Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 1 Chemistry: A Molecular Approach,...

Tro Chapter 4 + Burns 4/e Chapter 12

+ Timberlake Chapter 8

Slide 1

Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro

Chapter 5

Gases•Atmospheric Pressure, Pressure Units•Boyle’s Law: Gas Pressure and Volume•Charles’ Law: Gas Volume and Temperature•Avogadro’s Law: Gas Volume and Moles•Gay-Lussac’s Law: Gas Pressure and Temperature•Standard Temperature and Pressure•The Combined Gas Law•The Ideal Gas Law•Molar Volume and Gas Density at STP•Dalton’s Law of Partial Pressures•Gas Stoichiometry•The Kinetic Molecular Theory•Mean Free Path, Diffusion, and Effusion of Gasses•Real Gasses: The effects of size and intermolecular forces



Slide 2

Elements that exist as gases at 250C and 1 atmosphere



Slide 3

The Atmosphere

• The atmosphere is the thin layer of gases that surround the earth.

• Air is composed of a mixture of gases:– 78 % Nitrogen, N2

– 21 % Oxygen, O2

– 1 % Argon, Ar– 365 ppm carbon dioxide, CO2

– 0-4 % water, H2O

Dry Air



Slide 4

Physical Properties of Gases• No definite shape or volume:

– expand to fill container, take shape of container.

• Compressible– increase pressure, decrease volume.

• Low Density– air at room temperature and pressure:

0.00117 g/cm3.

• Exert uniform pressure on walls of container.

• Mix spontaneously and completely.– diffusion



Slide 5

Properties of Gases

• There are four basic properties that describe a gas.– Volume (V) – The space occupied by the gas.– Pressure (P) – The force that the gas exerts on

the walls of the container.– Temperature (T) – A measure of the kinetic

energy and rate of motion of a gas.– Amount (n) – The quantity of the present in the

container (moles or grams).



Slide 6

Pressure• Pressure is the force exerted per unit area:

• Pressure = Force/Area

• Atmospheric Pressure is the force per unit area exerted by the earth’s atmosphere.

• Atmospheric Pressure is measured with a barometer.

• Pressure can be measured by the height of a column of mercury that can be supported by a gas.



Slide 7

Barometer• A barometer measures

the pressure that is exerted by the atmosphere around us.

• The atmospheric pressure is measured as the height of a column of mercury, or sometimes as the height of a column of water.



Slide 8

Pressure Units• mm of mercury (mm Hg)• 1 mm Hg = 1 torr• 760 mm Hg = 1 atm• 1 atm is 1 atmosphere of pressure, sometimes

called standard pressure.• 1 Pascal, Pa, is the SI unit of pressure• 1 Pa = 1 N/m2 = 9.9 x 10-6 atm • Inches of mercury, similar to mm of mercury, is

the height of a column of mercury = 29.92 in Hg



Slide 9

Common Units of PressureUnit Average Air Pressure at

Sea Level

pascal (Pa), 101,325

kilopascal (kPa) 101.325

atmosphere (atm) 1 (exactly)

millimeters of mercury (mmHg) 760 (exactly)

inches of mercury (inHg) 29.92

torr (torr) 760 (exactly)

pounds per square inch (psi, lbs./in2) 14.7

2m

N 1 Pa 1

Example 5.1 – A high-performance bicycle tire has a pressure of 132 psi. What is the pressure in mmHg?

since mmHg are smaller than psi, the answer makes sense

1 atm = 14.7 psi, 1 atm = 760 mmHg

132 psi

mmHg

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

psi 14.7

atm 1

mmHg 10.826 atm 1

mmHg 760

psi 14.7

atm 1psi 132 3

atm 1

mmHg 760

psi atm mmHg



Slide 11

Kinetic Theory of Gases• 1. Gases move continuously, rapidly,

randomly in straight lines and in all directions.

• 2. Gas particles are extremely tiny and distances between them are great.

• 3. Gravitational forces and forces between molecules are negligible.

• 4. Collisions between gas molecules are elastic (no loss of energy in collision).– like billiard balls



Slide 12

Kinetic Energy• 5. The average kinetic energy of the gas particles

(molecules or atoms) is the same for all gases at the same temperature.– Kinetic Energy, K. E., is proportional to the

Kelvin temperature.• K. E. = ½mv2

– m = mass of the gas particle– v = velocity of particle– When temperature increases velocity of particles

increases.– At a fixed temperature, lighter particle move faster.



