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Unit 1 (formerly Module 2). Gases and Their Applications. Lesson 2-1 About Gases. 2. R. Gas is one of the three main states of matter Gas particles may be atoms or molecules , depending on the type of substance (ie, element or compound) - PowerPoint PPT Presentation
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Unit 1 (formerly Module 2) Gases and Their Applications
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Page 1: Unit 1 (formerly Module 2)

Unit 1(formerly Module 2)

Gases and Their Applications

Page 2: Unit 1 (formerly Module 2)

Lesson 2-1

About Gases

2

Page 3: Unit 1 (formerly Module 2)

Gas is one of the three main states of matter Gas particles may be atoms or molecules,

depending on the type of substance (ie, element or compound)

Gas particles have much more space between them than liquids or solids.

Gases are said to be an expanded form of matter, solids and liquids are condensed forms of matter.

3

R

Page 4: Unit 1 (formerly Module 2)

General Properties of a Gas

Gases do have mass (although it is sometimes difficult to measure).

Gases have no definite volume, Gases have no definite shape. Gases are compressible, meaning they can

be squeezed into smaller containers, or can expand to fill larger containers.– Because gases compress, the density of gases

can only be compared under specific conditions.

4

Page 5: Unit 1 (formerly Module 2)

Some Important Gases Oxygen (O2): clear, breathable, supports combustion. Ozone (O3): poisonous, unstable form of oxygen Nitrogen (N2): clear, low activity, most abundant gas in

the Earth’s atmosphere. Hydrogen (H2): clear, lighter than air,

flammable/explosive Carbon dioxide (CO2): clear, but turns limewater

cloudy. Does not support respiration but low toxicity. Heavier than air. Largely responsible for the greenhouse effect (global warming)

Sulphur dioxide (SO2): smelly gas. When it combines with oxygen and water vapour it can form H2SO4, responsible for acid rain.

5

Page 6: Unit 1 (formerly Module 2)

Some Important Gases Carbon monoxide (CO): clear, colourless, but very

toxic. It destroys the ability of blood to carry oxygen. About the same density as air.

Ammonia (NH3): toxic, strong smell, refrigerant . Very soluble in water, forms a basic solution called ammonia-water (NH4OH) which is found in some cleaners.

Freon® or CFC: Non-toxic (safe-to-inhale in moderation) refrigerant used in air-conditioners & freezers. Freon may catalyze ozone breakdown. The original Freon formula is now banned, but low chlorine versions are still in use.

Methane (CH4): flammable gas, slightly lighter than air, produced by decomposition. Found in natural gas. Methane is also a “greenhouse” gas.

Helium (He): inert, lighter than air. Used in balloons and in diver’s breathing mixtures. 6

Page 7: Unit 1 (formerly Module 2)

Acetylene (C2H2): AKA ethene, it is used as a fuel in welding, lanterns and other devices.

Propane (C3H8): used as a fuel in barbecues, stoves, lanterns and other devices.

Radon (Rn): A noble gas that is usually radioactive. It is heavier than air, and sometimes found in poorly ventilated basements.

Neon (Ne) and Xenon (Xe): Noble gases found in fluorescent light tubes, and as insulators inside windows. They glow more brightly than other gases when electrons pass through them. Neon is slightly lighter than air, Xenon is quite a bit heavier.

Compressed Air (78% N2, 21% O2): Not actually a pure gas, but a gas mixture that acts much like a pure gas. It is used by scuba divers (at shallow depths), and to run pneumatic tools, and for producing foam materials. 7

Page 8: Unit 1 (formerly Module 2)

Fun Gases(of no real importance)

Nitrous Oxide (N2O)– AKA: Laughing gas, Happy gas, Nitro, NOS– Once used as an anaesthetic in dentist offices, this

sweet-smelling gas reduces pain sensitivity and causes euphoric sensations. It is an excellent oxidizer, reigniting a glowing splint much like oxygen would. It is used in racing where it is injected into the carburetor to temporarily increase an engine’s horsepower.

Sulfur Hexafluoride– One of the densest gases in common use.

Fun with Sulfur hexafluoride

8

Page 9: Unit 1 (formerly Module 2)

Matchthe gas with the problem it causes

Gas Problem

Carbon Dioxide Ozone layer depletion

CFCs Global Warming Methane Toxic poisoning Carbon monoxide Noxious smell Sulfur dioxide Acid Rain

Next slide: Summary 9

Page 10: Unit 1 (formerly Module 2)

Some Gases Classified by Relative Density

Low Density gases Neutral Density Gases High Density gases“lighter than air”<25 g/mol “similar to air” 29±4 g/mol “Denser than air” (>34 g/mol)

Testable Property*: Balloon will float in air

Balloon drops slowly through air

Balloon drops quickly through air

Examples:Hydrogen (H2) 2Helium (He) 4Methane (CH4) 16Ammonia (NH3) 17Neon (Ne) 20Hydrogen Fluoride (HF) 21

Examples:“Cyanide“ (HCN) 27Acetylene (C2H4) 28Nitrogen (N2) 28Carbon monoxide 28Ethane (C2H6) 30Oxygen (O2) 32

Examples:Fluorine (F2) 38 Argon (Ar) 40Carbon dioxide (CO2) 44Propane (C3H8) 44Butane (C4H10) 58Sulphur Hexafluoride (SF6) 146

*balloon test: Fill a large, lightweight balloon with the gas, then release it from a height of about 1.8 m in a room with still air. If the gas is lighter than air the balloon will float upwards. If it is close to air, the balloon will fall very slowly. If the gas is heavier than air, the balloon will fall quickly.

Page 11: Unit 1 (formerly Module 2)

Some Gases Classified by Chemical Properties

Combustible gases(combustion /explosion)

Oxidizing (reactive) Gases (support combustion)

Non-Reactive gases (don’t support combustion)

Testable property:Burning splint produces “pop”

Testable property:Glowing splint reignites, burning splint grows brighter

Testable property:Burning splint is extinguished, glowing splint is dimmed

Other properties:Useful as fuels

Other properties:Cause metals and some other materials to corrode or oxidize. Can improve combustion.

Other properties:Can be used to preserve foods by slowing oxidation

Examples:Hydrogen (H2) Methane (CH4)Propane (C3H8)Acetylene (C2H4)

Examples:Oxygen (O2)Fluorine (F2)Chlorine (Cl2)Nitrous Oxide (NO2)

Examples:Carbon dioxide (CO2)Nitrogen (N2)Argon (Ar)Helium (He)

Page 12: Unit 1 (formerly Module 2)

Textbook Assignments

Read Chapter 1: pp. 37 to 50

Do the exercises on pages 51 and 52– Questions # 1 to 22

12

Page 13: Unit 1 (formerly Module 2)

Summary: • Know the properties of gases• Know the features of some important

gases, esp:• Oxygen• Hydrogen• Carbon dioxide

• Know the environmental problems associated with some gases, eg.

• Carbon dioxide• CFC’s• Sulfur dioxide

13

Page 14: Unit 1 (formerly Module 2)

Chapter 2Physical Properties of Gases

Includes: The Kinetic Theory

“Moving, moving, moving,Keep those atoms moving...”

The Gas Laws.“Jumping Jack Flash, It’s a gas, gas,

gas...”

14

Page 15: Unit 1 (formerly Module 2)

2.1 Kinetic Theory

• Overview:The kinetic theory of gases (AKA. kinetic-

molecular theory) tries to explain the behavior of gases, and to a lesser extent liquids and solids, based on the concept of moving particles or molecules.

