CHAPTER 14 THE IDEAL GAS LAW AND KINETIC THEORY

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

PV = REASONING AND SOLUTION According to the ideal gas law (Equation 14.1),

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 ( 0f vv − )=∆ mt , 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 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 − v 0

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: 2 31rms2 2mv kT= .

If the translational speed of each molecule in an ideal gas is tripled, then the root-mean-square 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, 2 31rms2 2mv kT= , relates the average kinetic energy per particle to the Kelvin temperature of

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 32U 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


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 temperature: 2 31

rms2 2mv kT= . Since the mass of an argon atom is greater than the mass of a 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

22

23 164 500 g/mol 1 kg = 1.07×10 kg

1000 g6.022×10 mol−

−

______________________________________________________________________________ 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:

moleculem Nm=

Using M to denote the molecular mass (in grams per mole) and recognizing that

moleculeA

MmN

= , where NA is Avogadro’s number and is the number of molecules per mole,

we have

moleculeA

Mm Nm NN

= =

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


( ) ( ) ( ) ( ) (Carbon Hydrogen Chlorine Nitrogen Oxygen

Molecular mass 22 12.011 u 23 1.00794 u 1 35.453 u 2 14.0067 u 2 15.9994 u

382.89 u

= + + + +

=

)

The mass of the active ingredient in the standard dosage is

( )19 223

A

382.89 g/mol1.572 10 molecules 1.00 10 g6.022 10 molecules/mol

Mm NN

− = = × = × ×

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)

Mass of 14 Mass of 18carbon atoms hydrogen atoms

Mass of 2 nitrogen atoms

Molecular mass 14 (12.011 u) 18 (1.00794 u)

+ 2 (14.0067 u) +5

= +

Mass of 5oxygen atoms

(15.9994) = 294.307 u

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

–2523 1

294.307 g/mol 1 kg 4.887 10 kg1000 g6.022 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


Molecular massof Tylenol = 8 (12.011 u)

Mass of 8carbon atoms

+ 9 (1.00794 u)Mass of 9

hydrogen atoms

+ 14.0067 uMass of

nitrogen atom

+ 2 (15.9994)Mass of 2

oxygen atoms

=151.165 u

The molecular mass of Advil (C13H18O2) is

Molecular massof Advil = 13 (12.011 u)

Mass of 13carbon atoms

+18 (1.00794 u)Mass of 18

hydrogen atoms

+ 2 (15.9994)Mass of 2

oxygen atoms

= 206.285 u

a. Therefore, the number of molecules of pain reliever in the standard dose of Tylenol is

( )

A A

323 1 21

Number of molecules =Mass per mole

325 10 g 6.022 10 mol 1.29 10151.165 g/mol

mn N N

−−

=

×= × =

×

b. Similarly, the number of molecules of pain reliever in the standard dose of Advil is

( )

A A

23 1 20

Number of molecules =Mass per mole

2.00 10 g 6.022 10 mol 5.84 10206.285 g/mol

mn N N

−1−

=

×= × =

×

______________________________________________________________________________ 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

23 1A

30.1×10 = 5.00 mol6.022×10 mol

NnN −= =

Since the sample has a mass of 135 g, the mass per mole is

135 g5.00 mol

= 27.0 g/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 .

______________________________________________________________________________

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 . AN n N=

SOLUTION

a. Starting with n = m/(Mass per mole) and substituting in the relation m = (0.71)W/g, we have

(0.71)

Mass per mole Mass per mole

Wm gn = = (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. Converting this value to kilograms per mole gives

ggMass per mole = 18.0153 = 18.0153mol 3

1 kgmol 10 g

Substituting this relation into Equation 1 gives

( )2

(0.71) (0.71) 580 N9.80 m/s

Mass per mole g 18.0153

Wgn = =

31 kg

mol 10 g

32.3 10 mol= ×

b. The number of water molecules in the runner’s body is

( )( )3 23 1A 2.3 10 mol 6.022 10 mol 1.4 10N n N −= = × × = × 27

____________________________________________________________________________________________


7. 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

Mass per mole Mass per molem Vn ρ

= =

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

( )2

Mass per mole Mass per mole

r hVnρ πρ

= =

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.

