________________________
8.
REASONING AND SOLUTION A slippery cork is being pressed into a very full bottle of
wine. When released, the cork slowly slides back out. If some of the wine is removed from
the bottle before the cork is inserted, the cork does not slide out. When the bottle is very
full, the volume of air in the bottle above the wine is relatively small. Therefore, pushing
the cork in reduces the volume of that air by an appreciable fraction. As a result, the
pressure of the air increases appreciably and becomes large enough to push the slippery cork
back out of the bottle. If some of the wine is removed, the volume of air above the wine is
much larger to begin with, and pushing the cork in reduces the volume of that air by a much
smaller fraction. Consequently, the pressure of the air increases by a much smaller amount
and does not become large enough to push the cork back out.
____________________________________________________________________________________________
9.
REASONING AND SOLUTION Packing material consists of "bubbles" of air trapped
between bonded layers of plastic. The packing material exerts normal forces on the packed
object at each place where the bubbles make contact with the object. The motion of the
packed object is thereby restricted. The magnitude of the normal force that any given
Chapter 14 Conceptual Questions
705
bubble can exert depends on the pressure of the air trapped inside the bubble. The
magnitude of the normal force is equal to the pressure of the air times the area of contact
between the bubble and the object. Since the air is trapped in the bubbles, the number of air
molecules and the volume of the gas are fixed. The pressure exerted by the bubbles, then,
depends on the temperature according to the ideal gas law: PV = nRT . On colder days, T is
smaller than on warmer days; therefore, on colder days, the pressure P of the air in the
bubbles is less than on warmer days. When the pressure is smaller, the magnitude of the
normal force that each bubble can exert on a packed object is smaller. Therefore, the
packing material offers less protection on cold days.
____________________________________________________________________________________________
10. SSM REASONING AND SOLUTION The kinetic theory of gases assumes that a gas
molecule rebounds with the same speed after colliding with the wall of a container. From
the impulse-momentum theorem, Equation 7.4, we know that F∆t = m(v f − v 0 ) , where the
magnitude of F is the magnitude of the force on the wall. This implies that in a time interval
t, a gas molecule of mass m, moving with velocity v before the collision and –v after the
collision, exerts a force of magnitude F = 2mv / t on the walls of the container, since the
magnitude of v f − v 0 is 2v. If the speed of the gas molecule after the collision is less than
that before the collision, the force exerted on the wall of the container will be less than
F = 2mv / t , since the magnitude of v f − v 0 is then less than 2v. Therefore, the pressure of
the gas will be less than that predicted by kinetic theory.
____________________________________________________________________________________________
11. REASONING AND SOLUTION The relationship between the average kinetic energy per
particle in an ideal gas and the Kelvin temperature T of the gas is given by Equation 14.6:
1 mv 2 = 3 kT
.
rms
2
2
If the translational speed of each molecule in an ideal gas is tripled, then the root-meansquare speed for the gas is also tripled. From Equation 14.6, the Kelvin temperature is
proportional to the square of the root-mean-square speed. Therefore, the Kelvin temperature
will increase by a factor of 9.
____________________________________________________________________________________________
12. REASONING AND SOLUTION If the temperature of an ideal gas is doubled from 50 to
100 °C, the average translational kinetic energy per particle does not double. Equation 14.6,
1 mv 2 = 3 kT , relates the average kinetic energy per particle to the Kelvin temperature of
rms
2
2
the gas, not the Celsius temperature. If the Celsius temperature increases from 50 to 100 °C,
the Kelvin temperature increases from 323.15 to 373.15 K. This represents a fractional
