Download
gases n.
Skip this Video
Loading SlideShow in 5 Seconds..
Gases PowerPoint Presentation

Gases

197 Views Download Presentation
Download Presentation

Gases

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

  1. Gases Properties of Gases Gas Pressure

  2. Gases What gases are important for each of the following: O2, CO2 and/or He? A. B. C. D.

  3. Gases What gases are important for each of the following: O2, CO2 and/or He? A. CO2 B. O2/CO2 C. O2 D. He

  4. Some Gases in Our Lives Air: oxygen O2 nitrogen N2 ozone O3 argon Ar carbon dioxide CO2 water H2O Noble gases: helium He neon Ne krypton Kr xenon Xe Other gases: fluorine F2 chlorine Cl2 ammonia NH3 methane CH4 carbon monoxide CO nitrogen dioxide NO2 sulfur dioxide SO2

  5. The Nature of Gases • Gases are compressible Why can you put more air in a tire, but can’t add more water to a glass full of water? • Gases have low densities Dsolid or liquid = 2 g/mL Dgas 2 g/L

  6. Nature of Gases 1. Why does a round balloon become spherical when filled with air? 2. Suppose we filled this room halfway with water. Where would pressure be exerted?

  7. Nature of Gases • Gases fill a container completely and uniformly • Gases exert a uniform pressure on all inner surfaces of their containers

  8. Kinetic Theory of Gases The particles in gases • Are very far apart • Move very fast in straight lines until they collide • Have no attraction (or repulsion) • Move faster at higher temperatures

  9. Barometers 760 mmHg atm pressure

  10. Learning Check G1 A.The downward pressure of the Hg in a barometer is _____ than (as) the weight of the atmosphere. 1) greater 2) less 3) the same B.A water barometer has to be 13.6 times taller than Hg barometer (DHg = 13.6 g/mL) because 1) H2O is less dense 2) Hg is heavier 3) air is more dense than H2O

  11. Solution G1 A.The downward pressure of the Hg in a barometer is 3) the same (as) the weight of the atmosphere. B.A water barometer has to be 13.6 times taller than Hg barometer (DHg = 13.6 g/mL) because 1) H2O is less dense

  12. Unit of Pressure One atmosphere (1 atm) • Is the average pressure of the atmosphere at sea level • Is the standard of pressure • P = Force Area 1.00 atm = 760 mm Hg = 760 torr

  13. Learning Check G2 When you drink through a straw you reduce the pressure in the straw. Why does the liquid go up the straw? 1) the weight of the atmosphere pushes it 2) the liquid is at a lower level 3) there is empty space in the straw Could you drink a soda this way on the moon? 1) yes 2) no 3) maybe Why or why not?

  14. Solution G2 When you drink through a straw you reduce the pressure in the straw. Why does the liquid go up the straw? 1) the weight of the atmosphere pushes it 3) there is empty space in the straw Could you drink a soda this way on the moon? 2) no Why or why not? Low atmospheric pressure

  15. Types of Pressure Units Pressure Used in 760 mm Hg or 760 torr Chemistry 14.7 lb/in.2 U.S. pressure gauges 29.9 in. Hg U.S. weather reports 101.3 kPa (kilopascals) Weather in all countries except U.S. 1.013 bars Physics and astronomy

  16. Learning Check G3 A. What is 475 mm Hg expressed in atm? 1) 475 atm 2) 0.625 atm 3) 3.61 x 105 atm B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 1) 2.00 mm Hg 2) 1520 mm Hg 3) 22,300 mm Hg

  17. Solution G3 A. What is 475 mm Hg expressed in atm? 485 mm Hg x 1 atm = 0.625 atm (B) 760 mm Hg B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 29.4 psi x 1.00 atm x 760 mmHg = 1.52 x 103 mmHg 14.7 psi 1.00 atm (B)

  18. Chapter 7Gases Pressure and Volume (Boyle’s Law) Temperature and Volume (Charles’ Law) Temperature and Pressure (Gay-Lussac’s Law)

  19. Pressure and Volume Experiment Pressure Volume P x V (atm) (L) (atm x L) 1 8.0 2.0 16 2 4.0 4.0 _____ 3 2.0 8.0 _____ 4 1.0 16 _____ Boyle's Law P x V = k (constant) when T remains constant P1V1= 8.0 atm x 2.0 L = 16 atm L P2V2= 4.0 atm x 4.0 L = 16 atm L P1V1 = P2V2 = k Use this equation to calculate how a volume changes when pressure changes, or how pressure changes when volume changes. new vol. old vol. x Pfactor new P old P x Vfactor V2 = V1 x P1 P2 = P1 x V1 P2 V2

  20. P and V Changes P1 P2 V1 V2

  21. Boyle's Law • The pressure of a gas is inversely related to the volume when T does not change • Then the PV product remains constant P1V1 = P2V2 P1V1= 8.0 atm x 2.0 L = 16 atm L P2V2= 4.0 atm x 4.0 L = 16 atm L

  22. PV Problem Freon-12, CCl2F2, is used in refrigeration systems. What is the new volume (L) of a 1.6 L sample of Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg at constant T?

  23. PV Calculation Prepare a data table DATA TABLE Initial conditions Final conditions P1 = 50 mm Hg P2 = 200 mm Hg V1 = 1.6 L V2 = ? ?