Page 16: Unit 1 (formerly Module 2)

The Kinetic Theory of Gases(AKA: The Kinetic Molecular Theory)

• The Kinetic Theory of Gases tries to explain the similar behaviours of different gases based on the movement of the particles that compose them.

• “Kinetic” refers to motion. The idea is that gas particles* are in constant motion.

* For simplicity, I usually call the gas particles “molecules”, although in truth, they could include atoms or ions.

2.1Page 54

16

Page 17: Unit 1 (formerly Module 2)

The Particle ModelNot in text

• The Kinetic Theory is part of the Particle Model of matter, which includes the following concepts:– All matter is composed of particles (ions, atoms or

molecules) which are extremely small and have a varying space between them, depending on their state or phase.

– Particles of matter may attract or repel each other, and the force of attraction or repulsion depends on the distance that separates them.

– Particles of matter are always moving.++

-17

R

Page 18: Unit 1 (formerly Module 2)

Kinetic Molecular Theory And Temperature

• The absolute temperature of a gas (Kelvins) is directly proportional to the average kinetic energy of its molecules.– In other words, when it is cold, molecules move

slowly and have lower kinetic energy.– When the temperature increases, molecules speed

up and have more kinetic energy!

18

Page 19: Unit 1 (formerly Module 2)

Particle Motion and Phases of Matter

• Recall that:• In solids, the particles (molecules) are moving

relatively slowly. They have low kinetic energy• In liquids, molecules move faster. They have

higher kinetic energy.• In gases, the particles move fastest, and have high

kinetic energy.• But, as we will find out later:

• Heavy particles moving slowly can have the same kinetic energy as light particles moving faster.

2.1.1Page 54

19

R

Page 20: Unit 1 (formerly Module 2)

Kinetic Theory Model of States

SolidParticles vibrate but don’t “flow”. Strong molecular attractions keep them in place.

LiquidParticles vibrate, move and “flow”, but cohesion (molecular attraction) keeps them close together.

GasParticles move freely through container. The wide spacing means molecular attraction is negligible.

20

Page 21: Unit 1 (formerly Module 2)

Kinetic Motion of Particles• Particles (ie. Molecules) can have 3 types

of motion, giving them kinetic energy– Vibrational kinetic energy (vibrating)– Rotational kinetic energy (tumbling)– Translational kinetic energy (moving)

2.1.1Page 55

21

Page 22: Unit 1 (formerly Module 2)

Kinetic Theory and Solids & Liquids• When it is cold, molecules move slowly• In solids, they move so slowly that they are held

in place and just vibrate (only vibrational energy)• In liquids they move a bit faster, and can tumble

and flow, but they don’t escape from the attraction of other molecules (more rotational energy, along with a little bit of vibration & translation)

• In gases they move so fast that they go everywhere in their container (more translational energy, with a little bit of rotation & vibration).

2.1.1Page 56

22

Page 23: Unit 1 (formerly Module 2)

Plasma, the “Fourth State”(extension material)

• When strongly heated, or exposed to high voltage or radiation, gas atoms may lose some of their electrons. As they capture new electrons, the atoms emit light—they glow. This glowing, gas-like substance is called “plasma”

23

Page 24: Unit 1 (formerly Module 2)

Kinetic Theory and the Ideal Gas• As scientists tried to understand how gas

particles relate to the properties of gases, they saw mathematical relationships that very closely, but not perfectly, described the behaviour of many gases.

• They have developed theories and mathematical laws that describe a hypothetical gas, called “ideal gas.”

2.1.3Page 61

24

Page 25: Unit 1 (formerly Module 2)

• To make the physical laws (derived from kinetic equations from physics) work, they had to make assumptions about how molecules work.

• Four of these assumptions are listed on page 61 of your textbook

• Other textbooks contain additional assumptions associated with the kinetic theory.

2.1.3Page 61

2

21: mvEequationkinetic k

25

Page 26: Unit 1 (formerly Module 2)

Kinetic Theory Hypotheses about an Ideal Gas

1. The particles of an ideal gas are infinitely small, so the size is negligible compared to the volume of the container holding the gas.

2. The particles of an ideal gas are in constant motion, and move in straight lines (until they collide with other particles)

3. The particles of an ideal gas do not exert any attraction or repulsion on each other.

4. The average kinetic energy of the particles is proportional to the absolute temperature.

2.1.3Page 61

26

Page 27: Unit 1 (formerly Module 2)

No Gas is Ideal• Some of the assumptions on the previous

page are clearly not true.• Molecules do have a size (albeit very tiny)• Particles do exert forces on each other (slightly)

• As a result, there is no such thing as a perfectly “ideal gas”

• However, the assumptions are very good approximations of the real particle properties.

• Real gases behave in a manner very close to “ideal gas”, in fact so close that we can usually assume them to be ideal for the purposes of calculations.

27

Page 28: Unit 1 (formerly Module 2)

Other “Imaginary Features” of Ideal Gas

• An ideal gas would obey the gas laws at all conditions of temperature and pressure

• An ideal gas would never condense into a liquid, nor freeze into a solid.

• At absolute zero an ideal gas would occupy no space at all.

2.1.3Page 61

28

Page 29: Unit 1 (formerly Module 2)

Please Notice:• Not all molecules move at exactly the same

speed. The kinetic theory is based on averages of a great many molecules.– Even if the molecules are identical and at a uniform

temperature, a FEW will be faster than the average, and a FEW will be slower.

– If there are two different types of molecules, the heavier ones will be slower than the light ones – ON THE AVERAGE! – but there can still be variations. That means SOME heavy molecules may be moving as fast as the slowest of the light ones.

• Temperature is based on the average (mean) kinetic energy of sextillions of individual molecules. 29

Page 30: Unit 1 (formerly Module 2)

“Slow” molecules

The range of kinetic energies can be represented as a sort of “bell curve.” Maxwell’s Velocity Distribution Curve.

Increasing kinetic energyAverage kinetic energy

Incr

easin

g #

mol

ecul

es

Most molecules

mod

em

ean

“Average”molecules

The mean & mode can help establish

“average” molecules

“Fast”Molecules

30

Page 31: Unit 1 (formerly Module 2)

So, Given two different gases at the same temperature…What is the same about them?• The AVERAGE kinetic energy is the same.

• Not the velocity of individual molecules• Not the mass of individual molecules.• In fact, the lighter molecules will move faster

• Ek = mv2 kinetic energy of molecules

2So, kinetic energy depends on both the

speed (v) and on the mass (m) of the molecules. 31

Page 32: Unit 1 (formerly Module 2)

Distribution of Particles Around Average Kinetic Energies.

Kinetic Energy of molecules(proportional to velocity of molecules)

Num

ber o

f mol

ecul

es

Aver

age

kine

tic

ener

gy o

f mol

ecul

es

Aver

age

kine

tic

ener

gy o

f war

mer

mol

ecul

es

Faster than average molecules

Slowerthan

average molecules

32

Aver

age

kine

tic

ener

gy o

f col

der

mol

ecul

es

Page 33: Unit 1 (formerly Module 2)

Kinetic Theory Trivia• The average speed of oxygen molecules at

20°C is 1656km/h.• At that speed an oxygen molecule could travel from Montreal

to Vancouver in three hours…If it travelled in a straight line. • Each air molecule has about 1010 (ten billion)

collisions per second• 10 billion collisions every second means they bounce around

a lot!• The number of oxygen molecules in a classroom

is about:• 722 400 000 000 000 000 000 000 000

– that’s more than there are stars in the universe!

