1 version 1. 2 properties of gases 3 may be compressed expand to fill container low density may be...

1Version 1

4

The Kinetic-The Kinetic-Molecular TheoryMolecular Theory

(KMT)(KMT)

5

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.

6

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

Principle Assumptions of the KMT

7

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. 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.

Principle Assumptions of the KMT

8

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 (high temperature & low pressure).

11

Parameters forParameters forDescribing GasesDescribing Gases

12

Variables that describe a gas:Variable Symbol Units Conversions

Pressure P

- mm Hg

- torr

- atm

torr = mm Hg

760 torr = 1 atm

Volume V L, mL 1000 mL = 1 L

Temperature T K K = oC + 273

Moles n moles n = g/mw

13

Measurement of Measurement of Pressure of GasesPressure of Gases

14

Pressure results from the collisions of gas molecules with the walls of the container, whether it is a flexible wall or not.

19

The Gas Laws

1 1 2 2P V = P V

1 2

1 2

V V =

T T

1 2

1 2

P P =

T T

1 1 2 2

1 2

P V P V =

T T

PV = nRT

Ptotal = Pa + Pb + Pc + Pd + ….

20

Boyle’s LawBoyle’s Law16621662

22

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

23

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

24The effect of pressure on the volume of a gas.

25

V1 = 8.00 LP1 = 500 torr

V2 = 3.00 LP2 = ?

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).

Step 1. Organize the given information:

26

1 12

2

P VP =

V

1 1 2 2P V = P V

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

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).

27

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

3 = 1.33 x 10 torr

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).

28

A tank of O2 contains 1500. ml of gas at a pressure of 350. torr. What volume would the gas occupy at 1000. torr?

Step 1. Organize the given information:

V1 = 1500. ml

P1 = 350. torr

P2 = 1000. torr

V2 = ?

29

A tank of O2 contains 1500. ml of gas at a pressure of 350. torr. What volume would the gas occupy at 1000. torr?

Step 2. Write and solve the equation for

the unknown.

1 1 2 2P V = P V

1 12

2

P VV =

P

30

A tank of O2 contains 1500. ml of gas at a pressure of 350. torr. What volume would the gas occupy at 1000. torr?

Step 3. Put the given information into

the equation and calculate.

1 12

2

P VV =

P(350. torr)(1500. ml)

= 1000. torr

= 525 ml

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Charles’ LawCharles’ Law17871787

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

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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.

38

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

V1 = 255 mL T1 = 75oC = 348 K

V2 = ? T2 = 250oC = 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?

39

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

1 22

1

V TV =

T1 2

1 2

V V =

T T

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?

40

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

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?

41

A tank of O2 contains 16.0 L at 500.K is allowed to expand to a volume of 20.0 L. Find the new temperature (constant pressure).

Step 1. Organize the given information:

V1 = 16.0 L

T1 = 500.K

V2 = 20.0 L

T2 = ?

42

A tank of O2 contains 16.0 L at 500.K is allowed to expand to a volume of 20.0 L. Find the new temperature (constant pressure).

Step 2. Write and solve the equation for

the unknown.

1 2

1 2

V V =

T T2 1

21

V TT =

V

43

A tank of O2 contains 16.0 L at 500.K is allowed to expand to a volume of 20.0 L. Find the new temperature (constant pressure).

Step 3. Put the given information into

the equation and calculate:

2 12

1

V TT =

V(20.0 L)(500. K)

= 16.0 L

= 625 K

44

Gay-Lussac’s LawGay-Lussac’s Law18021802

46

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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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 T∝

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

P1 = 21.5 atm T1 = 40oC = 313 K

P2 = ? T2 = 100oC = 373 K

At a temperature of 40.oC an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100.oC what will be the pressure of the oxygen?

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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

At a temperature of 40.oC an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100.oC what will be the pressure of the oxygen?

50

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

= 25.6 atm1 2

21

P TP =

T

At a temperature of 40.oC an oxygen container is at a pressure of 21.5 atmospheres. If the temperature of the container is raised to 100.oC what will be the pressure of the oxygen?

P1 = 21.5 atm T1 = 40oC = 313 K

P2 = ? T2 = 100oC = 373 K

(21.5 atm)(373 K) =

313 K

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Combined Gas LawsCombined Gas Laws

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• A combination of Boyle’s and Charles’ Laws or Charles’ and Gay-Lussac’s Laws.

• 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

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A sample of hydrogen occupies 465 ml at at 0oC and 760 torr. 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:

54

Step 1. Organize the given information:

P1 = 760 torr P2 = 950 torr

V1 = 465 mL V2 = ?

T1 = 273 K T2 = 258 K

A sample of hydrogen occupies 465 ml at 0oC and 760 torr. If the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?

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1 1 2 2

1 2

PV PVT T

=

A sample of hydrogen occupies 465 ml at 0oC and 760 torr. If the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?

Step 2. Write and solve the equation for

the unknown V2.