increase of 1.15; therefore, the average translational kinetic energy per particle increases
only by a factor of 1.15.
____________________________________________________________________________________________
13. REASONING AND SOLUTION According to Equation 14.7, the internal energy of a
sample consisting of n moles of a monatomic ideal gas at Kelvin temperature T is
U = 32 nR T ; therefore, the internal energy of such an ideal gas depends only on the Kelvin
temperature. If the pressure and volume of this sample is changed isothermally, the internal
706 THE IDEAL GAS LAW AND KINETIC THEORY
energy of the ideal gas will remain the same. Physically, this means that the experimenter
would have to change the pressure and volume in such a way, that the product PV remains
the same. This can be verified from the ideal gas law (Equation 14.1), PV = nRT . If the
values of P and V are varied so that the product PV remains constant, then T will remain
constant and, from Equation 14.7, the internal energy of the gas remains the same.
____________________________________________________________________________________________
14. SSM REASONING AND SOLUTION The atoms in a container of helium have the
same translational rms speed as the molecules in a container of argon. Equation 14.6 relates
the average translational kinetic energy per particle in an ideal gas to the Kelvin
2
= 23 kT . Since the mass of an argon atom is greater than the mass of a
temperature: 12 mvrms
helium atom, then, from Equation 14.6, the average translational kinetic energy per atom of
the argon atoms is greater than the average translational kinetic energy per atom of the
helium atoms. Therefore, the temperature of the argon atoms is greater than that of the
helium atoms.
____________________________________________________________________________________________
15. REASONING AND SOLUTION When an object is placed in a warm environment, its
temperature increases for two reasons: (1) the object gains electromagnetic radiation from
objects in the warm environment, and (2) the object gains heat from air molecules that
collide with the object.
Suppose that an astronaut were placed in the ionosphere, where the temperature of the
ionized gas is about 1000 K, and the density of the gas is on the order of 1011 molecules/m3.
Since the density is extremely low, the distances between gas molecules are very large.
Even though the temperature of the gas is large, there is an insufficient number of molecules
to radiate sufficient energy to heat the astronaut. The number of molecules that collide with
the astronaut per unit time is very small; therefore, even though the average kinetic energy
of the gas molecules is large, the amount of heat transferred by the small number of
collisions per unit time is low. Most of the surface area of the astronaut is in contact with
empty space. The tissues of the astronaut would radiate more electromagnetic radiation than
they absorb. Therefore, the astronaut would not burn up; in fact, if the astronaut remained
in the environment very long, he would freeze to death.
____________________________________________________________________________________________
16. REASONING AND SOLUTION Fick's law of diffusion relates the mass m of solute that
diffuses in a time t through a solvent contained in a channel of length L and cross-sectional
area A: m = (DA∆C)t / L , where ∆C is the concentration difference between the ends of the
channel and D is the diffusion constant.
In the lungs, oxygen in very small sacs (alveoli) diffuses into the blood. The walls of the
alveoli are thin, so the oxygen diffuses over a small distance L. Since the number of alveoli
is large, the effective area A across which diffusion occurs is very large. From Fick's law,
we see that the mass of oxygen that diffuses per unit time is directly proportional to the
effective cross-sectional area A and inversely proportional to the diffusion distance L. Since
A is large and L is small, the mass of oxygen per second that diffuses into the blood is large.
____________________________________________________________________________________________
Chapter 14 Problems
707
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 
−22
 6.022×1023 mol−1   1000 g  = 1.07×10 kg