• The average distance air molecules travel between collisions is about 60nm.

– 0.00000006m is about the width of a virus.

33

Page 34: Unit 1 (formerly Module 2)

Videos• Kinetic Molecular Basketball

– http://www.youtube.com/watch?v=t-Iz414g-ro&NR=1

• Average Kinetic Energies– http://www.youtube.com/watch?v=UNn_trajMFo&NR=1

• Thermo-chemistry lecture on kinetics

34

Page 35: Unit 1 (formerly Module 2)

Assignments

• Read pages 53 to 61• Do Page 62 # 1-11

Page 36: Unit 1 (formerly Module 2)

Chapter 2.2

• Behaviors of Gases– Compressibility– Expansion– Diffusion and Effusion– Graham’s Law

Page 37: Unit 1 (formerly Module 2)

37

• 2.2.1 Compressibility:– Because the distances between particles in a

gas is relatively large, gases can be squeezed into a smaller volume.

– Compressibility makes it possible to store large amounts of a gas compressed into small tanks

• 2.2.2 Expansion:– Gases will expand to fill any container they

occupy, due to the random motion of the molecules.

Page 38: Unit 1 (formerly Module 2)

2.2.3 Diffusion Diffusion is the tendency for molecules to

move from areas of high concentration to areas of lower concentration, until the concentration is uniform. They do this because of the random motion of the molecules.

Effusion is the same process, but with the molecules passing through a small hole or barrier

Next slide: 38

Page 39: Unit 1 (formerly Module 2)

Rate of Diffusion or Effusion

It has long been known that lighter molecules tend to diffuse faster than heavy ones, since their average velocity is higher, but how much faster?

39

heavy particle light particle

Page 40: Unit 1 (formerly Module 2)

Graham’s Law Thomas Graham (c. 1840)

studied effusion (a type of diffusion through a small hole) and proposed the following law:

“The rate of diffusion of a gas is inversely proportional to the square root of its molar mass.”

In other words, light gas particles will diffuse faster than heavy gas molecules, and there is a math formula to calculate how much faster.

Next slide: Example

1

2

2

1

MM

vv

40

Where: v1= rate of gas 1 v2= rate of gas 2 M1= molar mass of gas1 M2=molar mass of gas 2

Internet demo of effusion

Page 41: Unit 1 (formerly Module 2)

Graham’s LawVersion #1, based on Effusion Rate

• The relationship between the rate of effusion or diffusion and the molar masses is:

1

2

2

1

MM

vv

Where: v1 is the rate of diffusion of gas 1, in any appropriate rate units*v2 is the rate of diffusion of gas 2, in the same units as gas 1M1 is the molar mass of gas 1M2 is the molar mass of gas 2

*Rate units must be an amount over a time for effusion (eg: mL/s or L/min), or a distance over a time for diffusion (eg: cm/min or mm/s)

Note: See the inversion of the 1 and 2 in the 2nd ratio!

Page 42: Unit 1 (formerly Module 2)

Thomas Graham (1805-1869)

• Graham derived his law by treating gases as ideal, and applying the kinetic energy formula to them.

• Ek = ½ mv2

• All gases have the same kinetic energy at the same temperature,

• Therefore, mv2 for the first gas = mv2 for the second gas: m1v1

2 = m2v22.

• A bit of algebra then gave him his famous law.

And in my spare time I invented dialysis, which has saved the

lives of thousands of

kidney patients

Page 43: Unit 1 (formerly Module 2)

Graham’s LawVersion #2, Based on Effusion Time

• Sometimes it’s easier to measure the time it takes for a gas to effuse completely, rather than the rate. Graham’s law can be changed for this, but the relationship between time and molar mass is direct as the square root:

2

1

2

1

MM

tt

Where: t1 is the time it takes for the first gas to effuse completely.t2 is the time it takes for an equal volume of the 2nd gas to effuseM1 is the molar mass of the first gasM2 is the molar mass of the second gas.

Note: In this variant law, the relationship is not inverted!

Page 44: Unit 1 (formerly Module 2)

Example of Graham’s Law: How much faster does He diffuse than N2?

Nitrogen (N2) has a molar mass of 28.0 g/mol

Helium (He) has a molar mass of 4.0 g/mol

The difference between their diffusion rates is:

Notice the reversal of order!

So helium diffuses 2.6 times faster than nitrogen

He

N

N

He

M

M

vv 2

2

6.223.5

/4/28

molgmolg

MN2=2x14.0=28 g/mol

Next slide: 2.3 Pressure of Gases

MHe=1x4.0=4 g/mol

44

Page 45: Unit 1 (formerly Module 2)

Assignments

Read pages 63 to 67 Do Questions 1 to 10 on

page 68

Page 46: Unit 1 (formerly Module 2)

Chapter 2.3• Pressure of Gases

– What is Pressure– Atmospheric Pressure– Measuring Pressure

100 km < 0.003 kPa

40 km 1 kPa

20 km 6 kPa

10 km 25 kPa 5 km 55 kPa 0 km 101 kPa

Mt Everest 31 kPa

46

Highest Jet 4 kPa

Edge of SpaceX15 (1963)Spaceship 1 (2006)Outer Space (immeasurable)

Mr. Smith

Page 47: Unit 1 (formerly Module 2)

Pressure

• Pressure is the force exerted by a gas on a surface.

• The surface that we measure the pressure on is usually the inside of the gas’s container.

• Pressure and the Kinetic Theory• Gas pressure is caused by billions of particles

moving randomly, and striking the sides of the container.

• Pressure Formula:

Pressure = force divided by areaAFP

47

Page 48: Unit 1 (formerly Module 2)

Atmospheric Pressure• This is the force of a 100 km high

column of air pushing down on us.• Standard atmospheric pressure is

• 1.00 atm (atmosphere), or• 101.3 kPa (kilopascals), or• 760 Torr (mmHg), or• 14.7 psi (pounds per square inch)

• Pressure varies with:• Altitude. (lower at high altitude)

• Weather conditions. (lower on cloudy days)

48

Page 49: Unit 1 (formerly Module 2)

Pressure conversions

)()(

)()(

wantedunitsSPwantedunitsP

givenunitsSPgivenunitsP

Example 1: convert 540 mmHg to kilopascals

kPaP

mmHgmmHg

3.101760540

=72.0 kPa

Example 2: convert 155 kPa to atmospheres

atmP

kPakPa

00.13.101155

=1.53 atm

SP1.00 atm760 mmHg760 Torr101.3 kPa14.7 psi1013 mB29.9 inHg

Multiply

Divide

Page 50: Unit 1 (formerly Module 2)

Measuring Pressure

• Barometer: measures atmospheric pressure.– Two types:

• Mercury Barometer• Aneroid Barometer

• Manometer: measures pressure in a container (AKA. Pressure guage)

• Dial Type: Similar to an aneroid barometer• U-Tube: Similar to a mercury barometer• Piston type: used in “tire guage” 50

Page 51: Unit 1 (formerly Module 2)

• A tube at least 800 mm long is filled with mercury (the densest liquid) and inverted over a dish that contains mercury.

• The mercury column will fall until the air pressure can support the mercury.

• On a sunny day at sea level, the air pressure will support a column of mercury 760 mm high.

• The column will rise and fall slightly as the weather changes.

