Unit 1Unit 1(formerly Module 2)(formerly Module 2)

Gases and Their ApplicationsGases and Their Applications

Lesson 2-1Lesson 2-1

About GasesAbout Gases

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Gas is one of the three main states of matterGas is one of the three main states of matter Gas particles may be Gas particles may be atomsatoms or or moleculesmolecules, ,

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

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

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

Gas is one of the three main states of matterGas is one of the three main states of matter Gas particles may be Gas particles may be atomsatoms or or moleculesmolecules, ,

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

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

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

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General Properties of a GasGeneral Properties of a Gas

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

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

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

can only be compared under specific conditions.can only be compared under specific conditions.

44

Some Important GasesSome Important Gases Oxygen (OOxygen (O22): ): clear, breathable, supports combustion.

Ozone (OOzone (O33): ): poisonous, unstable form of oxygen

Nitrogen (NNitrogen (N22): ): clear, low activity, most abundant gas in the Earth’s atmosphere.

Hydrogen (HHydrogen (H22): ): clear, lighter than air, flammable/explosive

Carbon dioxide (COCarbon dioxide (CO22): ): 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 (SOSulphur dioxide (SO22): ): smelly gas. When it combines with oxygen and water vapour it can form H2SO4, responsible for acid rain.

55

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

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

Ammonia (NHAmmonia (NH33): ): toxic, strong smell, refrigerant . Very soluble in water, forms a basic solution called ammonia-water (NH4OH) which is found in some cleaners.

FreonFreon®® or CFC: or CFC: Non-toxic 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 (CHMethane (CH44): ): flammable gas, slightly lighter than air, produced by decomposition. Found in natural gas. Methane is also a “greenhouse” gas.

Helium (He): Helium (He): inert, lighter than air. Used in balloons and in diver’s breathing mixtures. 66

Acetylene (CAcetylene (C22HH22): ): AKA ethene, it is used as a fuel in welding, lanterns and other devices.

Propane (CPropane (C33HH88): ): used as a fuel in barbecues, stoves, lanterns and other devices.

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

Neon (Ne) Neon (Ne) and and Xenon (Xe)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 Compressed Air (78% N(78% N22, 21% O, 21% O22): ): 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.ials.

77

Fun GasesFun Gases(of no real importance)(of no real importance)

Nitrous OxideNitrous Oxide (N (N22O)O)– AKA: Laughing gas, Happy gas, Nitro, NOSAKA: Laughing gas, Happy gas, Nitro, NOS– Once used as an anaesthetic in dentist offices, this Once used as an anaesthetic in dentist offices, this

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

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

with Sulfur hexafluoridewith Sulfur hexafluoride

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MatchMatchthe gas with the problem it causesthe gas with the problem it causes

GasGas ProblemProblem

Carbon DioxideCarbon Dioxide Ozone layer Ozone layer depletiondepletion

CFCsCFCs Global WarmingGlobal Warming MethaneMethane Toxic poisoningToxic poisoning Carbon monoxideCarbon monoxide Noxious smellNoxious smell Sulfur dioxideSulfur dioxide Acid RainAcid Rain

Next slide: SummaryNext slide: Summary 99

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.

Some Gases Classified by Chemical Properties

Combustible gases(combustion /explosion)

Reactive- oxidizing Gases (support combustion)

Non-Reactive gases

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)

Textbook AssignmentsTextbook Assignments

Read Chapter 1: pp. 37 to 50Read Chapter 1: pp. 37 to 50

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

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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

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Chapter 2Chapter 2

• The Kinetic Theory“Moving, moving, moving,

Keep those atoms moving...”

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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.

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

2.12.1

Page 54Page 54

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The Particle ModelNot in textNot 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.++

++

--

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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!

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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.

2.1.12.1.1

Page 54Page 54

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Kinetic Theory Model of States

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

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

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

2020

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)

2.1.12.1.1

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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).

2.1.12.1.1

Page 56Page 56

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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”

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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.”

2.1.32.1.3

Page 61Page 61

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

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

• The fifth one is not.

2.1.32.1.3

Page 61Page 61

2

2

1: mvEequationkinetic k

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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.

2.1.32.1.3

Page 61Page 61

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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.

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Other “Imaginary Features” Other “Imaginary Features” of Ideal Gasof Ideal Gas

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

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

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

2.1.32.1.3

Page 61Page 61

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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.

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““Slow” Slow”

moleculesmolecules

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

Increasing kinetic energyIncreasing kinetic energyAverage Average

kinetic energykinetic energy

Incr

easi

ng #

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cule

sIn

creasi

ng #

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Most moleculesMost molecules

mod

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mean

mean

““Average”Average”

moleculesmolecules

The mean & mode The mean & mode can help establish can help establish

“average” molecules“average” molecules

““Fast”Fast”

MoleculesMolecules

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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.3131

Distribution of Particles Around Average Kinetic Energies.

Kinetic Energy of moleculesKinetic Energy of molecules(proportional to velocity of molecules)(proportional to velocity of molecules)

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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.

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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

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Assignments

• Read pages 53 to 61

• Do Page 62 # 1-11

Chapter 2.2Chapter 2.2

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

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• 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.

2.2.3 Diffusion2.2.3 Diffusion

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

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

Next slide:Next slide: 3838

Rate of Diffusion or EffusionRate of Diffusion or Effusion

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

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heavy particleheavy particle light particlelight particle

Graham’s LawGraham’s Law Thomas Graham (c. 1840) Thomas Graham (c. 1840)

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

““The rate of diffusion of a gas is The rate of diffusion of a gas is inversely proportional to the inversely proportional to the square root of its molar mass.”square root of its molar mass.” In other words, light gas particles In other words, light gas particles

will diffuse faster than heavy gas will diffuse faster than heavy gas molecules, and there is a math molecules, and there is a math formula to calculate how much formula to calculate how much faster.faster.

