GasesGasesChapter 13Chapter 13

Solids, liquids, and GasesSolids, liquids, and Gases

Chapter 13 2

Compare the position and motion of the Compare the position and motion of the three states of matter.three states of matter.

EnergyEnergy• Potential Energy

– Stored energy – due to position

– Particles are attracted to one another. More energy is required to keep particles farther apart.

– Which of the 3 states has the highest potential energy?

Chapter 13 3

• Kinetic Energy– Motion energy – related to temperature

– The faster the particles are moving, the higher the kinetic energy, the higher the temperature (average kinetic energy)

– Which of the 3 states has the highest kinetic energy?

Chapter 13 4

Kinetic-Molecular TheoryKinetic-Molecular Theory- Theory developed to explain gas behavior- To describe the behavior of a gas, we must first

describe what a gas is:– Gases consist of very small particles each of which have a

mass.

– The distance between gas particles are relatively large. Volume of individual molecules is negligible compared to volume of container.

– Gas particles in rapid, constant, random motion.

Chapter 13 5

– Collisions between gas particles are perfectly elastic.

Energy can be transferred between molecules, but total kinetic energy is constant at constant temperature.

Kinetic-Molecular Theory (ContKinetic-Molecular Theory (Cont’’d)d)

No energy is lost during collisions.

Chapter 13 6

Kinetic-Molecular Theory (ContKinetic-Molecular Theory (Cont’’d)d)The average kinetic energy of gas particles depends only on the temperature of the gas.

Gas particles exert no forces on each other. Intermolecular forces (forces between gas molecules) are negligible.

What happens as temperature increases?

Which statements in KMT are assumptions?

Chapter 13 7

LetLet’’s generate some gass generate some gas

- Expand to fill a volume (expandability)- Compressible- Takes shape of container- Diffuses and flows

Properties of a gasProperties of a gas

Chapter 13 8

Variables that can be measured for gasesVariables that can be measured for gases

– Temperature

– Volume

– Amount

– Pressure

Temperature (T)Temperature (T)

• Measured in Fahrenheit, Celsius, or Kelvin.• For this chapter, we have to use Kelvin.

◦C = K – 273 K = ◦C + 273

Chapter 13 9

Volume (V)Volume (V)

• Measured in Liters, cubic meters, gallons, etc…

Amount (n)Amount (n)

• Measured in moles

Chapter 13 10

Pressure (P) – Pressure (P) – what causes pressure?what causes pressure?

1 atm = 760 mmHg = 760 torr = 101.3 kPa

The pressure of a gas is measured using a manometer.

*Atomspheric

pressure is

measured

using a

Barometer.

Unit ConversionsUnit Conversions

22◦C = ______ K 37 K = ______ ◦C

18 mL = ______ L 4.3 L = ______ cm3

500 cm3= ______ mL 2.2 dm3 = _______ L

Chapter 13 11

Unit ConversionsUnit Conversions

2.8 g N2 = ______ mol 612 mol SO3 = _______ g

22 kPa = ______ atm 289 mmHg = _______ kPa

4.3 atm = ______ torr 518 kPa = _______ mmHg

Chapter 13 12

Chapter 13 13

The Gas LawsThe Gas Laws- There are four variables required to describe a gas:

- Amount of substance: moles (n)- Volume of substance: volume (V)- Pressures of substance: pressure (P)- Temperature of substance: temperature (T)

- The gas laws will hold two of the variables constant and see how the other two vary (n,V,P,T)

Chapter 13 14

TodayToday’’s Lab – Boyles Lab – Boyle’’s Laws Law

• We will maintain a constant temperature and number of moles of gas.

• So we will vary the Pressure and the Volume and see how they relate.

