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Chemistry I Honors

Chemistry I Honors. Unit 9: Gases. Objectives #1-4: Introduction to the Kinetic Theory of Gases. The Kinetic Theory Assumptions of the Kinetic Theory Gases are composed of tiny particles that are arranged far apart from each other

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Chemistry I Honors

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  1. Chemistry I Honors Unit 9: Gases

  2. Objectives #1-4: Introduction to the Kinetic Theory of Gases • The Kinetic Theory • Assumptions of the Kinetic Theory • Gases are composed of tinyparticlesthat 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

  3. Objectives #1-4: 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

  4. Objectives #1-4: 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 attractiveforces between the particles of a gas

  5. Objectives #1-4: 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! )

  6. Objectives #1-4: 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 directlyproportional to the velocity of the gas particles and inverselyproportional 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

  7. Diffusion v Effusion

  8. Objectives #5-8: Relationships Among Gas Characteristics • Common Units of Pressure:

  9. Objectives #5-8: Relationships among Gas Characteristics • 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

  10. Objectives #5-8: Relationships among Gas Characteristics Dalton’s Law of Partial Pressure • The total pressure of a mixture of gases is equal to the sum of the partial pressures of 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 subtractedfrom the measured pressure of the gas in order to obtain the true pressure of the gas being collected Pgas = Patm - Pwater

  11. Interpreting Vapor Pressure Charts

  12. Objectives #5-8: Relationships among Gas Characteristics • 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

  13. Relationships Among Gas Characteristics

  14. Objectives #5-8: Relationships among Gas Characteristics • Relationships Among Gas Characteristics • Amount of Gas vs. Pressure (assumes temperature and volume are held constant) • What is happening at the particle level: How can you tell the can is full? Gases expand to fit the space, so increasing the particles makes the can heavier !

  15. Objectives #5-8: Relationships among Gas Characteristics • 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 Empty can = less pressure, so the can is lighter weight

  16. Objectives #5-8: Relationships Among Gas Characteristics • Pressure vs. Volume of Gases (assumes constant temperature) • What is happening at the particle level: Why are balloons “over-filled” with helium?

  17. Objectives #5-8: Relationships among Gas Characteristics • As the pressureof a fixed amount of gas increases, its volume decreases • This illustrates an inverse relationship

  18. Objectives #5-8: Relationships among Gas Characteristics • Temperature vs. Volume (assumes constant pressure) • What is happening at the particle level: Why does tires tend to look like they are low in air on a cold winter morning? Cold night temperatures result in slower movement of the gas particles, so the volume of the gas SEEMS to decrease. As the car moves, friction of the tire on the road increases the temp of the air in the tire.

  19. Objectives #5-8: Relationships Among Gas Characteristics • As the Kelvin temperature of a fixed amount of gas increases, its volume increases • This illustrates a direct relationship:

  20. Objectives #5-8: Relationships Among Gas Characteristics • Temperature vs. Pressure (assuming constant volume) • What is happening at the particle level: Keeping & transporting unopened pop cans in the car in the summer is NOT advisable. Why ? Increasing the temperature in the car increases the pressure in the can. If the pressure of the gas is greater than the pressure exerted by the wall of the can, the can will explode!

  21. Objectives #5-8: Relationships Among Gas Characteristics • As the Kelvin temperature of a fixed amount of gas increases, its pressure increases • This illustrates a direct relationship

  22. Objectives #5-8: Relationships Among Gas Characteristics 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

  23. Objective #9 Solving Problems Involving the Gas Laws *The Gas Laws are mathematical formulas based on the relationships discussed in the previous section of notes… Combined Gas Law: P1V1 = P2V2 T1 T2 **Note that all temps must be in KELVINS!! Ideal Gas Law: (P)(V) = (n)(R)(T)

  24. I. Combined Gas Law • 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? Step 1: Can a variable be eliminated? Yes; temperature is not a factor… Step 2: Substitute & Solve… P1V1 = P2V2 P2 = P1V1 = (1.00 atm) (90.0 L) V2 9.00 L P2 = 10.0 atm

  25. 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? Step 1: Convert temps!! K = C + 273 = 100.5oC + 273 = 374 K, AND K = C + 273 = 11.3oC + 273 = 284 K

  26. Step 2: Can a variable be eliminated? Yes; pressure is not a factor… Step 3: Substitute and solve… V1 = V2 T1 T2 (V1)(T2) = (T1)(V2) V1T2 = V2 T1 (.89 L) (284 K) = V2 374 K .676 L = V2

  27. 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? Step 1: Standardize units!! 5.0 L = 5000 ml 650. torr = .855 atm 24oC = 297 K

  28. Step 2: Can a variable be eliminated? No, all components are used… Step 3: Substitute and solve… P1V1 = P2V2 T1 T2 P1V1T2 = T1P2V2 T2 = T1P2V2 = (297 K)(3.0 atm)(100. ml) P1V1 (.855 atm) (5000 ml) T2 = 21 K

  29. 4. A sample of gas occupies 2.00 L at STP. What volume will it occupy at 27oC and 200. mm Hg? P1V1 = P2V2 T1 T2 Can a variable be eliminated? (P1)(V1)(T2) = (T1)(P2)(V2) P1V1T2 = V2 T1P2 (760 mm Hg) (2.00 L) (300 K) = V2 (273 K) (200. mm Hg) 8.35 L = V2

  30. II. 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 = 12.320 L 1 mole

  31. Calculate the moles of gas contained in 350 L of gas at STP. 350 L x 1 mole = 16 moles 22.4 L • Calculate the mass in grams of 3.50 L of chlorine gas. 3.50 L x 71.0 g = 11.1 g 22.4 L

  32. Objective #11: The Ideal Gas Law and Gas Stoichiometry Ideal Gas Law: (P)(V)=(n)(R)(T) R = ideal gas constant, so… R = (P)(V) = (1 atm) (22.4 L) = .0821 L . atm. (n)(t) (1 mole) (273K) mole.K

  33. Examples: • What is the temperature of a .65 L sample of fluorine gas at 620. torr which contains 1.3 mol fluorine? T = PV (620 torr/760 torr/atm = 0.82 atm ) nR = (0.82atm ) (.65 L) (1.3 mol) (.0821 L.atm / mol.K) = 5.0 K

  34. 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.

  35. Gas Stoichiometry There are two types of gas stochiometry problems: 1) at STP (Standard Temp & Pressure) 2) non STP

  36. Gas Stoichmetry: 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 + 2 HCl H2 + ZnCl2 5.00 g Zn x 1 mole Zn x 1 mole H2 x 22.4 L H2 65.4 g Zn 1 mole Zn 1 mole H2 = 1.71 L H2

  37. 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. 2 KClO3 2 KCl + 3 O2 3.50 x 1024 formula units KClO3x 1 mole KClO3 x3 mole O2 x 22.4 L 6.02 X 1023 f. units 2 mole KClO3 1 mole = 195 L

  38. 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. (Ignore the vapor pressure of water) Ca + 2H2O --› H2 + Ca(OH)2

  39. *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

  40. *use moles of gas found in ideal gas law to calculate volume of gas: PV=nRT V = nRT / P = (.661 moles) (.0821) (292 K) 1.50 L = 10.6 L

  41. 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. 2 NaCl+ F2 2 NaF + Cl2 First, find the moles of Cl2… 5.00 g NaCl x 1 mole NaCl x 1 mole Cl2 58.5 g NaCl 2 mole NaCl = .0427 mole Cl2

  42. Now, use the ideal gas relationship… PV = nRT V = nRT P = (.0427 mole) (.0821) (298 K) 1.25 atm = .836 L

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