Slide 13

As P (h) increases V decreases

Apparatus for Studying the Relationship between Pressure and Volume of a Gas



Slide 14

Boyle’s Law• As the pressure of a gas is increased the

volume decreases:• V P

– This is an inverse proportion:

• V= k/P• PV = k

– P1V1 = k– P2V2 = k– P1V1 = P2V2

– V2 = P1V1/P2, P2 = P1V1/ V2



Slide 15

Insert figure 12.9

Boyle’s Law Graph

•Constant temperature

•Constant amount of gas



Slide 16

Inverse Volume vs Pressure of Air, Boyle's Expt.

0

20

40

60

80

100

120

140

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09

Inv. Volume, in-3

Pre

ss

ure

, in

Hg



Slide 17

Relation of Volume and Pressure

• As the container volume decreases at constant temperature, the smaller volume has shorter distances between gas molecules and the walls, so collisions are more frequent. Hence, the pressure increases at lower volume.

P1 ∙ V1 = P2 ∙ V2

Example 5.2 – A cylinder with a movable piston has a volume of 7.25 L at 4.52 atm. What is the volume at 1.21 atm?

since P and V are inversely proportional, when the pressure decreases ~4x, the volume should increase ~4x, and it does

V1 =7.25 L, P1 = 4.52 atm, P2 = 1.21 atm

V2, L

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

2

112 P

VPV

V1, P1, P2 V2

L 1.27

atm 1.21

L 7.25atm 4.52

P

VPV

2

112



Slide 19

A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?



Slide 20

Practice – A balloon is put in a bell jar and the pressure is reduced from 782 torr to 0.500 atm. If the volume of the balloon is now 2780 mL, what was it originally?



Slide 21

CHARLES’S LAW

• V T– Volume is directly

proportional to Kelvin Temperature.

• V= kT • V/T = k

• In a closed system at constant pressure, if you change the temperature, what will happen to the volume?



Slide 22

CHARLES’S LAW (CONT.)

'2

2 kT

V'

1

1 kT

V

1

212 T

TVV

1

122 V

TVT

2

2

1

1

T

V

T

V



Slide 23

Insert figure 12.11

Charles’s Law Graph

Temperature must be in Kelvin

T (K) = t (0C) + 273.15



Slide 24

Relation of Volume and Temperature

• As the temperature increases, the most probable molecular speed and average kinetic energy increase. Thus the molecules hit the walls more frequently and more energetically. If the pressure is to remain constant, the volume of the container must increase.

T(K) = t(°C) + 273.15, 2

2

1

1

T

V

T

V

Example 5.3 – A gas has a volume of 2.57 L at 0.00°C. What was the temperature at 2.80 L?

since T and V are directly proportional, when the volume decreases, the temperature should decrease, and it does

V1 =2.57 L, V2 = 2.80 L, t2 = 0.00°C

t1, K and °C

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

2

121 V

VTT

V1, V2, T2 T1

K 6.729

L 2.80

L 2.57K 273.15

V

VTT

2

121

K 273.15T

273.150.00T

2

2

C 42t

273.156.729t

273.15Tt

1

1

11



Slide 26

Practice – The temperature inside a balloon is raised from 25.0°C to 250.0°C. If the volume of cold air was 10.0 L, what is the volume of hot air?



Slide 27

A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant?



Slide 28

Gay-Lussac’s Law

• At constant volume, the pressure exerted by a gas is directly proportional to the Kelvin temperature:– P T

– P = kT

– k = P/T

– P1/T1 = P2/T2

– P2 = P1T2/T1

– T2 = T1P2/P1



Slide 29

Relation of Pressure and Temperature

• As the temperature increases, the most probable molecular speed and average kinetic energy increase. Thus the molecules hit the walls more frequently and more energetically. A higher frequency of collisions causes higher internal pressure.



Slide 30

A gas has a pressure of 2 atm at 18 oC. What is the new pressure when the temperature is 62 oC (volume and amount constant)?



Slide 31

Combined Gas Equation

Charles’ law: V T(at constant n and P)

Boyle’s law: V (at constant n and T)1P

2

22

1

11

T

VP

T

VP

Gay-Lussac’s Law: P T (at constant n and V)

kT

PVor

P

TV



Slide 32

A sample of Helium gas has a volume of 0.180 L, a pressure of 0.800atm and a temperature of 29 oC. At what temperature will the sample have a volume of 90mL and a pressure of 3.20 atm (n constant)?



