1 The Gaseous State of Matter Chapter 12 Hein and Arena Version 1.1.

1

The Gaseous State of Matter Chapter 12

The Gaseous State of Matter Chapter 12

Hein and Arena

Version 1.1

2

Chapter Outline12.2 The Kinetic MolecularTheory

12.3 Measurement of Pressure of Gases

12.4 Dependence of Pressure on Number of Molecules and Temperature

12.11 Avogadro’s Law

12.12 Mole-Mass-Volume Relationships of Gases

12.5 Boyle’s Law

12.6 Charles’ Law

12.10 Dalton’s Law of Partial Pressures

12.13 Density of Gases

12.14 Ideal Gas Equation

12.7 Gay Lussac’s Law

12.8 Standard Temperature and Pressure

12.9 Combined Gas Laws

12.15 Gas Stoichiometry

12.16 Real Gases

3

The Kinetic-Molecular Theory

• KMT is based on the motions of gas particles.

• A gas that behaves exactly as outlined by KMT is known as an ideal gas.

• While no ideal gases are found in nature, real gases can approximate ideal gas behavior under certain conditions of temperature and pressure.

4

Principle Assumptions of the KMT

1. Gases consist of tiny subatomic particles.

2. The distance between particles is large compared with the size of the particles themselves.

3. Gas particles have no attraction for one another.

5

4. Gas particles move in straight lines in all directions, colliding frequently with one another and with the walls of the container.

5. No energy is lost by the collision of a gas particle with another gas particle or with the walls of the container. All collisions are perfectly elastic.


6

6. The average kinetic energy for particles is the same for all gases at the same temperature, and its value is directly proportional to the Kelvin temperature.


7

Kinetic Energy

21KE =

2mv

8

• All gases have the same average kinetic energy at the same temperature.

• As a result lighter molecules move faster than heavier molecules.

mH2= 2 mO2= 32

vH2

vO 2

=14

Kinetic Energy

9

Diffusion

The ability of two or more gases to mix spontaneously until they form a uniform mixture.

Stopcock closed No diffusion occurs

Stopcock open Diffusion occurs

10

Effusion

A process by which gas molecules pass through a very small orifice from a container at higher pressure to one at lower pressure.

11

Graham’s Law of Effusion

The rates of effusion of two gases at the same temperature and pressure are inversely proportional to the square roots of their densities, or molar masses.

rate of effusion of gas Arate of effusion of gas B

dB=

dAmolar mass B

= molar mass A

12

2

effusion rate COeffusion rate CO

2molar mass CO=

molar mass CO44.0 g

= 1.2528.0 g

What is the ratio of the rate of effusion of CO to CO2?

13

Measurement of Measurement of Pressure of GasesPressure of Gases

14

Pressure equals force per unit area.

ForcePressure =

Area

15

The pressure resulting from the collisions of gas molecules with the walls of the balloon keeps the balloon inflated.

16

The pressure exerted by a gas depends on the

• Number of gas molecules present.

• Temperature of the gas.

• Volume in which the gas is confined.

17

Mercury Barometer

A tube of mercury is inverted and placed in a dish of mercury.

The barometer is used to measure atmospheric pressure.

18

19

Average Composition of Dry Air

N2 78.08%

O2 20.95%

Ar 0.93%

CO2 0.033%

0.0018%Ne

He 0.0005%

Gas Volume Percent

CH4 0.0002%

Kr

Xe, H2, and N2O Trace

0.0001%

Gas Volume Percent

20

• Pressure is produced by gas molecules colliding with the walls of a container.

• At a specific temperature and volume, the number of collisions depends on the number of gas molecules present.

• For an ideal gas the number of collisions is directly proportional to the number of gas molecules present.

Dependence of Pressure on Number of Dependence of Pressure on Number of Molecules and TemperatureMolecules and Temperature

21

V = 22.4 LT = OoC

The pressure exerted by a gas is directly proportional to the number of molecules present.

22

Dependence of Pressure on Temperature

• The pressure of a gas in a fixed volume increases with increasing temperature.

• When the pressure of a gas increases, its kinetic energy increases.

• The increased kinetic energy of the gas results in more frequent and energetic collisions of the molecules with the walls of the container.

