Warm-Up

1

Warm-UpWarm-Up What does the motion of gas What does the motion of gas

molecules look like?molecules look like?

Why does a balloon inflate when you Why does a balloon inflate when you blow it up? Why will soda explode blow it up? Why will soda explode from a bottle if opened after shaking from a bottle if opened after shaking it?it?

2

Chapter 5Chapter 5

The Gas The Gas LawsLaws

3

Section 5.1- PressureSection 5.1- Pressure Force per unit area (P = force/area).Force per unit area (P = force/area). Gas molecules fill container.Gas molecules fill container. Molecules move around and hit Molecules move around and hit

sides.sides. Collisions are the force.Collisions are the force. Container is the area.Container is the area. Measured with a barometer.Measured with a barometer.

4

How Does A Barometer Work?How Does A Barometer Work?

The pressure of the The pressure of the atmosphere at sea atmosphere at sea level will cause the level will cause the column of mercury to column of mercury to rise to 760 mm Hg.rise to 760 mm Hg.

1 atm = 760 mm Hg1 atm = 760 mm Hg

Pressure of atmosphere pushes on

Hg

760 mm Hg

Vacuum

As a result, Hg rises up

into the glass tube

*Hg stops rising when it’s equal to atmospheric pressure

5

Units of pressureUnits of pressure 1 atmosphere = 760 mm Hg1 atmosphere = 760 mm Hg 1 mm Hg = 1 torr1 mm Hg = 1 torr 1 atm = 101,325 Pascals = 101.325 kPa1 atm = 101,325 Pascals = 101.325 kPa

*The first two are provided on the AP *The first two are provided on the AP equation sheet. No need to memorize equation sheet. No need to memorize the third- I assume you’ll be given that the third- I assume you’ll be given that if you need to use it.if you need to use it.

6

THE GAS LAWS OF BOYLE, THE GAS LAWS OF BOYLE, CHARLES, AND AVOGADROCHARLES, AND AVOGADRO

Section 5.2Section 5.2

7

About the Laws…About the Laws… You should be aware of the following You should be aware of the following

laws, however we will not focus laws, however we will not focus heavily on them as they can be heavily on them as they can be derived from the ideal gas law.derived from the ideal gas law.

After briefly going through each of After briefly going through each of the following laws, we will see how the following laws, we will see how to derive each from the ideal gas law.to derive each from the ideal gas law.

8

BoyleBoyle’’s Laws Law Pressure and volume are Pressure and volume are inverselyinversely related at related at

constant temperature.constant temperature. PP11VV11 = P = P22VV22

• As one goes up, the other goes down.As one goes up, the other goes down. Ex: if P increases (at constant T), V must go Ex: if P increases (at constant T), V must go

down down Further studies show that BoyleFurther studies show that Boyle’’s Law is only s Law is only

true at very low Ptrue at very low P• This will be discussed more in 5.8This will be discussed more in 5.8

Gases that obey these laws are called Gases that obey these laws are called ideal ideal gasesgases..

9

CharlesCharles’’s Laws Law Volume of a gas varies Volume of a gas varies directlydirectly with with

the temperature at constant the temperature at constant pressure.pressure.

VV1 1 VV22

TT1 1 T T22 As one goes up/down, so does the As one goes up/down, so does the

other.other.

==

10

Avogadro's LawAvogadro's Law At constant temperature and At constant temperature and

pressure, the volume of gas is pressure, the volume of gas is directly related to the number of directly related to the number of moles.moles.

VV1 1 VV22

nn1 1 n n22

As one goes up/down, so does the As one goes up/down, so does the other.other.

==

11

Gay- Lussac LawGay- Lussac Law At constant volume, pressure and At constant volume, pressure and

temperature are directly related.temperature are directly related.

PP1 1 PP22

TT11 T T22

As one goes up/down, so does the As one goes up/down, so does the

other.other.

==

12

Combined Gas LawCombined Gas Law Combination of BoyleCombination of Boyle’’s Law, Charless Law, Charles’’

Law, and Gay-Lussac Law.Law, and Gay-Lussac Law. Moles of gas remain constant.Moles of gas remain constant.