( ) ( )( )( ) ( )22 31.00 g /cm 4.50 cm 12.0 cm

42.4 molMass per mole 18.0 g /mol

r hn

ρ π π= = =

______________________________________________________________________________ 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. 23 26 2346.069 g/mol 7.65 10 g = 7.65 10 kg

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

2526

1.61 kg 2.11 107.65 10 kg

N −= =×

×

______________________________________________________________________________ 9. SSM REASONING AND SOLUTION The number of moles of air that is pumped into

the tire is ∆ . According to the ideal gas law (Equation 14.1), nn = nf − ni = PV /(RT); therefore,

∆n = nf − ni =PfVfRTf

−PiViRTi

=VRT

Pf − Pi( )


Substitution of the data given in the problem leads to

( ) ( )( ) ( )

4 35 54.1 10 m 6.2 10 Pa 4.8 10 Pa 2.3 10 mol

8.31 J/ mol K 296 Kn

−−× ∆ = × − × = × ⋅

2

______________________________________________________________________________ 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

mass per mole M (in grams per mole), according to mnM

= . Using this 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 NitrogenHelium

Helium NitrogenNumber of moles Number of moles

of helium of nitrogen

mmM M

=

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

( )( )Nitrogen HeliumNitrogen

Helium

28.0 g/mol 0.16 g1.1 g

4.00 g/mol

M mm

M= = =

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

( )( )( ) ( )

5 31.72 10 Pa 2.81 m201 mol

8.31 J/ mol K 273.15 15.5 KPVnRT

×= = =

⋅ + (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


( ) ( ) ( )

( )5

3

201 mol 8.31 J/ mol K 273.15 28.2 K1.21 10 Pa

4.16 mnRTPV

⋅ + = = = × (14.1)

______________________________________________________________________________ 12. REASONING AND SOLUTION The ideal gas law gives

( )3 31 22 1

2 1

65.0 atm 297 K 1.00 m 67.0 m1.00 atm 288 K

P TV V

P T = = =

______________________________________________________________________________ 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,

5 3 ) 5(1.1 10 Pa)(5400 m 2.55 10 mol[8.31 J/(mol K)](280 K)

PVnRT

×= = = ×

⋅ 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,

( )( )5 31 kg2.55 10 mol 4.002 60 g/mol 1.0 10 kg1000 g

m = × = ×

______________________________________________________________________________ 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.


SOLUTION Using Boyle’s law we find that

i if

f

PVV

P=

The volume of helium available for filling balloons is

3 3i if

f0.0031 m 0.0031 m

PVV

P− = −

The maximum number of balloons that can be filled is

( )( )7 33i i 3

5fBalloons 3

Balloon

1.6 10 Pa 0.0031 m0.0031 m 0.0031 m

1.2 10 Pa 120.034 m

PVP

NV

×− −

×= = =

15. SSM REASONING According to the ideal gas law (Equation 14.1), PV = nRT . Since

n, the number of moles, is constant, n1R = n2R . Thus, according to Equation 14.1, we have

1 1 2 2

1 2

PV P VT T

=

SOLUTION Solving for T2, we have

2 2 1 12 1

1 1 1 1

48.5 /16 (305 K)= 925 K

P V P VT T

P V P V

= =

______________________________________________________________________________

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

Mass per molem RT

nRTPV 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

Mass per molem RT

V

P = =

18.0153 g( )[ ]( )

43

8.31 J/ mol K 388 K/mol

2.4 10 Pa0.030 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( )PEarth / kTEarth( ) =

PVenus

PEarth

TEarth

TVenus

=

9.0 ×106 Pa1.0 ×10 5 Pa

320 K740 K

= 39

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: /m Vρ = . 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 /m Vρ = , we obtain

(Mass per mole) m nV V

ρ = = (1)

Substituting n = (PV)/(RT) from the ideal gas law into Equation 1 gives the following result:

(Mass per mole) P V

nV

ρ = =(Mass per mole)

RTV

(Mass per mole) P

RT=

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


gMass per mole 28= 3

1 kgmol 10 g

The density of the nitrogen gas is

( )5 g2 1.013 10 Pa 28

(Mass per mole)PRT

ρ

×

= =

31 kg

mol 10 g

( )[ ]( )32.2 kg/m

8.31 J/ mol K 310 K

=

⋅

_____________________________________________________________________________________________ 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 1 2T cm T2∆ = ∆ ,

1 1 2 2( ) (cm T T cm T T )− = −

1 1 2 2( ) (m T T m T T )− = −

Solving for T gives

2 2 1 1

2 1

m T m TT

m m+

=+

(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

( )( )[ ]( )

5 321 1

11

5.0 10 Pa 2.0 m 5.5 10 mol8.31 J/(mol K) 220 K

PVn

RT×

= = = ×⋅

This corresponds to a mass ( )2 4

1 = (5.5×10 mol) 20.179 g/mol = 1.1×10 g = 1.1×10 kgm 1 .

