chapter 11: behavior of gases - glencoe...1.00 atm 760 mm hg because atmospheric pressure is...

11CHAPTER

370

Behavior of GasesBehavior of Gases11

CHAPTER

11.1 Gas PressureMiniLab 11.1 Relating Mass and

Volume of Gas

11.2 The Gas Laws

MiniLab 11.2 How Straws FunctionChemLab Boyle’s Law

Chapter PreviewSections Fly, Fly Away!

Ahot air balloon is a great example of gasesin action. As the burner heats up the air inthe balloon, the air expands. The more the

air expands, the higher the balloon rises.

Fly, Fly Away!

371

More Than Just Hot AirHow does a temperature change affect the air in a balloon?

Safety Precautions

Always wear goggles to protect eyes from broken balloons.

Materials

• 5-gal bucket• round balloon• ice• string

Procedure

1. Inflate a round balloon and tie it closed.2. Fill the bucket about half full of cold water and add ice.3. Use a string to measure the circumference of the

balloon. 4. Stir the water in the bucket to equalize the

temperature. Submerge the balloon in the ice water for 15 minutes.

5. Remove the balloon from the water. Measure the circumference.

Analysis

What happens to the size of the balloon when its temper-ature is lowered? What might you expect to happen to itssize if the temperature is raised?

Before reading the chapter, make alist of some ways we might use gasesor gas pressure in everyday life.Next, scan the chapter to discoverother ways we use gas. Note some ofthe chapter features, such as “Chem-istry and Technology” on page 390.

Reading Chemistry

Review the following conceptsbefore studying this chapter.Chapter 10: properties of gases; the kinetic theory of gases

What I Already Know

Preview this chapter’s content andactivities at chemistryca.com

Start-up ActivitiesStart-up Activities

http://chemistryca.com

11.1

You might be surprised tosee that an overturnedtractor trailer can be

uprighted by inflating several airbags placed beneath it. But,you’re not at all surprised to seethat an uprighted truck is sup-ported by 18 tires inflated with air. Air can lift the tractor and support thetruck because air is a mixture of gases, and gases exert pressure.

Defining Gas PressureUnless a ball has an obvious dent, you can’t tell whether it is under-

inflated by looking at it. You have to squeeze it. If it’s soft, you know itneeds to be pumped up with more air. The springiness of a fully inflatedball is the pressure of the air inside. Figure 11.1 shows how the pressure ofair inside a soccer ball rises as air is added. How can changes in gas pres-sure be explained by the kinetic theory?

Gas Pressure

SECTION

Objectives✓ Model the effects of changing the num-ber of particles, mass,temperature, pressure,and volume on a gasusing kinetic theory.

✓ Measure atmos-pheric pressure.

✓ Demonstrate theability to use the fac-tor label method toconvert pressure units.

Review VocabularyDiffusion: process bywhich particles of mat-ter fill in a spacebecause of randommotion.

New Vocabularybarometerstandard

atmosphere (atm)pascal (Pa)kilopascal (kPa)factor label method

SECTION PREVIEW

372 Chapter 11 Behavior of Gases

Figure 11.1Pumping up a Soccer BallMolecules in air are in constantmotion and exert pressure whenthey strike the walls of the ball. Thepressure they exert counterbalancestwo other pressures—atmosphericpressure on the ball plus the pres-sure exerted by the tough rubber ofthe ball itself.

Pressure of wallPressure of wall

Gas pressureGas pressure

Atmosphericpressure

Atmosphericpressure

Pump

Pumping more air into the soft ball(left) increases the number of mole-cules inside. As a result, moleculesstrike the inner wall of the ballmore often and the pressureincreases (right). The increase inpressure is counterbalanced byincreased pressure from the toughwall. As a result, the ball becomesfirmer and bouncier. �

11.1 Gas Pressure 373

How are number of particles and gas pressure related?Recall from Chapter 10 that the pressure of a gas is the force per unit area

that the particles in the gas exert on the walls of their container. As youwould expect, more air particles inside the ball mean more mass inside. InFigure 11.2, one basketball was not fully inflated. The other identical ballwas pumped up with more air. The ball that contains more air has higherpressure and the greater mass.

From similar observations and measurements, scientists from as longago as the 18th century learned that the pressure of a gas is directly pro-portional to its mass. According to the kinetic theory, all matter is com-posed of particles in constant motion, and pressure is caused by the forceof gas particles striking the walls of their container. The more often gasparticles collide with the walls of their container, the greater the pressure.Therefore, the pressure is directly proportional to the number of particles.For example, doubling the number of gas particles in a basketball doublesthe pressure.

Figure 11.2Two Basketballs at Unequal PressuresThe mass of the ball on the left is greater than the mass of theone on the right. The ball on the left has greater mass becauseit has more air inside and, therefore, is at a higher pressurethan the ball on the right.


How are temperature and gas pressure related?How does temperature affect the behavior of a gas? You know from

Chapter 10 that at higher temperatures, the particles in a gas have greaterkinetic energy. They move faster and collide with the walls of the contain-er more often and with greater force, so the pressure rises. If the volumeof the container and the number of particles of gas are not changed, thepressure of a gas increases in direct proportion to the Kelvin temperature.

To demonstrate how a gas at constant temperature can be used to dowork, let’s examine the action of a piston as shown in Figure 11.3. Thepiston inside the cylinder is like the cover of a jar because it makes an air-tight seal, but it also acts like a movable wall. When gas is added, as shownin the top cylinder, the gas particles push out the piston until the pressureinside balances the atmospheric pressure outside.

If more gas is pumped into the cylinder, the number of particlesincreases, and the number of collisions on the walls of the containerincreases. Because the force on the inside face of the piston is now greaterthan the force on the outside face, the piston moves outward. As the gasspreads out into the larger volume, the number of collisions per unit areaon the inside face falls. Then, the pressure of the gas inside the containerfalls until it equals the pressure of the atmosphere outside and the pistoncomes to rest in its new position, farther out. The atmospheric pressureremains constant while the piston moves and the gas expands.

Air at atmosphericpressure

Gas at atmosphericpressure

More gas added andvolume increased

Piston

Piston

Figure 11.3Expanding a Gas atConstant PressureThe constant bombardmentof molecules and atoms of theatmosphere exerts a constantpressure on the outside faceof the piston. This pressurebalances the pressure causedby the confined gas bombard-ing the inside face of thepiston (top). �

When gas is added to thecylinder, the piston is pushedout (bottom). When the pres-sure of the new volume insidethe cylinder balances atmo-spheric pressure, the pistonstops moving out. �


What if the volume, such as in the piston in Figure 11.3, is not heldconstant? How would a change in temperature affect the volume andpressure of a gas? If the temperature of the gas inside the piston chamberis raised, the pressure will momentarily rise. But, if the gas is permitted toexpand as the temperature is raised, the pressure remains equal to theatmospheric pressure and the volume expands. Thus, the volume of a gasat constant pressure is directly proportional to the Kelvin temperature.

When the cylinder is in an automobile engine, the other end of the pis-ton rod, shown in Figure 11.3, is attached to the crankshaft. When themixture of gasoline and air in the cylinder ignites and burns, gases areproduced. The gas inside the cylinder expands because the heat of burn-ing raises the temperature. The pressure of expanding gases drives the pis-ton outward. As the piston rod moves out and then back in, it turns thecrankshaft, delivering power to the wheels that propel the vehicle forward.

1

Relating Mass and Volume of a GasRecall from Chapter 4 that one property of carbon dioxide is that it

changes directly from a solid (dry ice) to a gas. In other words, it sublimes.Can you determine some properties of gases from sublimed dry ice?