PP11VV1 1 PP22VV22

TT11 T T22

= =

13

SummarySummary Boyle’s: PBoyle’s: P11VV11 = P = P22VV22

Charles’: VCharles’: V11/T/T11 = V = V22/T/T22

Avogadro’s: VAvogadro’s: V11/n/n11 = V = V22/n/n22

Gay-Lussac: PGay-Lussac: P11/T/T11 = P = P22/T/T22

Combined: PCombined: P11VV11/T/T11 = P = P22VV22/T/T22

That’s a lot of laws! Or we can just That’s a lot of laws! Or we can just use the Ideal Gas Law!use the Ideal Gas Law!

14

Combined Gas Law Cont.Combined Gas Law Cont. ExEx: A 2.3L sample of gas has a pressure of : A 2.3L sample of gas has a pressure of

1.2atm at 200.K. If the pressure is raised to 1.2atm at 200.K. If the pressure is raised to 1.4atm and the temperature is increased to 300.K, 1.4atm and the temperature is increased to 300.K, what is the volume of the gas?what is the volume of the gas?

VV22 = P = P11VV11TT22

TT11PP22

VV22 = 3.0 L = 3.0 L

15

PracticePractice ExEx: A 12.2L sample of gas has 0.50mol of O: A 12.2L sample of gas has 0.50mol of O22 at 1atm at 1atm

and 25°C. How many moles of Oand 25°C. How many moles of O22 would occupy would occupy

19.4L at the same temperature and pressure? 19.4L at the same temperature and pressure?

SolutionSolution: V: V11/n/n11 = V = V22/n/n22

(12.2L)/(0.50mol) = (19.4L)/(n(12.2L)/(0.50mol) = (19.4L)/(n22) )

nn22 =0.80mol =0.80mol

*In other words, 0.80mol of O*In other words, 0.80mol of O22 would be required to would be required to

fill 19.4L in order to keep the same pressure as fill 19.4L in order to keep the same pressure as 0.50mol of O0.50mol of O22 in 12.2L. in 12.2L.

16

AP Practice QuestionAP Practice Question A sample of argon gas is sealed in a A sample of argon gas is sealed in a

container. The volume of the container. The volume of the container is doubled. If the pressure container is doubled. If the pressure remains constant, what must happen remains constant, what must happen to the temperature?to the temperature?

a)a) It doesn’t change.It doesn’t change.

b)b) It is halved.It is halved.

c)c) It is doubled.It is doubled.

d)d) It is squared.It is squared.

17

Demonstration Warm-Up!Demonstration Warm-Up! Observe the demonstration.Observe the demonstration. Keep in mind the properties of gases we Keep in mind the properties of gases we

have discussed so far: P, V, T, and n.have discussed so far: P, V, T, and n. Think about these properties before and Think about these properties before and

after imploding the can. Why do you after imploding the can. Why do you think the can was crushed?think the can was crushed?

As temperature decreases, so does the As temperature decreases, so does the pressure and volume. pressure and volume.

Remind you of a law we looked at?Remind you of a law we looked at?

18

Sections 1&2 HomeworkSections 1&2 Homework Pgs. 217-218 #: 2, 6, 34, 35Pgs. 217-218 #: 2, 6, 34, 35

19

THE IDEAL GAS LAWTHE IDEAL GAS LAWSection 5.3Section 5.3

20

Ideal Gas LawIdeal Gas Law PV = nRTPV = nRT At standard temperature and pressure (STP): V = At standard temperature and pressure (STP): V =

22.4L at 1atm, 0ºC, and n = 1mol. These conditions 22.4L at 1atm, 0ºC, and n = 1mol. These conditions were used to determine R (ideal gas constant):were used to determine R (ideal gas constant):

»R = 0.08206 L atm/mol KR = 0.08206 L atm/mol K»= 8.314 J/mol K= 8.314 J/mol K»= 62.36 L torr/mol K= 62.36 L torr/mol K

Tells you about a gas NOW.Tells you about a gas NOW. The other laws tell you about a gas when it The other laws tell you about a gas when it

changes. changes.

Choose R value according to units of P

KNOW THIS!

21

Ideal Gas Law Cont.Ideal Gas Law Cont. Looking back at the possible values Looking back at the possible values

for R, you will notice that all units for for R, you will notice that all units for temperature are in K.temperature are in K.

–When using the ideal gas law for When using the ideal gas law for calculations, convert all calculations, convert all temperatures to K!temperatures to K!