Similarly, n2 = 2.4 × 102 mol and m2 = 4.9 × 103 g = 4.9 kg. Substituting these mass values into Equation (1) yields

( )( ) ( )( )

( ) ( )1

21

4.9 kg 580 K 1.1 10 kg 220 K3.3 10 K

4.9 kg 1.1 10 kgT

+ ×= =

+ ×

×


b. From the ideal gas law,

( ) ( ) ( )[ ]( )2 2 25

3 35.5 10 mol 2.4 10 mol 8.31 J/ mol K 3.3 10 K 2.8 10 Pa

2.0 m 5.8 mnRTPV

× + × ⋅ × = = = ×+

______________________________________________________________________________ 20. REASONING Since the temperature of the confined air is constant, Boyle's law applies,

and PsurfaceVsurface = PhVh , where Psurface and V are the pressure and volume of the air in the tank when the tank is at the surface of the water, and

surfacePh and V 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, V is equal to the volume V 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

h

surfacetank

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

5 3 3 2surface

5

(1.01 10 Pa) (1.00 10 kg/m )(9.80 m/s )(40.0 m)

4.93 10 Pa

hP P ghρ = + = × + ×

= ×

where we have used a value of for the density of water. The desired

volume fraction is

31.00 10 kg/mρ = × 3

Vh

Vtank=

Psurface

Ph=

1.01×10 5 Pa4.93 ×10 5 Pa

= 0.205

______________________________________________________________________________ 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


moles of gas as TotalPVnRT

= , because the ideal gas law does not distinguish 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:

Nitrogen Oxygen Nitrogen Nitrogen Oxygen Oxygenm m m n M n M= + = +

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:

Nitrogen Nitrogen Oxygen Oxygen Nitrogen Total Nitrogen Oxygen Total Oxygenm n M n M f n M f n M= + = +

According to the ideal gas law, we have TotalPVnRT

= . With this substitution, the mass of the

air becomes

( ) ( )Nitrogen Nitrogen Oxygen Oxygen Total Nitrogen Nitrogen Oxygen OxygenPVm f M f M n f M f MRT

= + = +

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,

( )

( ) ( )( ) ( )( )(

( ) ( ))

Nitrogen Nitrogen Oxygen Oxygen

5

4

1.01 10 Pa 2.5 m 4.0 m 5.0 m0.79 28.0 g/mol 0.21 32.0 g/mol

8.31 J/ mol K 295 K

5.9 10 g

PVm f M f MRT

= +

× = + ⋅

= ×

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,

1 2 1

2 1 1 11

V P P gh ghV P P P

ρ ρ+= = = +


Therefore,

3 3 21

52

(1.000 10 kg/m )(9.80 m/s )(0.200 m)1 11.01 10 Pa

VV

×= + =

×.02

______________________________________________________________________________ 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

pressure ofwater vapor

= Percent relativehumidity

Equilibrium vapor pressure ofwater at the existing temperature

× 1

100%

=55%( ) 4250 Pa( )

100%= 2.34 × 103 Pa

The ideal gas law then gives the number of moles of water vapor per cubic meter of air as

nV

=PRT

=(2.34 ×103 Pa)

[8.31 J/(mol ⋅ K)](303 K)= 0.93 mol/m3

(14.1)

______________________________________________________________________________ 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 ( 34

3 rπ , 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


( ) ( )

( ) ( )( )( ) ( )

3 34 4A A3 3Atoms

Gas

310 23 1 543

Percentage 100 100 100

2.0 10 m 6.022 10 mol 1.01 10 Pa100 0.090 %

8.31 J/ mol K 273 K

r nN r N PVnRTV RT

P

π π

π − −

= × = × = ×

× × ×= ×

⋅ =

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

( )0 to

Total weight Buoyant forceof balloon

V g m gρ = tal (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:

airPVm n MRT

= =

M

Thus, the total mass of the balloon is mtotal = mload + PVM/(RT) and Equation (1) becomes

0 loadPVV m MRT

ρ = +

Solving for T gives

( )

( )( )( )( )( ) ( )

0 load

5 3 –3

3 3 3

–

1.01 10 Pa 650 m 29 10 kg/mol440 K

1.29 10 kg/m 650 m – 320 kg 8.31 J/ mol K

PVMTV m Rρ

=

× ×= =

× ⋅

______________________________________________________________________________


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

( )

1 1 1 12 1 1 1 1 2

2 2 2 2

5 4

(since )

= 1.01 10 Pa 5.05 10 Pa2

V A L LP P P P A A

V A L L

LL

= = = =

× = ×

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

3397 N 1.98 10 N/m0.200 m

Fkx

= = = ×

______________________________________________________________________________ 27. SSM REASONING According to the ideal gas law (Equation 14.1), . Since n,

the number of moles of the gas, is constant, PV nRT=

1 2n R n R= . Therefore, , where and 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

1 1 1PV T = 2 2 2/ /P V T

1 273 KT = 2T

1 1 2 2

1 2

Pd P dT T

= (1)

where the pressures and 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 T .