Procedure1. Place a small, zipper-closure

plastic bag on the pan of azeroed balance. Wearing glovesand using tongs, insert a 20- to30-g piece of dry ice into thebag.

2. Measure and record the mass ofthe bag and its contents. Quick-ly press the air out of the bagand zip it closed.

3. Immediately put the bag into alarger, clear plastic bag so thatyou can observe the sublimationprocess. After the inner bag isfilled with gas, unzip the closurethrough the outer bag. Immedi-ately remove the bag with dryice from the larger bag. Pressout the gas and zip it closed.

4. Measure and record the mass ofthe bag and its contents.

5. Use tongs and gloves to removeand dispose of the dry ice fromthe bag.

6. Determine the volume of theinner bag by filling it withwater and pouring the waterinto a graduated cylinder.Record its volume.

7. Repeat steps 1-6 with asmaller zipper-closurebag and a smaller pieceof dry ice.

Analysis1. Calculate the mass of

carbon dioxide that hassublimed in each trial.

2. Calculate the mass ratioby dividing the mass ofsublimed CO2 in Trial 1by the mass that sub-limed in Trial 2. Calcu-late the volume ratio of CO2.

3. How do the volume and massratios compare?

4. What can you infer about therelationship between the vol-ume and mass of a gas?

See page 868 inAppendix F forCrushing Cans

Lab


Devices to Measure PressureAlthough you can compare a property such as gas pressure by touching

partially and fully inflated basketballs, this method does not give you anaccurate measure of the two pressures. What is needed is a measuring device.

The BarometerOne of the first instruments used to measure gas pressure was designed

by the Italian scientist Evangelista Torricelli (1608-1647). He invented thebarometer, an instrument that measures the pressure exerted by theatmosphere. His barometer was so sensitive that it showed the differencein atmospheric pressure between the top and bottom of a flight of stairs.Figure 11.4 explains how Torricelli’s barometer worked. The height of themercury column measures the pressure exerted by the atmosphere. Welive at the bottom of an ocean of air. The highest pressures occur at thelowest altitudes. If you go up a mountain, atmospheric pressure decreasesbecause the depth of air above you is less.

One unit used to measure pressure is defined by using Torricelli’sbarometer. The standard atmosphere (atm) is defined as the pressure thatsupports a 760-mm column of mercury. This definition can be represent-ed by the following equation.

1.00 atm � 760 mm Hg

Because atmospheric pressure is measured with a barometer, it is oftencalled barometric pressure.

Figure 11.4How Torricelli’s BarometerWorksA barometer consists of atube of mercury that stands ina dish of mercury. Because themercury stands in a column inthe closed tube, you can con-clude that the atmosphereexerts pressure on the opensurface of the mercury in thedish. This pressure, transmit-ted through the liquid in thedish, supports the column ofmercury. � � The sphygmomanometer attached to the wall

also uses a column of mercury to measureblood pressure.

barometer:baros (G) heavymetron (G) tomeasure

A barometer mea-sures the pressure(force per unitarea) of the atmo-sphere.

760 mmMercury

Atmosphericpressure

The Pressure GaugeUnfortunately, the barometer can measure only atmospheric pressure.

It cannot measure the air pressure inside a bicycle tire or in an oxygentank. You need a device that can be attached to the tire or tank. This pres-sure gauge must make some regular, observable response to pressurechanges. If you have ever measured the pressure of an inflated bicycle tire,you’re already familiar with such a device.

10

psi

15

20

25

30

60

70

80

90

100

110

120

10

psi

15

20

25

30

60

70

80

90

100

110

120

Airtightpiston

Pressureof tire

Inletpin

Spring

Calibratedscale

Tire valve


Thinking Critically

Tire-Pressure GaugeA tire-pressure gauge is a device that measures the pressure

of the air inside an inflated tire or a basketball. Because anuninflated tire contains some air at atmospheric pressure, atire-pressure gauge records the amount that the tire pressureexceeds atmospheric pressure.

The most familiar tire gauge is about the size and shape ofa ballpoint pen. It is a convenient way to check tire pressurefor proper inflation regularly. Proper inflation ensures tiremaintenance and safety.

1. The pin in the head ofthe gauge pushesdownward on the tire-valve inlet and allowsair from the tire toflow into the gauge.

2. The air flowinginto the gaugepushes against amovable pistonthat pushes asliding calibrat-ed scale.

3. The piston movesalong the cylinder,compressing thespring until theforce of air pressingon the surface ofthe piston is equalto the force of thecompressed springon the piston.

4. The pressure of the air inthe cylinder, which is thesame as the pressure ofthe air in the tire, is readfrom the scale.

1. What does the cali-bration of the scaleindicate about therelationship betweenthe compression ofthe spring and pres-sure?

2. Explain whether atire-pressure gaugeshould be recalibrat-ed for use at loca-tions where atmo-spheric pressure isless than that at sealevel.


When you measure tire pressure, you are measuring pressure aboveatmospheric pressure. The recommended tire inflation pressures listed bymanufacturers are gauge pressures; that is, pressures read from a gauge. Abarometer measures absolute pressure; that is, the total pressures exerted byall gases, including the atmosphere. To determine the absolute pressure of aninflated tire, you must add the barometric pressure to the gauge pressure.

Pressure UnitsYou have learned that atmospheric pressure is measured in mm Hg.

Recall from Chapter 10, atmospheric pressure is the force per unit areathat the gases in the atmosphere exert on the surface of Earth. Figure 11.5shows two additional units that are used to measure pressure of one stan-dard atmosphere.

The SI unit for measuring pressure is the pascal (Pa), named after theFrench physicist Blaise Pascal (1623-1662). Because the pascal is a smallpressure unit, it is more convenient to use the kilopascal. As you recallfrom Chapter 1, the prefix kilo- means 1000; so, 1 kilopascal (kPa) isequivalent to 1000 pascals. One standard atmosphere is equivalent to101.3 kilopascals.

The weight of apostage stamp exerts apressure of about onepascal on the surface ofan envelope.

Expressed in SI units, onestandard atmosphere is101 300 pascals. Note thatSI units are based on thesquare meter area and notthe square inch. �

Figure 11.5Atmospheric PressureThe weight of the air in eachcolumn pushing on the areabeneath it exerts a pressureof one standard atmo-sphere. Each column of airextends to the outer limitsof the atmosphere.

If the unit of area is the square inchand the unit of force is the pound,then the unit of pressure is thepound per square inch (psi).Expressed in these units, one stan-dard atmosphere is 14.7 pounds persquare inch or 14.7 psi. �

14.7 psi

1 in.1 in.

101 300 Pa

1 m

1 m


Table 11.1 presents the standard atmosphere in equivalent units.Because there are so many different pressure units, the international com-munity of scientists recommends that all pressure measurements be madeusing SI units, but pounds per square inch continues to be widely used inengineering and almost all nonscientific applications in the United States.

In weather reports, barometric pressure is often expressed in inchesof mercury. What is one standard atmosphere expressed in inches ofmercury?

Analyze • You know that one standard atmosphere is equivalent to 760 mm ofHg. What is that height expressed in inches? A length of 1.00 inchmeasures 25.4 mm on a meterstick.

Set Up • Select the appropriate equivalent values and units given in Table 11.1.Multiply 760 mm by the number of inches in each millimeter toexpress the measurement in inches.

760 mm � � �The factor on the right of the expression above is the conversion factor.

Solve • Notice that the units are arranged so that the unit mm will cancelproperly and the answer will be in inches.

Check • Because 1 mm is much shorter than 1 in., the number of mm, 760,should be much larger than the equivalent number of inches.

760 mm 1.00 in.25.4 mm � 29.9 in.