–Recall conversion: K = °C + 273 Recall conversion: K = °C + 273 (provided on AP equation sheet)(provided on AP equation sheet)

22

Ideal Gas Law Derivation PracticeIdeal Gas Law Derivation Practice May be asked to prove one of the May be asked to prove one of the

laws discussed before!laws discussed before! Strategy: get all constants in the Strategy: get all constants in the

ideal gas law on one side and ideal gas law on one side and changing variables on the other.changing variables on the other.

We will go several of these in class.We will go several of these in class.

23

AP Practice QuestionAP Practice QuestionA 1.15mol sample of carbon monoxide A 1.15mol sample of carbon monoxide gas has a temperature of 27°C and a gas has a temperature of 27°C and a pressure of 0.300atm. If the pressure of 0.300atm. If the temperature is lowered to 17°C at temperature is lowered to 17°C at constant volume, what is the new constant volume, what is the new pressure? pressure?

a) 0.290atma) 0.290atm c) 0.206atmc) 0.206atm

b) 0.519atmb) 0.519atm d) 0.338atmd) 0.338atm

24

Ideal Gas Law- Why Ideal Gas Law- Why ‘‘IdealIdeal’’?? Ideal gases are hypothetical Ideal gases are hypothetical

substances.substances.– Gases only Gases only approachapproach ideal behavior ideal behavior

at low pressure (< 1 atm) and high at low pressure (< 1 atm) and high temperature.temperature.– They do not behave exactly according They do not behave exactly according

to this law, but they behave closely to this law, but they behave closely enough.enough.– Law provides good estimates of gas Law provides good estimates of gas

behavior under these conditions.behavior under these conditions. Unless told otherwise, assume ideal Unless told otherwise, assume ideal

gas behavior and use the ideal gas law.gas behavior and use the ideal gas law.

25

AP Practice QuestionAP Practice Question A sample of aluminum metal is A sample of aluminum metal is

added to HCl. How many grams of added to HCl. How many grams of aluminum metal must be added to an aluminum metal must be added to an excess of HCl to produce 33.6L of excess of HCl to produce 33.6L of hydrogen gas at STP?hydrogen gas at STP?

a)a) 18.0g18.0g

b)b) 35.0g35.0g

c)c) 27.0g27.0g

d)d) 4.50g4.50g

26

Section 3 HomeworkSection 3 Homework Complete the gas laws worksheet Complete the gas laws worksheet

AND #33, 40, 43, 52 on pg. 219-221.AND #33, 40, 43, 52 on pg. 219-221.

27

GAS STOICHIOMETRYGAS STOICHIOMETRYSection 5.4Section 5.4

28

Gases and StoichiometryGases and Stoichiometry Reactions involve moles of Reactions involve moles of

substances.substances. Recall that at STP (0ºC and 1 atm) Recall that at STP (0ºC and 1 atm)

1mol of any gas occupies 22.4 L.1mol of any gas occupies 22.4 L.

–At STP this can be a conversion At STP this can be a conversion factor: 1mol/22.4L or 22.4L/1molfactor: 1mol/22.4L or 22.4L/1mol

If not at STP, use the ideal gas law to If not at STP, use the ideal gas law to calculate moles or volume of a calculate moles or volume of a substance.substance.

29

Section 4 ExampleSection 4 Example Quicklime (CaO) is produced by the thermal Quicklime (CaO) is produced by the thermal

decomposition of calcium carbonate. Calculate the decomposition of calcium carbonate. Calculate the volume of carbon dioxide produced at STP if 152g of volume of carbon dioxide produced at STP if 152g of calcium carbonate are completely decomposed.calcium carbonate are completely decomposed.

CaCOCaCO33 CaO + CO CaO + CO22 Convert to moles: 152g x 1mol = 1.52molConvert to moles: 152g x 1mol = 1.52mol

100.09g CaCO100.09g CaCO33

1:1 mole ratio of CaCO1:1 mole ratio of CaCO33 to CO to CO22 1.52mol CO 1.52mol CO22 Use STP conditions & stoichiometry: Use STP conditions & stoichiometry: – At STP 1mol = 22.4LAt STP 1mol = 22.4L

– 1.52mol x (22.4L/1mol) = 34.1L CO1.52mol x (22.4L/1mol) = 34.1L CO22 Can double check using

ideal gas law

30

Gas Density and Molar MassGas Density and Molar Mass

Recall: D = m/VRecall: D = m/V Let Let mmmolarmolar stand for molar mass stand for molar mass

mmmolarmolar = m/n so n = m/m = m/n so n = m/mmolarmolar

PV = nRT solve for n: n= PV/RTPV = nRT solve for n: n= PV/RT Thus m/mThus m/mmolarmolar = PV/RT = PV/RT

Solve for mSolve for mmolarmolar: m: mmolarmolar = mRT/VP = mRT/VP

Replace m/V with D: mReplace m/V with D: mmolarmolar = DRT/P = DRT/P If density, temperature, and pressure are If density, temperature, and pressure are

known, molar mass can be found.known, molar mass can be found.