______________________________________________________________________________
2. 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
, where NA is Avogadro’s number and is the number of molecules per mole,
mmolecule =
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
708 THE IDEAL GAS LAW AND KINETIC THEORY
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
3.

382.89 g/mol


19
−2
 = 1.572 ×10 molecules 
 = 1.00 × 10 g
23
 6.022 × 10 molecules/mol 

(
)
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 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
Chapter 14 Problems
709
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
m


Number of molecules = n N A = 
 NA
 Mass per mole 
 325 × 10−3 g 
−1
23
21
=
 ( 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


______________________________________________________________________________
5.
SSM 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
710 THE IDEAL GAS LAW AND KINETIC THEORY
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 .
______________________________________________________________________________
6.
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
(0.71)W
m
g
n=
=
(1)
Mass per mole Mass per mole
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. Converting this value to kilograms per mole gives
Mass per mole = 18.0153

g   1 kg 
g
=  18.0153

mol 
mol   103 g 
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
____________________________________________________________________________________________
Chapter 14 Problems
7.
711
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
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
______________________________________________________________________________
8.
REASONING AND SOLUTION The molecular mass of C2H5OH is
2(12.011 u) + 6(1.00794 u) + 15.9994 u = 46.069 u
a.
46.069 g/mol
= 7.65 ×10−23 g = 7.65 ×10−26 kg
23
6.022 × 10 /mol
b. 2.00 × 10–3 m3 of this substance (density = 806 kg/m3) has a mass of m = ρ V = 1.61 kg.
The number of molecules is, therefore, equal to
1.61 kg
= 2.11×1025
−26
7.65 × 10
kg
______________________________________________________________________________
N=
9.
SSM REASONING AND SOLUTION The number of moles of air that is pumped into
the tire is ∆ n = nf − ni . According to the ideal gas law (Equation 14.1), n = PV /( RT ) ;
therefore,
V
PV
PV
∆n = nf − ni = f f − i i =
(P − Pi )
RTf
RTi RT f
712 THE IDEAL GAS LAW AND KINETIC THEORY
Substitution of the data given in the problem leads to
−4
3
4.1 × 10 m
( 6.2 × 105 Pa ) − ( 4.8 × 105 Pa )  = 2.3 × 10−2 mol


(
)
8.31 J/ mol ⋅ K ( 296 K )
______________________________________________________________________________
∆n =
10. 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 grams) and the
m
. Using this expression in the
mass per mole M (in grams per mole), according to n =
M
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
mNitrogen
mHelium
=
M Helium
M Nitrogen
Number of moles
of helium
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
11. 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
(
)(
)
1.72 × 105 Pa 2.81 m3
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
Chapter 14 Problems
P=
nRT ( 201 mol ) 8.31 J/ ( mol ⋅ K )  ( 273.15 + 28.2 ) K 
=
= 1.21 × 105 Pa
3
V
4.16 m
(
)
713
(14.1)
______________________________________________________________________________
12. 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
P
T
1.00
atm
288
K




 2  1 
______________________________________________________________________________
13. 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 
______________________________________________________________________________
(
)
14. 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.
714 THE IDEAL GAS LAW AND KINETIC THEORY
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 − 0.0031 m3
Pf
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
15. 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 = P2V2
T1
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 
______________________________________________________________________________
16. 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
Chapter 14 Problems
715
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
 Mass per mole 
 18.0153 g /mol 


=
P=
= 2.4 × 104 Pa
3
V
0.030 m
____________________________________________________________________________________________
17. 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.
SOLUTION We have
( N / V ) Venus
(N / V ) Earth
=
PVenus / (kTVenus )  PVenus   TEarth   9.0 × 10 6 Pa   320 K 
 = 39
=

=

PEarth / (kTEarth )  PEarth   TVenus   1.0 × 10 5 Pa   740 K 
Thus, we can conclude that the atmosphere of Venus is 39 times "thicker" than that of Earth.
______________________________________________________________________________
18. 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
716 THE IDEAL GAS LAW AND KINETIC THEORY

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
_____________________________________________________________________________________________
19. 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) cm1∆T1 = cm2 ∆T2 ,
cm1 (T − T1 ) = cm2 (T2 − T )
m1 (T − T1 ) = m2 (T2 − T )
Solving for T gives
T=
m2T2 + m1T1
m2 + m1
(1)
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
Chapter 14 Problems
717
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=
20. 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
Vh
Psurface
1.01 × 10 5 Pa
=
=
= 0.205
Vtank
Ph
4.93 × 10 5 Pa
______________________________________________________________________________
21. 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 using the given percentages
and the total number of moles. To obtain the total number of moles, we 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 number nTotal of
718 THE IDEAL GAS LAW AND KINETIC THEORY
moles of gas as nTotal =
PV
, because the ideal gas law does not distinguish between types of
RT
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:
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 the
RT
air becomes
(
)
(
m = f Nitrogen M Nitrogen + f Oxygen M Oxygen nTotal = f Nitrogen M Nitrogen + f Oxygen 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
22. 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 P2 P1 + ρ gh
ρ gh
=
=
= 1+
V2 P1
P1
P1
Chapter 14 Problems
719
Therefore,
(1.000 × 103 kg/m3 )(9.80 m/s 2 )(0.200 m)
= 1.02
V2
1.01×105 Pa
______________________________________________________________________________
V1
= 1+
23. SSM REASONING The graph that accompanies Problem 68 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 68 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
Percent relative  Equilibrium vapor pressure of 
pressure of = 
×
humidity   water at the existing temperature 100%