PV = k or P/V = kinverse direct

BolyleBolyle’’s Law Lab – 16 pts totals Law Lab – 16 pts total• Heading – 1 pt• Purpose – 1 pt• Procedure – 1 pt• Data – 4 pts (make sure they have units for the 3rd and 4th

columns.• Graph – 3 pts (should have a title and labeled axis)• Questions – 6 pts (in complete sentences)

1) 1/2 2) doubled 3) inverse 4) PV=k

5) The pressure and volume of a gas at a constant temperature are inversely proportional to each other.

6) Source of error.

Chapter 13 15

Chapter 13 16

Variables for Gases Discussed BeforeVariables for Gases Discussed Before

• T – Temperature• V – Volume• P – Pressure• n – Amount of a substance (moles)

Chapter 13 17

BoyleBoyle’’s Law – s Law – peeps in bell jar demo peeps in bell jar demo

The Pressure-Volume Relationship Boyle’s Law - The volume of a fixed quantity of gas is

inversely proportional to its pressure at a constant temperature.

2211

kPVor k

) and (constant 1

VPVPV

P

TnV

P

Chapter 13 18

A gas occupies 22 L at 2.43 atm. What is the new volume if the pressure changed to 5.11 atm?

Chapter 13 19

Gay-Lussac’s Law – can crush demo

Gay-Lussac’s Law – As the temperature of an enclosed gas increases, the pressure increases if the volume is constant.

The Pressure-Temperature Relationship:

)constantVandnfor(2

2

1

1

T

P

T

P

kT

PorkTPorTP

Chapter 13 20

A gas has a pressure of 1.47 atm at 303 K. What is the new temperature if the pressure changed to 680 mmHg?

Chapter 13 21

Charles Law – Charles Law – ivory soap demoivory soap demo

Charles’s Law - the volume of a fixed quantity of gas at constant pressure increases as the temperature increases.

The Temperature-Volume Relationship:

)constantPandnfor(2

2

1

1

T

V

T

V

kT

VorkTVorTV

Chapter 13 22

A gas occupies 14 L at 275 K. What is the new volume if the temperature changed to 297 K?

Chapter 13 23

AvogadroAvogadro’’s Laws Law

Avogadro’s Law - The volume of gas at a given temperature and pressure is directly proportional to the number of moles of gas.

) and (constant TPnV

The Quantity-Volume Relationship:

2

2

1

1

n

V

n

V

kn

V

Chapter 13 24

A balloon contains 1.98 mol of a gas and has a volume of 4.2 L. Some of the gas was let out to give a volume of 3.1 L. What is the amount of gas left in the balloon?

Chapter 13 25

The Gas Laws SummaryThe Gas Laws Summary

Boyle’s Law

2

2

1

1V

) and (constant

n

V

n

TPnV

2

2

1

1

)and(constant

T

V

T

V

PnTV

2211

) and (constant 1

VPVP

TnV

P

Charles’ Law

Avogadro’s Law

2

2

1

1P

) and (constant

T

P

T

VnTP

Guy-Lussac’s Law

Chapter 13 26

The Ideal Gas EquationThe Ideal Gas Equation

Ideal gas equation:

PV = nRT P = pressure (atm or mmHg or kPa)

V = volume (L) n = amount (mol)

R = gas constant T = temperature (K)

Combine the gas laws (Boyle, Charles, Guy-Lussac, Avogadro) yields a new law or equation.

Chapter 13 27

Finding R with Dry Ice Lab

nT

PVR PV = nRT

What units are used for R?

Calculating R with Dry Ice Lab – 20 pts totalCalculating R with Dry Ice Lab – 20 pts total• Heading – 1 pt

• Purpose – 1 pt

• Procedure – 1 pt

• Data – 10 pts (make sure they have units)

• Calculations – 2pts

• Conclusion – 5 pts– Their value for R

– Literature value for R

– % error

– Sources of error

– How they affected their results (higher or lower and must be consistent with their result for R)

Chapter 13 28

Chapter 13 29

Calculating RCalculating RWe define STP (standard temperature and pressure) as

0C = 273 K, and 1 atm = 760 mmHg = 101.3 kPa

Volume of 1 mol of gas at STP is 22.4 L.