Next slide: ExampleNext slide: Example

1

2

2

1

M

M

v

v

4040

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 effusionInternet demo of effusion

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

M

M

v

v

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 1

M1 is the molar mass of gas 1

M2 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!

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 And in my spare time I spare time I

invented invented dialysis, which dialysis, which has saved the has saved the

lives of lives of thousands of thousands of

kidney patientskidney patients

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

M

M

t

t

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 effuse

M1 is the molar mass of the first gas

M2 is the molar mass of the second gas.

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

Example of Graham’s Law: Example of Graham’s Law: How much faster does He diffuse than NHow much faster does He diffuse than N22??

Nitrogen (NNitrogen (N22) has a molar ) has a molar

mass of 28.0 g/molmass of 28.0 g/mol Helium (He) has a molar Helium (He) has a molar

mass of 4.0 g/molmass of 4.0 g/mol The difference between The difference between

their diffusion rates is:their diffusion rates is: Notice the reversal of Notice the reversal of

order!order! So helium diffuses 2.6 So helium diffuses 2.6

times faster than nitrogentimes faster than nitrogen

He

N

N

He

M

M

v

v 2

2

6.22

3.5

/4

/28

molg

molg

MMNN22=2x14.0=28 g/mol=2x14.0=28 g/mol

Next slide: 2.3 Pressure of GasesNext slide: 2.3 Pressure of Gases

MMHeHe=1x4.0=4 g/mol=1x4.0=4 g/mol

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AssignmentsAssignments

Read pages 63 to 67Read pages 63 to 67 Do Questions 1 to 10 on Do Questions 1 to 10 on

page 68page 68

Chapter 2.3Chapter 2.3• Pressure of Gases

– What is Pressure– Atmospheric Pressure– Measuring Pressure

100 km < 0.003 kPa100 km < 0.003 kPa

40 km 1 kPa40 km 1 kPa

20 km 6 kPa20 km 6 kPa

10 km 25 kPa10 km 25 kPa

5 km 55 kPa5 km 55 kPa

0 km 101 kPa0 km 101 kPa

Mt Everest 31 kPaMt Everest 31 kPa

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Highest Jet 4 kPaHighest Jet 4 kPa

Edge of SpaceEdge of SpaceX15 (1963)X15 (1963)Spaceship 1 (2006)Spaceship 1 (2006)Outer Space (immeasurable)Outer Space (immeasurable)

Mr. SmithMr. Smith

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 area

A

FP

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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)

4848

Pressure conversions

)(

)(

)(

)(

wantedunitsSP

wantedunitsP

givenunitsSP

givenunitsP

Example 1: convert 540 mmHg to kilopascals

kPa

P

mmHg

mmHg

3.101760

540 =72.0 kPa

Example 2: convert 155 kPa to atmospheres

atm

P

kPa

kPa

00.13.101

155 =1.53 atm

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

Multiply

DivideDivide

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”5050

• 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

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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.

• 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 gaugeU-tube manometer Pressure gauge

(mercury manometer) (aneroid)(mercury manometer) (aneroid) You Tube manometerYou Tube manometer

Manometers (Pressure Gauges)

Tire gaugeTire gauge

(piston manometer)(piston manometer)

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 Must be in in

mmHg, mmHg, not cm not cm or kPa!or kPa!

Atm. Atm. pressurepressure

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

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

Hgmm

Hgmm3.53.101

760

40

Open: Pgas(mmHg)=P atm(mmHg) +h (mmHg)

Pgas = 763 + 60mm Hg =823 mm Hg

Pgas = kPakPaHgmm

Hgmm7.1093.101

760

823

Open: Pgas(mmHg)=P atm(mmHg) -h (mmHg)

Pgas = 763 - 60mm Hg =703 mm Hg

Pgas = kPakPammHg

mmHg7.933.101

760

703

6 cm Higher

6 cm Lower

4 cm

6

9h= 4 cm

AssignmentsAssignments

• Read pages 69 to 73.

• Do Page 74, Questions 1 to 4.

Chapter 2.4Chapter 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:– Cole’s Law Relates thinly sliced cabbage

to vinegar– Murphy’s Law Anything that can go wrong will. – Clarke’s Laws Relates possible and

impossible

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Clarke’s LawsClarke’s Lawsof the impossible*of the impossible*

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

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

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

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

Lesson 2.4.1Lesson 2.4.1

Boyle’s LawBoyle’s LawRobert Boyle (1662)Robert Boyle (1662)

Lesson 2.4.1Lesson 2.4.1

Boyle’s LawBoyle’s LawRobert Boyle (1662)Robert Boyle (1662)

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

varies inversely with pressure.”varies inversely with pressure.”

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

varies inversely with pressure.”varies inversely with pressure.”

For Pressure and Volume

VP

1

Next slide: Air in SyringeNext slide: Air in Syringe 5959

Robert BoyleRobert Boyle

BBorn: 25 January 1627orn: 25 January 1627Lismore, County Waterford, Ireland Died Lismore, County Waterford, Ireland Died 31 December 1691 (aged 64)31 December 1691 (aged 64)London, EnglandLondon, England

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

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

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

Notable awards: Fellow of the Royal Notable awards: Fellow of the Royal SocietySociety

6060

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

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

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

Standard PressureStandard PressureAtmospheres (atm)Atmospheres (atm) 1 atm1 atmKilopascals (kPa)or(N/mKilopascals (kPa)or(N/m22)) 101.3 kPa = 101.3 N/m101.3 kPa = 101.3 N/m22

Millibars (mB)Millibars (mB) 1013 mB1013 mBTorr (torr)Torr (torr) or mm mercuryor mm mercury 760 torr = 760 mmHg760 torr = 760 mmHgCentimetres of mercuryCentimetres of mercury 76 cmHg76 cmHgInches of mercury (inHg)Inches of mercury (inHg) 29.9 inHg29.9 inHg (USA only)(USA only)Pounds per sq. in (psi)Pounds per sq. in (psi) 14.7 psi14.7 psi (USA (USA

only)only)

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

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

Next slide: Inside syringeNext slide: Inside syringe 6262

Inside the syringe…Inside the syringe…

The harder you press, the smaller The harder you press, the smaller the volume of air becomes. the volume of air becomes. Increasing the pressure makes the Increasing the pressure makes the volume smaller!volume smaller!