Slide 33

Avogadro’s Law• The volume of a gas at

constant temperature and pressure is proportional to the number of moles of gas:– V n

– V = kn

– V1/n1 = V2/n2

– V2 = V1n2/n1Constant temperatureConstant pressure



Slide 34

Avogadro’s Law• volume directly proportional to

the number of gas molecules– V = constant x n

– constant P and T

– more gas molecules = larger volume

• count number of gas molecules by moles

• equal volumes of gases contain equal numbers of molecules– the gas doesn’t matter

2

2

1

1

n

V

n

V

mol added = n2 – n1, 2

2

1

1

n

V

n

V

Example 5.4 – A 0.225 mol sample of He has a volume of 4.65 L. How many moles must be added to give 6.48 L?

since n and V are directly proportional, when the volume increases, the moles should increase, and it does

V1 =4.65 L, V2 = 6.48 L, n1 = 0.225 mol

n2, and added moles

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

21

21 n

V

Vn

V1, V2, n1 n2

mol 314.0

L 4.65

L 6.48mol 0.225

V

Vnn

1

212

mol 089.0added moles

225.0314.0added moles



Slide 36

If 0.75 moles of Helium gas occupies a volume of 1.5 L, what volume will 1.2 moles of Helium occupy at the same temperature and pressure?



Slide 37

Standard Temperature and Pressure

• Reference conditions for gases are called standard conditions.

• Standard Temperature is 273 K or 0oC.• Standard Pressure is 1 atm or 760 torr.• Together 273 K and 1 atm is called:• Standard Temperature and Pressure• or STP.



Slide 38

Standard Molar Volume

• Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.

• This is called the standard molar volume.

• The volume of any gas at STP can be calculated if the number of moles is known:

• V = (moles) x 22.4



Slide 39

Molar Volume



Slide 40

What is the volume at STP of 4.00 grams of CH4?



Slide 41

Density at Standard Conditions

• density is the ratio of mass-to-volume

• density of a gas is generally given in g/L

• the mass of 1 mole = molar mass

• the volume of 1 mole at STP = 22.4 L

L 22.4

g Mass,Molar Density



Slide 42

DENSITY PROBLEM

• Calculate the density of CH4 at STP

• Assume 1 mole of CH4. The mass of one mol is the molar mass C + H4 = 12 + 4x1= 16 g/mol.

• V= 22.4 L (the molar volume at STP)

• density = mass/volume = 16/22.4=

• 0.714 g/L



Slide 43

Ideal Gas Equation

Charles’ law: V T(at constant n and P)

Avogadro’s law: V n(at constant P and T)

Boyle’s law: V (at constant n and T)1P

V nT

PR is the gas constant

PV = nRT

R = 0.082057 L • atm / (mol • K)

RnT

PV



Slide 44

IDEAL GAS LAWPV = nRT

– R= GAS CONSTANT, 0.0821 L Atm/mol K

– P = PRESSURE (in atm)

– V= VOLUME (in L)

– n = MOLES OF GAS

– T= TEMPERATURE (in K)

• Be consistent with units!



Slide 45

Deriving a Gas Constant

R = 0.082057 L • atm / (mol • K)

PV = nRT

At STP,P = 1 atm n = 1molV = 22.4 L T = 0 oC = 273 K

Kmoles

atmL

Kmole

Latm

nT

PVR

082057.02731

4.221

1 atm = 14.7 psi

T(K) = t(°C) + 273.15Kmol

Latm 0.08206 R nRT,PV

Example 5.6 – How many moles of gas are in a basketball with total pressure 24.3 psi, volume of 3.24 L at 25°C?

1 mole at STP occupies 22.4 L, since there is a much smaller volume than 22.4 L, we expect less than 1 mole of gas

V = 3.24 L, P = 24.3 psi, t = 25 °C,

n, mol

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

RT

PV n

P, V, T, R n

mol 219.0

K 9820.08206

L 24.3atm 3151.6TR

VPn

Kmol

Latm

atm 3151.6psi 14.7

atm 1psi 24.3

K 298T

273.15C25 T(K)



Slide 47

What is the volume (in liters) occupied by 49.8 g of HCl at STP?



Slide 48

Dinitrogen oxide (N2O), laughing gas, is used by dentists as an anesthetic. If a 20 L tank of laughing gas contains 2.8 moles of N2O at 23oC, what is the pressure (in mmHg) of the gas?



Slide 49

Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?



Slide 50

Dalton’s Law of Partial Pressure

• Dalton’s Law of Partial Pressure states that the total pressure of a mixture of gases is equal to the sum of the partial pressures of the gases:

• PT = P1 + P2 + P3 …

• Partial pressure is the pressure the gas would exert in the same volume in the absence of other gases.



