Chapter 5 Gases Doesn’t include AP PHall integration.

Chapter 5

Gases

Doesn’t include AP PHall integration.

GasesGases

3

Chapter 5

Table of Contents

Copyright © Cengage Learning. All rights reserved

5.1 Pressure

5.2 The Gas Laws of Boyle, Charles, and Avogadro

5.3 The Ideal Gas Law

5.4 Gas Stoichiometry

5.5 Dalton’s Law of Partial Pressures

5.6The Kinetic Molecular Theory of Gases

5.7Effusion and Diffusion

5.8Real Gases

5.9Characteristics of Several Real Gases

5.10Chemistry in the Atmosphere

4

Chapter 5


Why study gases?

• An understanding of real world phenomena.• An understanding of how science “works.”

5

Chapter 5


• 50 random questions out of 100 questions on ch. 5 gas laws with self-check answers back.

• http://www.kentchemistry.com/links/GasLaws/GasLawsQuiz.htm

• Don’t do these now; wait until you finish ch. 5 notes.

http://www.kentchemistry.com/links/GasLaws/GasLawsQuiz.htm

http://www.kentchemistry.com/links/GasLaws/GasLawsQuiz.htm

6

Section 5.1

Pressure


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• Uniformly fills any container.• Gases expand to fill their containers Gases are easily compressible• Mixes completely with any other gas.• Exerts pressure on its surroundings.

A Gas

Gases are fluid – they flowGases have low density1/1000 the density of the equivalent liquid or

solidGases effuse and diffuse

Ideal GasesIdeal Gases

Ideal gases are imaginary gases that perfectlyfit all of the assumptions of the kinetic molecular theory. (section 5.6 kmt) Gases consist of tiny particles that are far apart relative to their size.

Collisions between gas particles and between particles and the walls of the container are elastic collisions

No kinetic energy is lost in elastic collisions

Ideal Gases Ideal Gases (continued)

Gas particles are in constant, rapid motion. They therefore possess kinetic energy, the energy of motion

There are no forces of attraction between gas particles

The average kinetic energy of gas particles depends on temperature, not on the identity of the particle.

9

Section 5.1

Pressure


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• SI units = Newton/meter2 = 1 Pascal (Pa)• 1 standard atmosphere = 101,325 Pa• 1 standard atmosphere = 1 atm =

760 mm Hg = 760 torr

Pressure

PressurePressure

Unit Symbol Definition/Relationship

Pascal Pa SI pressure unit1 Pa = 1 newton/meter2

Millimeter of mercury

mm Hg Pressure that supports a 1 mm column of mercury in a barometer

Atmosphere atm Average atmospheric pressure at sea level and 0 C

Torr torr 1 torr = 1 mm Hg

PressurePressure is the force created by the collisions is the force created by the collisions

of molecules with the walls of a containerof molecules with the walls of a container

Standard Standard Temperature and Temperature and

PressurePressure“STP”“STP”

Either of these: 273 Kelvin (273 K)273 Kelvin (273 K) 0 0 CC

Either of these: 273 Kelvin (273 K)273 Kelvin (273 K) 0 0 CC

And any one of these: 1 atm1 atm 101.3 kPa101.3 kPa 14.7 lbs/in14.7 lbs/in2 2 (psi)(psi) 760 mm Hg760 mm Hg 760 torr760 torr

And any one of these: 1 atm1 atm 101.3 kPa101.3 kPa 14.7 lbs/in14.7 lbs/in2 2 (psi)(psi) 760 mm Hg760 mm Hg 760 torr760 torr

Measuring PressureMeasuring Pressure

The first device for measuring atmosphericpressure was developed by Evangelista Torricelli during the 17th century.

The device was called a “barometer”

Baro = weight Meter = measure

13

Section 5.1

Pressure


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Barometer

• Device used to measure atmospheric pressure.

• Mercury flows out of the tube until the pressure of the column of mercury standing on the surface of the mercury in the dish is equal to the pressure of the air on the rest of the surface of the mercury in the dish.

14

Section 5.1

Pressure


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• Device used for measuring the pressure of a gas in a container.

• Measuring blood pressure also uses a manometer.

Manometer

15

Section 5.1

Pressure


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The pressure of a gas is measured as 2.5 atm. Represent this pressure in both torr and pascals.

Pressure Conversions: An Example

Gas LawsGas Laws

Joseph Louis Gay-LussacAmadeo Avogadro

Robert Boyle Jacques Charles

Copyright © Houghton Mifflin Company. All rights reserved. 5–

Liquid Nitrogen and a BalloonClick here to watch video.

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http://college.cengage.com/chemistry/zumdahl/chemistry/7e/assets/students/protected/fae/index.html?layer=act&src=qtiworkflowflash_5_19.xml&w=750&h=434

18

Section 5.2

The Gas Laws of Boyle, Charles, and Avogadro


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Liquid Nitrogen and a Balloon

19

Section 5.2



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• What happened to the gas in the balloon?

• A decrease in temperature was followed by a decrease in the volume of the balloon.