water vapor
=
(55%)(4250 Pa )
100%
= 2.34 × 10 3 Pa
The ideal gas law then gives the number of moles of water vapor per cubic meter of air as
n
P
(2.34 × 10 3 Pa)
=
=
= 0.93 mol/m 3
(14.1)
V RT [8.31 J/(mol ⋅ K)](303 K)
______________________________________________________________________________
24. 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
720 THE IDEAL GAS LAW AND KINETIC THEORY
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 )
25. 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
( ρ0V ) g =
mtotal g
(1)
Total weight
of balloon
Buoyant force
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
ρ 0V = mload + 

M

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
______________________________________________________________________________
Chapter 14 Problems
721
26. 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
2
L


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=
27. 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
722 THE IDEAL GAS LAW AND KINETIC THEORY
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 ) 
 P2 d 2 
T2 = 
T =
(273 K) = 308 K
 P d  1  ( 2.02 × 105 Pa ) ( 0.760 m ) 
 1 1


______________________________________________________________________________
28. 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


T = rms =
= 750 K
−23
3k
3 1.38 × 10 J/K
(
)
______________________________________________________________________________
29. SSM REASONING AND SOLUTION Using the expressions for v 2 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 
2
m/s) 2 + (7.0 m/s) 2 + (9.0 m/s) 2  = 46.3 m 2 /s 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
30. 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,
Chapter 14 Problems
T=
723
( 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
m
7.31 × 10−26 kg
______________________________________________________________________________
vrms =
31. 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), according to
m
. Substituting this expression into our result for the kinetic-energy ratio gives
n=
M
m
nArgon
M Argon
M Krypton
KE Krypton
=
=
=
m
nKrypton
M Argon
KE Argon
M Krypton
724 THE IDEAL GAS LAW AND KINETIC THEORY
Taking the masses per mole from the periodic table on the inside of the back cover of the
text, we find
KE Krypton M Krypton 83.80 g/mol
=
=
= 2.098
39.948 g/mol
M Argon
KE Argon
32. REASONING The translational rms-speed vrms is related to the Kelvin temperature T by
1 mv 2
rms
2
= 32 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
( vrms )earth's surface =
and
3kTearth's surface
m
Dividing the first equation by the second gives
( vrms )ionosphere
=
( vrms )earth's surface
3kTionosphere
m
3kTearth's surface
m
=
Tionosphere
Tearth's surface
= 3 = 1.73
____________________________________________________________________________________________
33. REASONING The smoke particles have the same average translational kinetic energy as the
2
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
(
)
______________________________________________________________________________
34. REASONING The total average kinetic energy ( KE ) total of all the molecules in the gas is
equal to the number N of molecules times the average kinetic energy KE of each molecule:
( KE )total = N ( KE )
(1)
Chapter 14 Problems
725
The number of molecules is equal to the number n of moles times Avogadro’s number NA
(which is the number of molecules per mole), so N = nNA. Substituting this relation for N
into Equation (1) gives
( KE )total = ( n NA ) ( KE )
(2)
According to Equation 14.6, the average kinetic energy of an individual molecule is related
to the Kelvin temperature T of the gas by KE = 32 k T , where k is Boltzmann’s constant.
Substituting this result into Equation (2) yields
( KE )total = ( n NA ) ( KE ) = ( n NA ) ( 32 k T )
SOLUTION The total average kinetic energy of all the molecules is
( KE )total = n N A ( 32 k T )
= ( 3.0 mol ) ( 6.022 × 1023 mol−1 )  32 (1.38 × 10−23 J/K ) ( 320 K )  = 1.2 × 104 J
__________________________________________________________________________
35. SSM WWW REASONING AND SOLUTION Since we are treating the air as an ideal
gas, PV = nRT, so that U = 52 nRT = 52 PV = 52 ( 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