• Mercury barometers are very accurate, but have lost popularity due to the toxicity of mercury.

the Mercury Barometer

51

Page 52: Unit 1 (formerly Module 2)

The Aneroid Barometer

• In an aneroid barometer, a chamber containing a partial vacuum will expand and contract in response to changes in air pressure

• A system of levers and springs converts this into the movement of a dial.

Page 53: Unit 1 (formerly Module 2)

• Manometers work much like barometers, but instead of measuring atmospheric pressure, they measure the pressure difference between the inside and outside of a container.

• Like barometers they come in mercury and aneroid types. There is also a cheaper “piston” type used in tire gauges, but not in science.

U-tube manometer Pressure gauge(mercury manometer) (aneroid)

You Tube manometer

Manometers (Pressure Gauges)

Tire gauge(piston manometer)

Page 54: Unit 1 (formerly Module 2)

Reading U-tube manometers• When reading a mercury U-

tube manometer, you measure the difference in the heights of the two columns of mercury.

• If the tube is “closed” then the height (h) is the gas pressure in mmHg. P(mmHg)=h(mmHg)

• If the tube is “open” and h is positive (the pressure you are measuring is greater than the atmosphere) then you must add atmospheric pressure in mmHg. Pgas(mmHg) = Patm(mmHg)+h(mm)

Must be in

mmHg, not cm or kPa!

Atm. pressure

After you finish, you can convert your answer to kPa, or atm. Or whatever.

Page 55: Unit 1 (formerly Module 2)

Manometer Exampleson a day when the air pressure is 763mmHg (101.7 kPa)

Closed tube: Pgas(mm Hg)=h (mm Hg)Pgas = h = 4 cm = 40 mm HgPgas = kPakPa

HgmmHgmm 3.53.101

76040

Open: Pgas(mmHg)=P atm(mmHg) +h (mmHg)Pgas = 763 + 60mm Hg =823 mm Hg

Pgas = kPakPaHgmmHgmm 7.1093.101

760823

Open: Pgas(mmHg)=P atm(mmHg) -h (mmHg)Pgas = 763 - 60mm Hg =703 mm HgPgas = kPakPa

mmHgmmHg 7.933.101

760703

6 cm Higher

6 cm Lower

4 cm

6

9h= 4 cm

Page 56: Unit 1 (formerly Module 2)

Assignments

• Read pages 69 to 73. • Do Page 74, Questions 1 to 4.

Page 57: Unit 1 (formerly Module 2)

Chapter 2.4• The Simple Gas Laws

– Boyle’s Law Relates volume & pressure– Charles’ Law Relates volume & temperature– Gay-Lussac’s Law Relates pressure & temperature– Avogadro’s Law Relates to the number of moles

• Other Simple Laws that are a Gas:– Clarke’s Laws Relates possible and

impossible– Murphy’s Law Anything that can go

wrong will– Cole’s Law Relates thinly sliced cabbage

to vinegar.57

Page 58: Unit 1 (formerly Module 2)

Clarke’s Lawsof the impossible*

Clarke’s 1st Law: If an elderly and respected science teacher (like me) tells you that something is possible, he is probably right. If he tells you something is impossible, he’s almost certainly wrong.

Clarke’s 2nd Law: The only way to find the limits to what is possible is to go beyond them.

Clarkes 3rd Law: Any sufficiently advanced technology is indistinguishable from magic.

*these are slightly paraphrased, I quote them from memory. They were developed by science fiction writer Arthur C. Clarke

Page 59: Unit 1 (formerly Module 2)

Lesson 2.4.1Boyle’s Law

Robert Boyle (1662)

“For a given mass of gas at a constant temperature, the volume

varies inversely with pressure.”

For Pressure and Volume

VP 1

59

Page 60: Unit 1 (formerly Module 2)

Robert Boyle Born: 25 January 1627

Lismore, County Waterford, Ireland Died 31 December 1691 (aged 64)London, England

Fields: Physics, chemistry; Known for Boyle's Law. Considered to be the founder of modern chemistry

Influences: Robert Carew, Galileo Galilei, Otto von Guericke, Francis Bacon

Influenced: Dalton, Lavoisier, Charles, Gay-Lussack, Avogadro.

Notable awards: Fellow of the Royal Society

60

Page 61: Unit 1 (formerly Module 2)

Pressure Gas pressure is the force placed on the sides of a

container by the gas it holds Pressure is caused by the collision of trillions of

gas particles against the sides of the container Pressure can be measured many ways

Standard PressureAtmospheres (atm) 1 atmKilopascals (kPa)or(N/m2) 101.3 kPa = 101.3 N/m2

Millibars (mB) 1013 mBTorr (torr) or mm mercury 760 torr = 760 mmHgCentimetres of mercury 76 cmHgInches of mercury (inHg) 29.9 inHg (USA only)Pounds per sq. in (psi) 14.7 psi (USA

only)

61Next slide: Air in Syringe

Page 62: Unit 1 (formerly Module 2)

Example of Boyle’s Law:Air trapped in a syringe

If some air is left in a syringe, and the needle removed and sealed, you can measure the amount of force needed to compress the gas to a smaller volume.

Next slide: Inside syringe 62

Page 63: Unit 1 (formerly Module 2)

Inside the syringe… The harder you press, the smaller

the volume of air becomes. Increasing the pressure makes the volume smaller!

The original pressure was low, the volume was large. The new pressure is higher, so the volume is small.

Click Here for an internet demo using psi (pounds per square inch) instead of kilopascals (1kPa=0.145psi)

Next slide: PV

low

high

63

Page 64: Unit 1 (formerly Module 2)

This means that: As the volume of a contained gas

decreases, the pressure increases As the volume of a contained gas

increases, the pressure decreases This assumes that:

no more gas enters or leaves the container, and

that the temperature remains constant. The mathematical formula for this is given

on the next slideNext slide: Example 64

Page 65: Unit 1 (formerly Module 2)

Boyle’s LawRelating Pressure and Volume of a Contained Gas

• By changing the shape of a gas container, such as a piston cylinder, you can compress or expand the gas. This will change the pressure as follows:

2211 VPVP Where: P1 is the pressure* of the gas before the container changes shape.

P2 is the pressure after, in the same units as P1.V1 is the volume of the gas before the container changes, in L or mLV2 is the volume of the gas after, in the same units as V1

*appropriate pressure units include: kPa, mmHg, atm. Usable, but inappropriate units include psi, inHg.

Page 66: Unit 1 (formerly Module 2)

Example 1 You have 30 mL of air in a syringe at 100 kPa.

If you squeeze the syringe so that the air occupies only 10 mL, what will the pressure inside the syringe be?

P1 × V1 = P2 × V2, so.. 100 kPa × 30 mL = ? kPa × 10 mL 3000 mL·kPa ÷ 10 mL = 300 kPa The pressure inside the syringe will be 300

kPaNext slide: Graph of Boyle’s Law 66

Page 67: Unit 1 (formerly Module 2)

Graph of Boyle’s LawThe Pressure-Volume Relationship

Pressure (kPa)

Volu

me

(L)

100 200 300 400 500 600 700 800

1

2

3

4

5

6

7

8

Boyle’s Law produces an inverse relationship graph.

100 x 8 = 800200 x 4 = 800

400 x 2 = 800

800 x 1 = 800

P(kpa) x V(L)

Next slide: Real Life Data

300 x 2.66 = 800

500 x 1.6 = 800600 x 1.33 = 800700 x 1.14 = 800

67

Page 68: Unit 1 (formerly Module 2)

Example 2: Real Life Data

2 4 6 8 10 12 14 16 18

5

10

15

2

0

25

30

35

40

In an experiment Mr. Taylor and Tracy put weights onto a syringe of air.