The original pressure was low, the The original pressure was low, the volume was large. The new volume was large. The new pressure is higher, so the volume is pressure is higher, so the volume is small.small. Click Here for an internet demo using Click Here for an internet demo using

psi (pounds per square inch) instead of psi (pounds per square inch) instead of kilopascals (1kPa=0.145psi)kilopascals (1kPa=0.145psi)

Next slide: PVNext slide: PV

lowlow

highhigh

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This means that:This means that:

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

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

This assumes that:This assumes that: no more gas enters or leaves the container, and no more gas enters or leaves the container, and that the temperature remains constant.that the temperature remains constant.

The mathematical formula for this is given on The mathematical formula for this is given on the next slidethe next slide

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

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

This assumes that:This assumes that: no more gas enters or leaves the container, and no more gas enters or leaves the container, and that the temperature remains constant.that the temperature remains constant.

The mathematical formula for this is given on The mathematical formula for this is given on the next slidethe next slide

Next slide: ExampleNext slide: Example 6464

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 mL

V2 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.

Example 1Example 1

You have 30 mL of air in a syringe at 100 kPa. You have 30 mL of air in a syringe at 100 kPa. If you squeeze the syringe so that the air If you squeeze the syringe so that the air occupies only 10 mL, what will the pressure occupies only 10 mL, what will the pressure inside the syringe be?inside the syringe be?

PP11 ×× V V11 = P = P22 ×× V V22, so.. , so.. 100 kPa 100 kPa ×× 30 mL = 30 mL = ?? kPa kPa ×× 10 mL 10 mL 3000 mL3000 mL··kPa kPa ÷ ÷ 10 mL = 300 kPa 10 mL = 300 kPa The pressure inside the syringe will be 300 kPaThe pressure inside the syringe will be 300 kPa

Next slide: Graph of Boyle’s LawNext slide: Graph of Boyle’s Law 6666

Graph of Boyle’s LawGraph of Boyle’s LawThe Pressure-Volume RelationshipThe Pressure-Volume Relationship

Pressure (kPa) Pressure (kPa)

Volu

me (

L)

Volu

me (

L)

100 200 300 400 500 600 700 800100 200 300 400 500 600 700 800

1

2

3

4

5

6

7

8

1

2

3

4

5

6

7

8

Boyle’s Law produces Boyle’s Law produces an inverse an inverse relationship graph.relationship graph.

100 x 8 = 800100 x 8 = 800

200 x 4 = 800200 x 4 = 800

400 x 2 = 800400 x 2 = 800

800 x 1 = 800800 x 1 = 800

P(kpa) x V(L)P(kpa) x V(L)

Next slide: Real Life DataNext slide: Real Life Data

300 x 2.66 = 800300 x 2.66 = 800

500 x 1.6 = 800500 x 1.6 = 800

600 x 1.33 = 800600 x 1.33 = 800700 x 1.14 = 800700 x 1.14 = 800

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Example 2: Real Life DataExample 2: Real Life Data

2 4 6 8 10 12 14 16 182 4 6 8 10 12 14 16 18

5

10

15

2

0

25

30

35

40

5

10

15

2

0

25

30

35

40

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

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

0+4= 4kg : 29 mL (116)0+4= 4kg : 29 mL (116)

2+4= 6kg : 20 mL (120)2+4= 6kg : 20 mL (120)

4+4=8kg : 15 mL (120)4+4=8kg : 15 mL (120)

6+4=10kg: 12 mL (120)6+4=10kg: 12 mL (120)

8+4=12kg: 10.5 mL8+4=12kg: 10.5 mL (126)(126)

Next slide: Boyle’s Law Experiment or skip to: Lesson 2.3 Charles’ Law:Next slide: Boyle’s Law Experiment or skip to: Lesson 2.3 Charles’ Law: 6868

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)

Volu

me (

L)

Pressure (kPa) Pressure (kPa)

VP

1

PP11VV11=P=P22VV22

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Assignments on Boyle’s Law

• Read pages 75 to 79

• Do questions 1 to 10 on page 97

Boyle’s Law Lab Activity• We will use the weight of a column of

mercury to compress and expand air (a gas) sealed in a glass tube.

• Read the handout for details of the procedure. (Note: You may shorten the procedure section in your report by including and referring to this handout as part of a complete sentence.)

• You should still write all other report sections (purpose, materials, diagram, observations etc.) in full, as normal.