1.00 in.�25.4 mm

Converting Barometric Pressure UnitsSAMPLE PROBLEM 1

1.00 atm 760 mm Hg 14.7 psi 101.3 kPa

Table 11.1 Equivalent Pressures

Pressure ConversionsYou can use Table 11.1 to convert pressure measurements to other

units. For example, you can now find the absolute pressure of the air in abicycle tire. Suppose the gauge pressure is 44 psi. To find the absolutepressure, add the atmospheric pressure to the gauge pressure. Because thegauge pressure is given in pounds per square inch, use the value of thestandard atmosphere that is expressed in pounds per square inch. Onestandard atmosphere equals 14.7 psi.

44 psi � 14.7 psi � 59 psi

The following Sample Problems show how to use the values in Table 11.1to express pressure in other units.


The method used in the Sample Problem to convert measurement toother units is called the factor label method. Study the Sample Problemagain. The following equation gives the conversion factor because it con-tains both the given unit and the desired unit.

1.00 in. � 25.4 mm

You know that you can divide both sides of an equation by the samevalue and maintain equality. For example, you can divide both sides of theequation 1.00 in. � 25.4 mm by 25.4 mm, as shown below.

�

Now, simplify the right side. The conversion factor is on the left.

� 1

In the Sample Problem, the height of mercury in millimeters was beingmultiplied by this conversion factor. Because multiplying any quantity by 1doesn’t affect its value, the height of the mercury column isn’t changed—only the units are. The factor label method changes the units of the mea-surement without affecting its value. You will use the factor label method tosolve problems in this chapter and in several of the following chapters. Youwill find more information on the factor label method on pages 801-803 ofAppendix A, the Skill Handbook.

1.00 in.�25.4 mm

25.4 mm�25.4 mm

1.00 in.�25.4 mm

The reading of a tire-pressure gauge is 35 psi. What is the equivalentpressure in kilopascals?

Analyze • The given unit is pounds per square inch (psi), and the desired unit iskilopascals (kPa). According to Table 11.1, the relationship betweenthese two units is 14.7 psi � 101.3 kPa.

Set Up • Write the conversion factor with kPaunits as the numerator and psi units asthe denominator. Note that the psi unitswill cancel and only the kPa units willappear in the final answer.

35 psi � � �Solve • Multiply and divide the values and units.

Notice that the given units (psi) will cancel properly and the quantitywill be expressed in the desired unit (kPa) in the answer.

Check • Because 1 psi is a much greater pressure than 1 kPa, the number of psi,35, should be much smaller than the equivalent number of kilopascals.

35 psi 101.3 kPa 14.7 psi

35 � 101.3 kPa 14.7� � 240 kPa

101.3 kPa��

14.7 psi

Converting Pressure UnitsSAMPLE PROBLEM 2

In the factor label method, terms arearranged so that units will cancel out.

Prob lem -So l v i ngH I N T


Your skill in converting units will help you relate measurements ofgas pressure made in different units.

Use Table 11.1 and the equation 1.00 in. � 25.4 mm to convert thefollowing measurements.

1. 59.8 in. Hg to psi

2. 7.35 psi to mm Hg

3. 1140 mm Hg to kPa

4. 19.0 psi to kPa

5. 202 kPa to psi

PRACTICE PROBLEMS

SECTION REVIEWFor more practice

with solving problems,see Supplemental Practice Problems, Appendix B.

Understanding Concepts1. Compare and contrast how a barometer and a

tire-pressure gauge measure gas pressure.

2. At atmospheric pressure, a balloon contains2.00 L of nitrogen gas. How would the volumechange if the Kelvin temperature were only 75 percent of its original value?

3. A cylinder containing 32 g of oxygen gas is placedon a balance. The valve is opened and 16 g of gasare allowed to escape. How will the pressurechange?

Thinking Critically4. Interpreting Data Sketch graphs of pressure

versus time, volume versus time, and number ofparticles versus time for a large, plastic garbagebag being inflated until it ruptures.

Applying Chemistry 5. Overloaded Why do tire manufacturers rec-

ommend that tire pressure be increased if therecommended number of car passengers orload size is exceeded?

chemistryca.com/self_check_quiz

http://chemistryca.com/self_check_quiz

11.2

An uninflated airmattress doesn’tmake a comfort-

able bed because comfort depends upon the pressure of the air inside it. So,you inflate it by huffing and puffing or by using the exhaust feature of a vac-uum cleaner. Recall that when more air is added to the mattress, the pressureof the air inside increases. When the mattress is filled and its air valve isclosed, it is far more comfortable. The air inside supports the walls of themattress and pushes against the force exerted by atmospheric pressure onthe outside surface of the mattress. As you add your weight to the mattress,do you know how the air inside is now acting upon the mattress walls,against the force of the atmosphere on its surface, and on you?

Boyle’s Law: Pressure and VolumeAs you know from squeezing a balloon, a confined gas can be com-

pressed into a smaller volume. Robert Boyle (1627-1691), an Englishscientist, used a simple apparatus like the one pictured in Figure 11.6 to

SECTION

Objectives✓ Analyze data thatrelate temperature,pressure, and volumeof a gas.

✓ Model Boyle’s lawand Charles’s lawusing kinetic theory.

✓ Predict the effect ofchanges in pressureand temperature onthe volume of a gas.

✓ Relate how volumesof gases react in termsof the kinetic theoryof gases.

Review VocabularyPascal: SI unit formeasuring pressure.

New VocabularyBoyle’s lawCharles’s lawcombined gas lawstandard temperature

and pressure, STPlaw of combining

gas volumesAvogadro’s principle

SECTION PREVIEW


Figure 11.6Relation Between Pressure and Volume1. The pressure of the trapped air in the J tube balances the

atmospheric pressure, 1 atm or 760 mm Hg.

2. When mercury is added to a height of 760 mmabove the height in the closed end, the volumeof the trapped air is halved, and the pressurethe air exerts is now 2 atm.

3. When an additional 760 mm of mer-cury is added to the column, thepressure of the trapped air is tripled.

100 mLof air1 atm

0 mm Hg

50 mLof air2 atm

33 mLof air3 atm76

0 m

m H

g

1520

mm

Hg

1 2 3

The Gas Laws

11.2 The Gas Laws 383

compress gases. The weight of the mercury in the open end of the tubecompresses air trapped in the closed end.

After performing many experiments with gases at constant tempera-tures, Boyle had four findings.

a) If the pressure of a gas increases, its volume decreases proportionately.b) If the pressure of a gas decreases, its volume increases proportionately.c) If the volume of a gas increases, its pressure decreases proportionately.d) If the volume of a gas decreases, its pressure increases proportionately.

Because the changes in pressure and volume are always opposite and pro-portional, the relationship between pressure and volume is an inverse pro-portion. By using inverse proportions, all four findings can be included inone statement called Boyle’s law. Boyle’s law states that the pressure andvolume of a gas at constant temperature are inversely proportional.

Recall from Chapter 10 that the kinetic energy of an ideal gas is directlyproportional to its temperature and that the graph of a direct proportion is astraight line. The graph of an inverse proportion is a curve like that shown inFigure 11.7. Just as a road runs both ways, you can think of a gas followingthe curve A-B-C or C-B-A. The path A-B-C represents the gas being com-pressed, which forces its pressure to rise. As the gas is compressed by half,from 1.0 L to 0.5 L, the pressure doubles from 100 kPa to 200 kPa. As it iscompressed by half again, from 0.5 L to 0.25 L, the pressure again doublesfrom 200 kPa to 400 kPa. Look at Figure 11.9 on page 386 to see anotherexample of this relationship. The reverse path C-B-A represents what happenswhen the pressure of the gas decreases and the volume increases accordingly.