nT

PVR PV = nRT

Chapter 13 30

The Ideal Gas Equation-Finding RThe Ideal Gas Equation-Finding R

Kmol

LkPa

Kmol

LkPaR

31.8

2731

4.223.101

Kmol

Latm

Kmol

LatmR

0821.02731

4.221

Kmol

LmmHg

Kmol

LmmHgR

4.62

2731

4.22760

Chapter 13 31

Combined Gas LawCombined Gas Law• So in cases with changing conditions

11

11

Tn

VPR

22

22

Tn

VPR and

22

22

11

11

Tn

VP

Tn

VP

Chapter 13 32

Ideal vs. Combined Gas Law-When UseIdeal vs. Combined Gas Law-When Use• Ideal Gas Law: Conditions not changing

• Combined Gas Law: Conditions Changing

22

22

11

11

Tn

VP

Tn

VP

nRTPV

Chapter 13 33

Example 1Example 1

• How many moles of a gas at 100 degrees C does it take to fill a 1.0 L flask to a pressure of 1.50 atm?

Not a changing situation so use Ideal Gas Law

PV=nRT

At 60 Celsius a 0.10 L sample of a gas has a pressure of 75.6 kPa. What would its volume be at STP?

Chapter 13 34

Changing situation so use combined gas law.

22

22

11

11

Tn

VP

Tn

VP

Example 2Example 2

Chapter 13 35

• What pressure would 3.55 grams of argon gas be under in a 2.40 L cylinder at -35 Celsius?

Not a changing situation so use Ideal Gas Law

PV=nRT

Example 3Example 3

Chapter 13 36

• What is the density of bromine gas the gas fills a 52.5 L cylinder at 145 K and 583 mmHg?

Not a changing situation so use Ideal Gas Law

PV=nRT

Example 4Example 4

Chapter 13 37

Example 5Example 5A gas occupies 4430 mL at 30 Celsius. It was

transferred in a 3.5 L cylinder. What is the new temperature.

Changing situation so use combined gas law.

22

22

11

11

Tn

VP

Tn

VP

Chapter 13 38

DemosDemos

Egg in E-flask

Fountain

Tank CarTank Car

• This tank car was cleaned with steam then all the valves were shut and tank car was sealed. The workers went home and when they came back the next morning this is what they saw.

Chapter 13 39

Chapter 13 40

Chapter 13 41

Chapter 13 42

Chapter 13 43

Ideal vs. Real gasesIdeal vs. Real gasesIdeal gases behave “ideally” according to the kinetic molecular theory and follow the ideal gas law PV = nRT.

But kinetic molecular theory has several assumptions that work most of the time but not always.

Chapter 13 44

Kinetic Molecular Theory Kinetic Molecular Theory AssumptionsAssumptions

1) The volume of gas particles are so small and the spaces between particles are so large.

Therefore KMT assumes that gas particles have no volume.

2) The distance between gas particles are very large.

Therefore KMT assumes that there is no attraction (IMF) between gas particles.

Chapter 13 45

But Real gases have volumeBut Real gases have volumeUnder most conditions the volume of gas particles is negligible.

To compensate, we plug in the following equation for V in PV = nRT

nbVV

But at small volumes or if the gas particles are large, the volume of the gas particles become significant.

Chapter 13 46

And real gases have attractive And real gases have attractive forces (IMF)forces (IMF)

Under most conditions, the intermolecular forces between gas particles are negligible.

To compensate for the pressure difference caused by IMF we plug in the following equation instead of P in PV = nRT

)(2

V

naPP

But at high pressure and low temperatures, the attractive forces become significant.

Chapter 13 47

Ideal vs. Real gasesIdeal vs. Real gasesWhich gas behaves more ideally?

Ne or HCl?

Neon because it’s particles have a smaller volume and weaker intermolecular forces.