31

AP Practice QuestionAP Practice QuestionDetermine the formula for a gaseous silane Determine the formula for a gaseous silane (Si(SinnHH2n+22n+2) if it’s density is 5.47g/L at 0ºC and ) if it’s density is 5.47g/L at 0ºC and

1.00atm.1.00atm.

*There are several ways to solve!*There are several ways to solve!

SiHSiH44

SiSi22HH66

SiSi33HH88

SiSi44HH1010

32

Section 4 HomeworkSection 4 Homework Pg. 220-221 #51, 54, 57, 63, 64Pg. 220-221 #51, 54, 57, 63, 64

33

DALTONDALTON’’S LAW OF S LAW OF PARTIAL PRESSURESPARTIAL PRESSURES


34

DaltonDalton’’s Law of Partial Pressuress Law of Partial Pressures The total pressure in a container is the The total pressure in a container is the

sum of the pressure each gas would sum of the pressure each gas would exert if it were alone in the container.exert if it were alone in the container.

Total pressure = sum of partial Total pressure = sum of partial pressures.pressures.

PPtottot = P = P11 + P + P22 + P + P33 + ... + ...

–PP11, P, P22, P, P33 are individual gases are individual gases

From the ideal gas law: PFrom the ideal gas law: PTotalTotal = (n = (nTotalTotal)RT)RTV

35

Partial Pressures Cont.Partial Pressures Cont. What does DaltonWhat does Dalton’’s Law tell us about ideal s Law tell us about ideal

gases?gases? Total # of gas particles, not their identities, Total # of gas particles, not their identities,

is important.is important.– V of individual gas particles doesnV of individual gas particles doesn’’t t

affect the total P.affect the total P.– Forces between gas particles doesnForces between gas particles doesn’’t t

affect the total P.affect the total P. If these were important, the different If these were important, the different

identities of gas particles would affect the identities of gas particles would affect the total P differently.total P differently.

36

AP Practice QuestionAP Practice QuestionA gaseous mixture at 25°C contained 1mol CHA gaseous mixture at 25°C contained 1mol CH44 and and

2mol O2mol O22, and P = 2atm. The gases underwent the , and P = 2atm. The gases underwent the

following reaction: following reaction:

CHCH44(g) + 2O(g) + 2O22(g) (g) CO CO22(g) + 2H(g) + 2H22O(g)O(g)

What is the P in the container after the reaction goes What is the P in the container after the reaction goes to completion and the T is allowed to return to 25°C?to completion and the T is allowed to return to 25°C?

1atm1atm2atm2atm3atm3atm4atm4atm

37

AP Practice QuestionAP Practice Question A sealed, rigid container is filled with three A sealed, rigid container is filled with three

identical gases: A, B, and C. The partial identical gases: A, B, and C. The partial pressure of each gas is known as well as pressure of each gas is known as well as T and V. What additional information is T and V. What additional information is needed to find the masses of the gases in needed to find the masses of the gases in the container?the container?

a) average distance travelled between a) average distance travelled between molecular collisionsmolecular collisions

b) the intermolecular forcesb) the intermolecular forces

c) the molar masses of the gasesc) the molar masses of the gases

d) the total pressured) the total pressure

38

The mole fractionThe mole fraction Ratio of moles of a substance to the Ratio of moles of a substance to the

total moles.total moles.

symbol is Greek letter chi symbol is Greek letter chi

= n= n11 = P= P1 1

nntot tot PPtottot

Mole fractions have no units!Mole fractions have no units!