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The pressure of a gas in a fixed volume increases with increasing temperature.

Lower T

Lower P

Higher T

Higher P

Increased pressure is due to more frequent and more energetic collisions of the gas molecules with the walls of the container at the higher

temperature.

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At constant temperature (T), the volume (V) of a fixed mass of gas is inversely proportional to the Pressure (P).

1V

P

1 1 2 2P V = P V

Boyle’s LawBoyle’s Law

26

Graph of pressure versus volume. This shows the inverse PV relationship of an ideal gas.

27The effect of pressure on the volume of a gas.

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Method A. Conversion Factors

Step 1. Determine whether volume is being increased or decreased.

Initial volume = 8.00 L Final volume = 3.00 L

volume decreases pressure increases

An 8.00 L sample of N2 is at a pressure of 500 torr. What must be the pressure to change the volume to 3.00 L? (T is constant).

29

8.00 Lx

3.00 L

Step 2. Multiply the original pressure by a ratio of volumes that will result in an increase in pressure.

P = 500 torr = 1333 L 3= 1.33 x 10 L

new pressure = original pressure x ratio of volumes


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V1 = 8.00 LP1 = 500 torr

V2 = 3.00 LP2 = ?


Method B. Algebraic Equation

Step 1. Organize the given information:

31

1 12

2

P VP =

V

1 1 2 2P V = P V

Step 2. Write and solve the equation for the unknown.


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Step 3. Put the given information into the equation and calculate.

1 12

2

P VP =

V(500 torr)(8.00 L)

= 3.00 L

1 = 1.33 x 10 torr


33

Absolute Zero on the Kelvin Scale

• If a given volume of any gas at 0oC is cooled by 1oC the volume of the gas decreases by .1

273


273

Charles’ LawCharles’ Law

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Absolute Zero on the Kelvin Scale


273

• -273oC (more precisely –273.15oC) is the zero point on the Kelvin scale. It is the temperature at which an ideal gas would have 0 volume.

35Volume-temperature relationship of methane (CH4).

36

Charles’ Law

At constant pressure the volume of a fixed mass of gas is directly proportional to the absolute temperature.

V T

1 2

1 2

V V =

T T

37

Effect of temperature on the volume of a gas. Pressure is constant at 1 atm. When temperature increases at constant pressure, the volume of the gas increases.

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75oC + 273 = 348 K

250oC + 273 = 523 K

A 255 mL sample of nitrogen at 75oC is confined at a pressure of 3.0 atmospheres. If the pressure remains constant, what will be the volume of the nitrogen if its temperature is raised to 250oC?

Method A. Conversion Factors

Step 1. Change oC to K: oC + 273 = K

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Step 2: Multiply the original volume by a ratio of Kelvin temperatures that will result in an increase in volume:

= 383 mLV = (255mL)523K348K


40

Method B. Algebraic EquationStep 1. Organize the information (remember

to make units the same):

V1 = 255 mL T1 = 75oC = 348 K

V2 = ? T2 = 250oC = 523 K


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Step 2. Write and solve the equation for the unknown:

1 22

1

V TV =

T1 2

1 2

V V =

T T


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Step 3. Put the given information into the equation and calculate:

= 383 mL1 22

1

V TV =

T(255mL)(523K)

= 348K

V1 = 255 mL T1 = 75oC = 348 K

V2 = ? T2 = 250oC = 523 K


43

The pressure of a fixed mass of gas, at constant volume, is directly proportional to the Kelvin temperature.

1 2

1 2

P P =

T T

P = kT

Gay-Lussac’s LawGay-Lussac’s Law

44

40oC + 273 = 313 K

At a temperature of 40oC an oxygen container is at a pressure of 2.15 atmospheres. If the temperature of the container is raised to 100oC what will be the pressure of the oxygen?

Method A. Conversion FactorsStep 1. Change oC to K:

oC + 273 = K

100oC + 273 = 373 K

temperature increases pressure increases

Determine whether temperature is beingincreased or decreased.