1 1 22

2 1

V PTV

P T=

56

Step 3 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)

A sample of hydrogen occupies 465 ml at 0oC and 760 torr. If the pressure is increased to 950 torr and the temperature is decreased to –15oC, what would be the new volume?

57

Dalton’s Law ofDalton’s Law ofPartial PressuresPartial Pressures

58

John Dalton

59

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 + ….

60

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

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• 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

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Oxygen collected over water.

63

A sample of O2 was collected over water in a bottle at a temperature of 25oC when the atmospheric pressure was 760 torr. What is the pressure of the O2 alone? 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

64

A sample of nitrogen was collected over water and occupies 300. ml at 23oC and 750 torr. If the pressure is increased to 760 torr and the temperature is decreased to 0oC, what would be the new volume of the dry nitrogen? Vapor pressure of H20 @ 23oC is 21.0 torr.

0oC + 273 = 273 K23oC + 273 = 296 K

Step 1. Organize the given information, putting temperature in Kelvin and correcting for water vapor pressure:

2NP = 750 torr - 21.0 torr = 729 torr

65

A sample of nitrogen was collected over water and occupies 300. ml at 23oC and 750 torr. If the pressure is increased to 760 torr and the temperature is decreased to 0oC, what would be the new volume of the dry nitrogen? Vapor pressure of H20 @ 23oC is 21.0 torr.

Step 1. Organize the given information,

P1 = 729 torr P2 = 760 torr

V1 = 300. mL V2 = ?

T1 = 296 K T2 = 273 K

66

A sample of nitrogen was collected over water and occupies 300. ml at 23oC and 750 torr. If the pressure is increased to 760 torr and the temperature is decreased to 0oC, what would be the new volume of the dry nitrogen? Vapor pressure of H20 @ 23oC is 21.0 torr.

Step 2. Write and solve the equation for the unknown V2.

1 1 2 2

1 2

PV PVT T

= 1 1 22

2 1

V PTV

P T=

67

A sample of nitrogen was collected over water and occupies 300. ml at 23oC and 750 torr. If the pressure is increased to 760 torr and the temperature is decreased to 0oC, what would be the new volume of the dry nitrogen? Vapor pressure of H20 @ 23oC is 21.0 torr.

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

( )2

(300 ml) 729 torr (273 K)V = = 265 mL

(760 torr)(296 K)

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Avogadro’s LawAvogadro’s Law18111811

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Avogadro’s Law

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

V n∝

71

Standard Temperature Standard Temperature and Pressureand Pressure

72

Standard Temperature and Pressure

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.

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.

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75

Ultimate Combined Gas Ultimate Combined Gas LawLaw

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• A combination of Boyle’s , Charles’ and Avogadro’s Laws.

• Used when pressure, temperature, and moles change at the same time.

• Solve the equation for any one of the 8 variables

1 1 2 2

1 1 2 2

P V P V =

n T n T

77

A 8.74 L sample of He at 300 K and 9.7 atm containing 3.45 moles is heated to 310 K under a pressure of 15.0 atm. An additional 1.27 moles of He are added to the container. What is the new volume?

Step 1. Organize the given information:

P1 = 9.7 atm

V1 = 8.74 L

T1 = 300 K

n1 = 3.45 moles

P2 = 15.0 atm

V2 = ?

T2 = 310 K

n2 = 3.45 + 1.27 = 4.72

78

A 8.74 L sample of He at 300 K and 9.7 atm containing 3.45 moles is heated to 310 K under a pressure of 15.0 atm. An additional 1.27 moles of He are added to the container. What is the new volume?

Step 2. Write and solve the equation for the unknown V2.

1 1 2 2

1 1 2 2

=PV PVnT nT

1 1 2 22

2 1 1

=VPT nVPTn

79

A 8.74 L sample of He at 300 K and 9.7 atm containing 3.45 moles is heated to 310 K under a pressure of 15.0 atm. An additional 1.27 moles of He are added to the container. What is the new volume?

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

1 1 2 22

2 1 1

=VPT nVPTn

( )2

(8.74 L) 9.7 atm (310 K)(4.72 moles)V =

(15.0 atm)(300 K)(3.45 moles)

= 7.99 L

80

AVOGADRO'S LAWAVOGADRO'S LAW

Explained Gay Lussac's

Law of Combining Volumes

Explained Gay Lussac's

Law of Combining VolumesProvided a method

for

Provided a method

for

Served as a foundation

for the devolopment of the

Kinetic-Molecular Theory

Served as a foundation

for the devolopment of the

Kinetic-Molecular Theory

the determination of

mole weights of gases

the determination of

mole weights of gasescomparing densities of gases

of known mole weight

comparing densities of gases

of known mole weight

81

Mole-Mass-Volume Mole-Mass-Volume RelationshipsRelationships

82

Density of GasesDensity of Gases

83

massdensity =

volume

84

Density of Gases

md =

vliters

grams

85

Density of Gases

md =

vdepends

on T and P

86

The density of neon at STP is 0.900 g/L. What is the mole weight of neon?