39

AP Practice QuestionAP Practice Question A reaction makes a mixture of COA reaction makes a mixture of CO22, ,

CO, and HCO, and H22O. The gaseous products O. The gaseous products

contained 0.60mol COcontained 0.60mol CO22, 0.30mol CO, , 0.30mol CO,

and 0.10mol Hand 0.10mol H22O. If the total P is O. If the total P is

0.80atm, what is the partial P of CO?0.80atm, what is the partial P of CO?

a)a) 0.24atm0.24atm

b)b) 0.34atm0.34atm

c)c) 0.080atm0.080atm

d)d) 0.13atm0.13atm

40

Vapor PressureVapor Pressure Water evaporates!Water evaporates! When water evaporates, the resulting When water evaporates, the resulting

water vapor has a pressure.water vapor has a pressure.–Vapor pressure changes with T- Vapor pressure changes with T-

must be looked up.must be looked up. Gases are often collected over water Gases are often collected over water

so the vapor pressure of water must be so the vapor pressure of water must be subtracted from the total pressure.subtracted from the total pressure.

Vapor pressure must be given.Vapor pressure must be given.

41

AP Practice QuestionAP Practice QuestionA sample of methane gas was collected over A sample of methane gas was collected over water at 35°C. The sample had a total water at 35°C. The sample had a total pressure of 756mm Hg. Determine the partial pressure of 756mm Hg. Determine the partial pressure of methane gas in the sample. pressure of methane gas in the sample. (Vapor pressure of water at 35°C is 41mm Hg.)(Vapor pressure of water at 35°C is 41mm Hg.)

a)a)760mm Hg760mm Hg

b)b)41mm Hg41mm Hg

c)c)715mm Hg715mm Hg

d)d)797mm Hg797mm Hg

42

Section 5 HomeworkSection 5 Homework Pg. 221-222 #65, 67, 69, 72Pg. 221-222 #65, 67, 69, 72

43

Collapsing Can DemoCollapsing Can Demo Watch the demonstration.Watch the demonstration. Why did the can collapse?Why did the can collapse?

-The heat vaporized the water, which in turn increased -The heat vaporized the water, which in turn increased P and pushed air out of the can.P and pushed air out of the can.

-When the can was inverted the water vapor quickly -When the can was inverted the water vapor quickly cooled. This caused a quick drop in P (created a cooled. This caused a quick drop in P (created a partial vacuum because essentially no air was left to partial vacuum because essentially no air was left to maintain P).maintain P).

-The atmospheric P outside of the can was much -The atmospheric P outside of the can was much greater than P inside of the can, which allowed the greater than P inside of the can, which allowed the can to be crushed.can to be crushed.

44

THE KINETIC MOLECULAR THE KINETIC MOLECULAR THEORY OF GASESTHEORY OF GASES


45

Kinetic Molecular Theory (KMT)- Kinetic Molecular Theory (KMT)- Explains Behavior & Properties of GasesExplains Behavior & Properties of Gases

1.1. Gases are made up of molecules or atoms.Gases are made up of molecules or atoms.2.2. V of particles can be ignored (very small in V of particles can be ignored (very small in

comparison to distance b/t particles).comparison to distance b/t particles).3.3. Particles constantly move and collide with Particles constantly move and collide with

each other and the walls of the container. each other and the walls of the container. Collisions with the walls of the container Collisions with the walls of the container cause P of the gas. cause P of the gas.

4.4. Particles don’t attract or repel each other; Particles don’t attract or repel each other; when they collide, it’s elastic (no KE is lost- when they collide, it’s elastic (no KE is lost- it’s transferred).it’s transferred).

5.5. The The averageaverage KE is proportional to the Kelvin T. KE is proportional to the Kelvin T.

46

KMT Cont.KMT Cont. Assumes gases are ideal.Assumes gases are ideal. BUT no gases are truly ideal- they BUT no gases are truly ideal- they

only approach ideal behavior only approach ideal behavior (specifically nonpolar gases at low P (specifically nonpolar gases at low P and high T).and high T).

In reality, gases DO have V (although In reality, gases DO have V (although small), and they CAN interact with small), and they CAN interact with each other. each other.

Even so, assuming ideal behavior Even so, assuming ideal behavior gives us good enough answers gives us good enough answers about properties of gases.about properties of gases.

