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## Chemistry I Honors

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**Chemistry I Honors**Unit 9: Gases**Objectives #1-3: Introduction to the Kinetic Theory of Gases**• The Kinetic Theory • Assumptions of the Kinetic Theory • Gases are composed of tiny particlesthat are arranged far apart from each other • Gases are composed of individual atoms such as in the element neon or molecules such as in the element oxygen**Objectives #1-3: Introduction to the Kinetic Theory of**Gases • Collisions that may occur between gas particles are elastic with no net loss of energy • Gas particles are in constant, random motion • No attractiveforces exist between gas particles • The temperatureof a gas depends on the average kinetic energy of the particles • (video clip Ch. 3 “Common Gas Properties”)**Objectives #1-3: Introduction to the Kinetic Theory of**Gases • The Kinetic Theory and its Implications on the Properties of Gases • The temperature of a gas is directlyrelated to its kinetic energy • The temperature at which no kinetic energy is present is called absolute zero; this temperature is 0 on the Kelvin scale and -273oon the Celsius scale • Gases are able to expand freely due to their randommotion and the lack of attractive forces between the particles of a gas • Gas particles are tiny and can move past each other with ease due to the lack of attractive forces between gas particles**Objectives #1-3: Introduction to the Kinetic Theory of**Gases • The density of gases is generally less than other substances due to the ability of gases to movefreely • Gases can be compressed due to the great distances between gas particles • The ability of gases to spread out spontaneously or diffuseis due to the rapid motion of gas particles; particles with less mass and faster velocities spread out at a faster rate than heavier, slower gas particles (This is why you smell cookies baking! )**Objectives #1-3: Introduction to the Kinetic Theory of**Gases • Due to their small size, gas particles can be made to pass through smallopenings; this characteristic of effusion depends on the velocity and molar mass of the gas particles present; the rate of effusion is directly proportional to the velocity of the gas particles and inversely proportional to the molar mass of the gas particles • When gas particles collide against a surface they exhibit pressure; which is defined as the force per unit area**Objectives #4-9 : Properties of Gases**• Common Units of Pressure:**Objectives #4-9: Properties of Gases**• Pressure Conversion Examples: • Convert 2.5 atm to mmHg 2.5 atm x 760 mmHg = 1900 mmHg 1 atm • Convert 300. Pa to atm 300. Pa x 1 atm_____ = .00296 atm 101325 Pa**Objectives #4-9: Properties of Gases**Dalton’s Law of Partial Pressure • The total pressureof a mixture of gases is equal to the sum of the partial pressuresof each of the individual gases in the mixture PT = P1 + P2 + P3 + …. • This concept must be kept in mind when gases are collected over water in the laboratory; the vapor pressure of water must be subtracted from the measured pressure of the gas in order to obtain the true pressure of the gas being collected Pgas = Patm - Pwater**Objectives #4-9: Properties of Gases**• The vapor pressure due to water increases with increasing temperature (see attached chart) Molar Volume • at standard temperature and pressure (STP), which are defined as O o C and 1atm, 1 mole of any gas occupies 22.4 L; this is referred to as molar volume**Examples:**• Which substance has the highest vapor pressure? • Which substance has the lowest vapor pressure? • Boiling occurs when the vapor pressure of a substance equals the external atmospheric pressure. At what external pressure will ethanol boil at 40oC? • A substance is volatile when it readily evaporates. Which substance is the most volatile? 5. Which substance has the strongest intermolecular forces? 6. Which substance has the weakest intermolecular forces?**Objectives #4-9: Properties of Gases**• Relationships Among Gas Characteristics • Amount of Gas vs. Pressure (assumes temperature and volume are held constant) • What is happening at the particle level:**Objectives #4-9: Properties of Gases**• As the amount of gas in a container increases, the pressure increases • This illustrates a directrelationship between the amt. of the gas and the pressure of the gas**Objectives #4-9: Properties of Gases**• Pressure vs. Volume of Gases (assumes constant temperature) • What is happening at the particle level:**Objectives #4-9: Properties of Gases**• As the pressureof a fixed amount of gas increases, its volume decreases • This illustrates an inverse relationship**Objectives #4-9: Properties of Gases**• Temperature vs. Volume (assumes constant pressure) • What is happening at the particle level:**Objectives #4-9: Properties of Gases**• As the Kelvin temperature of a fixed amount of gas increases, its volume increases • This illustrates a direct relationship:**Objectives #4-9: Properties of Gases**• Temperature vs. Pressure (assuming constant volume) • What is happening at the particle level:**Objectives #4-9: Properties of Gases**• As the Kelvin temperatureof a fixed amount of gas increases, its pressure increases • This illustrates a direct relationship**Objectives #4-9: Properties of Gases**Ideal vs. Real Gases • Ideal gases always follow the kinetictheoryunder any conditions • Real gas particles do have attractiveforces among each other • Real gases no longer act as ideal gases under conditions of highpressure and extremely lowtemperature**Objective #10 Solving problems Involving the Gas Laws***the Gas Laws *the Gas Laws are mathematical formulas based on the relationships discussed in the previous section of notes P1V1 / T1 = P2V2 / T2 (Combined Gas Law) PV = nRT (Ideal Gas Law) (video clip “Ch. 4 Ideal Gas Law) • Combined Gas Law *examples:**To what pressure must a gas be compressed to get it into a**9.00 L tank if it occupies 90.0 L at 1.00 atm? Can a variable be eliminated? P1V1 = P2V2 P1V1 / V2 = P2 (1.00 atm) (90.0 L) / 9.00 L = P2 10.0 atm = P2**A container with a movable piston contains .89 L of methane**gas at 100.50C. If the temperature of the gas drops to 11.3oC, what is the new volume of the gas? K = C + 273 = 100.5oC + 273 = 374 K**K = C + 273**= 11.3oC + 273 = 284 K Can a variable be eliminated? V1 / T1 = V2 / T2 V1T2 = T1V2 V1T2 / T1 = V2 (.89 L) (284 K) / 374 K = V2 .676 L = V2**3. A sample of gas occupies a volume of 5.0 L at a pressure**of 650. torr and a temperature of 24oC . We want to put the gas in a 100. ml container that can only withstand a pressure of 3.0 atm. What temperature must be maintained so that the container does not explode?**5.0 L = 5000 ml**650. torr = .855 atm 24oC = 297 K P1V1 / T1 = P2V2 / T2 Can a variable be eliminated? P1V1T2 = T1P2V2 T2 = T1P2V2 / P1V1**T2 = T1P2V2 / P1V1**T2 = (297 K) (3.0 atm) (100. ml) / = (.855 atm) (5000 ml) = 21 K 4. A sample of gas occupies 2.00 L at STP. What volume will it occupy at 27oC and 200. mmHg?**P1V1 / T1 = P2V2 / T2**Can a variable be eliminated? P1V1T2 = T1P2V2 P1V1T2 / T1P2 = V2 (760 mm Hg) (2.00 L) (300 K) / (273 K) (200. mm Hg) = V2 8.35 L = V2**Additional Problems***Recall that at STP conditions, 1 mole of any gas = 22.4 L *examples: • Calculate the volume of .55000 moles of gas at STP. .55000 moles X 22.4 L / 1 mole = 12.320 L**Calculate the moles of gas contained in 350 L of gas at STP.**350 L X 1 mole / 22.4 L = 16 moles • Calculate the mass in grams of 3.50 L of chlorine gas. 3.50 L X 71.0 g / 22.4 L = 11.1 g**Objective #11 The Ideal Gas Law and Gas Stoichiometry**PV=nRT *ideal gas constant: R = PV / nt R = ( 1 atm) (22.4 L) / (1 mole) (273K) = .0821 L . atm. / mole.K *examples: • What is the temperature of a .65 L sample of fluorine gas at 620. torr which contains 1.3 mol fluorine? T = PV / nR***examples:**• What is the temperature of a .65 L sample of fluorine gas at 620. torr which contains 1.3 mol fluorine? T = PV / nR = (620 torr / 760 torr) (.65 L) / (1.3 mol) (.0821 L.atm / mol.K) = 5.0 K**A 25.0 gram sample of argon gas is placed inside a container**with a volume of 10.0 L at a temperature of 65oC. What is the pressure of argon at this temperature? PV=nRT P = nRT / V = (25.0 g / 39.9 g) (.0821) (338 K) / 10.0 L = 1.74 atm.***there are two types of gas stochiometry problems:**at STP non STP *examples:**Calculate the volume of hydrogen gas that can be produced**from the reaction of 5.00 g of zinc reacted in an excess of hydrochloric acid. Assume STP conditions. Zn + 2HCl --› H2 + ZnCl2 5.00 g Zn X 1 mole Zn / 65.4 g Zn X 1 mole H2 / 1 mole Zn X 22.4 L / 1 mole H2 = 1.71 L**Calculate the volume in liters of oxygen gas that can be**produced from the decomposition of 3.50 X 1024 formula units of potassium chlorate. Assume STP conditions. 2KClO3 --› 2KCl + 3O2 3.50 X 1024 formula units KClO3 X 1 mole KClO3 / 6.02 X 1023 f. units X 3 mole O2 / 2 mole KClO3 X 22.4 L / 1 mole = 195 L**Calculate the volume of hydrogen produced at 1.50 atm and**19oC by the reaction of 26.5 g of calcium metal with excess water. The vapor pressure of water is 16.5 mmHg. Ca + 2H2O --› H2 + Ca(OH)2***use amount of given reactant and stoichiometry to determine**moles of gas desired in problem: 26.5 g Ca X 1 mole Ca / 40.1 g Ca X 1 mole H2 / 1 mole Ca = .661 mole H2***determine net pressure of gas:**16.5 mmHg X 1atm/760 mmHg = .0217 atm 1.50 atm - .0217 atm = 1.48 atm *use moles of gas found in ideal gas law to calculate volume of gas: PV=nRT V = nRT / P = (.661 moles) (.0821) (292 K) / 1.48 atm = 10.7 L**Calculate the volume of chlorine gas produced at 1.25 atm at**25oC from the reaction of 5.00 g of sodium chloride and an excess of fluorine. 2NaCl + F2 --› 2NaF + Cl2 5.00 g NaCl X 1 mole NaCl / 58.5 g NaCl X 1 mole Cl2 / 2 mole NaCl = .0427 mole Cl2**PV = nRT**V = nRT / P = (.0427 mole) (.0821) (298 K) /1.25 atm = .836 L