______________________________________________________________________________
36. REASONING AND SOLUTION We know from the ideal gas law that PV = nRT and
U = 32 nRT = 32 PV = 9300 J . A 0.25-hp engine provides (0.25 hp)(746 W)/(1 hp) = 187 J/s.
In order to equal the internal energy, therefore, the engine must run for a time of
9300 J
= 5.0 × 101 s
187 J/s
______________________________________________________________________________
726 THE IDEAL GAS LAW AND KINETIC THEORY
37. SSM WWW REASONING AND SOLUTION The average force exerted by one
electron on the screen is, from the impulse-momentum theorem (Equation 7.4),
F = ∆p/ ∆t = m∆v / ∆t . Therefore, in a time ∆t , N electrons exert an average force
F = Nm∆v / ∆t = (N / ∆t)m∆v . Since the pressure on the screen is the average force per unit
area (Equation 10.19), we have
P=
F ( N / ∆t )m∆v
=
A
A
16
(6.2 × 10
–31
7
electrons/s)(9.11 × 10
kg)(8.4 × 10 m/s – 0 m/s)
= 4.0 × 101 Pa
–7
1.2 × 10 m 2
______________________________________________________________________________
=
38. REASONING AND SOLUTION
a. Assuming that the direction of travel of the bullets is positive, the average change in
momentum per second is
∆p/∆t = m∆v/∆t = (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 = ∆p/∆t. 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
______________________________________________________________________________
39. REASONING The mass m of oxygen that diffuses in a time t through a trachea is given by
Equation 14.8 as m = ( DA ∆C ) t / L , where ∆C 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 ∆C is the higher concentration C2
minus the lower concentration C1, or ∆C = C2 − C1 .
SOLUTION Substituting the relation ∆C = 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
Chapter 14 Problems
727