20

Section 5.2



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• This is an observation (a fact).• It does NOT explain “why,” but it does tell

us “what happened.”


21

Section 5.2



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• What law results from observations like these?

• The volume of a gas depends on the temperature of the gas (constant P, pressure, and n, # of moles of gas).

Volume and Temperature

22

Section 5.2



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• Pressure and volume are inversely related (constant T, temperature, and n, # of moles of gas).

• PV = k (k is a constant for a given sample of air at a specific temperature)

Boyle’s Law

23

Section 5.2



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http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/gaslaw/boyles_law_graph.html

Click on the above link to take you to the icon. Drag the piston to obtain data. After obtaining several points, graph the data to see the relationship between pressure and volume using the piston.Marshmallow in piston is good demo. for this also.




Copyright © by Holt, Rinehart and Winston. All rights reserved.

ResourcesChapter menu

Visual Concepts

Boyle’s Law

Chapter 11

QuickTime™ and aSorenson Video 3 decompressorare needed to see this picture.


Boyle’s LawClick here to watch animation.

Skip for nowlog-inrequired.


26

Section 5.2



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Exercise

A sample of helium gas occupies 12.4 L at 23°C and 0.956 atm. What volume will it occupy at 1.20 atm assuming that the temperature stays constant?

27

Section 5.2



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Exercise

A sample of helium gas occupies 12.4 L at 23°C and 0.956 atm. What volume will it occupy at 1.20 atm assuming that the temperature stays constant?

Answer: 9.88 L

Let me know if you got a different answer. You should show your work.

28

Section 5.2



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• Volume and Temperature (in Kelvin) are directly related (constant P and n).

• V=kT (k is a proportionality constant)• K = °C + 273• 0 K is called absolute zero.

Charles’s Law

Temperature MUST be in KELVIN!

29

Section 5.2



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http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/gaslaw/charles_law.swf

Click on the above link. Change the temperature and watch the piston and the data plots and/or graph to watch the relationship of temperature vs. volume. The balloon with heat is good example of this or the balloon and freezer.



QuickTime™ and aSorenson Video 3 decompressorare needed to see this picture.



Visual Concepts

Charles’s Law

Chapter 11


Charles’s LawClick here to watch visualization.

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32

Section 5.2



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Exercise

Suppose a balloon containing 1.30 L of air at 24.7°C is placed into a beaker containing liquid nitrogen at –78.5°C. What will the volume of the sample of air become (at constant pressure)?

33

Section 5.2



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Exercise

Suppose a balloon containing 1.30 L of air at 24.7°C is placed into a beaker containing liquid nitrogen at –78.5°C. What will the volume of the sample of air become (at constant pressure)?

Did you get 0.849 L for answer?

34

QuickTime™ and a decompressor

are needed to see this picture.

Section 5.1

Pressure


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Collapsing Can - Lussac’s Law

Gay Lussac’s LawGay Lussac’s LawThe pressure and temperature of a gas aredirectly related, provided that the volume remains constant.

Temperature MUST be in KELVINS!


Video: Collapsing CanClick here to watch video.

Log-inrequired now.Added videopreviously.


The Combined Gas LawThe Combined Gas LawThe combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas.



The Combined Gas Law, continued

Sample Problem AA helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C?

Chapter 5 Section 2 The Gas Laws



Sample Problem A SolutionGiven: V1 of He = 50.0 L

T1 of He = 25°C + 273 = 298 K

T2 of He = 10°C + 273 = 283 K

P1 of He = 1.08 atm

P2 of He = 0.855 atm

Unknown: V2 of He in L





V

2=

P1V1T2

P2T1

=(1.08 atm)(50.0 L He)(283 K)

(0.855 atm)(298 K) = 60.0 L He

Sample Problem A Solution, continuedSolution:

Rearrange the equation for the combined gas law to obtain V2.


V

2=

P1V1T2

P2T1

P1V

1

T1

=P2V2

T2

⎛

⎝⎜⎞

⎠⎟


Substitute the given values of P1, T1, and T2 into the equation to

obtain the final volume, P2:



Answer practice problems #1-2 in your notes and show your work.



ANSWERSPractice Problems

Answer practice problems #1-2 in your notes and show your work.

off a bit?

43

Section 5.2



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• Volume and number of moles are directly related (constant T and P).

• (k is a proportionality constant)• (n is the # of moles)

Avogadro’s Law

V = kn

44

Section 5.2



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Sample Problem

If 2.45 mol of argon gas occupies a volume of 89.0 L, what volume will 2.10 mol of argon occupy under the same conditions of temperature and pressure?

45

Section 5.2



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Sample Problem

If 2.45 mol of argon gas occupies a volume of 89.0 L, what volume will 2.10 mol of argon occupy under the same conditions of temperature and pressure?

Did you get 76.3 L?