Figure 11.7Boyle’s Law: An Inverse RelationshipAs you follow the curve from left to right, the pressure increases and thevolume decreases. Look at the change from points A to C. The volume isreduced from 1.0 L to 0.25 L; that is, the gas is compressed to one-fourthof its volume, and the pressure rises from 100 kPa to 400 kPa, which isfour times as high. The volume/pressure relationship is a two-way system.

1.0

0.8

0.6

0.4

0.2

0

50 100 150 200 250 300 350 400 450 500Pressure (kPa)

Vol

ume

(L)

A

B

C

1.2


Boyle’s LawThe quantitative relationship between the vol-

ume of a gas and the pressure of the gas, at con-stant temperature, is known as Boyle’s Law. Bymeasuring quantities directly related to the pres-sure and volume of a tiny amount of trapped air,you can deduce Boyle’s law.

ProblemWhat is the relationship between the volume

and the pressure of a gas at constant temperature?

Objectives

•Observe the length of a column of trapped airat different pressures.

•Examine the mathematical relationshipbetween gas volume and gas pressure.

Materialsthin-stem pipet matchesdouble-post metric ruler

screw clamp scissorsfine-tip marker small beakerfood coloring water

Safety Precautions

Use care in lighting matches and melting thepipet stem.

1. Cut off the stepped portion of the stem of thepipet with the scissors.

2. Place about 20 mL of water in a beaker, add afew drops of food coloring, and swirl to mix.

3. Draw the water into the pipet, completely fill-ing the bulb and allowing the water to extendabout 5 mm into the stem of the pipet.

4. Heat the tip of the pipet gently above a flameuntil it is soft. (CAUTION: If the stem acciden-tally begins to burn, blow it out.) Use a metallicor glass object to flatten the tip against thecountertop so that the water and air are sealedinside the pipet. Holding the pipet by thebulb, tap any water droplets in the stem downinto the liquid. You should observe a cylindri-cal column of air trapped in the stem of thepipet.

5. Center the bulb in a double-post screw clamp,and tighten the clamp until the bulb is justheld firmly. Mark the knob of the clamp witha fine-tip marker, and tighten the clamp threeor four turns so that the length of the air col-umn is 50 to 55 mm. Record this number ofturns T under Trial 1 in a data table like theone shown. Measure and record the length Lof the air column in millimeters for Trial 1.

6. Turn the clamp knob one complete turn, andrecord the trial number and the total numberof turns. Measure and record the length of theair column.

7. Repeat step 6 until the air column is reducedto a length of 25 to 30 mm.

SMALLSCALESMALLSCALE


1. Observing and Inferring Explain whether thevolume V of the air in the stem is directly pro-portional to the length L of the air column.What inferences can be made about the pres-sure P of the air in the column and the num-ber of turns T of the clamp screw?

2. Interpreting Data Calculate the product LTand the quotient L/T for each trial. Which cal-culations are more consistent? If L and T aredirectly related, L/T will yield nearly constantvalues for each trial. On the other hand, if Land T are inversely related, LT will yieldalmost constant values for each trial. Are Land T directly or inversely related?

3. Drawing Conclusions Explain whether thedata indicate that gas volume and gas pressureat constant temperature are directly related orinversely related.

1. For a mercury column barometer to measureatmospheric pressure as 760 mm Hg, the

column containing the mercury must be com-pletely evacuated. However, if the column isnot completely evacuated, the barometer canstill be used to correctly measure changes inbarometric pressure. How is the second state-ment related to Boyle’s law?

2. Using the kinetic theory, explain how adecrease in the volume of a gas causes anincrease in the pressure of the gas.

Pressure and Length Data

Trial Turns, T Length of Air Column, L, (mm) Numerical Value, LT Numerical Value, L /T

From Robert Boyle’s experiments, we knowthe relationship between the pressure andvolume of a gas at constant temperature.


The weather balloon in Figure 11.8 illustrates Boyle’s law. When heliumgas is pumped into the balloon, it inflates, just as a soccer ball does, untilthe pressure of gas inside equals the pressure of the air outside. Becausehelium is less dense than air, the mass of helium gas is less than the mass ofthe same volume of air at the same temperature and pressure. As a result,the balloon rises. As it climbs to higher altitudes, atmospheric pressurebecomes less. According to Boyle’s law, because the pressure on the heliumdecreases, its volume increases. The balloon continues to rise until the pres-sures inside and outside are equal, when it hovers and records weather data.

Kinetic Explanation of Boyle’s LawYou know that by compressing air in a tire pump, the air’s pressure is

increased and its volume is reduced. According to the kinetic theory, if thetemperature of a gas is constant and the gas is compressed, its pressuremust rise. Boyle’s law, based on volume and pressure measurements madeon confined gases at constant temperature, quantified this relationship bystating that the volume and pressure are inversely proportional. Figure 11.9shows how the kinetic theory explains Boyle’s observations of gas behavior.Figure 11.8

Weather Balloons andBoyle’s LawAs the balloon rises, theweight of the column of airabove it is shorter, and thepressure on the helium gas inthe balloon decreases. At analtitude of 15 km, atmospher-ic pressure is only 1/10 asgreat as at sea level. When aballoon that was filled withhelium at 1.00 atm reachesthis altitude, its volume isabout ten times as large as itwas on the ground.

Figure 11.9Modeling Boyle’s Law1. When the piston of the bicycle tire

pump is pulled all the way out, theair pressure inside balances the airpressure outside.

2. When the piston is forced halfway down, the average kinetic energy of the airparticles is unchanged because the temperature is unchanged. They strike thewall with the same average force but, because they have been compressedinto half the volume of the pump, the frequency of collisions with the wallsdoubles. The air pressure inside the pump is now 2 atm.

3. When the piston isforced down farther andthe air is compressedinto one-fourth the vol-ume of the pump, thefrequency of collisionswith the walls is fourtimes as great. The airpressure inside the pumpis 4 atm.

1 atm

1 atm

1 liter

2 atm

2 atm

0.5 liter

4 atm

4 atm0.25 liter

1 2 3


EARTH SCIENCE CONNECTION

Weather BalloonsIf you watch the sky any day at 1 P.M. EST,

you may see a weather balloon like the one inthe photograph. Every 12 hours, an instru-ment-packed helium or hydrogen balloon islaunched at each of 70 sites around the UnitedStates. The balloons provide the data used tomake the country’s weather forecasts.

Weather balloon system Since WorldWar II, meteorologists have used weatherballoons to provide profiles of temperature,pressure, relative humidity, and wind velocityin the upper atmosphere. Besides the 70launching sites in this country, there aremore than 700 other sites around the world.The balloons are released at the same time inall countries—at 6:00 and 18:00 hoursGreenwich Mean Time (GMT). Data fromweather stations all over the world reach theUnited States via a central computer inMaryland. From there, the data are sent toregional weather service stations throughoutthe country for release to newspapers andtelevision and radio networks.

Sending the balloons aloft The rubberweather balloons are inflated with either heli-um or hydrogen to a diameter of about 2 m.The inflated balloons become buoyant becausethe density of either gas is less than the densityof air. An instrument package is attached toeach balloon by a 15-m cord. As the atmo-spheric pressure decreases, the gas inside theballoon expands, carrying the balloon higher.When it reaches a height of 30 km, the vol-ume of the expanded gas is so great that theballoon bursts. Then a parachute carries thepackage containing the radio transmitter andmeasuring instruments safely to the ground.

Gathering data While aloft, devices in theinstrument box record and transmit temper-ature, relative humidity, and barometricpressure.

