Chemistry: A Molecular Approach Ed. Nivaldo Tro Chapter 5

Transcript Chemistry: A Molecular Approach Ed. Nivaldo Tro Chapter 5

Chemistry: A Molecular Approach

, 1 st 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 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 1

Elements that exist as gases atmosphere at 25 0 C and 1 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 2

The Atmosphere

• The

atmosphere

is the thin layer of gases that surround the earth.

• Air is composed of a mixture of gases: – 78 % Nitrogen, N 2 – 21 % Oxygen, O 2 – 1 % Argon, Ar – 365 ppm carbon dioxide, CO 2 – 0-4 % water, H 2 O

Dry Air

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 3

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/cm 3 .

• Exert uniform pressure on walls of container.

• Mix spontaneously and completely.

– diffusion Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 4

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).

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 5

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 barometer.

is measured with a

Pressure

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

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 6

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.

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 7

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/m 2 = 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 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 8

Common Units of Pressure

Unit

pascal (Pa), 1 Pa  1 N m 2 kilopascal (kPa) atmosphere (atm) millimeters of mercury (mmHg) inches of mercury (inHg) torr (torr) pounds per square inch (psi, lbs./in 2 ) Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8

Average Air Pressure at Sea Level

101,325 101.325

1 (exactly) 760 (exactly) 29.92

760 (exactly) 14.7

Slide 9

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

Given:

132 psi

Find: Concept Plan:

mmHg

Relationships:

psi 1 atm 14.7

psi atm 760 mmHg 1 atm mmHg 1 atm = 14.7 psi, 1 atm = 760 mmHg

Solution:

132 psi  1 atm 14.7

psi  760 mmHg 1 atm  6 .82

 10 3 mmHg

Check:

since mmHg are smaller than psi, the answer makes sense

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 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 11

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.,

Kelvin temperature.

is proportional to the • K. E. = ½mv 2 – 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.

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 12

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

(h) increases

decreases Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 13

•

Boyle’s Law

• As the pressure of a gas is increased the volume decreases: V 1/ P – This is an inverse proportion: • V= k/P • PV = k – P 1 V 1 – P 2 V 2 = k = k – P 1 V 1 – V 2 = P = P 1 V 2 1 V /P 2 2 , P 2 = P 1 V 1 / V 2 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 14

Boyle’s Law Graph

Insert figure 12.9

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 •Constant temperature •Constant amount of gas Slide 15

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

140 120 100 80 60 40 20 0 0 0.01

0.02

0.03

0.04

0.05

Inv. Volume, in -3

0.06

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 0.07

0.08

0.09

Slide 16

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.

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 17

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?

Given:

V 1 =7.25 L, P 1 = 4.52 atm, P 2 = 1.21 atm

Find:

V 2 , L

Concept Plan: Relationships: Solution:

V 1 , P 1 , P 2 V 2  P 1  V 1 P 2 P 1 ∙ V 1 = P 2 ∙ V 2 V 2 V 2   P 1  V 1 P 2  4.52

atm  7.25

L 1.21

atm    27 .

1 L

Check:

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

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?

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 19

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?

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 20

CHARLES’S LAW

•

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

• V  T – Volume is directly proportional to Kelvin Temperature.

• V= kT • V/T = k Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 21

CHARLES’S LAW (CONT.)

1 

'

2 

1 

2 

1 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 22

Charles’s Law Graph

Insert figure 12.11

Temperature

must

be in Kelvin

(K) =

( 0 C) + 273.15

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 23

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. Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 24

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

Given:

V 1 =2.57 L, V 2 = 2.80 L, t 2 = 0.00°C

Find:

t 1 , K and °C

Concept Plan: Relationships:

V 1 , V 2 , T 2 T 1  T 2  V V 2 1 T(K) = t(°C) + 273.15, T 1 V 1 T 1 T 2

Solution:

 0.00

 273.15

T 2  273.15

K T 1   T 2  V 1 V 2  273.15

K  2.80

L  2.57

L   29 7 .

6 K  V 2 T 2 t t 1 1   T 1  273.15

29 7 .