46

Section 5.3

The Ideal Gas Law


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• We can bring all of these laws together into one comprehensive law:

V = kT (constant P and n)V = kn (constant T and P)V = (constant T and n)

PV = nRT(where R = 0.08206 L·atm/mol·K, the universal gas constant)

47

Chapter FivePrentice Hall © 2005Hall © 2005General Chemistry 4th edition, Hill, Petrucci, McCreary, Perry

48

Section 5.3

The Ideal Gas Law


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5.3 Exercise 1

An automobile tire at 23°C with an internal volume of 25.0 L is filled with air to a total pressure of 3.18 atm. Determine the number of moles of air in the tire.

49

Section 5.3

The Ideal Gas Law


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5.3 Exercise 1

An automobile tire at 23°C with an internal volume of 25.0 L is filled with air to a total pressure of 3.18 atm. Determine the number of moles of air in the tire.

ANSWER: 3.27 mol

50

Section 5.3

The Ideal Gas Law


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5.3 Exercise 2

What is the pressure in a 304.0 L tank that contains 5.670 kg of helium at 25°C?

51

Section 5.3

The Ideal Gas Law


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5.3 Exercise 2

What is the pressure in a 304.0 L tank that contains 5.670 kg of helium at 25°C?

ANSWER: 114 atm

52

Section 5.3

The Ideal Gas Law


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5.3 Exercise 3

At what temperature (in °C) does 121 mL of CO2 at 27°C and 1.05 atm occupy a volume of 293 mL at a pressure of 1.40 atm?

53

Section 5.3

The Ideal Gas Law


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5.3 Exercise 3

At what temperature (in °C) does 121 mL of CO2 at 27°C and 1.05 atm occupy a volume of 293 mL at a pressure of 1.40 atm?

ANSWER: 696°C


Gas Volume, Pressure, and Concentration

Click here to watch video.

Log-inrequired.


55

Section 5.4

Gas Stoichiometry


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• For 1 mole of an ideal gas at 0°C and 1 atm, the volume of the gas is 22.42 L.

• STP = standard temperature and pressure

0°C and 1 atm Therefore, the molar volume is 22.42 L

at STP.

Molar Volume of an Ideal Gas

56

Section 5.4

Gas Stoichiometry


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5.4 Exercise 1

A sample of oxygen gas has a volume of 2.50 L at STP. How many grams of O2 are present?

57

Section 5.4

Gas Stoichiometry


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5.4 Exercise 1

A sample of oxygen gas has a volume of 2.50 L at STP. How many grams of O2 are present?

ANSWER: 3.57 g

58

Section 5.4

Gas Stoichiometry


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d = density of gas T = temperature in Kelvin P = pressure of gas R = universal gas constant

Molar Mass of a Gas

59

Section 5.4

Gas Stoichiometry


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Exercise 2

What is the density of F2 at STP (in g/L)?

60

Section 5.4

Gas Stoichiometry


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Exercise 2

What is the density of F2 at STP (in g/L)?

ANSWER: 1.70 g/L

Gas Stoichiometry Gas Stoichiometry RelationshipRelationship

If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios.

3 H2(g) + N2(g) 2NH3(g)

3 moles H2 + 1 mole N2 2 moles NH3 3 liters H2 + 1 liter N2 2 liters NH3

Gas Stoichiometry Exercise 3Gas Stoichiometry Exercise 3How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen?

3 H2(g) + N2(g) 2NH3(g)

12 L H2

L H2

= L NH3 L NH3

3

28.0

Gas Stoichiometry #4Gas Stoichiometry #4How many liters of oxygen gas, at STP, can be collected from the complete decomposition of 50.0 grams of potassium chlorate?

2 KClO3(s) 2 KCl(s) + 3 O2(g)

50.0 g KClO3 1 mol KClO3

122.55 g KClO3

3 mol O2

2 mol KClO3

22.4 L O2

1 mol O2

= 13.7 L O2

Gas Stoichiometry #5Gas Stoichiometry #5How many liters of oxygen gas, at 37.0C and 0.930 atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate?

2 KClO3(s) 2 KCl(s) + 3 O2(g)

50.0 g KClO3 1 mol KClO3

122.55 g KClO3

3 mol O2

2 mol KClO3

= mol O2

= 16.7 L

0.612

65

Section 5.5

Dalton’s Law of Partial Pressures


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• For a mixture of gases in a container,

PTotal = P1 + P2 + P3 + . . . • The total pressure exerted is the sum of

the pressures that each gas would exert if it were alone.

This is particularly useful in calculating the pressure of gases collected over water.

66

Section 5.5



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Exercise 1

Consider the following apparatus containing helium in both sides at 45°C. Initially the valve is closed.– After the valve is opened, what is the

pressure of the helium gas?

3.00 atm3.00 L

2.00 atm9.00 L

67

Section 5.5



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Exercise 2

27.4 L of oxygen gas at 25.0°C and 1.30 atm, and 8.50 L of helium gas at 25.0°C and 2.00 atm were pumped into a tank with a volume of 5.81 L at 25°C.• Calculate the new partial pressure of oxygen.• Calculate the new partial pressure of helium.• Calculate the new total pressure of both gases.

