Gases and Their Applications

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Presentation transcript:

Gases and Their Applications Module 2 Gases and Their Applications

Lesson 2-1 About Gases

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.

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.

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

Carbon monoxide (CO): Clear, odourless, but very toxic Carbon monoxide (CO): Clear, odourless, but very toxic. Destroys hemoglobin. 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.

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.

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

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 Next slide: Summary

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

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

Chapter 2.1 The Kinetic Theory Moving, moving, moving, Keep those atoms moving

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.

Review Not in text The Particle Model 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. + + -

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!

Particle Motion and Phases of Matter 2.1.1 Page 54 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.

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.

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)

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).

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”

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.”

2.1.3 Page 61 To make the physical laws (derived from kinetic equations) 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.

Kinetic Theory Assumptions about an Ideal Gas 2.1.3 Page 61 Kinetic Theory Assumptions 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. Collisions between particles are perfectly elastic, ie. No energy is lost in collisions.

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 “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.

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, or freeze into a solid. At absolute zero an ideal gas would occupy no space at all.

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.

The mean & median 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 & median can help establish “average” molecules Most molecules “Average” molecules Increasing # molecules “Slow” molecules “Fast” Molecules mode mean Average kinetic energy Increasing kinetic energy

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 KE = mv2 kinetic energy of molecules 2 So, kinetic energy depends on both the speed (v) and on the mass (m) of the molecules.

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

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.

Videos Kinetic Molecular Basketball Average Kinetic Energies 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

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

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

2.2.3 Diffusion Diffusion is the tendency for molecules to move from areas of high concentration to areas of lower concentration, until the concentration is uniform. They do this because of the random motion of the molecules. Effusion is the same process, but with the molecules passing through a small hole or barrier Next slide:

Rate or 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?

Graham’s Law Thomas Graham (c. 1840) studied effusion (a type of diffusion through a small hole) and proposed the following law: “The rate of diffusion of a gas is inversely proportional to the square root of its molar mass.” In other words, light gas particles will diffuse faster than heavy gas molecules, and there is a math formula to estimate how much faster. Next slide: Example

Example of Graham’s Law: How much faster does He diffuse than N2? Nitrogen (N2) has a molar mass of 28.0 g/mol Helium (He) has a molar mass of 4.0 g/mol The difference between their diffusion rates is: Notice the reversal of order! So helium diffuses 2.6 times faster than nitrogen Internet demo of effusion MHe=1x4.0=4 g/mol MN2=2x14.0=28 g/mol Next slide: Avogadro

Chapter 2.3 Pressure of Gases What is Pressure Atmospheric Pressure 100 km 0 kPa Pressure of Gases What is Pressure Atmospheric Pressure Measuring Pressure Graham’s Law 40 km 1 kPa 20 km 6 kPa 10 km 25 kPa Mt Everest 30 kPa 5 km 55 kPa 0 km 101 kPa

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

Atmospheric Pressure This is the force of a 100 km column of air pushing down on us. Standard atmospheric pressure is 1 atm (atmosphere) or 101.3 kPa (kilopascals), or 760 Torr (mmHg) 14.7 psi (pounds per square inch) Pressure varies with: Altitude Weather conditions.

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

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.

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

Chapter 2.4 The Simple Gas Laws Boyle’s Law Charles’ Law Gay-Lussac’s Law

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.” Next slide: Air in Syringe

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

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) 101.3 kPa Millibars (mB) 1013 mB Torr (torr) 760 torr Millimetres Hg (mmHg) 760 mmHg Inches {Hg} 29.9 inHg Pounds per sq. in (psi) 14.7 psi

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. Read- don't copy Next slide: Inside syringe

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 Next slide: PV

This means that: Remember! As the volume decreases, the pressure increases As the volume increases, the pressure decreases The formula for this is: P1 V1 = P2 V2 Where P1, V1 = pressure, volume before P2, V2 = pressure, volume after Remember! Next slide: Example

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 Next slide: Graph of Boyle’s Law

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)  1 2 3 4 5 6 7 8 300 x 2.66 = 800 400 x 2 = 800 500 x 1.6 = 800 600 x 1.33 = 800 700 x 1.14 = 800 100 200 300 400 500 600 700 800 800 x 1 = 800 Pressure (kPa)  Next slide: Real Life Data

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) 5 10 15 20 25 30 35 40 2 4 6 8 10 12 14 16 18 Next slide: Boyle’s Law Experiment or skip to: Lesson 2.3 Charles’ Law:

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)  Pressure (kPa) 

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.

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

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

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

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.

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.

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 1.00 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 = 213.53 kPa 473 mL 1 mark 1 mark Answer: the pressure will be about 214 kilopascals

In a thermonuclear device the pressure of 0. 050 L of gas reaches 4 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? formula P1 • V1 =P2 • V2 1 mark Known P1= 4.0x108kPa V1= 0.050 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 200 000L) of gas

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

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? 1 mark Known P1=5.00x103 kPa V1=0.0500 L P2= 1.00x102 kPa V2= Unknown Formula P1V1=P2V2 1 mark 5.0x103kPa • 0.0500L = 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

Lesson 2.4.2 Charles’ Law The Relationship between Temperature and Volume. “Volume varies directly with Temperature” Next slide: Jacques Charles

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” Formula for Charles’ Law: Where: V1= volume before change T1= temperature before change V2= volume after change T2 = temperature after change Next slide: The Evidence

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. Next slide: Diagram of Cylinder & Piston

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

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 -273°C Next slide: Example

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

Summary: Charles’ law The volume of a gas is directly proportional to its temperature Formula: Graph: Boyle’s law is usually represented by an direct relationship graph (straight line) Video1 Volume (L)  Absolute zero 0°C=273K Temp