Read,

Don

’t W

rite

Read,

Don

’t W

rite

#1. Horizontal#1. Horizontal

#2 Open end up#2 Open end up

#3 Open end down#3 Open end down

Diagram of Boyle’s Law Apparatus

Collecting Data

• You will need to find the length of the mercury column with the tube held horizontal:

• You also need this atmospheric information:

(a) Position of “right”side of mercury ___ mm

(b) Position of “left” of mercury column ___ mm

(c) Height of mercury column (a) – (b) (c) mm

(d) today’s temperature* ___ °C

(e) today’s barometric pressure (blackboard) (e) mmHg

*used to calibrate the barometer, not used in calculations*used to calibrate the barometer, not used in calculations

Collecting Data (continued)

(f) Position of “left” side of column mm

(g) Position of closure mm

(h) “volume” of gas (f) – (g) (h) Mm

Data set 2 - Open End Up:

(i) Position of bottom of column mm

(j) Position of closure should be same as (g) mm

(k) “volume” of gas (i) – (j) (k) mm

Data set 1 - Horizontal Tube:Data set 1 - Horizontal Tube:

Collecting Data (continued)

Data set 3 - Open End Down:

l) Position of Top of column mm

m) Position of closure should be same as (g) mm

n) “volume” of gas (l) – (m) mm

This concludes the collection of data, now This concludes the collection of data, now we must process it and calculate the PV we must process it and calculate the PV (pressure x volume) values at each of the (pressure x volume) values at each of the three conditions.three conditions.

Calculations

Barometric pressure

Item (e)

Column Height

Item (c)

“Pressure”

P

“Volume”

V

PV

PxV

Horizontal (e) (c) (e) (h)

Open End Up

(e) (c) (e)+(c) (k)

Open End Down

(e) (c) (e)- (c) (n)

Since we are using analogues for pressure & volume, the units don’t matter.

Conclusion and Discussion

• According to Boyle’s law, the PV values should all be identical. In the real world they will not be identical, but they should be very close.

• Analyze your results. While doing this you should find the percentage similarity between your largest and smallest result (smallest over largest x 100%). This can help you conclude if your results have supported Boyle’s Law or not.

• Discuss sources of error, and explain if they were significant in your results.

• Discuss the meaning of Boyle’s law as it relates to this activity.

Answers to Boyle’s Law Sheet

1. 1.00 L of a gas at standard temperature and pressure (101 kPa) is compressed to 473 mL. What is the new pressure of the gas?

formulaformula

PP11 • V • V11 =P =P22 • V • V22

KnownKnown

PP11= 101 kPa= 101 kPa

VV11= 1.00x10= 1.00x1033 mL mL

PP22= = unknownunknown

VV22= 473 mL= 473 mL

101kPa • 1000 mL = 101kPa • 1000 mL = PP22 kPa • 473 mL kPa • 473 mL

PP22 = = 101•1000 101•1000 kPa•mL = 213.53 kPakPa•mL = 213.53 kPa

473 mL473 mL

Answer: the pressure will be about 214 Answer: the pressure will be about 214 kilopascalskilopascals

1 mark 1 mark

1 mark

1 mark

2. In a thermonuclear device the pressure of 0.050 L of gas reaches 4.0x108kPa. When the bomb casing explodes, the gas is released into the atmosphere where it reaches a pressure of 1.00x102kPa. What is the volume of the gas after the explosion?

KnownKnown

PP11= 4.0x10= 4.0x1088kPakPa

VV11= 0.050 L= 0.050 L

PP22= 1x10= 1x1022kPakPa

VV22=unknown=unknown

formulaformula

PP11 • V • V11 =P =P22 • V • V22

4.0x104.0x1088kPa • 0.050L = 1x10kPa • 0.050L = 1x1022kPa • kPa • VV22LL

VV22 = = 4x104x1088•0.05 •0.05 kPa•L = 2.00x10kPa•L = 2.00x1055 L L

1x101x1022kPakPaAnswer: there will be 2.00x10Answer: there will be 2.00x1055Litres (or 200 000(or 200 000LL)) of gas of gas

1 mark

1 mark

1 mark

1 mark

3. synthetic diamonds can be manufactured at pressures of 6.00x104 atm. If we took 2.00L of gas at 1.00 atm and compressed it to 6.00x104 atm, what would the volume be?

KnownKnown

PP11= 1.00 atm= 1.00 atm

VV11= 2.00 L= 2.00 L

PP22= 6.0x10= 6.0x1044 atm atm

VV22= = unknownunknown

FormulaFormula

PP11VV11=P=P22VV22

1.00•2.00 = 6.0•101.00•2.00 = 6.0•1044 • • VV22 VV22 = 2.00 ÷ 6.0x10 = 2.00 ÷ 6.0x1044

V2 = 3.33 x10-5 L

1 mark1 mark

1 mark

orP1=1.01x102kPa,

P2=6.06x106kPa.

The volume would be 3.33x10The volume would be 3.33x10-5-5 LitresLitres1 mark

4. Divers get the bends if they come up too fast because gas in their blood expands, forming bubbles in their blood. If a diver has 0.0500L of gas in his blood at a depth of 50m where the pressure is 5.00x103 kPa, then rises to the surface where the pressure is 1.00x102kPa, what will the volume of gas in his blood be? Do you think this will harm the diver?

KnownKnown

PP11=5.00x10=5.00x1033 kPa kPa

VV11=0.0500 L=0.0500 L

PP22= 1.00x10= 1.00x1022 kPa kPa

VV22= = UnknownUnknown

FormulaFormula

PP11VV11=P=P22VV22

5.0x105.0x1033kPakPa • 0.0500 • 0.0500LL = 1x10 = 1x1022kPakPa • • VV22LL

VV22 = = 5x105x1033•0.05 •0.05 kPa•L = 2.50 LkPa•L = 2.50 L

1x101x1022kPakPaThe sudden appearance of 2½ The sudden appearance of 2½ litreslitres of gas in the of gas in the

diver’s bloodstream could be quite deadly.diver’s bloodstream could be quite deadly.

1 mark 1 mark

1 mark

1 mark

Lesson 2.4.2Lesson 2.4.2Charles’ LawCharles’ LawLesson 2.4.2Lesson 2.4.2Charles’ LawCharles’ Law

The Relationship between Temperature The Relationship between Temperature and Volume.and Volume.

