Ch14 the Ideal Gas Law and Kinetic Theory

7/31/2019 Ch14 the Ideal Gas Law and Kinetic Theory

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Ch14. The Ideal Gas Law and Kinetic Theory

Molecular Mass, the Mole, and Avogadro's Number

To set up atomic mass scale, a reference value (along with a

unit) is chosen for one of the elements. The unit is called theatomic mass unit(symbol: u). By international agreement, the

reference element is chosen to be the most abundant type or

isotope* of carbon, which is called carbon-12. Its atomic mass

* is defined to be exactly twelve atomic mass units, or 12 u.

The relationship between the atomic mass unit and the

kilogram is


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A portion of the periodic table showing the atomic number and

atomic mass of each element. In the periodic table it is

customary to omit the symbol u denoting the atomic mass

unit.


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The molecular mass of a molecule is the sum of the atomic masses

of its atoms.

Macroscopic amounts of materials contain large numbers of atoms

or molecules. Even in a small volume of gas, 1 cm3, for example,

the number is enormous. It is convenient to express such large

numbers in terms of a single unit, thegram-mole, or simply the

mole (symbol: mol). One gram-mole of a substance contains asmany particles (atoms or molecules) as there are atoms in 12 grams

of the isotope carbon-12.

12 grams of carbon-12 contain 6.022 1023 atoms. The number of

atoms per mole is known asAvogadros number NA, after theItalian scientist Amedeo Avogadro (17761856):


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The mass per mole (in g/mol) of a substance has thesame numerical value as the atomic or molecular mass

of the substance (in atomic mass units).


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Example 1. The Hope Diamond and the

Rosser Reeves Ruby

The Hope diamond (44.5 carats), which is almost pure carbon. The

Rosser Reeves ruby (138 carats), which is primarily aluminum

oxide (Al2O3). One carat is equivalent to a mass of 0.200 g.

Determine (a) the number of carbon atoms in the diamond and (b)

the number of Al2O3 molecules in the ruby.

(a) m = (44.5 carats)[(0.200 g)/(1 carat)] = 8.90 g


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(b)

m = (138 carats)[(0.200 g)/(1 carat)] = 27.6 g.

Calculations like those in part (a) reveal that the RosserReeves ruby contains 0.271 mol or

.


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Check Your Understanding 1

A gas mixture contains equal masses of the monatomic

gases argon (atomic mass = 39.948 u) and neon

(atomic mass = 20.179 u). They are the only gases in

the mixture. Of the total number of atoms, what

percentage is neon?

0.664


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The Ideal Gas Law

An ideal gas is an idealized model for real gases that havesufficiently low densities.


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The absolute pressure of an ideal gas is proportional to the number

of molecules or, equivalently, to the number of moles n of the gas (P

n).

P nT/V.

IDEAL GAS LAW

The absolute pressurePof an ideal gas is directly

proportional to the Kelvin temperature Tand the numberof moles n of the gas and is inversely proportional to the

volume Vof the gas:P=R(nT/V). In other words,

whereR is the universal gas constant and has the value of 8.31

J/(molK).


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The constant termR/NA is referred to asBoltzmanns

constant, in honor of the Austrian physicist Ludwig Boltzmann

(18441906), and is represented by the symbol k:


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Example 2. Oxygen in the Lungs

In the lungs, the respiratory membrane separates tiny sacs of air(absolute pressure = 1.00 105 Pa) from the blood in the

capillaries. These sacs are called alveoli, and it is from them that

oxygen enters the blood. The average radius of the alveoli is 0.125

mm, and the air inside contains 14% oxygen. Assuming that the

air behaves as an ideal gas at body temperature (310 K), find thenumber of oxygen molecules in one of the sacs.

.


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One mole of an ideal gas occupies a volume of 22.4 liters

at a temperature of 273 K (0 C) and a pressure of oneatmosphere (1.013 105 Pa). These conditions of

temperature and pressure are known as standard

temperature and pressure (STP).


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Conceptual Example 3.

Beer Bubbles on the Rise

The next time you get a chance, watch the bubbles rise in a

glass of beer. If you look carefully, youll see them grow in size

as they move upward, often doubling in volume by the time

they reach the surface. Why does a bubble grow as it ascends?

The number of moles does increase as the bubble rises.

Each bubble acts as a nucleation site for CO2 molecules,

so as a bubble moves upward, it accumulates carbon

dioxide from the surrounding beer and grows larger.


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Boyles law

A pressure-versus-volumeplot for a gas at a constant

temperature is called an

isotherm. For an ideal gas,

each isotherm is a plot of the

equationP = nRT/V =constant/V.