1P 2P

2 SOLUTION Using Equation 11.4, we obtain the following values for the pressures and

. Note that the initial height of the mercury is 1P

2P 11 2 (1.520 m) 0.760 mh = = , while the final

height of the mercury is 12 4 (1.520 m) 0.380 mh = = .

( )5 4 3 2

1 0 1 1.01 10 Pa (1.36 10 kg/m )(9.80 m/s )(0.760 m) 2.02 10 PaP P ghρ = + = × + × = × 5


( ) ( )( )5 4 3 22 0 2 1.01 10 Pa 1.36 10 kg/m 9.80 m/s (0.380 m) 1.52 10 PaP P ghρ = + = × + × = ×

5

3

In these pressure calculations, the density of mercury is . In

Equation (1) we note that d and

41.36 10 kg/mρ = ×

1 0.760 m= 2 1.14 md = . Solving Equation (1) for T and substituting values, we obtain

2

( )

( )( )

52 2

2 1 51 1

1.52 10 Pa (1.14 m (273 K) 308 K2.02 10 Pa 0.760 m

P dT T

Pd ×

= = = ×

)

______________________________________________________________________________ 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 31rms2 2mv 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

( ) ( )

( )

272

2rms

23

1.66 10 kg44.0 u 650 m /s1 u

750 K3 3 1.38 10 J/K

mvT

k

−

−

× = = =

×

______________________________________________________________________________

29. SSM REASONING AND SOLUTION Using the expressions for 2v and ( ) given in the statement of the problem, we obtain:

2v

a. 1 12 2 2 2 2 2 2 2 2

1 2 33 3( ) (3.0 m/s) (7.0 m/s) (9.0 m/s) 46.3 m /sv v v v

= + + = + + =

b. ( )2 2 2 2 21 11 2 33 3( ) (3.0 m/s 7.0 m/s 9.0 m/s) 40.1 m /sv v v v = + + = + + =

2v and ( ) are not equal, because they are two different physical quantities. 2

v______________________________________________________________________________ 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 12 m = 2

rmsv 32 kT, we can solve for the temperature to get T = m /(3k), where the mass of

an H2O molecule is (18.015 g/mol)/(6.022 × 1023 mol–1) = 2.99 × 10–23 g. Thus,

2rmsv


( )( )( )

226

232.99 10 kg 648 m/s

303 K3 1.38 10 J/K

T−

−

×= =

×

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

23

rms 263 3(1.38 10 J/K)(303 K) 414 m/s

7.31 10 kgkTvm

−

−×

= = =×

______________________________________________________________________________ 31. REASONING According to the kinetic theory of gases, the average kinetic energy of an

atom is 32 kT=KE (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 PVnR

=T , we find that

the desired ratio is

323

KryptonKrypton Krypton Argon23Argon KryptonArgon2 3

2Argon

KEKE

PVkn RkT n

nkT PVkn R

= = =

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

mnM

= . Substituting this expression into our result for the kinetic-energy ratio gives

Krypton Argon Argon Krypton

Argon Krypton Argon

Krypton

KEKE

mn M M

mn MM

= = =


Taking the masses per mole from the periodic table on the inside of the back cover of the text, we find

Krypton Krypton

Argon Argon

KE 83.80 g/mol 2.09839.948 g/molKE

M

M= = =

32. REASONING The translational rms-speed vrms is related to the Kelvin temperature T by

2 31rms2 2mv 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 rms 3 /v kT= 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

( ) ( )ionosphere earth's surfacerms rmsionosphere earth's surface

3 3 and

kT kTv v

m m= =

Dividing the first equation by the second gives

( )( )

ionosphererms ionosphere ionosphere

earth's surfaceearth's surfacerms earth's surface

3

3 1.733

kTv Tm

TkTvm

= = = =

____________________________________________________________________________________________ 33. REASONING The smoke particles have the same average translational kinetic energy as the

air molecules, namely, 2 31rms2 2mv 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

( )( )

( )

2315

2 23rms

3 1.38 10 J/K 301 K3 1.6 10 kg2.8 10 m/s

kTmv

−−

−

×= = = ×

×

______________________________________________________________________________

34. REASONING The total average kinetic energy ( )totalKE of all the molecules in the gas is equal to the number N of molecules times the average kinetic energy KE of each molecule:

( ) ( )totalKE KEN= (1)


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

( ) ( )( )total AKE KEn N= (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 32KE k T= , where k is Boltzmann’s constant.