6  273.15

t 1  2 4  C

Check:

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

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? Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 26

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

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 27

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 – P 1 /T 1 – P 2 = P = P 1 T 2 2 /T /T 1 2 – T 2 = T 1 P 2 /P 1 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 28

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. Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 29

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

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 30

Combined Gas Equation

Boyle’s law: V a (at constant

n P

Charles’ law:

(at constant

and

) ) Gay Lussac’s Law:

a T (at constant n and V)



T P or PV



k T P

1 

2 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 31

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

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 32

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 – V 1 /n 1 – V 2 = V = V 1 n 2 2 /n /n 1 2 Constant temperature Constant pressure Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 33

Avogadro’s Law

• volume directly proportional to the number of gas molecules –

= constant x

– 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

V 1  n 1 V 2 n 2 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 34

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?

Given:

V 1 =4.65 L, V 2 = 6.48 L, n 1 = 0.225 mol

Find:

n 2 , and added moles

Concept Plan: Relationships:

V 1 , V 2 , n 1 n 1  V 2 V 1 mol added = n 2  n 2 – n 1 , n 2 V 1 n 1  V 2 n 2

Solution:

n 2   n 1  V 1 V 2  0.225

mol    4.65

L  6.48

L  moles added  0 .

314  0 .

225 moles added  0 .

089 mol  0 .

314 mol

Check:

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

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?

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 36

Standard Temperature and Pressure

• Reference conditions for gases are called

standard conditions.

• Standard Temperature is 273 K or 0 o C.

• Standard Pressure is 1 atm or 760 torr.

• Together 273 K and 1 atm is called: •

Standard Temperature and Pressure

• or

STP.

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 37

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

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 38

Molar Volume

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 39

What is the volume at STP of 4.00 grams of CH 4 ?

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 40

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

Density



Molar Mass, 22.4

L g

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 41

DENSITY PROBLEM

•

Calculate the density of CH 4 at STP

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

• • V= 22.4 L (the molar volume at STP) • density = mass/volume = 16/22.4=

0.714 g/L

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 42

Ideal Gas Equation

Boyle’s law: V a (at constant

n P

Charles’ law:

(at constant

and

) ) Avogadro’s law: V a

(at constant

and

)

nT P PV



R nT R

is the

gas constant

nRT R

= 0.082057 L • atm / (mol • K) Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 43

IDEAL GAS LAW PV = 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!

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 44

Deriving a Gas Constant

nRT

At STP, P = 1 atm V = 22.4 L n = 1mol T = 0 o C = 273 K

R



PV nT



1 atm



22 .

4 L

1 mole



273 K



0 .

082057

L



atm moles



K

= 0.082057 L • atm / (mol • K) Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 45

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?

Given:

V = 3.24 L, P = 24.3 psi, t = 25 °C,

Find:

n, mol

Concept Plan: Relationships:

P, V, T, R 1 atm = 14.7 psi n  PV RT T(K) = t(°C) + 273.15

Solution:

24.3

psi  1 atm 14.7

psi  1.6

T(K)  25  C  273.15

T  298 K 5 31 atm n PV  nRT, R  0.08206

atm  L mol  K n   P  V R  T  1.6

5 31 atm  0.08206

atm  L mol  K   3 .

24 L  2 98 K    0 .

219 mol

Check:

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

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

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 47

Dinitrogen oxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If a 20 L tank of laughing gas contains 2.8 moles of N 2 O at 23 o C, what is the pressure (in mmHg) of the gas?

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 48

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 0 C is heated to 85 0 C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 49

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: •

P T = P 1 + P 2 + P 3 … Partial pressure

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

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 50

Dalton’s Law of Partial Pressures

and

are

constant

2 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8

total

= P

1 +

2 Slide 51

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

for two gases, A and B, mixed together P A  n A x R x T V P B  n B x R x T V the temperatu re and volume of everything in the mixture are the same n total  n A  n B P total  P A  P B  n total x R x T V Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 52

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.

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 53

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

A =

n A

V P

B =

n B

V P

T =

A +

A =

A is the number of moles of A

B is the number of moles of B

A =

A +

B =

B +

B =

P i

X i P

T Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8

mole fraction,

c Slide 55

A sample of natural gas contains 8.24 moles of CH 4 , 0.421 moles of C 2 H 6 , and 0.116 moles of C 3 H 8 . If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C 3 H 8 )?

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 56

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.

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 57

Bottle full of oxygen gas and water vapor

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 58

Vapor Pressure Problem

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

• P T = P gas + P water • P gas = P T - P water • At 27 o C, the vapor pressure of water is 26.74 mm Hg • The pressure of dry gas is P gas =760-26.74=733 mm Hg Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 59

Practice – 0.12 moles of H 2 is collected over water in a 10.0 L container at 323 K. Find the total pressure.

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 60

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.