  24. Find New Volume (V2) Solve for V2: P1V2 = P2V2 V2 = V1 x P1 /P2 V2 = 1.6 L x 50 mm Hg = 0.4 L 200 mm Hg

  25. Learning Check GL1 A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant) Explain. 1) 3.2 L 2) 6.4 L 3) 12.8 L

  26. Solution GL1 A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant) 6.4 L x 0.70 atm = 3.2 L (1) 1.40 atm Volume must decrease to cause an increase in the pressure

  27. Learning Check GL2 A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain. 1) 200. mmHg 2) 400. mmHg 3) 1200 mmHg

  28. Solution GL2 A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain. 600. mm Hg x 12.0 L = 200. mmHg (1) 36.0 L Pressure decrease when volume increases.

  29. Charles’ Law V = 125 mL V = 250 mL T = 273 K T = 546 K Observe the V and T of the balloons. How does volume change with temperature?

  30. Charles’ Law: V and T At constant pressure, the volume of a gas is directly related to its absolute (K) temperature V1 = V2 T1 T2

  31. Learning Check GL3 Use Charles’ Law to complete the statements below: 1. If final T is higher than initial T, final V is (greater, or less) than the initial V. 2. If final V is less than initial V, final T is (higher, or lower) than the initial T.

  32. Solution GL3 V1 = V2 T1 T2 1. If final T is higher than initial T, final V is (greater) than the initial V. 2. If final V is less than initial V, final T is (lower) than the initial T.

  33. V and T Problem A balloon has a volume of 785 mL on a Fall day when the temperature is 21°C. In the winter, the gas cools to 0°C. What is the new volume of the balloon?

  34. VT Calculation Complete the following setup: Initial conditions Final conditions V1 = 785 mL V2 = ? T1 = 21°C = 294 K T2 = 0°C = 273 K V2 = _______ mL x __ K = _______ mL V1 K Check your answer: If temperature decreases, V should decrease.

  35. Learning Check GL4 A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. What temperature (in °C) is needed to change the volume to 640 mL? 1) 443°C 2) 170°C 3) - 82°C

  36. Solution GL4 A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. What temperature (in °C) is needed to change the volume to 640 mL? 2) 170°C T2 = 291 K x 640 mL = 443 K 420 mL = 443 K - 273 K = 170°C

  37. Gay-Lussac’s Law: P and T The pressure exerted by a confined gas is directly related to the temperature (Kelvin) at constant volume. P (mm Hg) T (°C) 936 100 761 25 691 0

  38. Learning Check GL5 Use Gay-Lussac’s law to complete the statements below: 1. When temperature decreases, the pressure of a gas (decreases or increases). 2. When temperature increases, the pressure of a gas (decreases or increases).

  39. Solution GL5 1. When temperature decreases, the pressure of a gas (decreases). 2. When temperature increases, the pressure of a gas (increases).

  40. PT Problem A gas has a pressure at 2.0 atm at 18°C. What will be the new pressure if the temperature rises to 62°C? (V constant) T = 18°C T = 62°C

  41. PT Calculation ? P1 = 2.0 atm T1 = 18°C + 273 = 291 K P2 = ? T2 = 62°C + 273 = 335 K What happens to P when T increases? P increases (directly related to T) P2 = P1 x T2 T1 P2 = 2.0 atm x K = atm K

  42. Learning Check GL6 Complete with 1) Increases 2) Decreases 3) Does not change A. Pressure _____, when V decreases B. When T decreases, V _____. C. Pressure _____ when V changes from 12.0 L to 24.0 L (constant n and T) D. Volume _____when T changes from 15.0 °C to 45.0°C (constant P and n)

  43. Solution GL6 A. Pressure 1) Increases, when V decreases B. When T decreases, V 2) Decreases C. Pressure 2) Decreases when V changes from 12.0 L to 24.0 L (constant n and T) D. Volume 1) Increases when T changes from 15.0 °C to 45.0°C (constant P and n)

  44. Chapter 7Gases The Combined Gas Law Volume and Moles (Avogadro’s Law) Partial Pressures LecturePLUS Timberlake

  45. Combined Gas Law P1V1 = P2V2 T1 T2 Rearrange the combined gas law to solve for V2 P1V1T2 = P2V2T1 V2 = P1V1T2 P2T1 LecturePLUS Timberlake

  46. Combined Gas Law P1V1 = P2V2 T1 T2 Isolate V2 P1V1T2 = P2V2T1 V2 = P1V1T2 P2T1 LecturePLUS Timberlake

  47. Learning Check C1 Solve the combined gas laws for T2. LecturePLUS Timberlake

  48. Solution C1 Solve the combined gas law for T2. (Hint: cross-multiply first.) P1V1 = P2V2 T1 T2 P1V1T2 = P2V2T1 T2 = P2V2T1 P1V1 LecturePLUS Timberlake

  49. Combined Gas Law Problem A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm? LecturePLUS Timberlake

  50. Data Table Set up Data Table P1 = 0.800 atm V1 = 0.180 L T1 = 302 K P2 = 3.20 atm V2= 90.0 mL T2 = ?? ?? LecturePLUS Timberlake