Substituting this result into Equation (2) yields

( ) ( )( ) ( )( )3total A A 2KE KEn N n N k T= =

SOLUTION The total average kinetic energy of all the molecules is

( ) ( )( )( ) ( )( )

3total A 2

23 1 23 432

KE

3.0 mol 6.022 10 mol 1.38 10 J/K 320 K 1.2 10 J

n N k T

− −

=

= × × = ×

__________________________________________________________________________ 35. SSM WWW REASONING AND SOLUTION Since we are treating the air as an ideal

gas, PV = nRT, so that ( ) ( )6 55 5 52 2 2

37.7×10 Pa 5.6 10 m 1.1 10 JnRT PV= = = × = × 13U . The number of joules of energy consumed per day by one house is

3 8J h 3600 s30.0 kW h 30.0 10 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

( )13 58

1 home1.1×10 J = 1.0 ×10 homes1.08×10 J

______________________________________________________________________________ 36. REASONING AND SOLUTION We know from the ideal gas law that PV = nRT and

3 32 2 9300 JU nRT PV= = = . 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

19300 J 5.0 10 s187 J/s

= ×

______________________________________________________________________________


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 =FA

=(N / ∆t)m∆v

A

= (6.2 ×1016 electrons/s)(9.11 ×10–31 kg)(8.4 ×107 m/s – 0 m/s)1.2 ×10–7 m2 = 4.0 × 101 Pa

______________________________________________________________________________ 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 × 105 Pa ______________________________________________________________________________

39. REASONING The mass m of oxygen that diffuses in a time t through a trachea is given by Equation 14.8 as , 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

( ) /m DA C t= ∆ L

2 1C C C∆ = − . SOLUTION Substituting the relation 2C C C1∆ = − into Equation 14.8 and solving for C1

gives

1 2L mC C

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


( )( )( )3 1

1 5 2 9 21.9 10 m0.28 kg/m 1.7 10 kg/s 0.14 kg/m

1.1 10 m /s 2.1 10 mC

−3−

− −

×= − × =

× ×

2 3

______________________________________________________________________________ 40. REASONING AND SOLUTION Fick’s law of diffusion gives

( )( )( )5 2 4 2 2 3

38

4.2 10 m / s 4.0 10 m 3.5 10 kg/m7.0 10 m/s

8.4 10 kgD A CLv

t m

− − −−

−

× × ×∆= = = = ×

×

______________________________________________________________________________ 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

m =(DA ∆C)t

L=

(12.4 ×10 –10 m 2 /s)(4.00 × 10–4 m 2 )(1.50 kg/m3 )(3600 s)(0.0200 m)

= 1.34 ×10 –7 kg

______________________________________________________________________________ 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 Ct L

∆= (Equation 14.8), where D is the diffusion constant, A is the cross-sectional area

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

( )( )( )( )

lower higher

143 3 3

9 2 4 2

4.2 10 kg/s 0.020 m8.3 10 kg/m 3.0 10 kg/m

1.06 10 m /s 1.5 10 m

m LtC CDA

−− −

− −

= −

×= × − = ×

× ×3


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

13 33 3

–10 2 –4 2

( / )

(5.00×10 kg/s)(5.00×10 m) = 4.2×10 kg/m(20.0×10 m /s)(3.00×10 m )

mL m t LCDAt DA

− −−

∆ = =

=

Then,

–2 3 –3 3 –3A left end (1.00 10 kg/m ) (4.2 ×10 kg/m )= 5.8 ×10 kg/mC C C= − ∆ = × − 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

( )2 / 2t L 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, gives Fick's law of diffusion:

DA C tmL∆

= . Solving for the time t gives

mLtDA C

=∆

(1)


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:

RTPV nRT PM

ρ= ⇒ =

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,

( )( )

[ ]3

2 32

2.4 10 Pa 0.0180 kg/mol1.8 10 kg/m

8.31 J/(mol K) (293 K)PMCRT

ρ −×= = = = ×

⋅

Substituting values into Equation (1) gives

( )( )( )( )