Relative humidity readings are obtainedfrom a device containing a polymer thatswells when it becomes moist. The swellingcauses an increase in the electrical resistanceof a carbon layer in the polymer. Electricalmeasurements of the resistance of the mate-rial indirectly measure changes in relativehumidity.

Wind velocity is determined by trackingthe balloon by radar or by one of the radiolocation systems, known as LORAN-C,OMEGA, or VLF.

The instrument package transmits baro-metric, relative humidity, and temperaturedata to the ground by radio transmissionusing a multiplexing system. The systemtransmits the different kinds of data in rota-tion, for example, temperature first, followedby humidity, and then pressure.

Connecting to Chemistry

1. Thinking CriticallySome peoplethought that whenweather satelliteswere deployed,weather balloonswould become

obsolete. Why doyou think this didnot happen?

2. Inferring Whywould a meteorolo-gist have to under-stand the gas laws?

How Straws FunctionEach of us has used straws to sip soda, milk, or another liquid from a

container. How do they function?

Procedure1. Obtain a clean, empty food jar

with a screw-on lid. Fill the jarhalfway with tap water.

2. Put the ends of two soda strawsinto your mouth. Place theother end of one of the strawsinto the water in the jar andallow the end of the otherstraw to remain in the air. Tryto draw water up the straw bysucking on both straws simul-taneously. Record your obser-vations.

3. Using a hammer and nail,punch a hole of about the samediameter as that of a straw inthe lid of the jar. Push the

straw 2 to 3 cm into the holeand seal the straw in positionwith wax or clay.

4. Fill the jar to the brim withwater, and carefully screw thelid on so that no air enters thejar. Try to draw the water upthe straw. Record your observa-tions.

Analysis1. Explain your observations in

step 2.

2. Explain your observations instep 4.

3. Explain how a soda straw func-tions.

2


The maximum volume a weather balloon can reach without rup-turing is 22 000 L. It is designed to reach an altitude of 30 km. At thisaltitude, the atmospheric pressure is 0.0125 atm. What maximum vol-ume of helium gas should be used to inflate the balloon before it islaunched?

Analyze • At 30 km, the pressure exerted by the helium in the balloon equals theatmospheric pressure at that altitude, 0.0125 atm. What volume doesthis amount of helium occupy at 1 atm? The pressure is greater at highaltitude by the factor 1.0 atm/0.0125 atm, which is greater than 1. ByBoyle’s law, the volume at launch is smaller than at high altitude by thefactor 0.0125 atm/1.0 atm, which is less than 1.

Set Up • Multiply the maximum volume by the factor given by Boyle’s law.

V � 22 000 L � � �0.0125 atm��

1.0 atm

Boyle’s Law: Determining VolumeSAMPLE PROBLEM 3

Solve • Multiply and divide the values and units.

22 000 L 0.0125 atm 1.0 atm

22 000 L � 0.0125 1.0� � 275 L�V

Check • As expected, the volume is less at sea level because the gas is compressed.


Two liters of air at atmospheric pressure are compressed into the0.45-L canister of a warning horn. If its temperature remains constant,what is the pressure of the compressed air?

Analyze • The initial pressure of the air that was forced into the canister was, ofcourse, 1 atm. Because the volume of air is reduced, its pressure increas-es. Multiply the pressure by the factor with a value greater than 1.

Set Up • P � 1.00 atm � � �Solve •Check • Estimate and reason: Because the volume of the air was reduced from 2 L

to about half a liter, the volume changed by the factor �02.5� or about �

14

�.Thus, the pressure changed by a factor of about �

41

�. The final pressure,4.4 atm, is about four times as large as the initial pressure, 1.00 atm.

1.00 atm 2.0 L 0.45 L

1.00 atm � 2.0 0.45� � 4.4 atm�P

2.0 L�0.45 L

Boyle’s Law: Determining PressureSAMPLE PROBLEM 4

Assume that the temperature remains constant in the following problems.

6. Bacteria produce methane gas in sewage-treatment plants. This gas is often captured or burned. If a bacterial culture produces60.0 mL of methane gas at 700.0 mm Hg, what volume would beproduced at 760.0 mm Hg?

7. At one sewage-treatment plant, bacteria cultures produce 1000 Lof methane gas per day at 1.0 atm pressure. What volume tankwould be needed to store one day’s production at 5.0 atm?

8. Hospitals buy 400-L cylinders of oxygen gas compressed at 150 atm.They administer oxygen to patients at 3.0 atm in a hyperbaric oxy-gen chamber. What volume of oxygen can a cylinder supply at thispressure?

9. If the valve in a tire pump with a volume of 0.78 L fails at a pres-sure of 9.00 atm, what would be the volume of air in the cylinderjust before the valve fails?

10. The volume of a scuba tank is 10.0 L. It contains a mixture ofnitrogen and oxygen at 290.0 atm. What volume of this mixturecould the tank supply to a diver at 2.40 atm?

11. A 1.00-L balloon is filled with helium at 1.20 atm. If the balloon issqueezed into a 0.500-L beaker and doesn’t burst, what is the pres-sure of the helium?

PRACTICE PROBLEMS

For more practice with solvingproblems, see SupplementalPractice Problems,Appendix B.

&TECHNOLOGYC H E M I S T R Y


Hyperbaric Oxygen ChambersHyperbaric oxygen (HBO) chambers have been

used since the 1940s. They were used then by theNavy to treat divers with decompression sickness. Inthe last 20 years, researchers have shown that HBOtreatment has many more medical applications. Oneof these applications is healing bone and muscleinjuries.

Healing Sports InjuriesSeveral professional football teams have each

acquired HBO units. The units have been dubbedspace capsules because of their appearance. Aplayer with a severe sprain, ordinarily doominghim to warm the bench for several weeks, mayinstead spend three one-hour sessions in the

HBO. Thedevice com-presses the airto 3 atm ofpressurearound theinjured player,who is breath-ing pure oxy-gen through amask. Thehigher airpressure in theHBO forcesmore oxygento dissolve inthe blood-stream, speed-

ing up the flow of oxygen to 15 times the normalrate. The abnormal oxygen flow causes the bloodvessels in the injured muscles to constrict, limit-ing swelling in that area. In two days, the injuredplayer may walk without crutches. By the end ofthe week, the player could be back on the field.

After a Heart AttackHBO therapy is also being used along with

clot-dissolving drugs to save heart muscle andimprove the quality of life for patients who havesuffered heart attacks. In a study of heart-attackpatients, half were given t-PA, a drug that pre-vents clots. The other half received the samedrug, followed by two hours of HBO treatment.The patients who received the drug alone had lin-gering chest pains for an average of 10.75 hours.Those who had HBO therapy experienced painsfor only 4.5 hours. In addition, normal electricalactivity was restored sooner in the hearts of thepatients who were in the hyperbaric chamber.Their electrocardiograms were normal in ten to15 minutes—half the time it took for the othergroup to approach that goal.

Breath of Fresh Air for Premature BabiesOne of the most important uses of HBO is to

treat babies with hyaline-membrane disease.These babies suffer from respiratory distress soonafter birth because the alveoli in their lungs fail toinflate. HBO therapy has increased the chancesthat the lungs will normalize so that these babiescan survive.

1. Thinking Critically What two factors causethe remarkable effects of HBO? Explain how.

2. Hypothesizing How might HBO be effectivefor treating second- and third-degree burnsover large portions of the body?

DISCUSSING THE TECHNOLOGY


Charles’s Law: Temperature and VolumeYou may have observed the beautiful patterns and graceful gliding of a

hot-air balloon, but what happens to a balloon when it is cold? Figure 11.10shows some dramatic effects of cooling and warming a gas-filled balloon.