1.9 × 10−3 m
−12
3
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 ) 
______________________________________________________________________________
40. REASONING AND SOLUTION Fick’s law of diffusion gives
L D A ∆C ( 4.2 × 10 m / s )( 4.0 × 10 m )( 3.5 × 10 kg/m )
=
= 7.0 × 10−3 m/s
v= =
−8
t
m
8.4 × 10 kg
______________________________________________________________________________
−5
2
−4
2
−2
3
41. 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 ∆C)t (12.4 × 10 –10 m 2 /s)(4.00 × 10 –4 m 2 )(1.50 kg/m 3 )(3600 s)
=
= 1.34 × 10 –7 kg
L
(0.0200 m)
______________________________________________________________________________
m=
42. REASONING
The concentration difference between the ends of the channel is
DC = Chigher - Clower. Therefore, we can determine the lower concentration as
Clower = Chigher - DC, provided that we can obtain a value for DC. We can find DC by using
Fick’s law of diffusion. According to this law, the mass rate of diffusion is given by
m DA∆C
=
(Equation 14.8), where D is the diffusion constant, A is the cross-sectional area
t
L
of the diffusion channel, and L is the length of the channel.
SOLUTION The lower concentration is Clower = Chigher - DC. Solving Fick’s law for DC,
and substituting the result into this expression gives
Clower
m
 L
t
= Chigher −  
DA
= 8.3 × 10
−3
kg/m
3
4.2 ×10−14 kg/s ) ( 0.020 m )
(
−
= 3.0 × 10−3 kg/m3
−9
−4
2
2
(1.06 ×10 m /s )(1.5 ×10 m )
728 THE IDEAL GAS LAW AND KINETIC THEORY
43. SSM WWW 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, ∆C , 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 – ∆C.
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
mass flow rate of CCl4 at point A is 5.00 × 10–13 kg/s .
b. Solving Fick's law for ∆C , we obtain
∆C =
mL (m / t ) L
=
DAt
DA
=
(5.00×10−13 kg/s)(5.00×10−3 m)
= 4.2×10−3 kg/m3
–10
2
–4
2
(20.0×10 m /s)(3.00×10 m )
Then,
CA = Cleft end − ∆C = (1.00 × 10
–2
3
kg/m ) − (4.2 × 10
–3
3
kg/m )= 5.8 × 10
–3
kg/m
3
______________________________________________________________________________
44. 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 into this expression yields
t = (2.5 × 10–2 m)2/[2(1.0 × 10–5 m2/s)] = 31 s
______________________________________________________________________________
45. REASONING AND SOLUTION Equation 14.8,
DA ∆C t
m=
. Solving for the time t gives
L
mL
t=
DA ∆C
gives
Fick's
law
of
diffusion:
(1)
Chapter 14 Problems
729
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 ∆C = 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:
PV = nRT
⇒
P=
ρ RT
M
where M = 0.0180152 kg/mol is the mass per mole for water (H2O). From the figure that
accompanies Problem 68 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!
______________________________________________________________________________
46. 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
______________________________________________________________________________
47. SSM REASONING AND SOLUTION According to the ideal gas law (Equation 14.1),
the total number of moles n of fresh air in the sample 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 the sample is nNA , where NA is Avogadro's number.
Since the sample contains approximately 21% oxygen, the total number of oxygen
molecules in the sample is (0.21) nNA or
730 THE IDEAL GAS LAW AND KINETIC THEORY
(0.21)(1.94 × 10
−2
mol)(6.022 × 10
23
−1
mol )= 2.5 × 10
21
______________________________________________________________________________
48. REASONING AND SOLUTION From the drawing in the text we can see that the volume
of the cylinder is V = Ah, where A is the cross-sectional area of the piston. Assuming that
pressure is constant, V1/T1 = V2/T2 or A1h1/T1 = A2h2/T2. Since A1 = A2, we have
T 
 318 K 
h2 = h1  2  = ( 0.120 m ) 
 = 0.140 m
273
K
T


 1
______________________________________________________________________________
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
–5 2
D = 2.4 × 10 m /s , the time t required for the first water molecule to travel L = 0.010 m
is
L2
(0.010 m) 2
t=
=
= 2.1 s
2D 2(2.4 × 10 –5 m 2 / s)
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 two
hydrogen atoms
Mass of one
oxygen atom
The mass of a single molecule is
m=
18.0153 × 10 –3 kg/mol
= 2.99 × 10 –26 kg
23
–1
6.022 × 10 mol
The average translational kinetic energy of water molecules at 293 K is, according to
Equation 14.6,
3
3
KE = 2 kT = 2 (1.38 × 10 –23 J/K)( 293 K ) = 6.07 × 10 –21 J
Chapter 14 Problems
731
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 changes 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).
______________________________________________________________________________
The behavior of the molecules is described by Equation 14.5:
50. REASONING
2
2
1
PV = 3 N ( 2 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.
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
= (463 m/s)
=
= 327 m/s
2 N1
N2
2
______________________________________________________________________________
(vrms ) 2 = (vrms )1
732 THE IDEAL GAS LAW AND KINETIC THEORY
51. SSM REASONING The internal energy of the neon at any Kelvin temperature T is given
by Equation 14.7, U = (3 / 2)nRT . Therefore, when the temperature of the neon increases
from an initial temperature Ti to a final temperature Tf , the internal energy of the neon
increases by an amount
3
∆ U = Uf − Ui = nR (Tf − Ti )
2
In order to use this equation, we must first determine n. Since the neon is confined to a tank,
the number of moles n is constant, and we can use the information given concerning the
initial conditions to determine an expression for the quantity nR. According to the ideal gas
law (Equation 14.1),
PV
nR = i i
Ti
These two expressions can be combined to obtain an equation in terms of the variables that
correspond to the data given in the problem statement.
SOLUTION Combining the two expressions and substituting the given values yields
∆U =
3  PiVi 
(T − Ti )
2  Ti  f
3  (1.01 × 105 Pa)(680 m3 ) 
(294.3 K − 293.2 K) = 3.9 × 10 5 J