36

5 2 4 2 2 32.0×10 kg (0.15 m)

= 2.3 10 s2.4 10 m /s 3.0 10 m 1.8 10 kg/m

t−

− − −= ×

× × ×

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,

( )5

22 1 5

1

3.01 10 Pa284 K 304 K2.81 10 Pa

PT T

P ×

= = = ×

______________________________________________________________________________ 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 =PVRT

=(1.0 ×105 Pa)(5.0 ×10–4 m3)

[8.31 J/(mole ⋅ K)] (310 K)=1.94 ×10–2 mol

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


2 23 1(0.21)(1.94 ×10 mol)(6.022 ×10 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

( )22 1

1

318 K0.120 m 0.140 m273 K

Th h

T = = =

______________________________________________________________________________ 49. REASONING AND SOLUTION a. As stated, the time required for the first solute molecule to traverse a channel of length L

is t . Therefore, for water vapor in air at 293 K, where the diffusion constant is = L2 /(2D)= 2.4 ×10D –5 m2/s , the time t required for the first water molecule to travel L = 0.010 m

is

t =L2

2D=

(0.010 m)2

2(2.4 ×10–5 m2 / s)= 2.1 s

b. If a water molecule were traveling at the translational rms speed v for water, the time

t it would take to travel the distance rms

L = 0.010 m would be given by t = L / vrms , where,

according to Equation 14.6 (KE = 12 mvrms

2rms =), v 2 KE( )/ m . Before we can use the last

expression for the translation rms speed v , we must determine the mass m of a water molecule and the average translational kinetic energy KE

rms.

Using the periodic table on the inside of the text’s back cover, we find that the molecular

mass of a water molecule is

Mass of one Mass of two oxygen atomhydrogen atoms

2(1.00794 u) 15.9994 u 18.0153 u+ =

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,

KE = 32

kT = 32

(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 t =

Lvrms

=0.010 m637 m/s

= 1.6 ×10–5 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).

______________________________________________________________________________ 50. REASONING The behavior of the molecules is described by Equation 14.5:

22 1rms3 2(PV N mv= ) . Since the pressure and volume of the gas are kept constant, while the

number of molecules is doubled, we can write 1 2 2 2PV P V= , 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 1 2 12 2 2

1 rms 1 2 rms 2 1 rms 1 2 rms 23 2 3 2( ) ( ) or ( ) ( )N m v N m v N v N v = =

2

1N

The last expression can be solved for ( , the final translational rms speed. rms 2)v SOLUTION Since the number of molecules is doubled, 2 2N = . Solving the last

expression above for ( , we find rms 2)v

1 1rms 2 rms 1

2 1

463 m/s( ) ( ) (463 m/s) 327 m/s2 2

N Nv v

N N= = = =

______________________________________________________________________________


51. SSM REASONING The internal energy of the neon at any Kelvin temperature T is given by Equation 14.7, U . Therefore, when the temperature of the neon increases from an initial temperature

(3 / 2)n= RTTi to a final temperature Tf , the internal energy of the neon

increases by an amount ∆U = Uf −Ui = 3

2nR (Tf − Ti )

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),

nR =PiViTi

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 = 32

PiViTi

(Tf − Ti )

= 32

(1.01×105 Pa)(680 m3)(293.2 K)

(294.3 K − 293.2 K) = 3.9 ×105 J

______________________________________________________________________________

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

( ) ( )i f

Number of moles withdrawn

Mass per molem n n= − (1)

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

f if i and

P V PVn n

RT R= =

T

Note that the volume V and temperature T do not change. Dividing the first by the second

equation gives


f

f ff i

ii i or

P Vn PRT n nPVn P

RT

= = =

f

i

PP

Substituting this expression for nf into Equation 1 gives

( )( ) ( )fi f i i

iMass per mole Mass per mole

Pm n n n n

P

= − = −

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

( )

( )

fi i

iMass per mole

25 atm g0.85 mol 0.85 mol 28.0134 8.1 g38 atm mol

Pm n n

P

= −

= − =

______________________________________________________________________________ 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,

3

( )3 3

23 1 28 32700 ×10 g/m/ 6.022 ×10 mol = 6.0 ×10 atoms/m26.9815 g/mol

N V − =

Assuming that the volume of the solid contains many small cubes, with one atom at the

center of each, then there are (6 along each edge of a 1.00-m3 cube. Therefore, the spacing between the centers of neighboring atoms is

