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Gas Laws and Atmospheric Pressure

Learn about the properties of gases, including diffusion and effusion, and how gas volume, pressure, and temperature are related through the Combined Gas Law. Discover how to convert atmospheric pressure units and solve gas law problems step by step.

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Gas Laws and Atmospheric Pressure

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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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