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CHAPTER 12

CHAPTER 12. Gases. Diffusion. Particles of 2 or more substances mix spontaneously due to random motion How fast gases diffuse depends on: 1. speed of particles (KE) 2. Size of particles 3. attractive forces between particles. Visual Concepts. Chapter 12. Properties of Gases. Effusion.

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CHAPTER 12

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  1. CHAPTER 12 Gases

  2. Diffusion Particles of 2 or more substances mix spontaneously due to random motion How fast gases diffuse depends on: 1.speed of particles (KE) 2.Size of particles 3.attractive forces between particles

  3. Visual Concepts Chapter 12 Properties of Gases

  4. Effusion • Gas particles leaving a container through a small opening • Movement caused by pressure

  5. Pressure • Collisions on a surface • 1 atm = 760 mm Hg = 760 torr = 101.325 kPa • Can be used as conversion factors. 760 mm Hg 760 torr 1 atm 760 mm Hg 760 torr 101.325 kPa

  6. Visual Concepts Chapter 12 Atmospheric Pressure

  7. Sample Problem A The average atmospheric pressure in Denver, Colorado is 0.830 atm. Express this pressure in millimeters of mercury (mm Hg).

  8. Sample Problem A The average atmospheric pressure in Denver, Colorado is 0.830 atm. Express this pressure in millimeters of mercury (mm Hg). Step 1: Start with what you know 0.830 atm = X mm Hg

  9. Sample Problem A The average atmospheric pressure in Denver, Colorado is 0.830 atm. Express this pressure in millimeters of mercury (mm Hg). Step 2: Determine your ratio of new unit : old unit. 1 atm = 760 mm Hg

  10. Sample Problem A The average atmospheric pressure in Denver, Colorado is 0.830 atm. Express this pressure in millimeters of mercury (mm Hg). Step 3: Set-up so units cancel and solve. 0.830 atm = X mm Hg 1 atm 760 mm Hg X = 631 mm Hg

  11. Sample Problem B The average atmospheric pressure in Denver, Colorado is 631 mm Hg. Express this pressure in kPa.

  12. Sample Problem B The average atmospheric pressure in Denver, Colorado is 631 mm Hg. Express this pressure in kPa. Step 1: Start with what you know 631 mm Hg

  13. Sample Problem B The average atmospheric pressure in Denver, Colorado is 631 mm Hg. Express this pressure in kPa. Step 2: Determine your ratio of new unit : old unit. 101.32 kPa 760 mm Hg

  14. Sample Problem B The average atmospheric pressure in Denver, Colorado is 631 mm Hg. Express this pressure in kPa. Step 3: Set-up so units cancel and solve. 101.32 kPa 760 mm Hg 631 mm Hg X = 84.1 kPa

  15. Temperature • Measure of the average kinetic energy of the molecules • MUST ALWAYS be in Kelvin (K) • ALWAYS! • SERIOUSLY! K = ° C + 273

  16. Combined Gas Law

  17. Combined Gas Law • It’s a Before-After equation • Look for words like “starts”, “begins”, “initial”, etc… • Look for words like “ends”, “final”, “change”, etc… • Look for the units to be mentionned twice • Units don’t matter so long as they match • EXCEPT for Temperature = Kelvin (K)

  18. P1 V1 T1 P2 V2 T2 = Combined Gas Law P = Pressure (atm, mm Hg, torr, kPa) V = Volume (mL or L) T = Temperature (K)

  19. Sample Problem C A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C?

  20. Sample Problem C A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C? Step 1: Outline what you know. P1 = 1.08 atm V1 = 50.0 L T1 = 25 + 273 = 298 K P2 = 0.855 atm V2 = ? L T2 = 10 + 273 = 283 K

  21. Sample Problem C A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C? Step 2: Plug the #’s into the eqn. P1 V1 T1 P2 V2 T2 = (1.08 atm)(50.0 L) (298 K) (0.855 atm) V2 (283 K) =

  22. (1.08 atm)(50.0 L) (298 K) (0.855 atm) V2 (283 K) = Sample Problem C A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C? Step 3a: Simplify the expression. 0.1812 = 0.0029 V2

  23. Sample Problem C A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C? Step 3b: Solve for the unknown. 0.1812 = 0.0029 V2 0.0029 0.0029 62 L = V2

  24. Sample Problem D A balloon has a volume of 40.0 L at 20 °C and 0.95 atm. At what temperature will the balloon expand to 50.0 L and 1.00 atm?

