# GASES - PowerPoint PPT Presentation

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GASES

## GASES

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##### Presentation Transcript

1. GASES

2. The Kinetic-Molecular Theory for Ideal Gases The kinetic-molecular theory describes the behavior of IDEAL gases in terms of particles in motion. Real gases do not obey the kinetic molecular theory.

3. The Kinetic-Molecular Theory for Ideal Gases 1. Gas particles are much smaller than the spaces between them. The gas particles themselves have virtually no (negligible) volume.

4. The Kinetic-Molecular Theory for Ideal Gases 2. Gas particles are in constant, random motion. Particles move in a straight line until they collide with other particles or with the walls of their container.

5. The Kinetic-Molecular Theory for Ideal Gases 3. Gas particles do not attract or repel each other. Therefore ideal gases would never condense to form liquids.

6. The Kinetic-Molecular Theory for Ideal Gases 4. No kinetic energy is lost when gas particles collide with each other or with the walls of their container. Collisions are perfectly elastic.

7. The Kinetic-Molecular Theory for Ideal Gases 5. All gases have the same average kinetic energy at a given temperature.

8. Gas Solubility Gases are less soluble in warm liquids than in cooler liquids.

9. Gas Solubility Gases are more soluble when under pressure.

10. Vapor Pressure The vapor pressure of water increases as the temperature increases.

11. STP • STP stands for standard temperature and pressure.

12. Standard Pressure • Standard pressure is 1atm (atmosphere) which is equal to 760 mm Hg, 760 torr, or 101.3 kPa (kilopascals).

13. Standard Temperature • Standard temperature is 273 Kelvin.

14. Standard Atmospheric Pressure 1) Perform the following pressure conversions. a) 144 kPa = _____ atm (1.42) b) 795 mm Hg = _____ atm (1.05)

15. Standard Atmospheric Pressure Perform the following pressure conversions. c) 669 torr = ______ kPa (89.2) d) 1.05 atm = ______ mm Hg (798)

16. Standard Atmospheric Pressure • Air pressure at higher altitudes, such as on a mountaintop, is slightly lower than air pressure at sea level.

17. Standard Atmospheric Pressure • Air pressure is measured using abarometer.

18. The Gas Laws

19. Boyle’s Law • Boyle’s lawstates that the pressure and volume of a gas at constant temperature are inversely proportional. • Inversely proportional means as one goes up the other goes down.

20. 1 atm 4 Liters

21. Boyle’s Law • The P-V graph for Boyle’s law results in a hyperbola because pressure and volume are inversely proportional.

22. Boyle’s Law

23. Boyle’s Law • P x V = K (K is some constant) P1 V1 = P2 V2

24. Example • A balloon is filled with 25 L of air at 1.0 atm pressure. If the pressure is changed to 1.5 atm, what is the new volume?

25. Example • First, make sure the pressure or volume units in the question match. A balloon is filled with 25 L of air at 1.0 atm pressure. If the pressure is changed to 1.5 atm, what is the new volume? THEY DO!

26. Example P1 • A balloon is filled with 25 L of air at 1.0 atm pressure. If the pressure is changed to 1.5 atm, what is the new volume? V1 = P2 V2 V2 1.0 atm (25 L) 1.5 atm V2 = 17 L

27. Problem 2 • A balloon is filled with 73 L of air at 1.3 atm pressure. What pressure is needed to change the volume to 43 L? P2 = 2.2 atm

28. Problem 3 • A gas is collected in a 242 cm3 container. The pressure of the gas in the container is measured and determined to be 87.6 kPa. What is the volume of this gas at standard pressure? V2 = 209 cm3

29. Problem 4 • A gas is collected in a 24.2 L container. The pressure of the gas in the container is determined to be 756 mm Hg. What is the pressure of this gas if the volume increases to 30.0 L? P2 = 610. mm Hg

30. Charles’ Law • Charles’ Law states that the volume of a gas is directly proportional to the Kelvin temperature if the pressure is held constant. • Directly proportional means that as one goes up, the other goes up as well.

31. Charles’ Law • The V-T graph for Charles’ law results in a straight line because volume and temperature are directly proportional.

32. Charles’ Law

33. Charles’ Law • V / T = K (K is some constant) V1 V2 = T1 T2

34. Charles’ Law • In any gas law problem involving temperature, temperature must be in Kelvin. K = °C + 273

35. Example • What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure?

36. Example • First, make sure the volume units in the question match. • What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure? THEY DO!

37. Example • Second, make sure to convert degrees Celsius to Kelvin. • What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure? °C 25 K = + 273 K = 298 K

38. Example V1 • What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure? V2 2.5 L 4.1 L = T1 T2 298 K T2 = 489 K

39. Problem 5 • What is the final volume of a gas that starts at 8.3 L and 17 ºC and is heated to 96 ºC? V2 = 11 L

40. Problem 6 • A 225 cm3 volume of gas is collected at 57 ºC. What volume would this sample of gas occupy at standard temperature? V2 = 186 cm3

41. Problem 7 • A 225 cm3 volume of gas is collected at 42 ºC. If the volume is decreased to 115 cm3, what is the new temperature? T2 = 161 K

42. Gay-Lusaac’s Law • Gay-Lusaac’ Law states that the pressure of a gas is directly proportional to the Kelvin temperature if the volume is held constant.

43. Gay-Lusaac’s Law • At higher temperatures, the particles in a gas have greater kinetic energy. • They move faster and collide with the walls of the container more often and with greater force, so the pressure rises.

44. Gay-Lusaac’s Law • The P-T graph for Gay-Lusaac’s law results in a straight line because pressure and temperature are directly proportional.

45. Gay-Lussac’s Law • P / T = K (K is some constant) P1 P2 = T1 T2

46. Example • What is the pressure inside a 0.250 L can of deodorant that starts at 25 ºC and 1.2 atm if the temperature is raised to 100 ºC? Volume remains constant.

47. Example • First, make sure the pressure units in the question match. • What is the pressure inside a 0.250 L can of deodorant that starts at 25 ºC and 1.2 atm if the temperature is raised to 100 ºC? There is only one pressure unit!

48. Example • Second, make sure to convert degrees Celsius to Kelvin. • What is the pressure inside a 0.250 L can of deodorant that starts at 25 ºC and 1.2 atm if the temperature is raised to 100 ºC? °C 25 K = + 273 K = 298 K