g = 20.2

mol

0.900 g1 L

⎛ ⎞⎜ ⎟⎝ ⎠

22.4 L1 mol

⎛ ⎞⎜ ⎟⎝ ⎠

87

64.07 gd =

mol⎛ ⎞⎜ ⎟⎝ ⎠

1 mol22.4 L⎛ ⎞⎜ ⎟⎝ ⎠

g = 2.86

L

The mole weight 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

88

Ideal Gas EquationIdeal Gas Equation

89

V PnTnRT

V = P

PV = nRT

90

V PnTnRT

V = P

PV = nRT

atmospheres

91

V PnTnRT

V = P

PV = nRT

liters

92

V PnTnRT

V = P

PV = nRT

moles

93

V PnTnRT

V = P

PV = nRT

Kelvin

94

V PnTnRT

V = P

PV = nRT

Ideal Gas Constant

L-atm0.0821

mol-K

95

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

96

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

97

Mole-Weight CalculationsMole-Weight Calculations

98

Determination of Molecular Weights Using the Ideal Gas Equation

gmole weight =

mol

gRTM =

PV

gmol =

mole weight

M = mole weightg

n = mol = M

PV = nRTg

PV = RTM

99

Calculate the mole weight 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

100

Gas StoichiometryGas Stoichiometry

101

Definition

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

102

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

gases.

• A gas not at STP is converted to STP.

103

Gas Stoichiometry

Primary conversions involved in stoichiometry.

104

Mole-Volume CalculationsMole-Volume Calculations

Mass-Volume CalculationsMass-Volume Calculations

105

• 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?

106

2

3

3 mol O2 mol KClO

⎛ ⎞⎜ ⎟⎝ ⎠

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

3(0.500 mol KClO )

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

• 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

107

The problem can also be solved in one continuous calculation.

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

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

108

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.oC 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

109

• 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.

What volume of hydrogen, collected at 30.oC and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?

700 torr1 atm

760 torr⎛ ⎞⎜ ⎟⎝ ⎠

= 0.921 atm

110

What volume of hydrogen, collected at 30.oC and 700. torr, will be formed by reacting 50.0 g of aluminum with hydrochloric acid?

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

111

Volume-Volume Volume-Volume CalculationsCalculations

112

Gay Lussac’s Law of Combining Volumes 1809

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.

113

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

1 volume 1 volume 2 volumes

Y volume Y volume 2Y volumes

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

114

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

115

Molecular Formula Molecular Formula CalculationsCalculations

116

An unknown liquid contains 14.3% H and 85.7% C by weight. 29.4 grams were vaporized in a 3.60 liter container at 260C and 2.84 atm. What is the molecular formula of the unknown liquid?

Step 1. Determine the Mole Weight:

gRTM =

PV

117

An unknown liquid contains 14.3% H and 85.7% C by weight. 29.4 grams were vaporized in a 3.60 liter container at 260C and 2.84 atm. What is the molecular formula of the unknown liquid?

mass = 29.4 grams V = 3.60 LT = 260 + 273 = 533 K P = 2.84 atm

gRTM =

PV(29.4 g)

M =

(0.082 L ? atm/mol ? K)

(533 K)

(3.60 L)

(2.84 atm)

g = 126

mol

118

An unknown liquid contains 14.3% H and 85.7% C by weight. 29.4 grams were vaporized in a 3.60 liter container at 260C and 2.84 atm. What is the molecular formula of the unknown liquid?

Step 2. Use the mole weight and

percentages to find the formula:

126g comp'd 85.7g C 1 moleC9moleC

1molecomp'd 100g comp'd 12.0g

126g comp'd 14.3g H 1 mole H18mole H

1molecomp'd 100g comp'd 1.0g

× × =

× × =C9H18

119

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

120

Effusion

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

122

Graham’s Law of Effusion/Diffusion

The rates of effusion or diffusion of two gases at the same temperature and pressure are inversely proportional to the square roots of their mole weights.

rate of diffusion of gas Arate of diffusion of gas B

mole weight B=

mole weight A

123

2

diffusion rate COdiffusion rate CO

2mole weight CO=

mole weight CO44.0 g

= 1.2528.0 g

=

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

124

It takes 30.0 sec for 10. ml of O2 to travel down a 100. cm glass tube. At the same temperature & pressure, it takes 45.0 sec for 10. ml of an unknown gas to travel the same distance. Find the molecular weight of the unknown gas.

distance 100. cmrate = =

time time

2

100. cmunk.30.0 sec2

100. cm45.0 sec O

MWrate O= =

rate unk. MW

125

It takes 30.0 sec for 10. ml of O2 to travel down a 100. cm glass tube. At the same temperature & pressure, it takes 45.0 sec for 10. ml of an unknown gas to travel the same distance. Find the molecular weight of the unknown gas.

( )( )

2

unk.2

45.0 MW =

3230.0

( )( )( )

2g

unk. mole2

32 45.0MW = = 72

30.0

126

Real GasesReal Gases

127

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.

128

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.

129