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 61

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 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 62

Gas Stoichiometry What is the volume of CO 2 produced at 37 0 C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C 6 H 12 O 6 (

) + 6O 2 (

) 6CO 2 (

) + 6H 2 O (

) g C 6 H 12 O 6 mol C 6 H 12 O 6 mol CO 2

CO 2 5.60 g C 6 H 12 O 6 x 1 mol C 6 H 12 O 6 180 g C 6 H 12 O 6 x 6 mol CO 2 1 mol C 6 H 12 O 6 = 0.187 mol CO 2

nRT P

= L •atm 0.187 mol x 0.0821 x 310.15 K mol •K 1.00 atm = 4.76 L Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 63

Practice – What volume of O 2 at 0.750 atm and 313 K is generated by the thermolysis of 10.0 g of HgO?

2 HgO(

)  2 Hg(

) + O 2 (

) (MW HgO = 216.59 g/mol) Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 64

• • •

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

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 65

Effusion

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 66

Graham’s Law: Diffusion and Effusion of Gases

• When comparing the effusion (or diffusion) rates for two different gases: effusion rate for gas 1  effusion rate for gas 2 RMS velocity for gas 1  RMS velocity for gas 2 3

RT M

1 3

RT M

2 

rate

gas 1

rate

gas 2 

Molar Mass

gas 2

Molar Mass

gas 1 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 67

Graham’s Law: Diffusion and Effusion of Gases

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 CH 4 .

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 68

Ex 5.15 – Calculate the molar mass of a gas that

Given: Find:

effuses at a rate 0.462 times N 2 rate unknown gas rate N 2 MM, g/mol  0 .

462

Concept Plan: Relationships:

N 2 rate A /rate B , MM N2 Molar Mass unknown  = 28.01 g/mol Molar Mass N 2 2 rate unknown MM unknown rate N 2 rate gas A rate gas B  Molar Mass gas B Molar Mass gas A

Solution:

Molar Mass unknown  Molar Mass N 2 2 rate unknown rate N 2   28 .

01 g  0.462

mol  2   131 g mol

Ideal vs. Non-Ideal Gases

• Kinetic Theory Assumptions – Point Mass – No Forces Between Molecules – Molecules Exert Pressure Via Elastic Collisions With Walls Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 70 (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 71

Ideal vs. Non-Ideal Gases

• Non-Ideal Gas – Violates Assumptions • Volume of molecules • Attractive forces of molecules Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 72 (courtesy F. Remer)

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 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 73

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 –

is called a

van der Waals constant

and is different for every gas because their molecules are different sizes V  nRT P  n

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 74

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 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 75

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 –

is called a

van der Waals constant

and is different for every gas because their molecules are different sizes 2

P



nRT V



a



n V

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 76

Ideal vs. Non-Ideal Gases

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



an 2 V 2

  



 

nR * T a = constant b = constant

• Van der Waals Equation Accounts for – Volume of molecules – Attractive forces between molecules Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 77 (courtesy F. Remer)

Van der Waals’ Equation

• • used for real gases

a

and

b

are called van der Waal constants and are different for each gas   P 

n V 2     V n

  nRT Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 78

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 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 79

Pollutant Gases, SO

x • SO 2 and SO 3 , 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 SO 2 + O 2 SO 3 + 2 H 2 O  + H 2 O  H 2 2 H SO 4 2 SO 4 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 80

Pollutant Gases, NO

x • NO and NO 2 , oxides of nitrogen, come from burning of fossil fuels in cars, trucks, and power plants – as well as lightning storms • NO 2 causes the brown haze seen in some cities • lung and eye irritants • strong oxidizers • major contributor to acid rain 4 NO + 3 O 2 4 NO 2 + O 2 + 2 H 2 O  + 2 H 2 O  4 HNO 4 HNO 3 3 Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 81

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 O 2 • at high levels can cause sensory impairment, stupor, unconsciousness, or death Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 82

Pollutant Gases, O

3 • 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 • O 3 is present in the brown haze seen in some cities • lung and eye irritants • strong oxidizer Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 83

Ozone Holes

• satellite data over the past 3 decades reveals a marked drop in ozone concentration over certain regions Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 84

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: 5.46, 5.52, 5.58, 5.62, 5.67, 5.74, 5.77, 5.80, 5.96,

5.100

5.30, 5.36, 3.39, • Bonus:

5.100

Tro Chapter 4 + Burns 4/e Chapter 12 + Timberlake Chapter 8 Slide 85