  25. Sample Problem D A balloon has a volume of 40.0 L at 20 °C and 0.95 atm. At what temperature will the balloon expand to 50.0 L and 1.00 atm? Step 1: Outline what you know. P1 = 0.95 atm V1 = 40.0 L T1 = 20 + 273 = 293 K P2 = 1.00 atm V2 = 50.0 L T2 = ? K

  26. (0.95 atm)(40.0 L) (293 K) (1.00 atm)(50.0 L) T2 = Sample Problem D A balloon has a volume of 40.0 L at 20 °C and 0.95 atm. At what temperature will the balloon expand to 50.0 L and 1.00 atm? Step 2: Plug the #’s into the eqn. P1 V1 T1 P2 V2 T2 =

  27. 50.0 T2 (0.95 atm)(40.0 L) (293 K) (1.00 atm)(50.0 L) T2 0.1297 = = Sample Problem D A balloon has a volume of 40.0 L at 20 °C and 0.95 atm. At what temperature will the balloon expand to 50.0 L and 1.00 atm? Step 3a: Simplify the expression.

  28. Sample Problem D A balloon has a volume of 40.0 L at 20 °C and 0.95 atm. At what temperature will the balloon expand to 50.0 L and 1.00 atm? Step 3b: Cross multiply to solve for the denominator 50.0 T2 0.1297 = 0.1297 T2 = 50.0

  29. Sample Problem D A balloon has a volume of 40.0 L at 20 °C and 0.95 atm. At what temperature will the balloon expand to 50.0 L and 1.00 atm? Step 3c: Solve for the unknown. 0.1297 T2 = 50.0 0.1297 0.1297 T2 = 386 K

  30. Boyle’s Law A Variation

  31. Visual Concepts Chapter 12 Boyle’s Law

  32. Boyle’s Law • An equation • Temperature is constant P1 V1 = P2 V2 • When P increases, V decreases • When P decreases, V increases • Inversely proportional relationship

  33. Sample Problem E A sample of oxygen gas has a volume of 150.0 mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant?

  34. Sample Problem E A sample of oxygen gas has a volume of 150.0 mL when its pressure is 0.947 atm. That will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant? Step 1: Outline what you know. P1 = 0.947 atm V1 = 150.0 mL P2 = 0.987 atm V2 = ? mL

  35. Sample Problem E A sample of oxygen gas has a volume of 150.0 mL when its pressure is 0.947 atm. That will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant? Step 2: Plug into the equation. P1 V1 = P2 V2 (0.947 atm) (150.0 mL) = (0.987 atm) V2

  36. Sample Problem E A sample of oxygen gas has a volume of 150.0 mL when its pressure is 0.947 atm. That will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant? Step 3a: Simplify the equation. (0.947 atm) (150.0 mL) = (0.987 atm) V2 142.05 = 0.987 V2

  37. Sample Problem E A sample of oxygen gas has a volume of 150.0 mL when its pressure is 0.947 atm. That will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant? Step 3b: Solve. 142.05 = 0.987 V2 0.987 0.987 144 mL = V2

  38. Charles’ Law Another Variation

  39. Charles’ Law

  40. Charles’ Law • An equation • Pressure is constant • V1= V2 T1 T2 • When T increases, V increases • When T decreases, V decreases • Directly proportional

  41. Sample Problem F A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant?

  42. Sample Problem F A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant? Step 1: List what you know. V1 = 752 mL T1 = 25 + 273 = 298 K V2 = ? mL T2 = 50 + 273 = 323 K

  43. 752 mL 298 K V2 323 K = Sample Problem F A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant? Step 2: Plug into the equation: V1= V2 T1 T2

  44. 752 mL 298 K V2 323 K = Sample Problem F A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant? Step 3: Cross multiply and solve. 242896 = 298 V2 298 298

  45. Sample Problem F A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant? Step 3: Cross multiply and solve. 815 mL

  46. Gay-Lussac’s Law Yup, you guessed it … A 3rd Variation

  47. Gay-Lussac’s Law • An equation • Volume is constant • P1= P2 T1 T2 • When T increases, P increases • When T decreases, P decreases • Directly proportional

  48. Sample Problem G The gas in a container is at a pressure of 3.00 atm at 25°C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C?

  49. Sample Problem G The gas in a container is at a pressure of 3.00 atm at 25°C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C? Step 1: List what you know. P1 = 3.00 atm T1 = 25 + 273 = 298 K P2 = ? atm T2 = 52 + 273 = 325 K

  50. 3.00 atm 298 K P2 325 K = Sample Problem G The gas in a container is at a pressure of 3.00 atm at 25°C. Directions on the container warn the user not to keep it in a place where the temperature exceeds 52°C. What would the gas pressure in the container be at 52°C? Step 2: Plug into the equation: P1= P2 T1 T2

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