““Volume varies directly with Temperature”Volume varies directly with Temperature”

Next slide: Jacques CharlesNext slide: Jacques Charles

TV 8383

Jacques Charles (1787)Jacques Charles (1787)

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

Next slide: The Mathematical formula for this lawNext slide: The Mathematical formula for this law

Born: November 12, 1746 Born: November 12, 1746 (1746-11-12) Beaugency, (1746-11-12) Beaugency, Orléanais Orléanais

Died: April 7, 1823 (1823-Died: April 7, 1823 (1823-04-08) (aged 76), Paris 04-08) (aged 76), Paris

Nationality: France Nationality: France

Fields: physics, Fields: physics, mathematics, hot air mathematics, hot air ballooning ballooning

Institutions: Conservatoire Institutions: Conservatoire des Arts et Métiersdes Arts et Métiers

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

T

V

T

V

Where: T1 is Temperature of the gas before it is heated, in kelvins.

T2 is Temperature of the gas after it is heated, in kelvins

V1 is the volume of the gas before it was heated, in L or mL

V2 is the volume of the gas after it was heated, in the same units.

Charles Law EvidenceCharles Law Evidence

Charles used cylinders and pistons to Charles used cylinders and pistons to study and graph the expansion of study and graph the expansion of gases in response to heat.gases in response to heat.

See the next two slides for diagrams See the next two slides for diagrams of his apparatus and graphs.of his apparatus and graphs.

Lord Kelvin (William Thompson) used Lord Kelvin (William Thompson) used one of Charles’ graphs to discover one of Charles’ graphs to discover the value of absolute zero.the value of absolute zero.

Next slide: Diagram of Cylinder & PistonNext slide: Diagram of Cylinder & Piston 8686

Charles Law ExamplePistonPiston

CylinderCylinder

Trapped GasTrapped Gas

Next slide: Graph of Charles’ LawNext slide: Graph of Charles’ Law

Click Here for a simulated Click Here for a simulated internet experimentinternet experiment

8787

Graph of Charles Law

00°C°C 100100°C°C 200200°C°C150150°C°C5050°C°C 250250°C°C

1L1L

2L2L

3L3L

4L4L

5L5L

6L6L

-250-250°C°C -200-200°C°C -150-150°C°C -100-100°C°C -50-50°C°C

-273.15-273.15°C°C

Expansion of an “Ideal” Gas

Expansion of an “Ideal” Gas

Expansion of most real gases 2

73

27

3°° CC

Next slide: ExampleNext slide: Example

Liquid stateLiquid state

Solid stateSolid state condensa

tion

condensa

tion

freeze

freeze

Charles discovered the direct

relationship

Lord Kelvin traced it back

to absolute zero.

ExampleExample

If 2 Litres of gas at 27If 2 Litres of gas at 27°C are heated in a cylinder, °C are heated in a cylinder, and the piston is allowed to rise so that pressure is and the piston is allowed to rise so that pressure is kept constant, how much space will the gas take up kept constant, how much space will the gas take up at 327°C?at 327°C?

Convert temperatures to kelvins: 27°C =300k, Convert temperatures to kelvins: 27°C =300k, 327°C = 600k327°C = 600k

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

K

Litresx

K

Litresso

T

V

T

V

600300

2:,

2

2

1

1

Next slide: Lesson 2.4 Gay Lussac’s LawNext slide: Lesson 2.4 Gay Lussac’s Law

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-standing standard for comparing gases is called Standard Temperature and Pressure (STP)

• Standard Temperature =0°C = 273 K

• Standard Pressure =101.3 kPa

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)

(SATP)

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

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

T

V

T

V

Absolute zeroAbsolute zero 00°°C=273KC=273K TempTempVolu

me (

L)

Volu

me (

L)

TV

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˚C

T2=4˚C

V1=0.50 L

To find:V2= unknown

Temperatures must be converted to kelvinTemperatures must be converted to kelvin

=295K=295K

=277K=277K

2

2

1

1

T

V

T

V

So:V2=V1 x T2 ÷ T1

V2=0.5L x 277K

295KV2=0.469 L

The balloon will have a volume of 0.47 litres

multiply

divide

9494

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.

9595

DataV1= 0.40L

T1= 20°C

T2= 250°C

V2= ?

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

=293 K

=523 KUse Charles’ Law

K

V

K

L

T

V

T

V

523293

4.0... 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 litresAnswer: The balloon’s volume will be 0.71 litres

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 285mLAnswer: The bag will have a volume of 285mL

Data:V1=250 mL

T1= 19.0°C

T2=60.0°C

V2= ?

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

=292 K

=333 K

K

V

K

mL

T

V

T

V

333292

250... 2

2

2

1

1

Use Charles’ Law

250mL x 333 K ÷ 292 K = 285.10mL

285.10 mL

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

Be sure to show your known information

Change the temperature to Kelvins and show them.

Show the formula you used and your calculations

State 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!

Charles’ Law Assignments

• Read pages 80 to 84

• Do questions 11 to 21 on pages 97 and 98

Gay-Lussac’s LawGay-Lussac’s LawFor Temperature-Pressure changes.For Temperature-Pressure changes.

““Pressure varies directly with Temperature”Pressure varies directly with Temperature”

Lesson 2.4.3Lesson 2.4.3

Next slide:’Next slide:’

TP 9999

Joseph Gay-Lussac (1802)Joseph Gay-Lussac (1802)

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

Next slide:’Next slide:’ 101000

Born Born 6 December 1778 6 December 1778

Saint-Léonard-de-NoblaSaint-Léonard-de-Noblatt

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

Nationality:Nationality: FrenchFrench

Fields: Fields: ChemistryChemistry

Known for Known for Gay-Lussac's lawGay-Lussac's law

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

2

2

1

1

T

P

T

P

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.