The French scientist Jacques Charles (1746-1823) didn’t have liquidnitrogen, but he was a pioneer in hot-air ballooning. He investigated howchanging the temperature of a fixed amount of gas at constant pressureaffected its volume. The relationship Charles found can be demonstrated,as shown in Figure 11.11.

Figure 11.10Cooling and Warming a Gas-Filled BalloonAt 77 K, the nitrogen in thebeaker is so cold it is a liquid. Itrapidly cools the air-filled balloons,shrinking them by reducing thevolume of the balloons. When theballoons are removed from thebeaker and the temperature of theair inside rises to room tempera-ture, the balloons expand.

Figure 11.11Demonstrating Charles’s LawThe mercury plug is free to move back andforth in the horizontal tube because theend of the glass tube is open to the atmo-sphere. The pressure of the gas inside thebulb is always equal to the atmosphericpressure. The distance the mercury plugmoves right or left measures the increase ordecrease in the volume of gas as it is heatedor cooled.

Changein volumeof gas


Charles’s law states that at constant pressure, the volume of a gas isdirectly proportional to its Kelvin temperature, as shown in the graph inFigure 11.12. The straight lines for each gas indicate that volume andtemperature are in direct proportions. For example, if the Kelvin tempera-ture doubles, the volume doubles, and if the Kelvin temperature is halved,the volume is halved.

Kinetic Explanation of Charles’s LawWhy did the air in the balloons expand when heated and contract

when cooled? Study Figure 11.13 to learn how the kinetic theory ofmatter explains Charles’s law.

Figure 11.12Volume and Temperature for Three GasesThe three straight lines show that the volumeof each gas is directly proportional to its Kelvintemperature. The solid part of each line repre-sents actual data. Part of each line is dashedbecause, as you recall from Chapter 10, whenthe temperature of a gas falls below its boilingpoint, a gas condenses to a liquid.

100

80

60

40

20

100 200 300 400 500

Temperature (K)

Vol

ume

(L)

Gas A

Gas B

Gas C

0

Temperature decreasesTempera

ture incre

ases

Temperature decreasesTempera

ture incre

ases

Figure 11.13Modeling Charles’s Law

� When the balloon is heated, the tempera-ture of the air inside increases, and theaverage kinetic energy of the particles inthe air also increases. They exert moreforce on the balloon, but the pressureinside does not rise above the originalpressure because the balloon expands.

� When the balloon cools, the tempera-ture of the air inside falls and theaverage kinetic energy of the particlesin the air decreases. The particlesmove slower and strike the balloonless often and with less force. The bal-loon contracts and the pressure of theair inside the balloon continues to bal-ance the pressure of the atmosphere.


A balloon is filled with 3.0 L of helium at 22°C and 760 mm Hg. Itis then placed outdoors on a hot summer day when the temperature is31°C. If the pressure remains constant, what will the volume of theballoon be?

Analyze • Because the volume of a gas is propor-tional to its Kelvin temperature, youmust first express the temperatures inthis problem in kelvins. As in Chapter10, add 273 to the Celsius temperature to obtain the Kelvin temperature.

TK � TC � 273TK � 22 � 273 � 295 KTK � 31 � 273 � 304 K

Because the temperature of the helium increases from 295 K to 304 K, itsvolume increases in direct proportion. The temperature increases by thefactor 304 K/295 K. Therefore, the volume increases by the same factor.

Set Up • V � 3.0 L � � �Solve •Check • Does the answer have volume units? Does the volume increase as

expected? Because the answer to both questions is yes, this solution is reasonable. Check your calculations to make sure your answer iscorrect.

3.0 L 304 K 295 K

3.0 L � 304 295� � 3.1 L�V

304 K�295 K

Charles’s LawSAMPLE PROBLEM 5

PRACTICE PROBLEMS

Remember that gas volumes and pres-sures are proportional to temperatureonly if the temperature is expressed inkelvins.

Prob lem -So l v i ngH I N T

For more practicewith solving problems,

see Supplemental Practice Problems, Appendix B.

Assume that the pressure remains constant in the following problems.

12. A balloon is filled with 3.0 L of helium at 310 K and 1 atm. Theballoon is placed in an oven where the temperature reaches 340 K.What is the new volume of the balloon?

13. A 4.0-L sample of methane gas is collected at 30.0°C. Predict thevolume of the sample at 0°C.

14. A 25-L sample of nitrogen is heated from 110°C to 260°C. Whatvolume will the sample occupy at the higher temperature?

15. The volume of a 16-g sample of oxygen is 11.2 L at 273 K and1.00 atm. Predict the volume of the sample at 409 K.

16. The volume of a sample of argon is 8.5 mL at 15°C and 101 kPa.What will its volume be at 0.00°C and 101 kPa?

Figure 11.14Flying HighHot-air ballooning is a popularsport all around the world. Theheight of the balloon over theground is controlled by varyingthe temperature of the air withinthe balloon.


Combined Gas LawYou know that according to Boyle’s law, if you double the volume of a gas

while keeping the temperature constant, the pressure falls to half its initialvalue. You also know that according to Charles’s law, if you double theKelvin temperature of a gas while keeping the pressure constant, the volumedoubles. What do you think would happen to the pressure if you doubledthe volume and doubled the temperature? Suppose, like Robert Boyle andJacques Charles, you were to investigate how gases behave. You would con-duct experiments with gases just as they did. You would measure tempera-ture, pressure, and volume for a sample of gas; expand the gas to twice thevolume; raise the temperature to twice as high; and then measure the pres-sure. Following accepted principles of scientific methods, you would makemany such experiments. You might choose to triple the volume and thetemperature or change the volume and temperature by half or less to get awide range of data. You might then use a computer to help make a graph ofyour data and look for relationships among your three variables.

Another approach might be to first double the volume while keeping thetemperature constant or to double the volume and temperature and thenmeasure the pressure. In what other ways could you compare relationshipsamong temperature, pressure, and volume? Any one of the variables could bekept constant while varying another and measuring the effect on the third.You would find that doubling thevolume first and the temperaturesecond has exactly the same resultas doubling the temperature firstand the volume second. Theseresults should not be surprisingwhen you remember that you canread the curve in the graph forBoyle’s law in either direction.Whether the gas was expandingor contracting, if you knew itsvolume you could determine itspressure. If you knew its pressure,you could determine its volume.One application of these gas lawscan be seen at hot-air balloonevents, Figure 11.14.


Because Boyle’s law of gases and Charles’s law of gases are equally valid,it is possible to determine one of the three variables, regardless of the orderin which the other two are changed. If you doubled the volume and thetemperature at the same time, you would get exactly the same result as ifyou had doubled one first, then the other. For example, suppose you had 3L of gas at 200 K and 1 atm. If you double the volume, according to Boyle’slaw, the pressure falls to 0.5 atm. If you then double the temperature,according to Charles’s law, the pressure is increased to twice as high again,or 1 atm. Note that doubling the volume and then doubling the tempera-ture brings the sample back to its initial pressure. The effect on the pres-sure of doubling the volume and doubling the temperature offset eachother because pressure is inversely proportional to volume but directlyproportional to temperature.

The combination of Boyle’s law and Charles’s law is called the com-bined gas law. The factors are the same as in the previous sample prob-lems, but you may have more than one factor in a problem because morethan one quantity may vary. The set of conditions 0.00°C and 1 atm is sooften used that it is called standard temperature and pressure or STP.

A 154-mL sample of carbon dioxide gas is generated by burninggraphite in pure oxygen. If the pressure of the generated gas is 121 kPaand its temperature is 117°C, what volume would the gas occupy atstandard temperature and pressure, STP?