SEE NEXT SLIDE FOR ANSWERS AFTER YOU CALCULATE.

68

Section 5.5



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Exercise 2

27.4 L of oxygen gas at 25.0°C and 1.30 atm, and 8.50 L of helium gas at 25.0°C and 2.00 atm were pumped into a tank with a volume of 5.81 L at 25°C.• Calculate the new partial pressure of oxygen.

6.13 atm• Calculate the new partial pressure of helium.

2.93 atm• Calculate the new total pressure of both gases.

9.06 atm

69

Section 5.6

The Kinetic Molecular Theory of Gases


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• So far we have considered “what happens,” but not “why.”

• In science, “what” always comes before “why.”


Kinetic Molecular TheoryClick here to watch visualization.

Log-inrequired


Kinetic Molecular TheoryKinetic Molecular TheoryParticles of matter are

ALWAYS in motionVolume of individual

particles is ≈ zero.Collisions of particles with

container walls cause the pressure exerted by gas.

Particles exert no forces on each other.

Average kinetic energy is proportional to Kelvin temperature of a gas.

Kinetic Energy of Gas ParticlesKinetic Energy of Gas Particles

At the same conditions of temperature, all gases have the same average kinetic energy.

m = massv = velocity

∴ At the same temperature, small small moleculesmolecules move FASTERFASTER than large molecules

∴ At the same temperature, small small moleculesmolecules move FASTERFASTER than large molecules

The Meaning of TemperatureThe Meaning of Temperature

Kelvin temperature is an index of the random motions of gas particles (higher T means greater motion.)

74

Section 5.6



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R = 8.3145 J/K·mol

(J = joule = kg·m2/s2)

T = temperature of gas (in K)

M = mass of a mole of gas in kg

• Final units are in m/s.

Root Mean Square Velocity

75

Section 5.7

Effusion and Diffusion


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• Diffusion – the mixing of gases.• Effusion – describes the passage of a gas

through a tiny orifice into an evacuated chamber.

• Rate of effusion measures the speed at which the gas is transferred into the chamber.


Ammonia and HClclick below to watch video.

http://www.youtube.com/watch?v=Rf9j0ztzcs4#





DiffusionClick here to watch visualization.

Videoshownprevious

log-inrequired.


Diffusion describes the Diffusion describes the mixing of gases. The rate of mixing of gases. The rate of diffusion is the rate of gas diffusion is the rate of gas mixing.mixing.

Diffusion is the result of Diffusion is the result of random movementrandom movement of gas of gas moleculesmolecules

The rate of diffusion The rate of diffusion increases with temperatureincreases with temperature

Small molecules diffuse Small molecules diffuse faster than large moleculesfaster than large molecules

DiffusionDiffusion

EffusionEffusionEffusion: describes the passage of gas into an evacuated chamber.


EffusionClick here to watch videoclip.

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81

Section 5.7

Effusion and Diffusion


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• M1 and M2 represent the molar masses of

the gases.

Graham’s Law of Effusion



Graham’s Law of Effusion, continued

Sample Problem ACompare the rates of effusion of hydrogen and oxygen at the same temperature and pressure.

Chapter 11 Section 4 Diffusion and Effusion



Sample Problem A SolutionGiven: identities of two gases, H2 and O2

Unknown: relative rates of effusion

Solution: The ratio of the rates of effusion of two gases at the same temperature and pressure can be found from Graham’s law.

Chapter 11


Section 4 Diffusion and Effusion

rate of effusion of A

rate of effusion of B=

MB

MA



Sample Problem A Solution, continuedSubstitute the given values into the equation:

Hydrogen effuses 3.98 times faster than oxygen.

Chapter 11


Section 4 Diffusion and Effusion

rate of effusion of A

rate of effusion of B=

MB

MA

=32.00 g/mol

2.02 g/mol=

32.00 g/mol2.02 g/mol

=3.98



Practice Problems #1-3

Complete practice problems #1-3. Show work.

v



ANSWERSPractice Problems

Complete practice problems. Show work.

v

87

Section 5.8

Real Gases


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• Gases reach ideal behavior at high temperature and low pressure.

• We must correct for non-ideal gas behavior when:– Pressure of the gas is high.– Temperature is low.

• Under these conditions: – Concentration of gas particles is high.– Attractive forces become important.

88

Section 5.8

Real Gases


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Plots of PV/nRT Versus P for Several Gases (200 K)

89

Section 5.8

Real Gases


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Plots of PV/nRT Versus P for Nitrogen Gas at Three Temperatures

90

Section 5.8

Real Gases


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Real Gases (van der Waals Equation)

↑↑corrected pressurecorrected pressure

PPidealideal

↑↑corrected volumecorrected volume

VVidealideal

Real GasesReal Gases

↑↑ ↑↑corrected pressure corrected volume

PPidealideal VVidealideal

At high pressure (smaller volume) and low temperature (attractive forces become important) you must adjust for non-ideal gas behavior using van der Waal’s equation.