Gay-Lussac’s LawGay-Lussac’s Law

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

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

pressurepressure

101022

101033

ExampleExample A sealed can contains 310 mL of air A sealed can contains 310 mL of air

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

K

x

K

kPa

879293

100

2

2

1

1

T

P

T

P multiply

x x = 87900 = 87900 ÷ 293÷ 293

xx = 300 = 300Next slide: T vs P graphNext slide: T vs P graph

Data:P1= 100kPa

V1=310 mL

T1=20˚C

P2=unknown

T2=606˚C

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

Answer: the pressure Answer: the pressure will be 300 kPawill be 300 kPa

Remove irrelevant factRemove irrelevant fact

=293K =879K

divide

Formula:

Temperature & Pressure GraphTemperature & Pressure Graph

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

101044

Graph of Pressure-Temperature Graph of Pressure-Temperature RelationshipRelationship

(Gay-Lussac’s Law)(Gay-Lussac’s Law)

Temperature (K) Temperature (K) Pre

ssure

(kP

a)

Pre

ssure

(kP

a)

273K273KNext slide:’Next slide:’ 1010

55

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

T

P

T

P

Absolute zeroAbsolute zero 00°°C=273KC=273K TempTempPre

ssure

Pre

ssure

Assignment on Gay-Lussac’s Law

• Read pages 85 to 87

• Answer questions #22 to 30 on page 98

Avogadro’s LawAvogadro’s LawFor amount of gas.For amount of gas.

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

Lesson 2.4.4Lesson 2.4.4

Next slide: Next slide: Lorenzo Romano Amedeo Carlo Avogadro di QuaregnaLorenzo Romano Amedeo Carlo Avogadro di Quaregna

nV 101088

Lorenzo Romano Amedeo Carlo Lorenzo Romano Amedeo Carlo

Avogadro di QuaregnaAvogadro di Quaregna

““Equal volumes of gas at Equal volumes of gas at the same temperature the same temperature and pressure contain and pressure contain the same number of the same number of moles of particles.”moles of particles.” Amedeo AvogadroAmedeo Avogadro

Born: August 9, 1776 Born: August 9, 1776

Turin, ItalyTurin, Italy

Died: July 9, 1856Died: July 9, 1856

Field: PhysicsField: Physics

University of TurinUniversity of Turin

Known for Avogadro’s Known for Avogadro’s hypothesis, hypothesis, Avogadro’s number.Avogadro’s number.

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

particles (6.02 x 10particles (6.02 x 102323))– That a mole of gas at standard temperature and That a mole of gas at standard temperature and

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

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

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

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

110110

It’s mostly common sense…It’s mostly common sense…

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

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

111111

Avogadro’s LawsRelating Moles of Gas to Volume or Pressure

2

2

1

1

n

V

n

V

2

2

1

1

n

P

n

Por

Where: V1 = volume before, in appropriate volume units.

V2 = volume after, in the same volume units

P1=pressure before, in appropriate pressure units.

P2=pressure after, in the same pressure units.

n1 = #moles before

n2 = #moles after 112112

Assignments on Avogadro’s Law

• Read pages 92 to 96

• Do Questions 31 to 36 on page 98

113113

Lesson 2.5Lesson 2.5

The General Gas Law and the The General Gas Law and the Ideal Gas LawIdeal Gas Law

Next slide:Next slide: 114114

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)

115115

= General Gas Law= General Gas Law

The General Gas LawRelating all the Simple Laws Together

22

22

11

11

Tn

VP

Tn

VP

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 kelvins

n 1, n2 is the number of moles of the gas.

The neat thing about the General gas law is that it The neat thing about the General gas law is that it can replace the three original gas laws.can replace the three original gas laws.

Just cross out or cover the parts that don’t change, Just cross out or cover the parts that don’t change, and you have the other laws:and you have the other laws:

22

22

11

11

Tn

VP

Tn

VP

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

If the temperature is If the temperature is constant, then you have constant, then you have Boyle’s law.Boyle’s law.

If, instead, pressure If, instead, pressure remains constant, you remains constant, you have Charles’ Lawhave Charles’ Law

And finally, if the And finally, if the volume stays constant, volume stays constant, then you have Gay-then you have Gay-Lussac’s LawLussac’s Law

117117

The Ideal Gas LawThe Ideal Gas Law

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

Next slide:Next slide:

22

22

11

11

Tn

VP

Tn

VP

Remember Standard Temperature & PressureRemember Standard Temperature & Pressure(STP)(STP)

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

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

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

The Ideal Gas Law: Calculating the The Ideal Gas Law: Calculating the Ideal Gas Constant.Ideal Gas Constant.

We are going to We are going to calculate a new constant calculate a new constant by substituting in values by substituting in values for Pfor P22, V, V22, T, T22 and n and n22

At STP we know all the At STP we know all the conditions of the gas.conditions of the gas.

Substitute and solve to Substitute and solve to give us a constantgive us a constant

Kmol

LkPa

Tn

VP

2731

4.223.101

11

11

molKkPaLTn

VP /31.8

11

11

Next slide: R-- The Ideal Gas ConstantNext slide: R-- The Ideal Gas Constant

22

22

11

11

Tn

VP

Tn

VP

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

The Ideal Gas Constant has the symbol The Ideal Gas Constant has the symbol RR

R=8.31R=8.31 LL·· kPa kPa // KK··molmol

The Ideal Gas constant is 8.31 litre-The Ideal Gas constant is 8.31 litre-kilopascals per kelvin-mole.kilopascals per kelvin-mole.