Analyze • Reducing the pressure from 121 kPa to 101 kPa increases the volumeof the carbon dioxide gas. Therefore, Boyle’s law gives the factor121 kPa/101 kPa. Because this factor is greater than 1, when it is multi-plied by the volume, the volume increases. Cooling the gas from 117°Cto 0.00°C reduces the volume of gas. To apply Charles’s law, you mustexpress both temperatures in kelvins.

TK � TC � 273 TK � TC � 273� 117 � 273 � 0.00 � 273� 390 K � 273 K

Because the temperature decreases, the volume decreases. Charles’s lawgives the factor 273 K/390 K, which is less than 1. Therefore, multiply-ing the volume by this factor decreases the volume.

Set Up • Multiply the volume by these two factors.

V � 154 mL � � � � � �Solve • The combined gas law equation is solved in the following steps.

154 mL 121 kPa 273 K 101 kPa 390 K

154 mL � 121 � 273 101 � 390

�

� 129 mL

�V

273 K�390 K

121 kPa�101 kPa

Determining Volumes at STPSAMPLE PROBLEM 6


The Law of Combining Gas VolumesWhen water decomposes into its elements—hydrogen and oxygen

gas—the volume of hydrogen produced is always twice the volume ofoxygen produced. Because matter is conserved, in the reverse synthesisreaction, the volume of hydrogen gas that reacts is always twice the vol-ume of oxygen gas.

Experiments with many other gas reactions show that volumes of gasesalways react in ratios of small whole numbers. Figure 11.15 presents thecombining volumes for a synthesis reaction and a decomposition reac-tion. The observation that at thesame temperature and pressure,volumes of gases combine ordecompose in ratios of smallwhole numbers is called the lawof combining gas volumes.

17. A 2.7-L sample of nitrogen is collected at 121 kPa and 288 K. Ifthe pressure increases to 202 kPa and the temperature rises to303 K, what volume will the nitrogen occupy?

18. A chunk of subliming carbon dioxide (dry ice) generates a 0.80-Lsample of gaseous CO2 at 22°C and 720 mm Hg. What volumewill the carbon dioxide gas have at STP?

PRACTICE PROBLEMS

Check • Estimate to see whether the answer is reasonable. The reduction inpressure would expand the volume by a factor of about 12/10. Coolingthe gas would contract it by a factor of about 7/10. Both factorstogether would alter the volume by a factor of about 84/100, which isless than 1. The final volume should be less than the initial volume.The final volume, 129 mL, is less than 154 mL, the initial volume.

2 liters HCl

Figure 11.15Comparing Volumes in Gas ReactionsWhen two liters of hydrogen chloride gas decompose toform hydrogen gas and chlorine gas, equal volumes ofhydrogen gas and chlorine gas are formed—1 L of each.The ratio of hydrogen to chlorine is 1 to 1, and the ratioof volumes of hydrogen to hydrogen chloride is 1/2 to 1or 1 to 2, the same as the ratio of chlorine to hydrogenchloride. Consider the reverse reaction—the composi-tion of hydrogen gas with chlorine gas to form hydro-gen chloride gas. What is the ratio of the reactants andthe ratio of the product to each reactant?

For more practice with solvingproblems, see SupplementalPractice Problems,Appendix B.


Popping CornYou recognize the smell, and that

characteristic popping noise is adead giveaway. Popcorn! Whatcauses the kernels of popcorn togo through explosive changes?

History of popcorn The ancestor of thepopcorn you eat was cultivated in theNew World by Native Americansmore than 7000 years ago. Popcornwas used as food, as decoration, andin religious ceremonies.

Popcorn kernels Popcorn kernels are extremelysmall and hard. They are wrapped in a tough,shell-like covering called the hull. The hull protectsthe embryo and its food supply. This food supplyis starch located within the endosperm. Each ker-nel also contains a small amount of water.

Exploding the kernel When heated to about204°C, the water in the kernel turns to steam. Theexpansion of the steam rips the remarkably toughhull with an explosive force, and the popcornbursts open to 30 to 40 times its original size. Theheat released by the steam bakes the starch into thefluffy product people like to eat.

Water content The amount of water in the kernelis an important aspect of popping. Food chemists

have found that the kernel must contain about13.5 percent water by mass to pop properly.

Chemistry

Exploring Further

1. Hypothesizing Suggest why too much or toolittle water present in the kernel greatly increas-es the number of unpopped kernels.

2. Applying Why is popcorn better stored in thefreezer or refrigerator rather than on the shelfat room temperature?

3. Acquiring Information More than 1000 vari-eties of corn are grown in the world today.Write a report about as many uses for corn andcorn products as you can.

�

1 liter H2 1 liter Cl2

To learn more about the mechanics of popcorn cook-ing, visit the Chemistry Web site at chemistryca.com

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In the synthesis of water, two volumes of hydrogen reactwith one of oxygen, but only two volumes of water vaporare formed. You might expect three. In the reverse reaction,which is the decomposition of water, two volumes of watervapor yield one volume of oxygen gas and two volumes ofhydrogen gas. That mysterious third volume appears again.How can we account for it? Amedeo Avogadro (1776-1856),an Italian physicist, made the same observations and askedthe same question. He first noticed that when water formsoxygen and hydrogen, two volumes of gas become threevolumes of gas, and he hypothesized that water vapor con-sisted of particles, as shown in Figure 11.16. Each particlein the water vapor broke up into two hydrogen parts andone oxygen part. Two volumes of hydrogen and one volumeof oxygen formed because there were twice as many hydro-

gen particles as oxygen particles. Today, we can show this by doing some-thing that Avogadro could not do—write the chemical equation for theformation of water.

2H2(g) � O2(g) ˇ 2H2O(g)

Avogadro was the first to interpret the law of combining volumes interms of interacting particles. He reasoned that the volume of a gas at agiven temperature and pressure must depend on the number of gas parti-cles. Avogadro’s principle states that equal volumes of gases at the sametemperature and pressure contain equal numbers of particles.

Connecting IdeasKnowing that equal volumes of gases at the same temperature and

pressure have the same number of particles is the first step toward know-ing how many particles are present in a sample of a substance. Knowingthe number of particles in a reactant and the ratio in which the particlesinteract enables us to predict the amount of a product. This importantnumber has been estimated.

Understanding Concepts1. The gas laws deal with the following variables—

the number of gas particles, temperature, pres-sure, and volume. Which of these are held con-stant in Boyle’s law, in Charles’s law, and in thecombined gas law?

2. Explain how an air mattress supports theweight of a person lying on it.

3. Why is it important to determine the volume ofa gas at STP?

Thinking Critically4. Analyzing If the volume of a helium-filled bal-

loon increases by 20 percent at constant tem-perature, what will the percentage change in thepressure be?

Applying Chemistry5. Aerosol Cans Use the kinetic theory to explain

why pressurized cans carry the message “Do notincinerate.”

SECTION REVIEW

Because of his observa-tions of reactionsinvolving oxygen, nitro-gen, and hydrogengases, Avogadro wasthe first to suggest thatthese gases were madeup of diatomic mole-cules.

Figure 11.16A Particle of WaterAvogadro determined thatwater is composed of particles.

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Chapter 11 Assessment 399

11.1 Gas Pressure■ The pressure of a gas at constant temperature

and volume is directly proportional to thenumber of gas particles.

■ The volume of a gas at constant temperatureand pressure is directly proportional to thenumber of gas particles.

■ At sea level, the pressure exerted by gases ofthe atmosphere equals one standard atmo-sphere (1 atm).

■ Gas pressure can be expressed in the follow-ing units: atmospheres (atm), millimeters ofmercury (mm Hg), inches of mercury (in.Hg), pascals (Pa), and kilopascals (kPa).