Kinetic Molecular Theory vs. Theory of the ideal GasEver see a fast food commercial. Don't those burgers look great. Then you get one, and you are disappointed because it is not the perfect one on the TV. So you lie to yourself. No matter how many burgers you order in your life, none will be as perfect at the one on TV. But if you think it is perfect, who really cares. Ideal gas are like the commercials and real gases are what you get. Nothing is ever perfect, but close is good enough.

Ideal Gas Rules (the Kinetic Molecular Theory)

1.Gases consist of large numbers of molecules (or atoms, in the case of the noble gases) that are in continuous, random straight line motion

2.The volume of all the molecules of the gas is negligible compared to the total volume in which the gas is contained (This is equivalent to stating that the average distance separating the gas particles is large compared to their size).

3.Attractive and repulsive forces between gas molecules is negligible (they can never group together and form drops and liquefy)

4. Energy can be transferred between molecules during collisions, but the collisions are perfectly elastic and the total energy remains constant.

Here is what you really need to know.

1. The most ideal gas in nature is hydrogen then helium. It is small and has the least mass.

2. How real gases deviate from Ideal gases. They have mass, volume and attraction. this causes the molecules to be drawn to each other, which cause the actual volume of a gas to be smaller then its ideal gas calculation. Watch this simulation of how a real molecules would wobble as it travels.

AP Chemistry-The Van der Waals equation accounts for these shortcomings of the Theory of the Ideal Gas.

3. Also real gases can liquefy under the conditions of High Pressure and Low Temperature. These are the least ideal conditions.

Just like ideal gas conditions, Ideal Vacations conditions High Temperature and Low Pressure.

QuickTime™ and a decompressor

are needed to see this picture.

93

Section 5.9

Characteristics of Several Real Gases


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• For a real gas, the actual observed pressure is lower than the pressure expected for an ideal gas due to the intermolecular attractions that occur in real gases.

94

Section 5.9



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Values of the van der Waals Constants for Some Gases• The value of a reflects

how much of a correction must be made to adjust the observed pressure up to the expected ideal pressure.

• A low value for a reflects weak intermolecular forces among the gas molecules.

95

Section 5.9



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Example

96

Section 5.9



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97

Section 5.9



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Section 5.10 will not be tested & isn’t in the notes

• Feel free to read section 5.10 which are good applications with air pollution but will not be tested.

98

Section 5.10

Chemistry in the Atmosphere


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• Two main sources: Transportation Production of electricity

• Combustion of petroleum produces CO, CO2, NO, and NO2, along with unburned molecules from petroleum.

Air Pollution

99

Section 5.10



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• At high temperatures, N2 and O2 react to form NO, which oxidizes to NO2.

• The NO2 breaks up into nitric oxide and free oxygen atoms.

• Oxygen atoms combine with O2 to form ozone (O3).

Nitrogen Oxides (Due to Cars and Trucks)

100

Section 5.10



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Concentration for Some Smog Components vs. Time of Day

101

Section 5.10



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• Sulfur produces SO2 when burned.

• SO2 oxidizes into SO3, which combines with water droplets in the air to form sulfuric acid.

Sulfur Oxides (Due to Burning Coal for Electricity)

102

Section 5.10



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• Sulfuric acid is very corrosive and produces acid rain.

• Use of a scrubber removes SO2 from the exhaust gas when burning coal.

Sulfur Oxides (Due to Burning Coal for Electricity)

103

Section 5.10



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A Schematic Diagram of a Scrubber

104

Section 5.10



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END OF CH. 5 NOTES

105

Section 5.6



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Concept Check

Sketch a graph of:

I. Pressure versus volume at constant temperature and moles.


Molecular View of Boyle’s LawClick here to watch visualization.

Log-inrequired now.


107

Section 5.6



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Concept Check

Sketch a graph of:

II. Volume vs. temperature (C) at constant pressure and moles.

108

Section 5.6



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Concept Check

Sketch a graph of:

III. Volume vs. temperature (K) at constant pressure and moles.


Molecular View of Charles’s LawClick here to watch visualization.


110

Section 5.6



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Concept Check

Sketch a graph of:

IV. Volume vs. moles at constant temperature and pressure.


Molecular View of The Ideal Gas LawClick here to watch visualization.


112

Section 5.6



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Concept Check

Which of the following best represents the mass ratio of Ne:Ar in the balloons?

1:1

1:2

2:1

1:3

3:1

Ne

Ar

VNe = 2VAr

113

Section 5.6



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Concept Check

• You have a sample of nitrogen gas (N2) in a container fitted with a piston that maintains a

pressure of 6.00 atm. Initially, the gas is at 45C in a volume of 6.00 L.

• You then cool the gas sample.

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Section 5.6



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Concept Check

Which best explains the final result that occurs once the gas sample has cooled?

a) The pressure of the gas increases.

b) The volume of the gas increases.

c) The pressure of the gas decreases.

d) The volume of the gas decreases.

e) Both volume and pressure change.