.0 ×10 28 )1/3 atoms/m = 3.9 ×10 9 atoms/m

d =

13.9 ×109 /m

= 2.6 ×10–10 m

______________________________________________________________________________

54. REASONING The rms-speed vrms of the sulfur dioxide molecules is related to the Kelvin

temperature T by 2 31rms2 2mv kT= (Equation 14.6), where m is the mass of an SO2 molecule

and k is Boltzmann’s constant. Solving this equation for the rms-speed gives

rms3kTvm

= (1)


The temperature can be found from the ideal gas law, Equation 14.1, as ( )/V nR=T P ,

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

rms

3 3PVk k PVnRvm nm

= =

R

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

Mass of a single Mass of two

sulfur atom oxygen atoms

32.07 u 2 (15.9994 u) 64.07 u+ =

Since 1 u = 1.6605 × 10−27 kg (see Section 14.1), the mass of a sulfur dioxide molecule is

64.07 um = ( )27 1.6605 10 kg

1 u

−× 251.064 10 kg− = ×

The translational rms-speed of the sulfur dioxide molecules is

( )( )( )( )( ) ( )

rms

23 4 3

25

3

3 1.38 10 J/K 2.12 10 Pa 50.0 m 343 m/s421 mol 1.064 10 kg 8.31 J/ mol K

k PVvnm R

−

−

=

× ×= =

× ⋅

______________________________________________________________________________ 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 21

02 mv is equal to the average translational kinetic energy.

Therefore, 2 3102 KEmv kT= = 2 , 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: 2 21 1

f f 0 02 2Final mechanical energy Initial mechanical energy

mv mgh mv mgh+ = +

In this expression, we know that 2 31

02 KEmv kT= = 2 and that 21f2 0 Jmv = (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:

3

f f23 or 2

kTmgh kT hmg

= =

Using M to denote the molecular mass (in kilograms per mole) and recognizing that

A

MmN

= , where NA is Avogadro’s number and is the number of xenon atoms per mole, we

have A

f

A

33 32 2

2

kN TkT kThmg MgM g

N

= = =

Recognizing that kNA = R and that M = 131.29 g/mol = 131.29 × 10-3 kg/mol, we find

( ) ( )( )( )

Af 3 2

3 8.31 J/ mol K 291 K3 3 2820 m2 2 2 131.29 10 kg/mol 9.80 m/s

kN T RThMg Mg −

⋅ = = = =×

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

perspiration vm LQEN N

= =

But the mass of perspiration is equal to the mass m of a single water molecule

times the number N of water molecules. The mass of a single water molecule is equal to its 2H O molecule


molecular mass (18.0 u), converted into kilograms. The average energy given to a single water molecule is

( ) ( )

2

2

H O molecule vperspiration v

276 2

H O molecule v1.66 10 kg18.0 u 2.42 10 J/kg 7.23 10 J

1 u

m N Lm LE

N N

E m L−

−

= =

×= = × = ×

0

______________________________________________________________________________ 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

( ) ( ) ( )33 24

b 31.19 kg/m 1.50 m 9.80 m/s 164.9 NF mg Vgρ π = = = =

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:

313.8 kg 3450 mol

4.0026 10 kg/molmnM −= = =

×

Using the ideal gas law to find the pressure, we have

( )[ ]( )

( )5

343

3450 mol 8.31 J/(mol K) 305 K6.19 10 Pa

1.50 m

nRTPV π

⋅= = = ×

______________________________________________________________________________ 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:


( )27

251.6605 10 kg196.967 u 3.2706 10 kg1 u

m−

− ×= =

×

c. The number of moles of atoms is equal to the mass m divided by the mass per mole:

285 g 1.45 mol

Mass per mole 196.967 g /molmn = = =

______________________________________________________________________________ 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

1 21 2and

nRT nRTP P

V V= =

Taking the ratio P2/P1 of the final pressure to the initial pressure gives

2

2 2

11 1

70.0 K 2.0035.0 K

nRTP TV

nRTP TV

= = = =

b. The ratio of the pressures at the temperatures of 35.0 °C and 70.0 °C is

( )( )

2 2

1 1

273.15 70.0 K1.11

273.15 +35.0 KP TP T

+= = =

______________________________________________________________________________ 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 , 150 , 100 3Pa m⋅ 3Pa m⋅ 3Pa m⋅ , and 150 3Pa m⋅ . 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:

( )( )( ) ( )

( )( )( ) ( )

( )( )( ) ( )

( )( )( ) ( )