Next slide: Ideal Gas FormulaNext slide: Ideal Gas Formula

So, if So, if

Then, by a bit of algebra: PThen, by a bit of algebra: P11VV11=n=n11RTRT11

Since we are only using one set of Since we are only using one set of subscripts here, we might as well remove subscripts here, we might as well remove them: PV=nRTthem: PV=nRT

RTn

VP

11

11

The Ideal Gas LawRelating Conditions to the Ideal Gas Constant

nRTPV Where: P=Pressure, in kPa

V=Volume, in Litres

n= number of moles, in mol

R= Ideal Gas constant, 8.31 LkPa/Kmol

T = Temperature, in kelvins

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

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

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

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

2.0x102.0x1022 kPa kPa (ie: 200 Kpa w. 2 sig fig)(ie: 200 Kpa w. 2 sig fig) and a and a temperature of 15temperature of 15°°C. How many litres of C. How many litres of oxygen are there? Formula: oxygen are there? Formula: PV = nRTPV = nRT

Variables: Variables: P=P=20200 0 kPakPa V=V=?? (our unknown)= (our unknown)= xx n= 8.0n= 8.0gg ÷÷ 32 32 g/molg/mol = =0.25 0.25

molmol R=R=8.31 8.31 LL·kPa/K·mol ·kPa/K·mol (ideal gas (ideal gas

constant)constant)

T= 15T= 15°°C + 273 = C + 273 = 288288KK 200 200 xx = (0.25)(8.31)(288) , = (0.25)(8.31)(288) , thereforetherefore xx= (0.25)(8.31)(288) = (0.25)(8.31)(288) ÷ ÷ 200=200=2.92.99 L9 L There are 3.0 L of oxygen There are 3.0 L of oxygen (rounded to 2 S.D.)(rounded to 2 S.D.)

Sample problemSample problem

molmolg

g

M

mn 25.0

/32

8

8.0 g of oxygen gas is at a pressure of 2.0x102 kPa (ie: 8.0 g of oxygen gas is at a pressure of 2.0x102 kPa (ie: 200KPa) and a temperature of 15200KPa) and a temperature of 15°°C. How many litres C. How many litres of oxygen are there? (assume 2 significant digits)of oxygen are there? (assume 2 significant digits)

Data:Data:

P=P=20200 0 kPakPa

V=V=unknown = unknown = XX

n= not givenn= not given

R=R=8.31 8.31 LL·kPa/K·mol·kPa/K·mol

T= 15T= 15°°C + 273 = C + 273 = 288288KK

------

m (Om (O22) = 8g) = 8g

M (OM (O22) = 32.0 g/mol) = 32.0 g/mol

0.25 0.25 molmol

Temperature has been converted to Temperature has been converted to kelvinskelvins

Calculate the value of n using the mole Calculate the value of n using the mole formula:formula:

nRTPV 200 200 xx = (0.25)(8.31)(288) , = (0.25)(8.31)(288) , thereforetherefore

xx= (0.25)(8.31)(288) = (0.25)(8.31)(288) ÷ ÷ 200=200=2.92.99 L9 L

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

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 g

M(O2)= 32.0 g/mol

n =

To find:V

molmolg

g25.0

/32

8

Formula:

Work:

nRTPV

KmolK

kPaLmolVkPa 28831.825.0200

kPa

KmolKkPaL

molV

200

28831.825.0

LV 99.2

Next slide: Ideal vs. RealNext slide: Ideal vs. Real

Ideal vs. Real GasesIdeal vs. Real GasesThe gas laws were worked out by assuming that gases are The gas laws were worked out by assuming that gases are idealideal, that , that is, that they obey the gas laws at all temperatures and pressures. In 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 reality gases will condense or solidify at low temperatures and/or high pressures, at which point they stop behaving like gases. Also, pressures, at which point they stop behaving like gases. Also, attraction forces between molecules may cause a gas’ behavior to vary attraction forces between molecules may cause a gas’ behavior to vary slightly from ideal.slightly from ideal.

A gas is ideal if its particles are extremely small (true for most gases), 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 the distance between particles is relatively large (true for most gases near room temperature) and there are no forces of attraction between near room temperature) and there are no forces of attraction between the particles (not always true)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: SummaryNext slide: Summary

Testing if a gas is idealTesting if a gas is ideal

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

ExampleExample

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

Convert to kelvins: Convert to kelvins: 2727°°C+273=300KC+273=300K PV=nRTPV=nRT (ideal gas law formula)(ideal gas law formula) 100100kPakPa2525LL==11molmolRR300300KK, so…, so… R=R=100100kPakPa2525LL÷÷((300300KK11molmol)) R=R=8.33 8.33 kPakPaL L //KKmolmol expected value: 8.31 expected value: 8.31 kPa kPaL L //KKmolmol

So the gas is So the gas is notnot ideal, but it is fairly close to an ideal, but it is fairly close to an ideal gas, ideal gas,

It varies from ideal by only 0.24%It varies from ideal by only 0.24%%24.0%100

31.8

)31.833.8(

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.

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

Tn

VP

Tn

VP

nRTPV

VP

1

2

2

1

1

T

V

T

V

2211 VPVP 2

2

1

1

T

P

T

P

TV TP

R=8.31 Lkpa/Kmol

Video

• Simple gas laws

Assignments on the Simple Gas Laws

• Finish Exercises p. 99 #37 to 52

Extra Assignments

• Old text References: – Textbook Chapter 10: pp. 221 to 240– Student Study Guide pp. 2-4 to 2-11

• Old Textbook: page 241 # 25 to 30– Do these in your assignments folder.

Extra practice:• 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

Exercise Answers

14) The pressure will double, since there is twice as much gas occupying the same space. (I answered this using logic and Avagadro’s hypothesis rather than math. It stands to reason that twice as much gas in the same space will increase the pressure.)

