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Unit 1 (formerly Module 2)

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1 Unit 1 (formerly Module 2)
Gases and Their Applications

2 Lesson 2-1 About Gases 2

3 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

4 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

5 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

6 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 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

7 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

8 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

9 Match the 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 9 Next slide: Summary

10 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) 2 Helium (He) 4 Methane (CH4) 16 Ammonia (NH3) 17 Neon (Ne) 20 Hydrogen Fluoride (HF) 21 “Cyanide“ (HCN) 27 Acetylene (C2H4) 28 Nitrogen (N2) 28 Carbon monoxide 28 Ethane (C2H6) 30 Oxygen (O2) 32 Fluorine (F2) 38 Argon (Ar) 40 Carbon dioxide (CO2) 44 Propane (C3H8) 44 Butane (C4H10) 58 Sulphur 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.

11 Some Gases Classified by Chemical Properties
Combustible gases (combustion /explosion) Reactive- oxidizing Gases (support combustion) Non-Reactive gases Testable property: Burning splint produces “pop” Glowing splint reignites, burning splint grows brighter Burning splint is extinguished, glowing splint is dimmed Other properties: Useful as fuels Cause metals and some other materials to corrode or oxidize. Can improve combustion. Can be used to preserve foods by slowing oxidation Examples: Hydrogen (H2) Methane (CH4) Propane (C3H8) Acetylene (C2H4) Oxygen (O2) Fluorine (F2) Chlorine (Cl2) Nitrous Oxide (NO2) Carbon dioxide (CO2) Nitrogen (N2) Argon (Ar) Helium (He)

12 Textbook Assignments Read Chapter 1: pp. 37 to 50
Do the exercises on pages 51 and 52 Questions # 1 to 22 12

13 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. CFC’s Sulfur dioxide 13

14 Chapter 2 The Kinetic Theory “Moving, moving, moving,
Keep those atoms moving...” 14

15 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.

16 The Kinetic Theory of Gases (AKA: The Kinetic Molecular Theory)
2.1 Page 54 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. 16

17 R The Particle Model Not 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

18 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

19 Particle Motion and Phases of Matter
2.1.1 Page 54 Particle Motion and Phases of Matter R 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. 19

20 Kinetic Theory Model of States
Liquid Particles vibrate, move and “flow”, but cohesion (molecular attraction) keeps them close together. Gas Particles move freely through container. The wide spacing means molecular attraction is negligible. Solid Particles vibrate but don’t “flow”. Strong molecular attractions keep them in place. 20

21 Kinetic Motion of Particles
2.1.1 Page 55 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) 21

22 Kinetic Theory and Solids & Liquids
2.1.1 Page 56 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). 22

23 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

24 Kinetic Theory and the Ideal Gas
2.1.3 Page 61 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.” 24

25 2.1.3 Page 61 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. 25

26 Kinetic Theory Hypotheses about an Ideal Gas
2.1.3 Page 61 Kinetic Theory Hypotheses about an Ideal Gas The particles of an ideal gas are infinitely small, so the size is negligible compared to the volume of the container holding the gas. The particles of an ideal gas are in constant motion, and move in straight lines (until they collide with other particles) The particles of an ideal gas do not exert any attraction or repulsion on each other. The average kinetic energy of the particles is proportional to the absolute temperature. 26

27 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

28 Other “Imaginary Features” of Ideal Gas
2.1.3 Page 61 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. 28

29 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

30 The mean & mode can help establish “average” molecules
The range of kinetic energies can be represented as a sort of “bell curve.” Maxwell’s Velocity Distribution Curve. The mean & mode can help establish “average” molecules Most molecules “Average” molecules Increasing # molecules “Slow” molecules “Fast” Molecules mode mean Average kinetic energy Increasing kinetic energy 30

31 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 2 So, kinetic energy depends on both the speed (v) and on the mass (m) of the molecules. 31