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Consider equal masses of the three monatomic gases argon

(atomic mass = 39.948 u), krypton (atomic mass = 83.80 u), and

xenon (atomic mass = 131.29 u). The pressure and volume of

each is the same. Which gas has the greatest and which the

smallest temperature?

Xenon has the greatest and argon the

smallest temperature.


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Example 4. Scuba Diving

In scuba diving, a greater water pressure acts on a diver at

greater depths. The air pressure inside the body cavities (e.g.,

lungs, sinuses) must be maintained at the same pressure as that

of the surrounding water; otherwise they would collapse. A

special valve automatically adjusts the pressure of the airbreathed from a scuba tank to ensure that the air pressure

equals the water pressure at all times. The scuba gear consists of

a 0.0150-m3 tank filled with compressed air at an absolute

pressure of 2.02 107 Pa. Assuming that air is consumed at a

rate of 0.0300 m3 per minute and that the temperature is the

same at all depths, determine how long the diver can stay under

seawater at a depth of (a) 10.0 m and (b) 30.0 m.


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(a)


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(b) The calculation here is like that in part (a).

Absolute water pressure is 4.02 105 Pa

Vf = 0.754 m3


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Charles law

Frenchman Jacques Charles (17461823) discovered that at

a constant pressure, the volume of a fixed mass (fixed

number of moles) of a low-density gas is directly

proportional to the Kelvin temperature (V T).


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Kinetic Theory of Gases


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THE DISTRIBUTION OF

MOLECULAR SPEEDS


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KINETIC THEORY

The pressure that a gas exerts is

caused by the collisions of itsmolecules with the walls of the

container.


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A gas particle is shown

colliding elastically with

the right wall of the

container andrebounding from it.


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C t l E l 5


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Does a Single Particle Have a

Temperature?Each particle in a gas has kinetic energy. Furthermore, the

equation establishes the relationship between

the average kinetic energy per particle and the temperature of

an ideal gas. Is it valid, then, to conclude that a single particlehas a temperature?

A single gas particle does

not have a temperature.


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The pressure of a monatomic ideal gas is doubled, while its

volume is reduced by a factor of four. What is the ratio of the

new rms speed of the atoms to the initial rms speed?

707.0

,

,=

initialrms

newrms

v

v


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Example 6.

The Speed of Molecules in Air

Air is primarily a mixture of nitrogen N2 (molecular mass = 28.0

u) and oxygen O2 (molecular mass = 32.0 u). Assume that each

behaves as an ideal gas and determine the rms speeds of thenitrogen and oxygen molecules when the temperature of the air is

293 K.


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THE INTERNAL ENERGY OF A

MONATOMIC IDEAL GASThe internal energy of a substance is the sum of the various kinds of

energy that the atoms or molecules of the substance possess. A

monatomic ideal gas is composed of single atoms. These atoms are

assumed to be so small that the mass is concentrated at a point, with

the result that the moment of inertiaIabout the center of mass isnegligible.


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Diffusion

The process in which molecules move from a region of higherconcentration to one of lower concentration is called diffusion.

The host medium, such as the air or water, is referred to as the

solvent, while the diffusing substance, like the perfume

molecules, is known as the solute. Relatively speaking,

diffusion is a slow process, even in a gas.


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Why Diffusion Is Relatively Slow

In Example 6 we have seen that

a gas molecule has a

translational rms speed of

hundreds of meters per second

at room temperature. At such aspeed, a molecule could travel

across an ordinary room in just

a fraction of a second. Yet, it

often takes several seconds, and

sometimes minutes, for the

fragrance of a perfume to reach

the other side of a room. Why

does it take so long?


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When a perfume molecule diffuses through air, it

makes millions of collisions each second with airmolecules. The speed and direction of motion change

abruptly as a result of each collision. Between

collisions, the perfume molecule moves in a straight

line at a constant speed. Although a perfume

molecule does move very fast between collisions, it

wanders only slowly away from the bottle because of

the zigzag path resulting from the collisions. It would

take a long time for a molecule to diffuse in this

manner across a room. Usually, however, convectioncurrents are present and carry the fragrance across

the room in a matter of seconds or minutes.


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Using diffusion, a transdermal patch delivers a drug directly into the

skin, where it enters blood vessels. The backing contains the drugwithin the reservoir, and the control membrane limits the rate of

diffusion into the skin. Another way to control the diffusion is to

adjust the concentration of the drug in the reservoir by dissolving it

in a neutral material.


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(a) Solute diffuses through the channel from the region of higher

concentration to the region of lower concentration. (b) Heat is

conducted along a bar whose ends are maintained at different

temperatures.


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FICKS LAW OF DIFFUSION

The mass m of solute that diffuses in a time tthrough a

solvent contained in a channel of lengthL and cross-sectional

areaA is

where Cis the concentration difference between the ends ofthe channel andD is the diffusion constant.