Slide 51

V and T are

constant

P1 P2 Ptotal = P1 + P2

Dalton’s Law of Partial Pressures



Slide 52

The partial pressure of each gas in a mixture can be calculated using the ideal gas law

V

T x R x n P P P

n n n

same theare mixture in the

everything of volumeand re temperatutheV

T x R x n P

V

T x R x n P

togethermixed B, andA gases, for two

totalBAtotal

BAtotal

BB

AA



Slide 53

Practice – Find the partial pressure of neon in a mixture with total pressure 3.9 atm, volume 8.7 L, temperature 598 K, and 0.17 moles Xe.



Slide 55

Consider a case in which two gases, A and B, are in a container of volume V.

PA = nART

V

PB = nBRT

V

nA is the number of moles of A

nB is the number of moles of B

PT = PA + PB XA = nA

nA + nB

XB = nB

nA + nB

PA = XA PT PB = XB PT

Pi = Xi PTmole fraction,



Slide 56

A sample of natural gas contains 8.24 moles of CH4, 0.421 moles of C2H6, and 0.116 moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)?



Slide 57

Gases are often collected over water

• Dalton’s Law of Partial Pressure is often used to correct for the vapor pressure of water, which is a function of temperature but not volume or amount. The vapor pressure of water can be looked up in standard reference books such as the Handbook of Chemistry and Physics.



Slide 58

Bottle full of oxygen gas and water vapor



Slide 59

Vapor Pressure Problem

• PT = Pgas + Pwater

• Pgas = PT - Pwater

• At 27oC, the vapor pressure of water is 26.74 mm Hg

• The pressure of dry gas is Pgas=760-26.74=733 mm Hg

A gas is collected over water at 300K (27 oC), at 1.00 atm(760 Torr). Calculate the pressure of the dry gas.



Slide 60

Practice – 0.12 moles of H2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.



Slide 61

Gas Volume Stoichiometry

• Do stoichiometry problems using gas laws.

• Law of Combining Volumes: In chemical reactions, volumes of gases combine in small whole number ratios.

• The ratio of combination of volumes follow the moles This puts this unit together with the stoichiometry unit.



Slide 62

Reactions Involving Gases• the principles of reaction stoichiometry from Chapter 4

can be combined with the gas laws for reactions involving gases

• in reactions of gases, the amount of a gas is often given as a volume– instead of moles– as we’ve seen, must state pressure and temperature

• the ideal gas law allows us to convert from the volume of the gas to moles; then we can use the coefficients in the equation as a mole ratio

• when gases are at STP, use 1 mol = 22.4 L

P, V, T of Gas A mole A mole B P, V, T of Gas B



Slide 63

Gas Stoichiometry

What is the volume of CO2 produced at 370 C and 1.00 atm when 5.60 g of glucose are used up in the reaction:

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)

g C6H12O6 mol C6H12O6 mol CO2 V CO2

5.60 g C6H12O6

1 mol C6H12O6

180 g C6H12O6

x6 mol CO2

1 mol C6H12O6

x = 0.187 mol CO2

V = nRT

P

0.187 mol x 0.0821 x 310.15 KL•atmmol•K

1.00 atm= = 4.76 L



Slide 64

Practice – What volume of O2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO?

2 HgO(s) 2 Hg(l) + O2(g)(MW HgO = 216.59 g/mol)



Slide 65

Graham’s Law: Diffusion and Effusion of Gases

• Diffusion the process whereby a gas spreads out through another gas to occupy the space with uniform partial pressure.

• Effusion the process in which a gas flows through a small hole in a container.

• Graham’s law of Effusion the rate of effusion of gas molecules through a hole is inversely proportional to the square root of the molecular mass of the gas at constant temperature and pressure.

Rate=k

MW



Slide 66

Effusion



Slide 67


• When comparing the effusion (or diffusion) rates for two different gases:

1

2

32

31

effusion rate for gas 1 RMS velocity for gas 1

effusion rate for gas 2 RMS velocity for gas 2

RTM

RTM

M

M

1 gas

2 gas

2 gas

1 gas

MassMolar

MassMolar

rate

rate



Slide 68


E.g. determine the molecular mass of an unknown compound if it effused through a small orifice if it effused 3.55 times slower than CH4.