At the beginning, Mr. Taylor calculated the equivalent of 4 kgf of atmospheric pressure were exerted on the syringe.

0+4= 4kg : 29 mL (116)2+4= 6kg : 20 mL (120)4+4=8kg : 15 mL (120)6+4=10kg: 12 mL (120)8+4=12kg: 10.5 mL (126)

Next slide: Boyle’s Law Experiment or skip to: Lesson 2.3 Charles’ Law: 68

Page 69: Unit 1 (formerly Module 2)

Summary: Boyle’s law• The volume of a gas is

inversely proportional to its pressure

• Formula: P1V1=P2V2

• Graph: Boyle’s law is usually represented by an inverse relationship graph (a curve)

Volu

me

(L)

Pressure (kPa)

VP 1

P1V1=P2V2

69

Page 70: Unit 1 (formerly Module 2)

70

Page 71: Unit 1 (formerly Module 2)

Assignments on Boyle’s Law

• Read pages 75 to 79• Do questions 1 to 10 on page 97

Page 72: Unit 1 (formerly Module 2)

Lesson 2.4.2Charles’ Law

The Relationship between Temperature and Volume.

“Volume varies directly with Temperature”

Next slide: Jacques Charles

TV 72

Page 73: Unit 1 (formerly Module 2)

Jacques Charles (1787)“The volume of a fixed mass of gas is directly proportional to its temperature (in kelvins) if the pressure on the gas is kept constant”This assumes that the container can expand, so that the pressure of the gas will not rise.

Next slide: The Mathematical formula for this law

Born: November 12, 1746 (1746-11-12) Beaugency, Orléanais Died: April 7, 1823 (1823-04-08) (aged 76), Paris Nationality: France Fields: physics, mathematics, hot air ballooning Institutions: Conservatoire des Arts et Métiers

Page 74: Unit 1 (formerly Module 2)

Charles’ LawRelating Volume and Temperature of a Gas

• If you place a gas in an expandable container, such as a piston or balloon, as you heat the gas its volume will increase, as you cool it the volume will decrease.

2

2

1

1

TV

TV

Where: T1 is Temperature of the gas before it is heated, in kelvins.T2 is Temperature of the gas after it is heated, in kelvinsV1 is the volume of the gas before it was heated, in L or mLV2 is the volume of the gas after it was heated, in the same units.

Page 75: Unit 1 (formerly Module 2)

Charles Law EvidenceCharles used cylinders and pistons to study and graph the expansion of gases in response to heat.See the next two slides for diagrams of his apparatus and graphs.Lord Kelvin (William Thompson) used one of Charles’ graphs to discover the value of absolute zero.

Next slide: Diagram of Cylinder & Piston 75

Page 76: Unit 1 (formerly Module 2)

Charles Law ExamplePiston

Cylinder

Trapped Gas

Next slide: Graph of Charles’ Law

Click Here for a simulated internet experiment

76

Page 77: Unit 1 (formerly Module 2)

Graph of Charles Law

0°C 100°C 200°C150°C50°C 250°C

1L

2L

3L

4L

5L

6L

-250°C -200°C -150°C -100°C -50°C

-273.15°C

Expansion of an “Ideal” Gas

Expansion of most real gases 27

3°C

Next slide: Example

Liquid stateSolid state co

nden

satio

n

freez

e Charles discovered

the direct relationship

Lord Kelvin traced it back

to absolute zero.

Page 78: Unit 1 (formerly Module 2)

-273.15°C is called absolute zero. It is the coldest possible temperature.

ABSOLUTE ZERO IS REAL COOL!

ABSOLUTE ZERO

At absolute zero, molecules stop moving and even vibrating. Since temperature is based on the average kinetic energy of molecules, temperature cannot be said to exist if there is no kinetic energy (movement)

Page 79: Unit 1 (formerly Module 2)

William Thompson, The right honourable

Lord Kelvin1st Baron of Largs

1824-1907

Born in Belfast IrelandDied in Largs, Scotland

Worked at the University of Glasgow

Experimented in Thermodynamics.

"There is nothing new to be discovered in physics now. All that remains is more and more precise measurement"

Page 80: Unit 1 (formerly Module 2)

Kelvin’s Scale

To convert from Celsius to Kelvin, simply add 273 to the Celsius temperature. To convert back, subtract 273Note: Temperature readings are always assumed to have at least 3 significant digits. For example, 6°C is assumed to mean 279 K with 3 sig.fig., even though the data only showed 1 sig.fig.

In 1848 Lord Kelvin suggested using a temperature scale based on absolute zero, but with divisions exactly the same as the Celsius scale. For many years this was called the “absolute temperature scale” but long after his death it was renamed to honour Lord Kelvin

Page 81: Unit 1 (formerly Module 2)

Example If 2 Litres of gas at 27°C are heated in a cylinder,

and the piston is allowed to rise so that pressure is kept constant, how much space will the gas take up at 327°C?

Convert temperatures to kelvins: 27°C =300k, 327°C = 600k

Use Charles’ Law: (see below) Answer: 4 Litres

KLitresx

KLitresso

TV

TV

6003002:,

2

2

1

1

Next slide: Lesson 2.4 Gay Lussac’s Law

Page 82: Unit 1 (formerly Module 2)

Summary: Charles’ law• The volume of a gas is

directly proportional to its temperature

• Formula:• Graph: Charles’ law is

usually represented by a direct relationship graph (straight line)

• Video1

2

2

1

1

TV

TV

Absolute zero 0°C=273K TempVolu

me

(L)

TV

Page 83: Unit 1 (formerly Module 2)

Charles’ Law Assignments

• Read pages 80 to 84• Do questions 11 to 21 on pages 97 and 98

Page 84: Unit 1 (formerly Module 2)

Charles’ Law Worksheet1. The temperature inside my fridge is about 4˚C, If I place a

balloon in my fridge that initially has a temperature of 22˚C and a volume of 0.50 litres, what will be the volume of the balloon when it is fully cooled? (for simplicity, we will assume the pressure in the balloon remains the same)

Data:T1=22˚CT2=4˚CV1=0.50 LTo find:V2= unknown

Temperatures must be converted to kelvin=295K=277K

2

2

1

1

TV

TV

So:V2=V1 x T2 ÷ T1

V2=0.5L x 277K 295K

V2=0.469 LThe balloon will have a volume of 0.47 litres

multiplydivide

84

Page 85: Unit 1 (formerly Module 2)

2. A man heats a balloon in the oven. If the balloon has an initial volume of 0.40 L and a temperature of 20.0°C, what will the volume of the balloon be if he heats it to 250°C.

85

DataV1= 0.40LT1= 20°CT2= 250°CV2= ?

Convert temperatures to kelvin20+273= 293K, 250+273=523k

=293 K=523 K

Use Charles’ Law

KV

KL

TV

TV

5232934.0... 2

2

2

1

1

0.40L x 523 K ÷ 293 K = 0.7139L

0.7139L

Answer: The balloon’s volume will be 0.71 litres

Page 86: Unit 1 (formerly Module 2)

3. On hot days you may have noticed that potato chip bags seem to inflate. If I have a 250 mL bag at a temperature of 19.0°C and I leave it in my car at a temperature of 60.0°C, what will the new volume of the bag be?

Answer: The bag will have a volume of 285mL

Data:V1=250 mLT1= 19.0°CT2=60.0°CV2= ?