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 0.5 litres, what will be the volume of the balloon when it is fully cooled? Known T1=22˚C T2=4˚C V1=0.5 L 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 The balloon will have a volume of 0.47 litres

The balloon will have a volume of 0.71 litres. Although only the answers are shown here, in order to get full marks you need to show all steps of the solution! The balloon will have a volume of 0.71 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. The bag will have a volume of 285mL The volume of air in my lungs will be 2.35 litres

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

Joseph Gay-Lussac (1802) “The pressure of a gas is directly proportional to the temperature (in kelvins) if the volume is kept constant.” Formula for Gay-Lussac’s Law: Born 6 December 1778 Saint-Léonard-de-Noblat Died 9 May 1850 @ Saint-Léonard-de-Noblat NationalityFrench FieldsChemistry Known forGay-Lussac's law Next slide:’

Gay-Lussac’s Law As the gas in a sealed container is heated, the pressure increases. For calculations, you must use Kelvin temperatures: K=°C+273 pressure

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 Known P1= 100kPa V1=310 mL T1=20˚C P2=unknown T2=606˚C ˚Celsius must be converted to kelvins 20˚C = 293 K 606˚C = 879 K =293K =879K Formula: multiply divide x = 87900 ÷ 293 x = 300 Answer: the pressure will be 300 kPa Next slide: T vs P graph

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.

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

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

Avogadro’s Hypothesis Lesson 2.45 Avogadro’s Hypothesis For amount of gas. “The volume of a gas is directly related to the number of moles of gas” Next slide: Lorenzo Romano Amedeo Carlo Avogadro di Quaregna

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.

You already know most of the facts that relate to Avogadro’s hypothesis: 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 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).

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.

Mathematical statements of Avogadro’s Hypothesis Where V1 = volume before V2 = volume after P1=pressure before P2=pressure after n1 = #moles before n2 = #moles after

Reference & Assignment Reference: Textbook pp. 226 to 231 Textbook Assignment: page 241 # 14 – 24 Do these in your assignments folder Jump to next lesson: The combined gas laws Next slide: answers to exercises

Lesson 2.5 1. The Combined Gas Law A.K.A. General Gas Law or Universal Gas Law 2. The Ideal Gas Law Created by combining Boyle’s Law, Charles’ Law and Gay-Lussac’s Law, and adding Avogadro’s mole concept Next slide:

The Combined Gas Law Formula: Memorize #moles before #moles after Next slide: Ideal Gas Law

The neat thing about the combined 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, instead, pressure remains constant, you have Charles’ Law If the temperature is constant, then you have Boyle’s 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.

FYI: Deriving a Formula, The Ideal Gas Law The Ideal Gas Law is derived from the Combined Gas Law in several mathematical steps. First, start with the combined 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:

Remember Standard Temperature & Pressure (STP) Standard Temperature is 0°C (or more to the point, 273K) Standard Pressure is 101.3 kPa (one atmospheric pressure at sea level) At STP one mole of an ideal gas occupies exactly 22.4 Litres OK, You should already know this part. If you don't, record it now!

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

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

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

The Ideal Gas Law Formula Memorize Where P=Pressure (in kPa) V=Volume (in Litres) n= number of moles R= Ideal Gas constant (8.31 LkPa/Kmol) T = Temperature (in Kelvins) Next slide: Sample Problem

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.

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) Formula: PV = nRT Variables: P=200 kPa V=? (our unknown)= x n= 8.0g ÷ 32 g/mol =0.25 mol R=8.31 L·kPa/K·mol (ideal gas constant) T= 15°C + 273 = 288K 200 x = (0.25)(8.31)(288) , therefore x= (0.25)(8.31)(288) ÷ 200=2.99 L There are 3.0 L of oxygen (rounded to 2 S.D.) Next slide: Ideal vs. Real

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 Known P=200 kPa V=unknown = X n= not given R=8.31 L·kPa/K·mol T= 15°C + 273 = 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.)

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

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.

Example A sample of gas contains 1 mole of particles and occupies 25L., its pressure 100kPa is and its temperature is 27°C. Is the gas ideal, nearly ideal, or not even close to 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=8.33kPaL/Kmol (actual value: R=8.31) so the gas is very close to ideal (but not perfectly ideal).

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: Summary R=8.31 Lkpa/Kmol Original 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

Video Simple gas laws

References and Assignments Textbook Chapter 10: pp. 221 to 240 Student Study Guide pp. 2-4 to 2-11 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 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

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).

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 91.2kpa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

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)

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 127.8 kPa Multiply these together Then divide by these

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: 52.7kPaxL=0.375mol8.31Lk•Pa/Kmol298K 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 32g/mol 0.375mol 298K

#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 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

Other laws dealing with gases: Lesson 2.6 Other laws dealing with gases: Dalton’s Law (partial pressure) Graham’s Law (diffusion) Avogadro’s Hypothesis (moles of gas)

The Law of Partial Pressures Dalton’s Law The Law of Partial Pressures 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. P total = P1 + P2 + ... The total pressure exerted by a mixture of gases is equal to the sum of the partial pressures. 101.3 kPa (Pair) = 79.1 kPa (PN2)+ 21.2 kPa (PO2) + 1.0 kPa(P3) Next slide:

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.

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.

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

Avogadro’s Hypothesis “At the same temperature and pressure, equal volumes of gases contain the same number of particles (molecules)” You already know some facts that came from Avogadro’s hypothesis: Avogadro’s Number: 6.02 x 1023 particles per mole We now know that at STP a mole of gas will occupy 22.4 L

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

Exercises Assignment, do in your assignment folder: Page 241 # 31-34 (If you have not shown me the previously assigned exercises; p241 #14-30; hand them in at the same time.) 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 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.

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

Assigned Activities References: Practice problems: 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

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