32 Distribution of Particles Around Average Kinetic Energies.
Average kinetic energy of molecules Average kinetic energy of warmer molecules Average kinetic energy of colder molecules Number of molecules Slower than average molecules Faster than average molecules Kinetic Energy of molecules (proportional to velocity of molecules) 32

33 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: that’s more than there are stars in the universe! The average distance air molecules travel between collisions is about 60nm. m is about the width of a virus. 33

34 Videos Kinetic Molecular Basketball Average Kinetic Energies
Average Kinetic Energies Thermo-chemistry lecture on kinetics 34

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

36 Chapter 2.2 Behaviors of Gases Compressibility Expansion
Diffusion and Effusion Graham’s Law

37 2.2.1 Compressibility: 2.2.2 Expansion:
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. 37

38 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 38 Next slide:

39 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?  heavy particle  light particle 39

40 Graham’s Law Internet demo of effusion
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. Where: v1= rate of gas 1 v2= rate of gas 2 M1= molar mass of gas1 M2=molar mass of gas 2 40 Next slide: Example

41 Graham’s Law Version #1, based on Effusion Rate
The relationship between the rate of effusion or diffusion and the molar masses is: Note: See the inversion of the 1 and 2 in the 2nd ratio! 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)

42 And in my spare time I invented dialysis, which has saved the lives of thousands of kidney patients
Thomas Graham ( ) 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: m1v12 = m2v22. A bit of algebra then gave him his famous law.

43 Graham’s Law Version #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: Note: In this variant law, the relationship is not inverted! 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.

44 Example of Graham’s Law: How much faster does He diffuse than N2?
MN2=2x14.0=28 g/mol 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 MHe=1x4.0=4 g/mol 44 Next slide: 2.3 Pressure of Gases

45 Assignments Read pages 63 to 67 Do Questions 1 to 10 on page 68

46 Chapter 2.3 Pressure of Gases What is Pressure Atmospheric Pressure
Outer Space (immeasurable) Spaceship 1 (2006) X15 (1963) 100 km < kPa Chapter 2.3 Edge of Space Pressure of Gases What is Pressure Atmospheric Pressure Measuring Pressure 40 km 1 kPa Highest Jet 4 kPa 20 km 6 kPa 10 km 25 kPa Mt Everest 31 kPa 5 km 55 kPa 46 Mr. Smith 0 km kPa

47 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 47

48 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

49 Pressure conversions SP =72.0 kPa =1.53 atm
760 mmHg 760 Torr 101.3 kPa 14.7 psi 1013 mB 29.9 inHg Pressure conversions Example 1: convert 540 mmHg to kilopascals Multiply =72.0 kPa Divide Example 2: convert 155 kPa to atmospheres =1.53 atm

50 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

51 the Mercury Barometer 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. 51

52 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.

53 Manometers (Pressure Gauges)
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. Tire gauge (piston manometer) U-tube manometer Pressure gauge (mercury manometer) (aneroid) You Tube manometer

54 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) Atm. pressure Must be in mmHg, not cm or kPa! After you finish, you can convert your answer to kPa, or atm. Or whatever.

55 Manometer Examples on a day when the air pressure is 763mmHg (101
Manometer Examples on 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 Hg Pgas = h= 4 cm 4 cm Open: Pgas(mmHg)=P atm(mmHg) +h (mmHg) Pgas = mm Hg =823 mm Hg Pgas = 6 cm Higher 6 Open: Pgas(mmHg)=P atm(mmHg) -h (mmHg) Pgas = mm Hg =703 mm Hg Pgas = 9 6 cm Lower

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

57 Chapter 2.4 The Simple Gas Laws Other Simple Laws that are a Gas:
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 57

58 Clarke’s Laws of 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

59 Lesson 2.4.1 Boyle’s Law Robert Boyle (1662)
For Pressure and Volume “For a given mass of gas at a constant temperature, the volume varies inversely with pressure.” 59 Next slide: Air in Syringe