Convert temperatures to kelvin19+273= 292K, 60+273=333K

=292 K=333 K

KV

KmL

TV

TV

333292250... 2

2

2

1

1

Use Charles’ Law

250mL x 333 K ÷ 292 K = 285.10mL

285.10 mL

Page 87: Unit 1 (formerly Module 2)

4. The volume of air in my lungs will be 2.35 litres

Be sure to show your known informationChange the temperature to Kelvins and show them.Show the formula you used and your calculationsState the answer clearly.

5.6. The temperature is 279.7 K, which corresponds to 6.70 C. A

jacket or sweater would be appropriate clothing for this weather.

Although only the answers are shown here, in order to get full marks you need to show all steps of the solution!

Page 88: Unit 1 (formerly Module 2)

Gay-Lussac’s LawFor Temperature-Pressure changes.

“Pressure varies directly with Temperature”

Lesson 2.4.3

Next slide:’

TP 88

Page 89: Unit 1 (formerly Module 2)

Joseph Gay-Lussac (1802)“The pressure of a gas is directly proportional to the temperature (in kelvins) if the volume is kept constant.”

Next slide:’ 89

Born 6 December 1778 Saint-Léonard-de-Noblat

Died 9 May 1850 @Saint-Léonard-de-Noblat

Nationality: FrenchFields: ChemistryKnown for Gay-Lussac's law

Page 90: Unit 1 (formerly Module 2)

Gay-Lussac’s LawRelating Pressure and Temperature of a Gas

2

2

1

1

TP

TP

Where: P1 is the pressure* of the gas before the temperature change.P2 is the pressure after the temperature change, in the same units.T1 is the temperature of the gas before it changes, in kelvins.T2 is the temperature of the gas after it changes, in kelvins.

*appropriate pressure units include: kPa, mmHg, atm.

Page 91: Unit 1 (formerly Module 2)

Gay-Lussac’s Law

As the gas in a sealed container that cannot expand is heated, the pressure increases.

For calculations, you must use Kelvin temperatures: K=°C+273

pressure

91

Page 92: Unit 1 (formerly Module 2)

92

Example A sealed can contains 310 mL of air

at room temperature (20°C) and an internal pressure of 100 kPa. If the can is heated to 606 °C what will the internal pressure be?

Kx

KkPa

879293100

2

2

1

1

TP

TP

multiply

x = 87900 ÷ 293x = 300Next slide: T vs P graph

Data:P1= 100kPaV1=310 mLT1=20˚CP2=unknownT2=606˚C

˚Celsius must be converted to kelvins20˚C = 293 K 606˚C = 879 K

Answer: the pressure will be 300 kPa

Remove irrelevant fact

=293K =879K

divide

Formula:

Page 93: Unit 1 (formerly Module 2)

Temperature & Pressure Graph

The graph of temperature in Kelvin vs. pressure in kilopascals is a straight line. Like the temperature vs. volume graph, it can also be used to find the value of absolute zero.

93

Page 94: Unit 1 (formerly Module 2)

Graph of Pressure-Temperature Relationship

(Gay-Lussac’s Law)

Temperature (K) Pres

sure

(kPa

)

273KNext slide:’ 94

Page 95: Unit 1 (formerly Module 2)

Summary: Gay-Lussac’s law• The pressure of a gas is

directly proportional to its temperature

• Formula:• Graph: Gay-Lussac’s law is

usually represented by an direct relationship graph (straight line)

2

2

1

1

TP

TP

Absolute zero 0°C=273K TempPres

sure

Page 96: Unit 1 (formerly Module 2)

Assignment on Gay-Lussac’s Law

• Read pages 85 to 87• Answer questions #22 to 30 on page 98

Page 97: Unit 1 (formerly Module 2)

Avogadro’s LawFor amount of gas.

“The volume of a gas is directly related to the number of moles of gas”

Lesson 2.4.4

Next slide: Lorenzo Romano Amedeo Carlo Avogadro di Quaregna

nV 97

Page 98: Unit 1 (formerly Module 2)

Lorenzo Romano Amedeo Carlo

Avogadro di Quaregna “Equal volumes of gas at

the same temperature and pressure contain the same number of moles of particles.” Amedeo Avogadro

Born: August 9, 1776 Turin, Italy

Died: July 9, 1856Field: PhysicsUniversity of TurinKnown for Avogadro’s

hypothesis, Avogadro’s number.

Page 99: Unit 1 (formerly Module 2)

You already know most of the facts that relate to Avogadro’s Law:– That a mole contains a certain number of

particles (6.02 x 1023)– That a mole of gas at standard temperature and

pressure will occupy 22.4 Litres (24.5 at SATP) The only new thing here, is how changing

the amount of gas present will affect pressure or volume.– Increasing the amount of gas present will

increase the volume of a gas (if it can expand), – Increasing the amount of gas present will

increase the pressure of a gas (if it is unable to expand).

99

Page 100: Unit 1 (formerly Module 2)

It’s mostly common sense…

If you pump more gas into a balloon, and allow it to expand freely, the volume of the balloon will increase.

If you pump more gas into a container that can’t expand, then the pressure inside the container will increase.

100

Page 101: Unit 1 (formerly Module 2)

Avogadro’s LawsRelating Moles of Gas to Volume or Pressure

2

2

1

1

nV

nV

2

2

1

1

nP

nP

or

Where: V1 = volume before, in appropriate volume units.V2 = volume after, in the same volume unitsP1=pressure before, in appropriate pressure units.P2=pressure after, in the same pressure units.n1 = #moles beforen2 = #moles after 101

Page 102: Unit 1 (formerly Module 2)

Assignments on Avogadro’s Law

• Read pages 88 to 92• Do Questions 31 to 36 on page 98

102

Page 103: Unit 1 (formerly Module 2)

Lesson 2.45

Standard Conditions and Molar Volume

Next slide: 103

Page 104: Unit 1 (formerly Module 2)

Standard Temperature & Pressure(STP)

• Since the volume of a gas can change with pressure and temperature, gases must be compared at a specific temperature and pressure. The long-held standard for comparing gases is called Standard Temperature and Pressure (STP)

• Standard Temperature =0°C = 273 K• Standard Pressure =101.3 kPa

Page 105: Unit 1 (formerly Module 2)

Ambient Temperature • Some chemists prefer to compare gases at 25°C rather

than 0°C. At zero it is freezing, a temperature difficult to maintain inside the lab. This alternate set of conditions is known as Standard Ambient Temperature and Pressure (SATP). Although not widely used, you should be aware of it, and always watch carefully in case a question uses AMBIENT temperature instead of STANDARD temperature.

• Ambient Temperature = 25°C = 298 K

• Standard Pressure = 101.3 kPa(SATP)

Page 106: Unit 1 (formerly Module 2)

Molar Volume

• The volume of 1 mole of an ideal gas depends on the conditions:– At STP one mole of an ideal gas has a

volume of 22.4 litres– AT SATP one mole of an ideal gas has a

volume of 24.5 litres• Since all common gases are very near

ideal at these temperatures, we can use these as standard molar volumes for ANY common gas.