2 
(293.2 K)

______________________________________________________________________________
=
52. 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 )
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
(1)
Chapter 14 Problems
Pf V
nf
P
RT
=
= f
PV
ni
Pi
i
RT
or
733
P 
nf = ni  f 
 Pi 
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

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
53. SSM 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/m3 
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
= 2.6 × 10 –10 m
9
3.9 × 10 /m
______________________________________________________________________________
d=
54. REASONING The rms-speed vrms of the sulfur dioxide molecules is related to the Kelvin
2
= 32 kT (Equation 14.6), where m is the mass of an SO2 molecule
temperature T by 12 mvrms
and k is Boltzmann’s constant. Solving this equation for the rms-speed gives
vrms =
3kT
m
(1)
734 THE IDEAL GAS LAW AND KINETIC THEORY
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
vrms
 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
Mass of a single
sulfur atom
+ 2 (15.9994 u) = 64.07 u
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 ) 
______________________________________________________________________________
55. 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, 12 mv02 = 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.
Chapter 14 Problems
735
SOLUTION Equation 6.9b gives the principle of conservation of mechanical energy:
1 mv 2
f
2
+ mghf
=
Final mechanical energy
1 mv 2
0
2
+ mgh0
Initial mechanical energy
In this expression, we know that 12 mv02 = KE = 32 kT and that 12 mvf2 = 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 energyconservation equation as follows:
mghf = 32 kT
or
hf =
3kT
2mg
Using M to denote the molecular mass (in kilograms per mole) and recognizing that
M
, where NA is Avogadro’s number and is the number of xenon atoms per mole, we
m=
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
(
)(
)
56. 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.
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
736 THE IDEAL GAS LAW AND KINETIC THEORY
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


______________________________________________________________________________
(
)
57. REASONING AND SOLUTION We need to determine the amount of He inside the
balloon. 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 balloon 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 = 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
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
3
4 π (1.50 m )
V
3
______________________________________________________________________________
58. CONCEPT QUESTIONS
a. The mass of one of its atoms (in atomic mass units) has the same numerical value as the
mass per mole (in units of g/mol).
b. Dividing the mass of the sample by the mass per mole gives the number of moles of
atoms in the sample.
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. We can convert the mass from atomic mass units to kilograms by noting that 1 u =
1.6605 × 10–27 kg:
Chapter 14 Problems
737
 1.6605 × 10−27 kg 
−25
m = (196.967 u ) 
 = 3.2706 × 10 kg
1u