60 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

61 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 Pressure Atmospheres (atm) 1 atm Kilopascals (kPa)or(N/m2) kPa = N/m2 Millibars (mB) mB Torr (torr) or mm mercury 760 torr = 760 mmHg Centimetres of mercury 76 cmHg Inches of mercury (inHg) 29.9 inHg (USA only) Pounds per sq. in (psi) psi (USA only) 61

62 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. 62 Next slide: Inside syringe

63 Inside the syringe… Read- don't copy
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) low high 63 Next slide: PV

64 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 slide 64 Next slide: Example

65 Boyle’s Law Relating 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: 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.

66 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 kPa 66 Next slide: Graph of Boyle’s Law

67 Graph of Boyle’s Law The Pressure-Volume Relationship
Boyle’s Law produces an inverse relationship graph. P(kpa) x V(L) 100 x 8 = 800 200 x 4 = 800 Volume (L)  300 x 2.66 = 800 400 x 2 = 800 500 x 1.6 = 800 600 x 1.33 = 800 700 x 1.14 = 800 800 x 1 = 800 67 Pressure (kPa)  Next slide: Real Life Data

68 Example 2: Real Life Data
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) 68 Next slide: Boyle’s Law Experiment or skip to: Lesson 2.3 Charles’ Law:

69 Summary: Boyle’s law P1V1=P2V2
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) P1V1=P2V2 Volume (L)  69 Pressure (kPa) 

70 70

71 Assignments on Boyle’s Law
Read pages 75 to 79 Do questions 1 to 10 on page 97

72 Boyle’s Law Lab Activity
Read, Don’t Write 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.

73 Diagram of Boyle’s Law Apparatus
#1. Horizontal #2 Open end up #3 Open end down

74 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 (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

75 Collecting Data (continued)
Data set 1 - Horizontal Tube: (f) Position of “left” side of column mm (g) Position of closure (h) “volume” of gas (f) – (g) (h) Mm Data set 2 - Open End Up: (i) Position of bottom of column (j) Position of closure should be same as (g) (k) “volume” of gas (i) – (j) (k) mm

76 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) n) “volume” of gas (l) – (m) This concludes the collection of data, now we must process it and calculate the PV (pressure x volume) values at each of the three conditions.

77 Calculations (e) (c) (h) (e)+(c) (k) (e)- (c) (n) Barometric pressure
Item (e) Column Height Item (c) “Pressure” P “Volume” V PV PxV Horizontal (e) (c) (h) Open End Up (e)+(c) (k) Open End Down (e)- (c) (n) Since we are using analogues for pressure & volume, the units don’t matter.

78 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.

79 Answers to Boyle’s Law Sheet
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? 1 mark formula P1 • V1 =P2 • V2 1 mark Known P1= 101 kPa V1= 1.00x103 mL P2= unknown V2= 473 mL 101kPa • 1000 mL = P2 kPa • 473 mL P2 = 101•1000 kPa•mL = kPa 473 mL 1 mark 1 mark Answer: the pressure will be about 214 kilopascals

80 In a thermonuclear device the pressure of 0. 050 L of gas reaches 4
In a thermonuclear device the pressure of 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? formula P1 • V1 =P2 • V2 1 mark Known P1= 4.0x108kPa V1= L P2= 1x102kPa V2=unknown 1 mark 4.0x108kPa • 0.050L = 1x102kPa • V2L V2 = 4x108•0.05 kPa•L = 2.00x105 L 1x102kPa 1 mark 1 mark Answer: there will be 2.00x105Litres (or L) of gas

81 The volume would be 3.33x10-5 Litres
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? Known P1= 1.00 atm V1= 2.00 L P2= 6.0x104 atm V2= unknown 1 mark Formula P1V1=P2V2 1 mark 1.00•2.00 = 6.0•104 • V2 V2 = 2.00 ÷ 6.0x104 V2 = 3.33 x10-5 L or P1=1.01x102kPa, P2=6.06x106kPa. 1 mark The volume would be 3.33x10-5 Litres 1 mark