115

Section 5.6



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Concept Check

The gas sample is then cooled to a temperature of 15C. Solve for the new condition. (Hint: A moveable piston keeps the pressure constant overall, so what condition will change?)

5.43 L

116

Section 5.6



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Concept Check

You are holding two balloons of the same volume. One contains helium, and one contains hydrogen. Complete each of the following statements with “different” or “the same” and be prepared to justify your answer.

He

H2

117

Section 5.6



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Concept Check

• Complete the following statement with

“different” or “the same” and be prepared to justify your answer.

• The pressures of the gas in the two balloons

are __________.

He

H2

118

Section 5.6



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Concept Check



• The temperatures of the gas in the two

balloons are __________.

He

H2

119

Section 5.6



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Concept Check



• The numbers of moles of the gas in the two

balloons are __________.

He

H2

120

Section 5.6



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Concept Check



• The densities of the gas in the two balloons are __________.

He

H2


• Gases are said to exert “pressure.” Provide a molecular level explanation for this.

• What is meant by the term “volume” of a gas? Give a molecular level explanation for this.

• What is meant by the term “temperature”? What does temperature measure?

React 1


• You are holding two balloons. One contains helium, and one contains hydrogen. Complete each of the following statements with “different” or “the same” and be prepared to justify your answer.

He

H2

React 2


• Complete the following statement with “different” or “the same” and be prepared to justify your answer.

• The pressures of the gas in the two balloons are __________.

He

H2

React 2



• The temperatures of the gas in the two balloons are __________.

He

H2

React 2



• The numbers of moles of the gas in the two balloons are __________.

He

H2

React 2



• The densities of the gas in the two balloons are __________.

He

H2

React 2


Discuss what happens to the density of gas as you increase the temperature if the gas is in:

a) a rigid steel container.

b) a container fitted with a moveable piston.

React 3


Sketch a graph of:

I. Pressure versus volume at constant temperature and moles.

React 4


Sketch a graph of:

II. Volume vs. temperature (C) at constant pressure and moles.

React 4


Sketch a graph of:

III. Volume vs. temperature (K) at constant pressure and moles.

React 4


Sketch a graph of:

IV. PV vs. P at constant temperature for 1 mole of gas.

React 4


Sketch a graph of

V. Volume vs. moles at constant temperature and pressure.

React 4


Which of the following best represents the mass ratio of Ne:Ar in the balloons? 1:1

1:2

2:1

1:3

3:1

VNe = 2VArNe

Ar

React 5


• On a beautiful autumn day in Champaign, IL, you are holding two balloons, each with 15 mol of gas.

• Which balloon is larger? – The one filled with neon

gas.– The one filled with argon

gas.

React 6


• On a beautiful autumn day in Champaign, IL, you are holding two balloons, each with 15 mol of gas.

• Estimate the volume of each balloon.

React 6


• What mass of neon is in the balloon? <1.0g

1.0g

>1.0g

VHe = VNe Ne

He1 gram

React 7


• Determine the mass of neon in the balloon.

VHe = VNeNe

He1 gram

React 7


• Consider the following container of helium. Initially the valve is closed.

• The total pressure in the container after the valve is opened is:

a) <5.0 atm

b) 5.0 atm

c) >5.0 atm

React 8

3.00 atm3.00 L

2.00 atm9.00 L


• Consider the following container of helium. Initially the valve is closed.

• After the valve is opened, what is the pressure of the helium gas?

React 8

3.00 atm3.00 L

2.00 atm9.00 L


React 9

• You have a sample of nitrogen gas (N2) in a container fitted with a piston that is free to move. Initially, the gas is under 6.00 atm of pressure at 45C in a volume of 6.00 L.

• You then cool the gas sample.


• Which best explains the final result that occurs once the gas sample has cooled?

a) The pressure of the gas increases.

b) The volume of the gas increases.

c) The pressure of the gas decreases.

d) The volume of the gas decreases.

e) Both volume and pressure change.

React 9


• The gas sample is then cooled to a temperature of 15C.

a) Solve for the new condition.

React 9


Consider two containers at the same pressure and temperature. How do each of the following compare?

a) volume of containers

1 moleO2 (g)

1 moleH2 (g)

React 10



b) number of molecules

React 10

1 moleO2 (g)

1 moleH2 (g)



c) density of samples

React 10

1 moleO2 (g)

1 moleH2 (g)



d) average kinetic energy

React 10

1 moleO2 (g)

1 moleH2 (g)



e) number of collisions per second

React 10

1 moleO2 (g)

1 moleH2 (g)


You have two containers with the same volume at the same temperature, each containing two moles of gas.

a) In which container is the partial pressure of neon greater?

React 11

#1 #2


What is the ratio of partial pressure of neon in container 1 to partial pressure of neon in container 2?