47

KMTKMT #3 describes motion; let’s quantify it:#3 describes motion; let’s quantify it:

uurmsrms = √(3RT/m = √(3RT/mmolarmolar) )

– uurmsrms is root mean square velocity is root mean square velocity

– R value used is 8.314J/molKR value used is 8.314J/molK–molar mass in molar mass in kg/mol kg/mol (b/c J = kgm(b/c J = kgm22/s/s22))

#5: KE per mole (average KE) = 3/2 RT#5: KE per mole (average KE) = 3/2 RT- Recall definition of T! Directly related!Recall definition of T! Directly related!- Units: J/molUnits: J/mol

KE per molecule = ½ mvKE per molecule = ½ mv2 2 this is the only equation this is the only equation given on AP exam!given on AP exam!-- Units: JUnits: J

Large! For H2 at 20°C = 2,000m/s

48

Root Mean Square Velocity Root Mean Square Velocity ExampleExample

What is the root mean square What is the root mean square velocity for the atoms in a sample of velocity for the atoms in a sample of He gas at 25°C?He gas at 25°C?

Convert T to K: 25 + 273 = 298KConvert T to K: 25 + 273 = 298K M = 4.00g/mol M = 4.00g/mol 0.004000kg/mol 0.004000kg/mol uurms rms = 136m/s = 136m/s

49

Range of velocitiesRange of velocities The average distance a molecule travels The average distance a molecule travels

between collisions with another gas between collisions with another gas particle is called the mean free path and particle is called the mean free path and is small (near 10is small (near 10-7-7))–Results in a range of velocities.Results in a range of velocities.

Temperature is an average. There are Temperature is an average. There are molecules of many speeds in the average.molecules of many speeds in the average.

This is shown on a graph called a velocity This is shown on a graph called a velocity distribution.distribution.

50

num

ber

of p

arti

cles

Molecular Velocity

273 K

1273 K

2273 K

Notice that with higher T, average velocities increase and so does the velocity range.

Maxwell-Boltzmann Distribution

51

AP Practice QuestionAP Practice QuestionTwo balloons are at the same T and P. Two balloons are at the same T and P. One contains 14g of nitrogen and the One contains 14g of nitrogen and the other contains 20.0g of argon. Which of other contains 20.0g of argon. Which of the following is true?the following is true?

a)a)D of ND of N22 > D of Ar > D of Ar

b)b)Average speed of NAverage speed of N22 > average speed of > average speed of

Ar moleculesAr molecules

c)c)Average KE of NAverage KE of N22 molecules > average molecules > average

KE of Ar moleculesKE of Ar molecules

d)d)V of NV of N22 container < V Ar container < V Ar

52

AP Practice QuestionAP Practice QuestionIncreasing the T of an ideal gas from 50°C Increasing the T of an ideal gas from 50°C to 75°C at constant V will cause which of to 75°C at constant V will cause which of the following to increase for the gas?the following to increase for the gas?

a)a)average molecular mass of the gasaverage molecular mass of the gas

b)b)average distance between moleculesaverage distance between molecules

c)c)average speed of the moleculesaverage speed of the molecules

d)d)density of the gasdensity of the gas

53

Section 6 HomeworkSection 6 Homework Pg. 222-223 #78, 79, 82, 83 Pg. 222-223 #78, 79, 82, 83

54

EFFUSION AND DIFFUSIONEFFUSION AND DIFFUSIONSection 5.7Section 5.7

55

EffusionEffusion Passage of gas through a small hole, Passage of gas through a small hole,

into a vacuum.into a vacuum. Effusion rate = speed at which the Effusion rate = speed at which the

gas is transferred into the vacuum.gas is transferred into the vacuum. GrahamGraham’’s Law - the relative rates of s Law - the relative rates of

effusion are inversely proportional to effusion are inversely proportional to the square roots of the molar masses the square roots of the molar masses of the gas particles.of the gas particles.

Rate of effusion for gas 1

Rate of effusion for gas 2

M

M

2

1

56

DiffusionDiffusion The spreading of a gas through a room The spreading of a gas through a room

(mixing of gases).(mixing of gases). Slow considering molecules move at Slow considering molecules move at

hundreds of meters per second.hundreds of meters per second.

–Slower movement is caused by Slower movement is caused by collisions with other molecules in the air. collisions with other molecules in the air.

Best estimate is GrahamBest estimate is Graham’’s Law.s Law.

–Ratio is actually less.Ratio is actually less.

–More complex analysis required.More complex analysis required.

57

Section 7 HomeworkSection 7 Homework Pg. 223 #86, 88Pg. 223 #86, 88

58

REAL GASESREAL GASESSections 5.8 & 5.9Sections 5.8 & 5.9

59

Real GasesReal Gases Real molecules do take up space and Real molecules do take up space and

they do interact with each other they do interact with each other (especially polar molecules).(especially polar molecules).