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Step 2: Multiply the original pressure by a ratio of Kelvin temperatures that will result in an increase in pressure:

= 25.6 atmP = (21.5 atm)373K313K


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= 25.6 atmP = (21.5 atm)373K313K

A temperature ratio greater than 1 will

increase the pressure


47

Method B. Algebraic Equation

Step 1. Organize the information (remember to make units the same):

P1 = 21.5 atm T1 = 40oC = 313 K

P2 = ? T2 = 100oC = 373 K


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Step 2. Write and solve the equation for the unknown:

1 22

1

P TP =

T1 2

1 2

P P =

T T


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Step 3. Put the given information into the equation and calculate:

= 25.6 atm1 2

21

P TP =

T


P1 = 21.5 atm T1 = 40oC = 313 K

P2 = ? T2 = 100oC = 373 K

(21.5 atm)(373 K) =

313 K

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Standard ConditionsStandard Temperature and Pressure

STP

273.15 K or 0.00oC1 atm or 760 torr or 760 mm Hg

Selected common reference points of temperature and pressure.

Standard Temperature and Standard Temperature and PressurePressure

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• A combination of Boyle’s and Charles’ Law.• Used when pressure and temperature change

at the same time.• Solve the equation for any one of the 6

variables

1 1 2 2

1 2

P V P V =

T T

Combined Gas LawsCombined Gas Laws

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final volume =ratio of

pressuresratio of

temperaturesinitial volume

calculate final volume

54


pressuresratio of


increases or decreases volume

55


pressuresratio of


increases or decreases volume

56

A sample of hydrogen occupies 465 ml at STP. If the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?

oC + 273 = K

0oC + 273 = 273 K

-15oC + 273 = 258 K

Step 1. Organize the given information, putting temperature in Kelvins:

57

Step 1. Organize the given information, putting temperature in Kelvins:

P1 = 760 torr P2 = 950 torr

V1 = 465 mL V2 = ?

T1 = 273 K T2 = 258 K


58

Method A Conversion Factors

Step 2. Set up ratios of T and P

760 torrP ratio =

950 torr

258 KT ratio =

273 K


(decrease in T decreases V)

(increase in P decreases V)

59

Step 3. Multiply the original volumes by the ratios:

= 352 mL


2V = (465 ml)760 torr950 torr

258K273K

P1 = 760 torr P2 = 950 torr

V1 = 465 mL V2 = ?

T1 = 273 K T2 = 258 K

60

1 1 2 2

1 2

PV PVT T

Method B Algebraic Equation

2

2

TP

2

2

TP


Step 2. Write and solve the equation for

the unknown V2.

2

2

TP

1 1

1

PVT

2V

1 1 22

2 1

V PTV

P T

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Step 2 Put the given information into the equation and calculate.

1 1 22

2 1

V PTV

P T

2

(465 ml) 760 torr (258 K)V = = 352 mL

(950 torr)(273 K)


62

Each gas in a mixture exerts a pressure that is independent of the other gases present.The total pressure of a mixture of gases is the sum of the partial pressures exerted by each of the gases in the mixture.

Ptotal = Pa + Pb + Pc + Pd + ….

Dalton’s Law ofDalton’s Law ofPartial PressuresPartial Pressures

63

A container contains He at a pressure of 0.50 atm, Ne at a pressure of 0.60 atm, and Ar at a pressure of 1.30 atm. What is the total pressure in the container?

Ptotal = PHe + PNe+ PAr

Ptotal = 0.5 atm + 0.6 atm + 1.30 atm = 2.40 atm

64

• The pressure in the collection container is equal to the atmospheric pressure.

• The pressure of the gas collected plus the pressure of water vapor at the collection temperature is equal to the atmospheric pressure.

2total atm gas H OP = P = P + P

Collecting a Gas Sample Over Water

65

Oxygen collected over water.

66

A sample of O2 was collected in a bottle over water at a temperature of 25oC when the atmospheric pressure was 760 torr. The vapor pressure of water at 25oC is 23.8 torr.

2 2O H OP = 760 torr - P

totalP = 760 torr2 2

O H O = P +P

2OP = 760 torr - 23.8 torr = 736 torr

67

Gay Lussac’s Law of Combining Volumes

N2

1 volume

+

+

3 H2

3 volumes

→

→

2 NH3

2 volumes

2

2

H

N

V 3=

V 13

2

NH

H

V 2=

V 33

2

NH

N

V 2=

V 1

When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.