82 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 L 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? 1 mark Known P1=5.00x103 kPa V1= L P2= 1.00x102 kPa V2= Unknown Formula P1V1=P2V2 1 mark 5.0x103kPa • L = 1x102kPa • V2L V2 = 5x103•0.05 kPa•L = 2.50 L 1x102kPa 1 mark The sudden appearance of 2½ litres of gas in the diver’s bloodstream could be quite deadly. 1 mark

83 Lesson Charles’ Law The Relationship between Temperature and Volume. “Volume varies directly with Temperature” 83 Next slide: Jacques Charles

84 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. Born: November 12, 1746 ( ) Beaugency, Orléanais Died: April 7, ( ) (aged 76), Paris Nationality: France Fields: physics, mathematics, hot air ballooning Institutions: Conservatoire des Arts et Métiers Next slide: The Mathematical formula for this law

85 Charles’ Law Relating 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. 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.

86 Charles Law Evidence Charles 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. 86 Next slide: Diagram of Cylinder & Piston

87 Charles Law Example Piston Cylinder Trapped Gas
Click Here for a simulated internet experiment 87 Next slide: Graph of Charles’ Law

88 Graph of Charles Law Expansion of an “Ideal” Gas
Charles discovered the direct relationship 6L Lord Kelvin traced it back to absolute zero. 5L Expansion of an “Ideal” Gas 4L 3L 2L Expansion of most real gases condensation freeze Liquid state 1L 273°C Solid state -250°C -200°C -150°C -100°C -50°C 0°C 50°C 100°C 150°C 200°C 250°C °C Next slide: Example

89 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 Next slide: Lesson 2.4 Gay Lussac’s Law

90 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

91 (SATP) Ambient Temperature Ambient Temperature = 25°C = 298 K
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 = kPa (SATP)

92 Comparison Standard and Ambient Conditions
Standard Temperature & Pressure (STP) Ambient Temperature & Pressure (SATP) Pressure 101.3 kPa Temperature °C 0 °C 25 °C Temperature K K K Molar Volume 22.4 L/mol 24.5 L/mol

93 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 Volume (L)  Absolute zero 0°C=273K Temp

94 Charles’ Law Worksheet
1. 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 kelvin =295K =277K So: V2=V1 x T2 ÷ T1 V2=0.5L x 277K 295K V2=0.469 L multiply divide 94 The balloon will have a volume of 0.47 litres

95 Answer: The balloon’s volume will be 0.71 litres
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. Data V1= 0.40L T1= 20°C T2= 250°C V2= ? Convert temperatures to kelvin 20+273= 293K, =523k =293 K Use Charles’ Law =523 K 0.7139L 0.40L x 523 K ÷ 293 K = L Answer: The balloon’s volume will be 0.71 litres 95

96 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? Convert temperatures to kelvin 19+273= 292K, =333K Data: V1=250 mL T1= 19.0°C T2=60.0°C V2= ? Use Charles’ Law =292 K =333 K mL 250mL x 333 K ÷ 292 K = mL Answer: The bag will have a volume of 285mL

97 4. The volume of air in my lungs will be 2.35 litres
Although only the answers are shown here, in order to get full marks you need to show all steps of the solution! 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 K, which corresponds to 6.70 C. A jacket or sweater would be appropriate clothing for this weather.