c. The number 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=
59. CONCEPT QUESTIONS
a. According to the ideal gas law, PV = nRT , the absolute pressure P is directly
proportional to the temperature T, provided the temperature is measured on the Kelvin scale.
Therefore, if the temperature on the Kelvin scale doubles, the pressure also doubles.
b. The pressure is not proportional to the temperature when it is measured on the Celsius
scale. (The pressure is proportional to the temperature when measured on the Kelvin scale.)
Thus, the pressure does not double when the temperature on the Celsius scale doubles.
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. The ratio of the pressures at the temperatures of 35.0 °C and 70.0 °C is
P2
T ( 273.15 + 70.0 ) K
= 2 =
= 1.11
P1
T1 ( 273.15 +35.0 ) K
______________________________________________________________________________
60. CONCEPT QUESTION 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.
738 THE IDEAL GAS LAW AND KINETIC THEORY
SOLUTION The temperature of each gas can be found from the ideal gas law,
Equation 14.1:
( 25.0 Pa ) 4.0 m3
PAVA
TA =
=
= 120 K
nR
( 0.10 mol ) 8.31 J /( mol ⋅ K )
(
)
(
)
(
)
(
)
( 30.0 Pa ) 5.0 m3
PBVB
TB =
=
= 180 K
nR
( 0.10 mol ) 8.31 J /( mol ⋅ K )
( 20.0 Pa ) 5.0 m3
PCVC
TC =
=
= 120 K
nR
( 0.10 mol ) 8.31 J /( mol ⋅ K )
( 2.0 Pa ) 75 m3
PDVD
TD =
=
= 180 K
nR
( 0.10 mol ) 8.31 J /( mol ⋅ K )
______________________________________________________________________________
61. CONCEPT QUESTION According to the kinetic theory of gases, the average kinetic
2
= 32 kT . We see that
energy of an atom is related to the temperature of the gas by 12 mvrms
the temperature is proportional to the product of the mass and the square of the rms-speed.
2
has the greatest temperature. Using the
Therefore, the tank with the greatest value of mvrms
information from the table given with the problem statement, we see that the values of
2
mvrms
for each tank are:
Tank
2
mvrms
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:
Chapter 14 Problems
(
)
739
2
3.32 × 10 kg (1223 m /s )
mvrms
=
= 1200 K
TA =
−23
3k
3 1.38 × 10 J /K
TB =
TC
(
m 2vrms
3k
−26
(
TD =
(
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
______________________________________________________________________________
62. CONCEPT QUESTIONS
a. According to the kinetic theory of gases, the average kinetic energy KE of an atom 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 hydrogen and deuterium atoms have the same
average kinetic energy.
2
b. The average kinetic energy is related to the rms-speed by KE = 12 mvrms
. Since both
gases have the same average kinetic energy and hydrogen has the smaller mass, it has the
greater rms-speed. Therefore, hydrogen has the greater diffusion rate.
SOLUTION For a fixed temperature, the ratio RH/RD of the diffusion rates for hydrogen
and deuterium is equal to the ratio vrms, H/vrms, D of the rms-speeds. According to Equation
14.6, it follows that vrms = 2KE/m , so that
v
RH
= rms, H =
RD
vrms, D
2KE
mH
2KE
mD
=
mD
=
mH
2.0 u
= 1.4
1.0 u
______________________________________________________________________________
63. CONCEPT QUESTIONS
a. 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), according to
m
.
n=
M
740 THE IDEAL GAS LAW AND KINETIC THEORY
b. To calculate the percentage, we divide the number n of moles of that 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.
c. 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.
d. 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
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
20.179 g/mol
=
×100 = 13.5 %
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
Chapter 14 Problems
pHelium =
nHelium
nArgon + nNeon + nHelium
× 100 =
mHelium
M Helium
mArgon
M Argon
m
m
+ Neon + Helium
M Neon M Helium
741
×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
These results are consistent with our answers to Concept Questions (c) and (d).
64. CONCEPT QUESTIONS
a. 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
(1)
where k is the spring constant.
b. 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
F0 = P0 A
Force applied by
initial pressure
and
Ff = Pf A
(2)
Force applied by
final pressure
c. The ideal gas law is PV = nRT. Since the number of moles is constant, this equation can
PV
PV
= nR = constant . Thus, the value of
be written as
is the same initially and finally,
T
T
and we can write
P0V0 Pf Vf
=
(3)
T0
Tf
d. 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)
742 THE IDEAL GAS LAW AND KINETIC THEORY
SOLUTION The final temperature can be obtained by rearranging Equation (3) to show that
 PV 
Tf =  f f  T0
 PV 
 0 0
(5)
Into this result we can now substitute expressions for P0 and Pf. These expressions can be
obtained by using Equations (2) in Equation (1) as follows (and recognizing that, for the
initial and final forces, P0A = FxApplied and Pf A = FxApplied):
P0 A
Force applied to
spring by initial
pressure
= kx0
and
Pf A
= kxf
(6)
Force applied to
spring by final
pressure
In addition, we can substitute Equation (4) for the final volume into Equation (5). With these
substitutions Equation (5) becomes
 kxf
 A
 Pf Vf 
Tf = 
T =
 P V  0
 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 )
= 3.70 ×102 K
( 0.0800 m ) ( 6.00 ×10−4 m3 )