React 11

#1 #2

Chapter 5Questions

GasesGases

Copyright © Houghton Mifflin Company. All rights reserved. CRS Question, 5–

QUESTION #1Four bicycle tires are inflated to the following pressures. Which one has the highest pressure? Tire A 3.42 atm; Tire B 48 lbs/sq in; Tire C 305 kPa; Tire D 1520 mmHg. (Recall; 1.00 atm = 760 mmHg = 14.7 lb/sq in = 101.3 kPa)

1. Tire A2. Tire B3. Tire C4. Tire D


ANSWERChoice 1 Even though it has the smallest number, it represents the highest pressure of the four. When all four are changed to a common label (use conversion factors found on page 181 and dimensional analysis) 3.42 atm is a higher pressure than the others.

Section 5.1: Pressure


QUESTION #2Why is it critical that the temperature be held constant when applying Boyle’s law to changing the pressure of a trapped gas?

1. Gas molecules may expand at higher temperatures; this wouldchange the volume.

2. Changing the temperature causes the gas to behave in non-idealfashion.

3. Changing the temperature affects the average particle speed,which could affect the pressure.

4. Allowing the temperature to drop below 0°C would cause thetrapped gas to no longer follow Boyle’s Law.


ANSWERChoice 3 provides the connection between temperature and pressure that would introduce another variable when studying the pressure-volume relationship. Boyle’s law shows that if the molecules of a trapped gas maintain the same average velocity when their volume is changed, the pressure will be inversely related to the volume change.

Section 5.2: The Gas Laws of Boyle, Charles, and Avogadro


QUESTION #3After examining this figure, which of the following statements offers accurate, consistent statements about the gases shown at constant pressure?


QUESTION (continued)1. At –273°C all gases occupy nearly the same volume; the

different slopes are because of differences in molar masses. 2. At zero Celsius the gases have different volumes because the

larger the molecule, the larger the volume.3. Since the pressure is constant, the only difference that could

cause the different slopes is in the number of moles of gas present in each sample. For example, the N2O sample must have the fewest moles.

4. Although it does appear to do so, the volumes do not reach zero but if the graph used K instead of °C the volume would reach zero for all the gases.


ANSWERChoice 3 is the correct conclusion. If all the gases had the same number of moles in their samples, the volume should be the same for all four at any temperature (with constant pressure). This is reflected in Avogadro’s law.



QUESTION #4Each of the balloons hold 1.0 L of different gases. All four are at 25°C and each contains the same number of molecules. Of the following which would also have to be the same for each balloon? (obviously not their color)

1. Their density2. Their mass3. Their buoyancy4. Their pressure


ANSWERChoice 4 is consistent with Avogadro’s Law. The temperature, pressure, number of moles, and volume are related for a sample of trapped gas. If two samples have three variables out of the four the same, the fourth variable must be the same as well.



QUESTION #5If a 10.0 L sample of a gas at 25°C suddenly had its volume doubled, without changing its temperature what would happen to its pressure? What could be done to keep the pressure constant without changing the temperature?

1. The pressure would double; nothing else could be done toprevent this.

2. The pressure would double; the moles of gas could be doubled.3. The pressure would decrease by a factor of two; the moles of gas

could be halved.4. The pressure would decrease by a factor of two; the moles could

be doubled.


ANSWERChoice 4 describes two opposing changes. When the volume increases, the pressure of a trapped gas will decrease (at constant temperature and constant moles of gas). However, if the pressure drops, more collisions could be restored by adding more particles of gas in the same ratio as the pressure decline.



QUESTION #6The typical capacity for human lungs is approximately 5,800 mL. At a temperature of 37°C (average body temperature) and pressure of 0.98 atm, how many moles of air do we carry inside our lungs?

1. 1.9 mol2. 0.22 mol3. 230 mol4. No one wants moles in their lungs.


ANSWERChoice 2 is correct. The units for temperature must be in K, pressure in atm, and volume in L. Then using the universal constant 0.08206 L atm/ K mol in the PV = nRT equation moles may be solved.

Section 5.3: The Ideal Gas Law


QUESTION #7The aroma of fresh raspberries can be attributed, at least in part, to 3-(para-hydroxyphenyl)-2-butanone. What is the molar mass of this pleasant smelling compound if at 1.00 atmosphere of pressure and 25.0°C, 0.0820 grams has a volume of 12.2 mL?

1. 13.8 g/mol2. 164 g/mol3. 40.9 g/mol4. 224 g/mol


ANSWERChoice 2 is correct. Using PV = nRT and properly substituting0.0122 L for V, 298 K for T, using 0.08206 for R and solving for n the number of moles represented by 0.0820 grams can be obtained. Then the grams in one mole can be found.



QUESTION #8If a person exhaled 125 mL of CO2 gas at 37.0°C and 0.950 atm of pressure, what would this volume be at a colder temperature of 10.0°C and 0.900 atm of pressure?

1. 3.12 mL2. 0.130 L3. 0.120 L4. 22.4 L


ANSWERChoice 3 correctly accounts for the changing conditions. The combination of Charles and Boyle’s laws provides the mathematical basis for the calculation. After converting the temperature to K units, the equation P1V1/T1 = P2V2/T2 may be used by solving for V2.