98 Charles’ Law Assignments
Read pages 80 to 84 Do questions 11 to 21 on pages 97 and 98

99 Gay-Lussac’s Law Lesson 2.4.3 For Temperature-Pressure changes.
“Pressure varies directly with Temperature” 99 Next slide:’

100 Joseph Gay-Lussac (1802) “The pressure of a gas is directly proportional to the temperature (in kelvins) if the volume is kept constant.” Born 6 December 1778 Saint-Léonard-de-Noblat Died 9 May Saint-Léonard-de-Noblat Nationality: French Fields: Chemistry Known for Gay-Lussac's law 100100 Next slide:’

101 Gay-Lussac’s Law Relating Pressure and Temperature of a Gas
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.

102 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 102102

103 Remove irrelevant fact
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? Remove irrelevant fact Data: P1= 100kPa V1=310 mL T1=20˚C P2=unknown T2=606˚C ˚Celsius must be converted to kelvins 20˚C = 293 K ˚C = 879 K =293K =879K Formula: multiply divide x = ÷ 293 x = 300 Answer: the pressure will be 300 kPa 103103 Next slide: T vs P graph

104 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 be used to find the value of absolute zero. 104104

105 Graph of Pressure-Temperature Relationship (Gay-Lussac’s Law)
Pressure (kPa)  Temperature (K)  105105 273K Next slide:’

106 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) Pressure  Absolute zero 0°C=273K Temp

107 Assignment on Gay-Lussac’s Law
Read pages 85 to 87 Answer questions #22 to 30 on page 98

108 Avogadro’s Law Lesson 2.4.4 For amount of gas.
“The volume of a gas is directly related to the number of moles of gas” 108108 Next slide: Lorenzo Romano Amedeo Carlo Avogadro di Quaregna

109 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, 1856 Field: Physics University of Turin Known for Avogadro’s hypothesis, Avogadro’s number.

110 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). 110

111 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. 111

112 Avogadro’s Laws Relating Moles of Gas to Volume or Pressure
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 112

113 Assignments on Avogadro’s Law
Read pages 92 to 96 Do Questions 31 to 36 on page 98 113

114 The General Gas Law and the Ideal Gas Law
Lesson 2.5 The General Gas Law and the Ideal Gas Law 114 Next slide:

115 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) = General Gas Law 115

116 The General Gas Law Relating all the Simple Laws Together
Memorize 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.

117 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: If the temperature is constant, then you have Boyle’s law. If, instead, pressure remains constant, you have Charles’ Law And finally, if the volume stays constant, then you have Gay-Lussac’s Law Most of the time, the number of moles stays the same, so you can remove moles from the equation. 117

118 FYI: Deriving a Formula,
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) . FYI: Deriving a Formula, no need to copy all of it Next slide:

119 Remember Standard Temperature & Pressure (STP)
Standard Temperature is 0°C or more to the point, 273K = 25°C = 298K) Standard Pressure is kPa (one atmospheric pressure at sea level) At STP one mole of an ideal gas occupies exactly 22.4 Litres = 24.5 L) OK, You should already know this part. If you don't, record it now!

120 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 Next slide: R-- The Ideal Gas Constant

121 The Ideal Gas Constant is the proportionality constant that makes the ideal gas law work
The Ideal Gas Constant has the symbol R R=8.31 L· kPa / K·mol The Ideal Gas constant is 8.31 litre-kilopascals per kelvin-mole. Memorize Next slide: Ideal Gas Formula

122 FYI: Deriving a Formula,
no need to copy all of it 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

123 The Ideal Gas Law Relating Conditions to the Ideal Gas Constant
Where: P=Pressure, in kPa V=Volume, in Litres n= number of moles, in mol R= Ideal Gas constant, 8.31 LkPa/Kmol T = Temperature, in kelvins

124 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.

125 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 = 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.)

126 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? (assume 2 significant digits) Temperature has been converted to kelvins Data: P=200 kPa V=unknown = X n= not given R=8.31 L·kPa/K·mol T= 15°C = 288K --- m (O2) = 8g M (O2) = 32.0 g/mol Calculate the value of n using the mole formula: 0.25 mol 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.)

127 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·mol T = = 288K m(O2)= 8.0 g M(O2)= 32.0 g/mol n = To find: V Formula: Work: Next slide: Ideal vs. Real

128 Ideal vs. Real Gases The 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

129 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.