3

A AA

3

B BB

3

C CC

3

D DD

25.0 Pa 4.0 m120 K

0.10 mol 8.31 J / mol K

30.0 Pa 5.0 m180 K

0.10 mol 8.31 J / mol K

20.0 Pa 5.0 m120 K

0.10 mol 8.31 J / mol K

2.0 Pa 75 m180 K

0.10 mol 8.31 J / mol K

P VT

nR

P VT

nR

P VT

nR

P VT

nR

= = = ⋅

= = = ⋅

= = = ⋅

= = = ⋅

______________________________________________________________________________ 61. CONCEPT QUESTION According to the kinetic theory of gases, the average kinetic

energy of an atom is related to the temperature of the gas by 2 312 2rmsmv kT= . We see that

the temperature is proportional to the product of the mass and the square of the rms-speed. Therefore, the tank with the greatest value of has the greatest temperature. Using the information from the table given with the problem statement, we see that the values of

for each tank are:

2rmsmv

2rmsmv

Tank A

2rmsmv

B ( )2 2rms rms2 4m v mv=

C ( ) 2 2rms rms2 2m v mv=

D ( ) 2rms rms2 2 8m v mv=

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:


( )( )( )

( ) ( )( )( )

( ) ( )( )( )

( ) ( )( )

2262rms

A 23

2 226rms

B 23

2262rms

C 23

2 226rms

D

3.32 10 kg 1223 m /s1200 K

3 3 1.38 10 J /K

2 3.32 10 kg 2 1223 m /s4800 K

3 3 1.38 10 J /K

2 3.32 10 kg 1223 m /s22400 K

3 3 1.38 10 J /K

2 2 2 3.32 10 kg 2 1223 m /s

3 3 1

mvT

k

m vT

k

m vT

k

m vT

k

−

−

−

−

−

−

−

×= = =

×

× ×= = =

×

×= = =

×

× ×= =

( )239600 K

.38 10 J /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 32KE 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.

b. The average kinetic energy is related to the rms-speed by 21

rms2KE mv= . 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 rms 2KE/v m= , so that

rms, H HH D

D rms, D H

D

2KE2.0 u 1.41.0 u2KE

v mR mR v m

m

= = = = =

______________________________________________________________________________ 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 mnM

= .


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

Argon

Argon ArgonArgon

ArgonArgon Neon Helium Neon Helium

Argon Neon Helium

100 100

1.20 g39.948 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

mn M

p mn n n m mM M M

= × =+ +

+ +

= ×+ +

×

=

The percentage of neon is

Neon

Neon NeonNeon


Argon Neon Helium

100 100

2.60 g20.179 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

mn M

p mn n n m mM M M

= × =+ +

+ +

= ×+ +

×

=

The percentage of helium is


Helium

Helium HeliumHelium


Argon Neon Helium

100 100

3.20 g4.0026 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

mn M

p mn n n m mM M M

= × =+ +

+ +

= ×+ +

×

=

These results are consistent with our answers to Concept Questions (c) and (d). 64. CONCEPT QUESTIONS a. The force Fx

Applied 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

AppliedxF 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

0 0 f f

Force applied byForce applied byfinal pressureinitial pressure

and F P A F P A= = (2)

c. The ideal gas law is PV = nRT. Since the number of moles is constant, this equation can

be written as constantPV nRT

= = . Thus, the value of PVT

is the same initially and finally,

and we can write 0 0 f f

0 f

P V P VT T

= (3)

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 ( )f 0A x x− . Therefore, we have

( )f 0 f 0V V A x x= + − (4)


SOLUTION The final temperature can be obtained by rearranging Equation (3) to show that

f ff

0 0

P VT

P V

=

0T

f

(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 = Fx

Applied and Pf A = FxApplied):

0 0 f

Force applied toForce applied tospring by finalspring by initial

pressurepressure

and P A kx P A kx= = (6)

In addition, we can substitute Equation (4) for the final volume into Equation (5). With these

substitutions Equation (5) becomes

( ) ( )

( ) ( )( )( )( )

( )

f0 f 0 0

f 0 f 0 0f ff 0

00 0 0 00

4 3 3 2

4 3

2

0.1000 m 6.00 10 m 2.50 10 m 0.1000 m 0.0800 m 273 K

0.0800 m 6.00 10 m

3.70 10 K

kxV A x x T x V A x x TP V AT T

kxP V x VVA

− −

−

+ − + − = = =

× + × − =×

= ×

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CHAPTER 14 THE IDEAL GAS LAW AND KINETIC THEORY