Page 107: Unit 1 (formerly Module 2)

ComparisonStandard and Ambient Conditions

Standard Temperature & Pressure

(STP)Ambient Temperature &

Pressure

(SATP)Pressure 101.3 kPa 101.3 kPaTemperature °C 0 °C 25 °CTemperature K 273.15 K 298.15 KMolar Volume 22.4 L/mol 24.5 L/mol

Page 108: Unit 1 (formerly Module 2)

Assignments on Molar Volume

• Read pages 92 to 96• Do Questions 37 to 52 on page 98

108

Page 109: Unit 1 (formerly Module 2)

Lesson 2.5 & 2.6

The General Gas Law and the Ideal Gas Law

Next slide: 109

Page 110: Unit 1 (formerly Module 2)

The Combined or General Gas Law

• The general (or combined) gas law replaces the four simple gas laws. It puts together:

• Boyle’s Law• Charles’ Law• Gay-Lussac’s Law• Avogadro’s Law

• Advantages of the Combined Gas Law:• It is easier to remember one law than four.• It can handle changing more than one variable at a

time (eg. Changing both temperature and pressure)

110

= General Gas Law

Page 111: Unit 1 (formerly Module 2)

The General Gas LawRelating all the Simple Laws Together

22

22

11

11

TnVP

TnVP

Where: P1 P2 are the pressure of the gas before and after changes.

V1, V2 are the volume of the gas before and after changes.T1 T2 are the temperatures, in kelvinsn 1, n2 is the number of moles of the gas.

Page 112: Unit 1 (formerly Module 2)

The neat thing about the General gas law is that it can replace the three original gas laws.Just cross out or cover the parts that don’t change, and you have the other laws:

22

22

11

11

TnVP

TnVP

Most of the time, the number of moles stays the same, so you can remove moles from the equation.

If the temperature is constant, then you have Boyle’s law.If, instead, pressure remains constant, you have Charles’ LawAnd finally, if the volume stays constant, then you have Gay-Lussac’s Law

112

Page 113: Unit 1 (formerly Module 2)

The Ideal Gas Law

The Ideal Gas Law is derived from the General Gas Law in several mathematical steps. First, start with the general gas law, including P, V, T, and the amount of gas in moles (n) .

Next slide:22

22

11

11

TnVP

TnVP

Page 114: Unit 1 (formerly Module 2)

Remember Standard Temperature & Pressure(STP)

Standard Temperature is 0°C or more to the point, 273K (@SATP = 25°C = 298K)

Standard Pressure is 101.3 kPa (one atmospheric pressure at sea level)

At STP one mole of an ideal gas occupies exactly 22.4 Litres (@SATP = 24.5 L)

Page 115: Unit 1 (formerly Module 2)

The Ideal Gas Law: Calculating the Ideal Gas Constant.

We are going to calculate a new constant by substituting in values for P2, V2, T2 and n2

At STP we know all the conditions of the gas.

Substitute and solve to give us a constant

KmolLkPa

TnVP

27314.223.101

11

11

molKkPaLTnVP

/31.811

11

Next slide: R-- The Ideal Gas Constant

22

22

11

11

TnVP

TnVP

Page 116: Unit 1 (formerly Module 2)

The Ideal Gas Constantis the proportionality constant that makes the ideal gas law work

The Ideal Gas Constant has the symbol R

R=8.31 L· kPa / K·molThe Ideal Gas constant is 8.31 litre-

kilopascals per kelvin-mole.

Next slide: Ideal Gas Formula

Page 117: Unit 1 (formerly Module 2)

So, if

Then, by a bit of algebra: P1V1=n1RT1 Since we are only using one set of

subscripts here, we might as well remove them: PV=nRT

RTnVP

11

11

Page 118: Unit 1 (formerly Module 2)

The Ideal Gas LawRelating Conditions to the Ideal Gas Constant

nRTPV Where: P=Pressure, in kPa

V=Volume, in Litresn= number of moles, in molR= Ideal Gas constant, 8.31 LkPa/Kmol

T = Temperature, in kelvins

Page 119: Unit 1 (formerly Module 2)

The Ideal gas law is best to use when you don’t need a “before and after” situation.

Just one set of data (one volume, one pressure, one temperature, one amount of gas)

If you know three of the data, you can find the missing one.

Page 120: Unit 1 (formerly Module 2)

Sample Problem 8.0 g of oxygen gas is at a pressure of

2.0x102 kPa (ie: 200 Kpa w. 2 sig fig) and a temperature of 15°C. How many litres of oxygen are there? Formula: PV = nRT

Variables: P=200 kPa V=? (our unknown)= x n= 8.0g ÷ 32 g/mol =0.25

mol R=8.31 L·kPa/K·mol (ideal gas

constant) T= 15°C + 273 = 288K 200 x = (0.25)(8.31)(288) , therefore x= (0.25)(8.31)(288) ÷ 200=2.99 L There are 3.0 L of oxygen (rounded to 2 S.D.)

Page 121: Unit 1 (formerly Module 2)

Sample problem

molmolg

gMmn 25.0

/0.320.8

8.0 g of oxygen gas is at a pressure of 2.0x102 kPa (ie: 200KPa) and a temperature of 15°C. How many litres of oxygen are there? (assume 2 significant digits)

Data:P=200 kPaV=unknown = X n= not givenR=8.31 L·kPa/K·molT= 15°C + 273 = 288K---m (O2) = 8gM (O2) = 32.0 g/mol

0.25 mol

Temperature has been converted to kelvins

Calculate the value of n using the mole formula:

nRTPV 200 x = (0.25)(8.31)(288) , thereforex= (0.25)(8.31)(288) ÷ 200=2.99 L

There are 3.0 L of oxygen (rounded to 2 S.D.)

Page 122: Unit 1 (formerly Module 2)

Sample Problem• 8.0 g of oxygen gas is at a pressure of 2.0x102 kPa (ie:

200KPa) and a temperature of 15°C. How many litres of oxygen are there? (give answer to 2 significant digits)

Data:P = 200 kPa R = 8.31 L·kPa/K·molT = 15+273 = 288Km(O2)= 8.0 gM(O2)= 32.0 g/moln =

To find:V

molmolg

g 25.0/32

8

Formula:

Work:

nRTPV

KmolKkPaLmolVkPa 28831.825.0200

kPa

KmolKkPaLmol

V200

28831.825.0

LV 99.2

Next slide: Ideal vs. Real

Page 123: Unit 1 (formerly Module 2)

Ideal vs. Real GasesThe gas laws were worked out by assuming that gases are ideal, that is, that they obey the gas laws at all temperatures and pressures. In reality gases will condense or solidify at low temperatures and/or high pressures, at which point they stop behaving like gases. Also, attraction forces between molecules may cause a gas’ behavior to vary slightly from ideal.A gas is ideal if its particles are extremely small (true for most gases), the distance between particles is relatively large (true for most gases near room temperature) and there are no forces of attraction between the particles (not always true)At the temperatures where a substance is a gas, it follows the gas laws closely, but not always perfectly. For our calculations, unless we are told otherwise, we will assume that a gas is behaving ideally. The results will be accurate enough for our purposes!

Next slide: Summary

Page 124: Unit 1 (formerly Module 2)

Testing if a gas is ideal If you know all the important properties of a

gas (its volume, pressure, temperature in kelvin, and the number of moles) substitute them into the ideal gas law, but don’t put in the value of R. Instead, calculate to see if the value of R is close to 8.31, if so, the gas is ideal, or very nearly so. If the calculated value of R is quite different from 8.31 then the gas is far from ideal.

Page 125: Unit 1 (formerly Module 2)

Example

A sample of gas contains 1 mole of particles and occupies 25L., its pressure 100 kPa is and its temperature is 27°C. Is the gas ideal?