130 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) 100kPa25L=1molR300K, so… R=100kPa25L÷(300K1mol) R=8.33 kPaL /Kmol expected value: 8.31 kPaL /Kmol So the gas is not ideal, but it is fairly close to an ideal gas, It varies from ideal by only 0.24%

131 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.

132 Gas Laws: Summary R=8.31 Lkpa/Kmol Simple gas laws Boyle’s Law:
Charles’ Law: Gay-Lussac’s Law: Combined gas law: Ideal gas law: The ideal gas constant: R=8.31 Lkpa/Kmol

133 Video Simple gas laws

134 Assignments on the Simple Gas Laws
Finish Exercises p. 99 #37 to 52

135 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

136 Exercise Answers 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.) 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) 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

137 The gas cannot expand, so it exerts force on its container
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 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 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).

138 20) PCO2 = 3.33 kPa, since all the partial pressures will add up to the total pressure ( =33.3) 21) Use Boyle’s law: P1V1=P2V2, therefore 91.2kpa4.0L=20.3kpaxL so therefore x=91.2x4÷20.3 the new volume is 17.9 L 22) Use Boyle’s law: P1V1=P2V2 ,so x=100kPa6L÷25.3kPa. The new volume will be 23.7L

139 23) Use Charles’Law: V1/T1=V2/T2, convert the temperature from °CK, so -50°C223K and 100 °C373K so… 5L/223K = x/373K so… x=5373÷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°C300K. So… 200kPa/300K=223kPa/x. The new temperature will be 61.5°C (converted from 334.5K)

140 ANSWERS 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 kPa Multiply these together Then divide by these

141 27) Data given: need to find: m=12g(O2) M(O2) P=52.7kPa V=x L
R=8.31LkPa/Kmol n in mol T= 25°C T in kelvin Find the number of moles of O2: n=m/M M(O2)=32g/mol so: 12g ÷ 32g/mol = 0.375mol. Convert CK, 25°C+273=298K formula: PV=nRT so: kPaxL=0.375mol8.31Lk•Pa/Kmol298K so: x = (0.375 mol  8.31L•kPa  298 K) • __1_ K mol kpa Answer: The volume will be about 17.6 L 32g/mol 0.375mol 298K

142 #28-30, answers (with brief explanation) (see me at lunch if you need more explanation)
28) Litres at STP 56 L b) 6.72 L c) 7.84 L (remember: each mole of 29) Answer: The pressure will be 1714 kPa (use the formula PV=nRT) 30) Answer: The volume will be 16.8 L

143 Dalton’s Law of partial pressures
Lesson 2.8 Dalton’s Law of partial pressures

144 John Dalton 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 Born 6 September Eaglesfield, Cumberland, England Died 27 July 1844 Manchester, England Notable students James Prescott Joule Known for Atomic Theory, Law of Multiple Proportions, Dalton's Law of Partial Pressures, Daltonism Influences John Gough

145 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) kPa (O2) kPa(Other) Next slide:

146 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.

147 Dalton’s Law The 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...

148 Variant of Dalton’s Law (used for finding partial pressure of a gas in a mixture)
Where: PA=Pressure of gas A nA = moles of gas A nT= total moles of all gases PT= Total Pressure of all gases

149 Uses of Dalton’s Law Story Don’t copy
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.

150 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.

151 Crew of Apollo 1 Gus Grissom, Ed White, Roger Chaffee

152 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

153 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

154 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.

155 Gases are made of particles Particles move randomly!
Pressure Energy of a particle: KE = ½ mV 2 Pressure is the result of particles colliding with the container walls. P = F /A

156 Assigned Activities References: Practice problems:
Read Textbook pp Practice problems: Textbook: p199 #1-3 Student study guide: pp 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.)

157 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

158 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 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

159 The end of module 2

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