68

Gay Lussac’s Law of Combining Volumes

When measured at the same temperature and pressure, the ratio of the volumes of reacting gases are small whole numbers.

69

Equal volumes of different gases at the same temperature and pressure contain the same number of molecules.

Avogadro’s LawAvogadro’s Law

70

E xp la ine d G ay Lu ssa c 'sL a w o f Co m b in in g V o lu m es

th e d e te rm in a tio n o fm o la r m a sse s o f ga ses

co m p ar in g d en sit ies o f g aseso f kn o wn m o la r m a ss

P rov id ed a m e th odfo r

S e rv ed as a fou nd ationfo r th e d ev o lop m e nt o f the K in e tic-M o lecu la r T h eo ry

AVOGADR O'S LAW

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There are 2 molecules of hydrogen chloride.

1 volume 1 volume 2 volumes

1 molecule 1 molecule 2 molecules

1 mol 1 mol 2 mol

hydrogen + chlorine hydrogen chloride

Each molecule of hydrogen chloride contains at least 1 atom of hydrogen and 1 atom of chlorine.

72


1 molecule 1 molecule 2 molecules

1 mol 1 mol 2 mol

hydrogen + chlorine → hydrogen chloride

Each molecule of hydrogen and each molecule of chlorine must contain at least 2 atoms.

H2 + Cl2 → 2 HCl

73

• Volume of one mole of any gas at STP = 22.4 L.

• 22.4 L at STP is known as the molar volume of any gas.

Mole-Mass-Volume Mole-Mass-Volume RelationshipsRelationships

74

• 22.4 L at STP is known as the molar volume of any gas.

75

76

The density of neon at STP is 0.900 g/L. What is the molar mass of neon?

g = 20.2

mol

0.900 g1 L

22.4 L1 mol

77

massdensity =

volume

Density of GasesDensity of Gases

78

Density of Gases

md =

vliters

grams

79

Density of Gases

md =

vdepends

on T and P

80

64.07 gd =

mol

1 mol22.4 L

g = 2.86

L

The molar mass of SO2 is 64.07 g/mol. Determine the density of SO2 at STP.

1 mole of any gas occupies 22.4 L at

STP

81

V PnTnRT

V = P

PV = nRT

Ideal Gas EquationIdeal Gas Equation

82

V PnTnRT

V = P

PV = nRT

atmospheres

83

V PnTnRT

V = P

PV = nRT

liters

84

V PnTnRT

V = P

PV = nRT

moles

85

V PnTnRT

V = P

PV = nRT

Kelvin

86

V PnTnRT

V = P

PV = nRT

Ideal Gas Constant

L-atm0.0821

mol-K

87

A balloon filled with 5.00 moles of helium gas is at a temperature of 25oC. The atmospheric pressure is 750. torr. What is the balloon’s volume?

Step 1. Organize the given information. Convert temperature to kelvins.

K = oC + 273

K = 25oC + 273 = 298K

Convert pressure to atmospheres.

P = 750. torr 1 atm

x 760 torr

= 0.987 atm

88

Step 2. Write and solve the ideal gas equation for the unknown.

Step 3. Substitute the given data into the equation and calculate.

A balloon filled with 5.00 moles of helium gas is at a temperature of 25oC. The atmospheric pressure is 750. torr. What is the balloon’s volume?

nRTV =

PPV = nRT

(0.987 atm)

(5.00 mol)V =

(0.0821 L×atm/mol×K)

(298 K)

= 124 L

89

Determination of Molecular Weights Using the Ideal Gas Equation

gmolar mass =

mol

gRTM =

PV

gmol =

molar mass

M = molar massg

n = mol = M

PV = nRTg

PV = RTM

90

Calculate the molar mass of an unknown gas, if 0.020 g occupies 250 mL at a temperature of 305 K and a pressure of 0.045 atm.

gRTM =

PV

V = 250 mL = 0.250 L g = 0.020 g

T = 305 K P = 0.045 atm

(0.020 g)M =

(0.082 L × atm/mol × K)

(305 K)

(0.250 L)

(0.045 atm)g

= 44 mol

91

• All calculations are done at STP.• Gases are assumed to behave as ideal

gases.