Chapter 13 48

Ideal vs. Real gasesIdeal vs. Real gasesIn PV = nRT , plugging in

We get the Van der Waals Equation

)(2

V

naPP

nbVV

ExampleExample• Use the Van der Waals equation to calculate the

temperature of 3.6 moles of nitrogen gas in a 4.6 L cylinder at 2.5 atm if a and b for nitrogen are 1.390 and 0.03910 respectively.

Chapter 13 49

Chapter 13 50

Gas Mixtures and Partial PressuresGas Mixtures and Partial Pressures

Dalton’s Law - In a gas mixture the total pressure is given by the sum of partial pressures of each component:

Pt = P1 + P2 + P3 + …

- The pressure due to an individual gas is called a partial pressure.

Dalton’s Law

Chapter 13 51

Partial Pressure of Dry AirPartial Pressure of Dry Air

At top of Mt. Everest, total atmospheric pressure is 33.73 kPa (about 1/3 that of sea level). So partial pressure of O2 is 7.06 kPa. To support respiration the partial pressure must be 10.76 kPa or higher so oxygen tanks are needed.

ExampleExample

Chapter 13 52

What is the total pressure of a mixture of gases containing oxygen, hydrogen, and water vapor if their partial pressures are 120 kPa, 94 kPa, and 137 kPa respectively?

Chapter 13 53

Molecular Effusion and DiffusionMolecular Effusion and DiffusionGraham’s Law of Diffusion or Effusion

Effusion – The escape of gas through a small opening.

Diffusion – The movement of gas particles through another gas.

Chapter 13 54

Molecular Effusion and DiffusionMolecular Effusion and DiffusionGraham’s Law of Effusion or Diffusion

Chapter 13 55

Kinetic-Molecular TheoryKinetic-Molecular Theory

• As kinetic energy increases, the velocity of the gas molecules increases.

• Average kinetic energy, KE, is related to root mean square speed:

KE= ½mv2

Molecular Speed

Chapter 13 56

GrahamGraham’’s Law of Effusion or Diffusions Law of Effusion or Diffusion

• Gases at same temperature – tell me about them– They have the same kinetic energy (KE)

• Which would move faster? CO2 or O2

– O2 would move faster because KE=1/2 mv2 and O2 has the lower mass

• Thomas Graham noticed how rate of diffusion (or effusion) related to molar mass

Chapter 13 57

Molecular Effusion and DiffusionMolecular Effusion and Diffusion

Graham’s Law of Effusion – The rate of effusion of a gas is inversely proportional to the square root of its molecular mass.

Graham’s Law of Effusion or Diffusion

1

2

2

1

22

21

1

2

222

211

222

211

21

2

1

2

1

KEKE : tempsameat gases 2for

m

m

v

v

v

v

m

m

vmvm

vmvm

m1 and m2 are molar mass and v1 and v2 are effusion/diffusion rates

Chapter 13 58

Molecular Effusion and DiffusionMolecular Effusion and Diffusion

Graham’s Law of Effusion/Diffusion – The rate of effusion/diffusion of a gas is inversely proportional to the square root of its molecular mass.

Graham’s Law of Effusion/Diffusion

1

2

2

1MM

rr

ExampleExample

Chapter 13 59

Find the relative rate of diffusion for the gases krypton and bromine.

Chapter 13 60

Gas Laws & StoichiometryGas Laws & Stoichiometry

PV = nRT calculations in reactions

with molar ratios

Link?Moles!!

Chapter 13 61

Example 1Example 1- What mass of NaCl could be produced form excess

sodium and 10.0 L of chlorine gas at 23oC and 1.02 atm?

Chapter 13 62

Example 2Example 2- Calcium carbonate decomposes to form carbon

dioxide and calcium oxide. What volume of carbon dioxide at 25oC and 99.8 kPa is produced if 25.0 g of calcium carbonate is decomposed?