Need to add correction factors to the Need to add correction factors to the ideal gas law to account for these.ideal gas law to account for these.

a = correction factor for pressurea = correction factor for pressure b = correction factor for volumeb = correction factor for volume

60

Volume CorrectionVolume Correction The actual volume free to move in is less The actual volume free to move in is less

because particles do take up some of the because particles do take up some of the volume.volume.

More molecules will have more effect More molecules will have more effect (taking up more space).(taking up more space).

Corrected volume VCorrected volume V’’ = V - nb = V - nb– b is a constant that differs for each gas.b is a constant that differs for each gas.

PP’’ = nRT = nRT

(V-nb) (V-nb)

61

Pressure CorrectionPressure Correction Molecules are attracted to each other- Molecules are attracted to each other-

pressure on the container will be less pressure on the container will be less than ideal gases.than ideal gases.

Size of correction factor depends on the Size of correction factor depends on the # of molecules per liter (conc. of gas).# of molecules per liter (conc. of gas). More molecules = closer together and More molecules = closer together and

more likely to interact/attract.more likely to interact/attract. Since Since twotwo molecules interact, the effect molecules interact, the effect

must be squared.must be squared.

Pobserved = P’ - a

2( )Vn a =

proportionality constant

62

All TogetherAll Together PPobsobs= nRT - a n = nRT - a n 22

V-nb VV-nb V

Called the Van der WaalCalled the Van der Waal’’s equation if s equation if

rearranged:rearranged:

Corrected Corrected Corrected Corrected Pressure Volume Pressure Volume

( )

P + an

V x V - nb nRTobs

2

NOT given on AP Equation sheet!

63

Graphing Real GasesGraphing Real Gases For ideal For ideal

gases PV/nRT gases PV/nRT should be 1 should be 1 (since both (since both are equal are equal according to according to ideal gas law).ideal gas law).

Not seen for Not seen for real gases.real gases.

Notice the Notice the effect of T on effect of T on ideal gas ideal gas behavior.behavior.

64

Graphing Real GasesGraphing Real Gases Deviation from Deviation from

ideal behavior ideal behavior depends on depends on identity of the identity of the gas too.gas too.

Smaller, Smaller, nonpolar gases nonpolar gases exhibit more exhibit more ideal behavior.ideal behavior.

65

Where Do Constants Come From?Where Do Constants Come From? a and b are experimentally a and b are experimentally

determined.determined. Different for each gas.Different for each gas. Bigger molecules have larger b.Bigger molecules have larger b. a depends on both size and polarity.a depends on both size and polarity. Note: table of constants for some Note: table of constants for some

gases is on pg. 210 in the book.gases is on pg. 210 in the book.

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Graphing Real GasesGraphing Real Gases

Take a closer look at HTake a closer look at H22 on the graph. on the graph.

– Most ideal behavior, so it has lowest Most ideal behavior, so it has lowest ‘‘aa’’ value value of the gases shown for Van der Waals of the gases shown for Van der Waals equation.equation.

– Lower a means less correction needed.Lower a means less correction needed.– Thus it must have weak intermolecular forces.Thus it must have weak intermolecular forces.

Real gas behavior can tell us how big of a Real gas behavior can tell us how big of a role intermolecular forces play in role intermolecular forces play in attraction between gas molecules.attraction between gas molecules.

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AP Practice QuestionAP Practice QuestionThe true volume of a real gas is smaller The true volume of a real gas is smaller than that calculated from the ideal gas than that calculated from the ideal gas equation. This occurs because the ideal equation. This occurs because the ideal gas equation does not consider which of gas equation does not consider which of the following?the following?

a)a)Attraction between moleculesAttraction between molecules

b)b)Shapes of moleculesShapes of molecules

c)c)Volume of moleculesVolume of molecules

d)d)Mass of moleculesMass of molecules

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AP Practice QuestionAP Practice Question

Which of the following gases probably Which of the following gases probably shows the greatest deviation from shows the greatest deviation from ideal gas behavior?ideal gas behavior?

a)a)HeHe

b)b)OO22

c)c)SFSF44

d)d)SiHSiH44

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Sections 8&9 HomeworkSections 8&9 Homework Pg. 223 #89, 90Pg. 223 #89, 90

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