15) The pressure will be four times as high, since the volume is one quarter what it was before: P1V1 = P2V2 so… P1V1 = 4P1 x ¼V1 (again, although you can do it with math, logic works better)

16) The pressure will be one third as great as it was before, since there is three times the volume: P1V1 = P2V2, so = 1/3 P1 x 3V1

17) The gas cannot expand, so it exerts force on its container. As the temperature increases, the gas particles move faster, hitting the container sides more frequently and with more force. This causes greater pressure. You can also explain this using Gay-Lussac’s law; P1/T1 = P2/T2

18) Make sure you use the KELVIN temperatures. The formula is P1/T1 = P2/T2 or 300 kpa/300K = xkPa/100K, so the pressure will be 100 kPa

19) An ideal gas obeys the gas laws at all temperatures and pressures (no real gas is perfectly ideal. More ideal properties will be discussed in the next section).

20) PCO2 = 3.33 kPa, since all the partial pressures will add up to the total pressure (3.33+23.3+6.67=33.3)

21) Use Boyle’s law: P1V1=P2V2, therefore 91.2kpa4.0L=20.3kpaxL so therefore x=91.2x4÷20.3 the new volume is 17.9 L

22) Use Boyle’s law: P1V1=P2V2 ,so x=100kPa6L÷25.3kPa. The new volume will be 23.7L

• 23) Use Charles’Law: V1/T1=V2/T2, convert the temperature from °CK, so -50°C223K and 100 °C373K so… 5L/223K = x/373K so… x=5373÷223. The new volume will be about 8.36 L

• 24) Use Gay-Lussack’s law: P1/T1=P2/T2, don’t forget to change 27°C300K. So… 200kPa/300K=223kPa/x. The new temperature will be 61.5°C (converted from 334.5K)

ANSWERS

25) The combined gas laws:(this answer is straight from the lesson)

26) Convert the temperatures to kelvin, set up equation, leaving out n1 and n2 (moles don’t change), cross multiply:

Answer:

The new pressure is 127.8 kPa

22

22

11

11

Tn

VP

Tn

VP

K

LP

K

LkPa

373

7

223

5107 2

Multiply these togetherMultiply these together

Then divide by theseThen divide by these

27) Data given: need to find: m=12g(O2) M(O2)

P=52.7kPa V=x LR=8.31LkPa/Kmol n in molT= 25°C T in kelvin

Find the number of moles of O2: n=m/MM(O2)=32g/mol so: 12g ÷ 32g/mol = 0.375mol. Convert CK, 25°C+273=298K

formula: PV=nRTso: 52.7kPaxL=0.375mol8.31Lk•Pa/Kmol298K

so: x = (0.375 mol 8.31L•kPa 298 K) • __1_ K mol 52.7 kpa

Answer: The volume will be about 17.6 L

3232g/molg/mol

0.3750.375molmol298298KK

#28-30, answers (with brief explanation)(see me at lunch if you need more explanation)

28) Litres at STPa) 56 L b) 6.72 L c) 7.84 L(remember: each mole of gas @STP=22.4L)

29) Answer: The pressure will be 1714 kPa(use the formula PV=nRT)

30) Answer: The volume will be 16.8 L (use the formula PV=nRT)

Lesson 2.8Lesson 2.8

Dalton’s Law of partial pressuresDalton’s Law of partial pressures

John Dalton

Born 6 September 1766 Eaglesfield, Cumberland, England

Died 27 July 1844Manchester, England

Notable students James Prescott Joule

Known 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

Partial PressurePartial 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:Next slide:

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.

Dalton’s LawThe Law of Partial Pressures of Gases

Where: PT is the total pressure of mixed gases

P1 is the pressure of the 1st gas

P2 is the pressure of the 2nd gas

etc...

...21 PPPT

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

TT

AA P

n

nP

Where: PA=Pressure of gas A

nA = moles of gas A

nT= total moles of all gases

PT= Total Pressure of all gases

Uses of Dalton’s Law Uses of Dalton’s Law

In the 1960s NASA used the law of partial In the 1960s NASA used the law of partial pressures to reduce the launch weight of their pressures to reduce the launch weight of their spacecraft. Instead of using air at 101 kPa, they spacecraft. Instead of using air at 101 kPa, they used pure oxygen at 20kPa.used pure oxygen at 20kPa.

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

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

StoryStory

Don’t copy

Don’t copy

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. hatch.

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. hatch.

Gus Grissom, Ed White, Roger ChaffeeGus Grissom, Ed White, Roger ChaffeeGus Grissom, Ed White, Roger ChaffeeGus Grissom, Ed White, Roger Chaffee

Crew of Apollo 1Crew of Apollo 1Crew of Apollo 1Crew of Apollo 1

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

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

M

M

Rate

Rate

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.

Energy of a Energy of a particle:particle:

KE = ½ KE = ½ mV mV 22

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

P = P = F /AF /A

Gases are made of particlesGases are made of particlesParticles move randomly!Particles move randomly!

PressurePressure

Assigned Activities

• References: – Read Textbook pp.197-203

• Practice problems: – Textbook: p199 #1-3– Student study guide: pp. 2-19 to 2-20(practice problems are for self-correction)

• Assignments (to be collected in your folder):– Page 241: all questions from 25 to 34– Handout #1: “combined gas law” #52-58– Handout #2: “gases & gas laws” 5 questions

(on the back.)

Answers (sheet 1)• 52: The volume of gas will be 36.5 L

• 53: The temperature will be 908K or 635C

• 54: The volume will be 250 mL or 0.25L

• 55: The pressure will be 251 kPa

• 56: The pressure will stay the same

• 57: The pressure will be 42.2 kPa

• 58: The volume will be 10.2 L

Answers (sheet 2)

• 1: The volume is about 32.5 L

• 2: The mass is about 1.53 x 10-7 g

• 3: The pressure is about 61909 kPa• 4: The pressure will increase by 168 kPa

(tricky: most students say 268kPa, but that’s what it ends at, NOT how much it changes!)

• 5: The total pressure is about 172kPa

• The end of module 2