Convert to kelvins: 27°C+273=300K PV=nRT (ideal gas law formula) 100kPa25L=1molR300K, so… R=100kPa25L÷(300K1mol) R=8.33 kPaL /Kmol expected value: 8.31 kPaL /Kmol So the gas is not perfectly ideal, but it is very

close to an ideal gas, It varies from ideal by only 0.24%

%24.0%10031.8

)31.833.8(

Page 126: Unit 1 (formerly Module 2)

Gas Laws Overview• When using gas laws, remember that

temperatures are given in Kelvins (K)– Based on absolute zero: –273°C

• The three original gas laws can be combined, and also merged with Avogadro’s mole concept to give us the Combined Gas Law.

• Rearranging the Combined Gas Law and doing a bit of algebra produces the Ideal Gas Law.

• Substituting in the STP conditions we can find the Ideal Gas Constant.

• “Ideal gases” are gases that obey the gas laws at all temperatures and pressures. In reality, no gas is perfectly ideal, but most are very close.

Page 127: Unit 1 (formerly Module 2)

Gas Laws: SummarySimple gas laws– Boyle’s Law: – Charles’ Law:– Gay-Lussac’s Law:

– Combined gas law:

– Ideal gas law:

– The ideal gas constant:

22

22

11

11

TnVP

TnVP

nRTPV

VP 1

2

2

1

1

TV

TV

2211 VPVP 2

2

1

1

TP

TP

TV TP

R=8.31 Lkpa/Kmol

Page 128: Unit 1 (formerly Module 2)

Video• Simple gas laws

Page 129: Unit 1 (formerly Module 2)

Assignments on the Ideal Gas Law

• Read pages 100 to 104in textbook• Do Exercises p. 104 #1 to 16

Page 130: Unit 1 (formerly Module 2)

Lesson 2.7

Stoichiometry of Gases

Page 131: Unit 1 (formerly Module 2)

Stoichiometry of Gases

• When using stoichiometry with gases it is important to remember Avogadro’s hypothesis: that equal volumes of gas under the same conditions of temperature and pressure contain equal number of particles – Ie. At same pressure and temperature

• Same volumes have same # moles• Volumes are proportional to numbers of moles

Page 132: Unit 1 (formerly Module 2)

Simple question 1• How many litres of hydrogen will react with

3 Litres of oxygen to form water if both gases are at the same pressure and temperature?

• 2 H2 + O2 2 H2O• 2 : 1 : 2 molar ratio• ?L : 3L : ?L volume ratio• 6L : 3L : XX proportion solutionAnswer: Six litres of hydrogen gas will react.

In theory, this reaction would produce 6L of water vapour, but because the reaction is highly exothermic, the temperature would go up, so the result for water would be meaningless (hence the XX)

Page 133: Unit 1 (formerly Module 2)

Assignments on the Stoichiometry of Gases

• Read pages 108 to 109 in textbook• Do Exercises p. 110 #1 to 9

Page 134: Unit 1 (formerly Module 2)

Lesson 2.8

Dalton’s Law of partial pressures

Page 135: Unit 1 (formerly Module 2)

John Dalton

Born 6 September 1766 Eaglesfield, Cumberland, England

Died 27 July 1844Manchester, England

Notable students James Prescott JouleKnown for Atomic Theory, Law of

Multiple Proportions, Dalton's Law of Partial Pressures, Daltonism

Influences John Gough

Besides being the founder of modern atomic theory, John Dalton experimented on gases. He was the first to reasonably estimate the composition of the atmosphere at 21% oxygen, 79% Nitrogen

Page 136: Unit 1 (formerly Module 2)

Partial Pressure�Many gases are mixtures, �eg. Air is 78% nitrogen, 21% Oxygen, 1% other gases�Each gas in a mixture contributes a partial

pressure towards the total gas pressure.

�The total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of the individual gases in the mixture.

�101.3 kPa (Pair) = 79.1 kPa (N2)+ 21.2 kPa (O2) + 1.0 kPa(Other)

Next slide:

Page 137: Unit 1 (formerly Module 2)

Kinetic Theory Connection

• Hypothesis 3 of the kinetic theory states that gas particles do not attract or repel each other.

• Dalton established that each type of gas in a mixture behaved independently of the other gases.

• The pressure of each gas contributes towards the total pressure of the mixture.

Page 138: Unit 1 (formerly Module 2)

Dalton’s LawThe Law of Partial Pressures of Gases

Where: PT is the total pressure of mixed gasesP1 is the pressure of the 1st gasP2 is the pressure of the 2nd gasetc...

...21 PPPT

Page 139: Unit 1 (formerly Module 2)

Variant of Dalton’s Law(used for finding partial pressure of a gas in a mixture)

TT

AA P

nnP

Where: PA=Pressure of gas AnA = moles of gas AnT= total moles of all gasesPT= Total Pressure of all gases

Page 140: Unit 1 (formerly Module 2)

Uses of Dalton’s Law In the 1960s NASA used the law of partial

pressures to reduce the launch weight of their spacecraft. Instead of using air at 101 kPa, they used pure oxygen at 20kPa.

Breathing low-pressure pure oxygen gave the astronauts just as much “partial pressure” of oxygen as in normal air.

Lower pressure spacecraft reduced the chances of explosive decompression, and it also meant their spacecraft didn’t have to be as strong or heavy as those of the Russians (who used normal air).. This is one of the main reasons the Americans beat the Russians to the moon.

Story

Don’t copy

Page 141: Unit 1 (formerly Module 2)

Carelessness with pure oxygen, however, lead to the first major tragedy of the American space program…

At 20 kPa, pure oxygen is very safe to handle, but at 101 kPa pure oxygen makes everything around it extremely flammable, and capable of burning five times faster than normal.

On January 27, 1967, during a pre-launch training exercise, the spacecraft Apollo-1 caught fire. The fire spread instantly, and the crew died before they could open the hatch.

Page 143: Unit 1 (formerly Module 2)

Gus Grissom, Ed White, Roger Chaffee

Crew of Apollo 1

Page 144: Unit 1 (formerly Module 2)

Exercises:• Page 113 in new textbook, # 1 to 8

Extra practice (if you haven’t already started):• Study guide: pp 2.12 to 2.17 # 1 to 22

– There is an answer key in the back for these– Do these on your own as review

Page 145: Unit 1 (formerly Module 2)

Summary:• Dalton’s Law: The total pressure of a

gas mixture is the sum of the partial pressures of each gas.

PT = P1 + P2 + …• Graham’s Law: light molecules diffuse

faster than heavy ones

• Avogadro’s hypothesis– A mole of gas occupies 22.4L at STP and

contains 6.02x1023 particles

1

2

2

1

MM

RateRate

Page 146: Unit 1 (formerly Module 2)

Summary of Kinetic Theory• Hypotheses (re. Behaviour of gas molecules):

1. Gases are made of molecules moving randomly 2. Gas molecules are tiny with lots of space between.3. They have elastic collisions (no lost energy).4. Molecules don’t attract or repel each other (much)

• Results:• The kinetic energy of molecules is related to their

temperature (hot molecules have more kinetic energy because they move faster)

– Kinetic theory is based on averages of many molecules (graphed on the Maxwell distribution “bell” curve)

– Pressure is caused by the collision of molecules with the sides of their containers.

– Hotter gases and compressed gases have more collisions, therefore greater pressure.

Page 147: Unit 1 (formerly Module 2)

Energy of a particle:

KE = ½ mV 2

Pressure is the result of particles colliding with the container walls.

P = F /A

Gases are made of particlesParticles move randomly!

Pressure

Page 148: Unit 1 (formerly Module 2)

• The end of module 2


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