■ The factor label method is a simple tool forconverting measurements from one unit toanother.

11.2 The Gas Laws■ Boyle’s law states that the pressure and vol-

ume of a confined gas are inversely propor-tional.

■ Charles’s law states that the volume of anysample of gas at constant pressure is directlyproportional to its Kelvin temperature.

■ According to the combined gas law, whenpressure and temperature are both changing,both Boyle’s and Charles’s laws can be appliedindependently.

■ The law of combining gas volumes states thatin chemical reactions involving gases, theratio of the gas volumes is a small wholenumber.

■ Avogadro’s principle states that equal volumesof gases at the same temperature and pressurecontain equal numbers of particles.

VocabularyFor each of the following terms, write a sentence that showsyour understanding of its meaning.

Avogadro’s principlebarometerBoyle’s lawCharles’s lawcombined gas lawfactor label methodkilopascal (kPa)law of combining gas volumespascal (Pa)standard atmosphere (atm)standard temperature and pressure, STP

UNDERSTANDING CONCEPTS1. Name three ways to increase the pressure

inside an oxygen tank.

2. What Celsius temperature corresponds toabsolute zero?

3. What physical conditions are specified by theterm STP?

4. What factors affect gas pressure?

5. Convert the pressure measurement 547 mm Hgto the following units.

a) psi b) in. of Hgc) kPa d) atm

APPLYING CONCEPTS6. The passenger cabin of an airplane is pressur-

ized. Explain what this term means and why itis done.

7. Explain why balloons filled with helium rise inair but balloons filled with carbon dioxide sink.

8. At 1250 mm Hg and 75°C, the volume of asample of ammonia gas is 6.28 L. What volumewould the ammonia occupy at STP?

9. The volume of a gas is 550 mL at 760 mm Hg.If the pressure is reduced to 380 mm Hg andthe temperature is constant, what will be thenew volume?

REVIEWING MAIN IDEAS

CHAPTER 11 ASSESSMENT

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10. A 375-mL sample of air at STP is heated atconstant volume until its pressure increases to980 mm Hg. What is the new temperature ofthe sample?

11. A sample of neon gas has a volume of 822 mLat 160°C. The sample is cooled at constantpressure to a volume of 586 mL. What is itsnew temperature?

12. The volume of a gas is 1.5 L at 27°C and 1 atm.What volume will the gas occupy if the temper-ature is raised to 127°C at constant pressure?

13. A 50-mL sample of krypton gas at STP iscooled to �73°C at constant pressure. Whatwill be the volume?

14. How many liters of hydrogen will be needed toreact completely with 10 L of oxygen in thecomposition reaction of hydrogen peroxide?

Chemistry and Technology15. Compare the amount of oxygen you would

receive in an HBO chamber with the amountyou would receive breathing ordinary air.

Everyday Chemistry16. Explain why the density of a popped kernel of

popcorn is less than that of an unpopped kernel.

Earth Science Connection17. Would chlorine gas be suitable for weather bal-

loons? Explain.

THINKING CRITICALLYComparing and Contrasting18. ChemLab How are inferences made about the

relationship between volume and pressure inthis chapter’s ChemLab similar to inferencesmade from observing the apparatus describedin Figure 11.6?

Observing and Inferring19. MiniLab 1 How might you account for the

possible popping of the bag containing thesubliming dry ice?

Applying Concepts20. MiniLab 2 Explain how air pressure helps you

transfer a liquid with a pipet.

Making Predictions21. Use Boyle’s law to predict the change in density

of a sample of air if its pressure is increased.

CUMULATIVE REVIEW22. What are allotropes? Name three elements that

occur as allotropes. (Chapter 5)

23. Can an atom of an element have six electronsin the 2p sublevel? (Chapter 7)

24. Classify the following bonds as ionic, covalent,or polar covalent using electronegativity val-ues. (Chapter 9)

a) NH c) BHb) BaCl d) NaI

SKILL REVIEW25. Factor Label Method In a laboratory, 400 L of

methane are stored at 600 K and 1.25 atm. Ifthe methane is forced into a 200-L tank andcooled to 300 K, how does the pressure change?

WRITING IN CHEMISTRY26. Modern manufacturing applies the properties

of gases in many ways. List three examples ofthings you see and use every day that apply gasproperties. Include at least one householdproduct. Write a short explanation of theproperty used in each example.

PROBLEM SOLVING27. A sample of argon gas occupies 2.00 L at

–33.0°C and 1.50 atm. What is its volume at207°C and 2.00 atm?

28. A welding torch requires 4500 L of acetylene gasat 2 atm. If the acetylene is supplied by a 150-Ltank, what is the pressure of the acetylene?

CHAPTER 11 ASSESSMENT

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1. Which of the following units is equivalent to1.00 atm?

a) 760 mm Hg c) 101.3 kPab) 14.7 psi d) all of the above

2. 18.7 psi equals

a) 0.36 kPa. c) 128.9 kPa.b) 2.7 kPa. d) 966.8 kPa.

Interpreting Graphs: Use the graph to answerquestions 3–4.

3. The graph shows that

a) as temperature increases, pressure decreases.b) as pressure decreases, volume decreases.c) as temperature decreases, moles decrease.d) as pressure decreases, temperature decreases.

4. The predicted pressure of Gas B at 310 K is

a) 260 kPa. c) 1000 kPa.b) 620 kPa. d) 1200 kPa.

5. A sample of argon gas is compresses into avolume of 0.712 L by a piston exerting a pres-sure of 3.92 atm of pressure. The piston isslowly released until the pressure of the gas is1.50 atm. The new volume of the gas is

a) 0.272 L. c) 1.8 L.b) 3.67 L. d) 4.19 L.

6. Charles’s Law states that the volume of a gas is

a) inversely proportional to the temperature ofthe gas.

b) directly proportional to the temperature ofthe gas.

c) inversely proportional to the pressure of thegas.

d) directly proportional to the pressure of thegas.

7. While it is on the ground, a blimp is filled with5.66 � 106 L of He gas. The pressure inside thegrounded blimp, where the temperature is 25�C,is 1.10 atm. Modern blimps are non-rigid,which means their volume is changeable. If thepressure inside the blimp remains the same,what will be the volume of the blimp at a heightof 2300 m, where the temperature is 12�C?

a) 5.66 � 106 L c) 5.40 � 106 Lb) 2.72 � 106 L d) 5.92 � 106 L

8. When the temperature of the air inside a basketball is increased, the

a) pressure of the air inside the ball increases.b) kinetic energy of the air particles decreases.c) the volume of the air increases.d) pressure of the air inside the ball decreases.

9. Which of the following conditions is STP?

a) 0.00�C c) 273 Kb) 760 mm Hg d) all of the above

10. A volume of gas has a pressure of 1.4 atm.What would the pressure be if you doubled thevolume and reduced the temperature in kelvinsby half?

a) 0.35 atm c) 2.8 atmb) 1.4 atm d) 5.6 atm

Standardized Test Practice

Standardized Test Practice 401

Maximize Your Score If possible, find outhow your standardized test will be scored. In orderto do your best, you need to know if there is apenalty for guessing, and if so, what the penalty is.If there is no random-guessing penalty at all, youshould always fill in an answer, even if you haven’tread the question!

Test Taking Tip

chemistryca.com/standardized_test

250 260 270 280 290 300

Pres

ure

(kP

a)

1200

1000

800

600

400

200

0

Gas C

Gas A

Gas B

Temperature (K)

Pressures of Three Gasesat Different Temperatures

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chapter 11: behavior of gases - glencoe...1.00 atm 760 mm hg because atmospheric pressure is...

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