QUESTION #9Under STP conditions what is the density of O2 gas?

1. Not enough information is given to solve this.2. 1.31 g/L3. 1.43 g/L4. 0.999 g/L


ANSWERChoice 3 is the correct density for O2 at STP. At STP conditions the volume of one mole of a gas is approximately 22.4 L. One mole of oxygen = 32.0 grams. The density is calculated by dividing mass by volume.

Section 5.4: Gas Stoichiometry


QUESTION #10The primary way your body produces exhaled CO2 is by the

combustion of glucose, C6H12O6 (molar mass = 180. g/mol.). The

balanced equation is shown here:

C6H12O6 (aq) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l)

If you oxidized 5.42 grams of C6H12O6 at STP conditions, how many

liters of O2 did you use?

1. 0.737 L2. 0.672 L3. 4.05 L4. 22.4 L


ANSWERChoice 3, assuming you were not holding your breath, is correct. The number of moles of glucose must first be determined (5.42/180. = 0.0301 moles), then this is multiplied by 6 to account for the stoichiometric ratio between glucose and oxygen. From this, PV = nRT may be used with the appropriate substitutions.

Section 5.4: Gas Stoichiometry


QUESTION #11Freon-12 had been widely used as a refrigerant in air conditioning systems. However, it has been shown to be related to destroying Earth’s important ozone layer. What is the molar mass of Freon-12 if 9.27 grams was collected by water displacement, in a 2.00 liter volume at 30.0°C and 764 mmHg. Water’s vapor pressure at this temperature is approximately 31.8 mmHg.

1. 120. g/mol2. 12.0 g/mol3. 115 g/mol4. 92.7 g/mol


ANSWERChoice 1 is the molar mass of Freon-12. The pressure must be corrected for the presence of water by subtracting 31.8 mmHg from the total pressure. This should also be converted to atm. The temperature must be converted to K. Then PV = nRT can be used if n is written as g/mm and solved for mm.

Section 5.5: Dalton’s Law of Partial Pressures


QUESTION #12Radon (element 86) is a radioactive noble gas that can be found in some homes. In those situations radon can diffuse in the air and end up inhaled by the occupants. What is the ratio of the velocity of radon gas compared with O2 in the same room at the same

conditions?

1. 2.6 times faster2. 0.38 times as fast3. 0.14 times as fast4. 6.9 times faster


ANSWERChoice 2 shows the approximate ratio for the more massive radon velocity compared to the lighter oxygen molecules. This is an application of Graham’s law. Velocity of Rn/Velocity of O2 = (mm O2).5/(Rn).5

Section 5.7: Effusion and Diffusion


QUESTION #13If the mole fraction of O2 in our atmosphere at standard conditions is approximately 0.209, what is the partial pressure of the oxygen in every breath you take?

1. 1.00 atm2. 4.78 atm3. 159 torr4. 3640 mmHg


ANSWERChoice 3 shows the correct pressure. This calculation is based on the application of Dalton’s law of partial pressures. For one atmosphere of pressure 0.209 is caused by oxygen so 760 .209 = 159 torr.

Section 5.5: Dalton’s Law of Partial Pressures


QUESTION #14Kinetic molecular theory helps explain the definition of temperature based on molecular motion. Which statement describes an important aspect of this connection?

1. Temperature is inversely related to the kinetic energy of thegas particles.

2. At the same temperature, more massive gas particles will bemoving faster than less massive gas particles.

3. As the temperature of a gas sample increases, the average velocity of the gas particles increases.4. Kinetic energy is directly related to temperature. This is valid

for any units of temperature.


ANSWERChoice 3 states a correct connection between temperature and velocity. The important relationship is summarized in the following equation: ½ mass velocity2 = 3/2 RT (T must be in K).

Section 5.6: The Kinetic Molecular Theory of Gases


QUESTION #15As the temperature of a gas increases, which statement best correlates to information about molecular velocity?

1. The average molecular velocity will increase, but thedistribution of molecular velocities will stay the same.

2. The average molecular velocity will stay the same, but themolecular velocity distribution will spread.

3. The average molecular velocity will increase, and thedistribution of the molecular velocities will spread.

4. The average molecular velocity will stay the same, and thedistribution of the molecular velocities will stay the same.


ANSWERChoice 3 accurately reflects the connection between molecular velocity and an increase in temperature. Figure 5.21 on page 206 visually shows the relationship. Kelvin temperature is directly related to molecular velocity - with greater temperature the distribution of available velocities also increases.

Section 5.6: The Kinetic Molecular Theory of Gases


QUESTION #16Real gases exhibit their most “ideal” behavior at which relative conditions?

1. Low temperatures and low pressures2. High temperatures and high pressures3. High temperatures and low pressures4. Low temperatures and high pressures


ANSWERChoice 3 provides the correct choice. At those conditions gas molecules are farthest apart and exert their least influence on each other thereby permitting their behavior to follow mathematical predictions.

Section 5.8: Real Gases

Chapter 5 Gases Doesn’t include AP PHall integration.

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