• A gas not at STP is converted to STP.

Gas StoichiometryGas Stoichiometry

92

Definition

Stoichiometry: The area of chemistry that deals with the quantitative relationships among reactants and products in a chemical reaction.

93

Gas Stoichiometry

Primary conversions involved in stoichiometry.

94

Mole-Volume CalculationsMole-Volume Calculations

Mass-Volume CalculationsMass-Volume Calculations

95

• Step 1 Write the balanced equation

2 KClO3 2 KCl + 3 O2

• Step 2 The starting amount is 0.500 mol KClO3. The conversion is

moles KClO3 moles O2 liters O2

What volume of oxygen (at STP) can be formed from 0.500 mol of potassium chlorate?

96

2

3

3 mol O2 mol KClO

• Step 3. Calculate the moles of O2, using the mole-ratio method.

3(0.500 mol KClO )


• Step 4. Convert moles of O2 to liters of O2

2 = 0.750 mol O

2(0.750 mol O )22.4 L1 mol

2= 16.8 L O

2 KClO3 2KCl + 3 O2

97

The problem can also be solved in one continuous calculation.


2 KClO3 2KCl + 3 O2

3(0.500 mol KClO ) 2

3

3 mol O2 mol KClO

22.4 L1 mol

2= 16.8 L O

98

2 Al(s) + 6 HCl(aq) 2AlCl3(aq) + 3 H2(g)

Step 1 Calculate moles of H2.

grams Al moles Al moles H2

What volume of hydrogen, collected at 30.0oC and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?

50.0 g Al1 mol Al

26.98 g Al

23 mol H2 mol Al

2 = 2.78 mol H

99

• Convert oC to K: 30.oC + 273 = 303 K

• Convert torr to atm:

2 Al(s) + 6 HCl(aq) 2AlCl3(aq) + 3 H2(g)

Step 2 Calculate liters of H2.


700 torr1 atm

760 torr

= 0.921 atm

100


PV = nRT

nRTV =

P

• Solve the ideal gas equation for V

(0.921 atm)

2(2.78 mol H )V =

(0.0821 L-atm)

(303 K)

(mol-K) 2 = 75.1 L H

101

Volume-Volume Volume-Volume CalculationsCalculations

102

H2(g) + Cl2(g) 2HCl(g)

1 mol H2 1 mol Cl2 2 mol HCl

22.4 LSTP

22.4 LSTP

2 x 22.4 LSTP


Y volume Y volume 2Y volumes

For reacting gases at constant temperature and pressure: Volume-volume relationships are the same as mole-mole relationships.

103

What volume of nitrogen will react with 600. mL of hydrogen to form ammonia? What volume of ammonia will be formed?

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

2600. ml H 2

2

1 vol N3 vol H

2= 200. mL N

2600. ml H 3

2

2 vol NH3 vol H

3= 400. mL NH

104

Ideal Gas

• An ideal gas obeys the gas laws.

– The volume the molecules of an ideal gas occupy is negligible compared to the volume of the gas. This is true at all temperatures and pressures.

– The intermolecular attractions between the molecules of an ideal gas are negligible at all temperatures and pressures.

Real GasesReal Gases

105

Real Gases• Deviations from the gas laws occur at

high pressures and low temperatures.– At high pressures the volumes of the real

gas molecules are not negligible compared to the volume of the gas

– At low temperatures the kinetic energy of the gas molecules cannot completely overcome the intermolecular attractive forces between the molecules.

106

Key Concepts12.2 The Kinetic MolecularTheory

12.3 Measurement of Pressure of Gases

12.4 Dependence of Pressure on Number of Molecules and Temperature

12.11 Avogadro’s Law

12.12 Mole-Mass-Volume Relationships of Gases

12.5 Boyle’s Law

12.6 Charles’ Law

12.10 Dalton’s Law of Partial Pressures

12.13 Density of Gases

12.14 Ideal Gas Equation

12.7 Gay Lussac’s Law

12.8 Standard Temperature and Pressure

12.9 Combined Gas Laws

12.15 Gas Stoichiometry

12.16 Real Gases

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1 The Gaseous State of Matter Chapter 12 Hein and Arena Version 1.1.

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