SI Unit for the Diffusion Constant: m2/s


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Example 8.

Water Given Off by Plant Leaves


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Large amounts of water can be given off by plants. It has been

estimated, for instance, that a single sunflower plant can lose up

to a pint of water a day during the growing season. Figure 14.17shows a cross-sectional view of a leaf. Inside the leaf, water

passes from the liquid phase to the vapor phase at the walls of

the mesophyll cells. The water vapor then diffuses through the

intercellular air spaces and eventually exits the leaf through

small openings, called stomatal pores. The diffusion constant for

water vapor in air isD = 2.4 105 m2/s. A stomatal pore has a

cross-sectional area of aboutA = 8.0 1011 m2 and a length of

aboutL = 2.5 105 m. The concentration of water vapor on the

interior side of a pore is roughly C2 = 0.022 kg/m3

, while that onthe outside is approximately C1 = 0.011 kg/m

3. Determine the

mass of water vapor that passes through a stomatal pore in one

hour.


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The same solute is diffusing through the same solvent in eachcase referred to in the table below, which gives the length and

cross-sectional area of the diffusion channel. In each case, the

concentration difference between the ends of the diffusion

channel is the same. Rank the diffusion rates (in kg/s) in

descending order (largest first). Length Cross-Sectional

Area

(a) L A

(b) L A

(c) L 2A

c, b, a


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Concepts & Calculations Example 9.

The Ideal Gas Law and Springs


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There are three identical chambers containing a piston and a

spring whose spring constant is k= 5.8 104 N/m. The chamber

in part a is completely evacuated, and the piston just touches itsleft end. In this position, the spring is unstrained. In part b of the

drawing, 0.75 mol of ideal gas 1 is introduced into the chamber,

and the spring compresses byx1 = 15 cm. In part c, 0.75 mol of

ideal gas 2 is introduced into the chamber, and the spring

compresses byx2 = 24 cm. Find the temperature of each gas.

P = F/A F = kx P = kx/A


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Concepts & Calculations Example 10.

Hydrogen Atoms in Outer Space

In outer space the density of matter is extremely low, about one

atom per cm3. The matter is mainly hydrogen atoms (m = 1.67

1027 kg) whose rms speed is 260 m/s. A cubical box, 2.0 m on a

side, is placed in outer space, and the hydrogen atoms are allowed

to enter. (a) What is the magnitude of the force that the atoms

exert on one wall of the box? (b) Determine the pressure that the

atoms exert. (c) Does outer space have a temperature, and, if so,

what is it?


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(a)


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(b)

(c)


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Problem 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 + gh. We

also know that P1V1 = P2V2. Then,

V1

V2=

P2

P1=

P1

+ gh

P1= 1+

gh

P1

V1

V2

= 1+(1.000103 kg/m3)(9.80 m/s2)(0.200 m)

(1.01105 Pa)= 1.02


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Problem 41

REASONING AND SOLUTION

a. As stated, the time required for the first solute molecule to

traverse a channel of lengthL is . Therefore, for

water vapor in air at 293 K, where the diffusion constant is

D=2.4*10-5

m2

/s , the time trequired for the first water molecule totravel is

t= L2

/(2D)

L = 0.010 m

( )

( ) ssmm

D

L

t 1.2/104.22

01.0

2 25

22

=

==


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b. If a water molecule were traveling at the translational rms speed

for water, the time tit would take to travel the distance

would be given by , where, according to Equation 14.6 (), . Before we can use the last

expression for the translation rms speed vrms, we must determine the

mass m of a water molecule and the average translational kinetic

energy

L = 0.010 m

t= L / vrmsKE =1

2mvrms

2vrms = 2 KE( )/ m

KE

Using the periodic table on the inside of the texts back cover,

we find that the molecular mass of a water molecule is

Mass of oneMass of twooxygen atomhydrogen atoms

2(1.00794 u) 15.9994 u 18.0153 u+ =1 4 2 431 4 4 2 4 43


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The average translational kinetic energy of water molecules at 293

K is, according to Equation 14.6,

kgmol

molkg

m

26

123

3

1099.210022.6

/100153.18

=

=

( )( ) JKKJ 2123 1007.6293/1038.1 =

( )sm

kg

J/637

1099.2

1007.6226

21

=


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Thus, the time trequired for a water molecule to travel the

distance at this speed is

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 endof 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)

s5106.1


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Problem 42

REASONING AND SOLUTION Ficks 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

8.4 10 kg

D A CL

vt m

= = =


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Problem 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 tyields

( )2 / 2t L D=

b. Substituting into this expression yields

t= (2.5 * 102 m)2/[2(1.0 * 105 m2/s)] = 31 s

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Ch14 the Ideal Gas Law and Kinetic Theory