Ex 5.15 – Calculate the molar mass of a gas that effuses at a rate 0.462 times N2

MM, g/mol

Solution:

Concept Plan:

Relationships:

Given:

Find:

rateA/rateB, MMN2 MMunknown

462.0rate

rate

2N

gasunknown

N2 = 28.01 g/mol A gas

B gas

B gas

A gas

MassMolar

MassMolar

rate

rate

2

N

unknown

Nunknown

2

2

rate

rate

MassMolar MassMolar

mol

g2

molg

2

N

unknown

Nunknown 131

0.462

01.28

rate

rate

MassMolar MassMolar

2

2



Slide 70

Ideal vs. Non-Ideal Gases

• Kinetic Theory Assumptions– Point Mass– No Forces Between Molecules– Molecules Exert Pressure Via Elastic Collisions

With Walls

(courtesy F. Remer)

Ideal vs. Real Gases• Real gases often do not behave like ideal

gases at high pressure or low temperature

• at low temperatures and high pressures these assumptions are not valid



Slide 72


• Non-Ideal Gas– Violates Assumptions

• Volume of molecules• Attractive forces of molecules

(courtesy F. Remer)



Slide 73

Real Gas Behavior

• because real molecules take up space, the molar volume of a real gas is larger than predicted by the ideal gas law at high pressures



Slide 74

The Effect of Molecular Volume• at high pressure, the amount of space occupied by the

molecules is a significant amount of the total volume• the molecular volume makes the real volume larger

than the ideal gas law would predict• van der Waals modified the ideal gas equation to

account for the molecular volume– b is called a van der Waals constant and is different for

every gas because their molecules are different sizes

bnP

nRTV



Slide 75

Real Gas Behavior

• because real molecules attract each other, the molar volume of a real gas is smaller than predicted by the ideal gas law at low temperatures



Slide 76

The Effect of Intermolecular Attractions

• at low temperature, the attractions between the molecules is significant

• the intermolecular attractions makes the real pressure less than the ideal gas law would predict

• van der Waals modified the ideal gas equation to account for the intermolecular attractions– a is called a van der Waals constant and is different for

every gas because their molecules are different sizes2

V

n

V

nRTP

a



Slide 77


• Van der Waals Equation Accounts for– Volume of molecules– Attractive forces between molecules

TnRnbVV

anp *

2

2

a = constanta = constant

b = constantb = constant

(courtesy F. Remer)

combining the equations to account for molecular volume and intermolecular attractions we get the following equation



Slide 78

Van der Waals’ Equation

• used for real gases

• a and b are called van der Waal constants and are different for each gas

nRTn-VV

nP

2

ba



Slide 79

Air Pollution• air pollution is materials added to the atmosphere that

would not be present in the air without, or are increased by, man’s activities– though many of the “pollutant” gases have natural sources as

well• pollution added to the troposphere has a direct effect on

human health and the materials we use because we come in contact with it– and the air mixing in the troposphere means that we all get a

smell of it!• pollution added to the stratosphere may have indirect

effects on human health caused by depletion of ozone– and the lack of mixing and weather in the stratosphere means

that pollutants last longer before “washing” out



Slide 80

Pollutant Gases, SOx

• SO2 and SO3, oxides of sulfur, come from coal combustion in power plants and metal refining– as well as volcanoes

• lung and eye irritants

• major contributor to acid rain

2 SO2 + O2 + 2 H2O 2 H2SO4

SO3 + H2O H2SO4



Slide 81

Pollutant Gases, NOx

• NO and NO2, oxides of nitrogen, come from burning of fossil fuels in cars, trucks, and power plants– as well as lightning storms

• NO2 causes the brown haze seen in some cities• lung and eye irritants• strong oxidizers• major contributor to acid rain

4 NO + 3 O2 + 2 H2O 4 HNO3

4 NO2 + O2 + 2 H2O 4 HNO3



Slide 82

Pollutant Gases, CO

• CO comes from incomplete burning of fossil fuels in cars, trucks, and power plants

• adheres to hemoglobin in your red blood cells, depleting your ability to acquire O2

• at high levels can cause sensory impairment, stupor, unconsciousness, or death



Slide 83

Pollutant Gases, O3

• ozone pollution comes from other pollutant gases reacting in the presence of sunlight– as well as lightning storms– known as photochemical smog and ground-level

ozone

• O3 is present in the brown haze seen in some cities

• lung and eye irritants• strong oxidizer



Slide 84

Ozone Holes

• satellite data over the past 3 decades reveals a marked drop in ozone concentration over certain regions



Slide 85

Homework

• You should examine and be able to answer all of the ‘Problems’…some of them (or similar) may be on the test

• To be handed in for grading: 5.30, 5.36, 3.39, 5.46, 5.52, 5.58, 5.62, 5.67, 5.74, 5.77, 5.80, 5.96, 5.100

• Bonus: 5.100

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 1 Chemistry: A Molecular Approach,...

Documents

Transcript of Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 1 Chemistry: A Molecular Approach,...