1. Kinetic Theory of Gases 1. Gas particles do not attract or repel one another 2. Gas particles are much smaller than the distances between them

2. Kinetic Theory of Gases 3. Gas particles are in constant random motion 4. Gas particle collisions with other particles or their container are elastic (no kinetic energy lost)

3. Kinetic Theory of Gases 5. All gases have the same kinetic energy at a given temperature

4. Boyle’s Law States that pressure and volume are inversely proportional to one another. –As one variable increases the other decreases

5. Boyle’s Law P 1 V 1 = P 2 V 2 P 1 V 1 are the initial set of conditions P 2 V 2 are the new set of conditions

6. Boyle’s Law Gas in an engine cylinder is at a pressure of 3 atm and has a volume of.65 L. The piston is then pushed down reducing the volume to.34 L, what is the new pressure in the balloon?

7. Boyle’s Law P 1 V 1 = P 2 V 2 (3atm)(.65L) = P 2 (.34L) P 2 = 5.74 atm * As you can see the pressure increases as the volume decreases

8. Boyle’s Law Pressure VOLUMEVOLUME

9. Charles’s Law States that as temperature increases, volume increases There is a directly proportional relationship between temperature (in K) and pressure. –As one increases the other increases

10. Charles’s Law V 1 = V 2 T 1 T 2 If you know three of the four values you can calculate the other

11. Charles’s Law A sample of CO 2 at 293 o K occupies a 4.52 L container, what volume would the container have to be reduced to in order to bring the temperature down to the freezing point of 194.5 o K?

12. Charles’s Law V 1 = V 2 T 1 T 2 4.52L = V 2 293 o K 194.5 o K V1 = 3.00L

13. Charles’s Law Temperature VOLUMEVOLUME

14. Gay-Lussac’s Law States that temperature and pressure are directly proportional to one another –As temperature increases so does the pressure

15. Gay-Lussac’s Law P1 = P2 T1 T2 If you know three of the four values you can calculate the other

16. Gay-Lussac’s Law The pressure of a gas is raised from 4.65 atm at 30 o C to 32.56 atm? What is the new temperature?

17. Gay-Lussac’s Law 4.65atm = 32.56atm 303 o K T 2 T 2 = 2121.65 K

18. Gay-Lussac’s Law Temperature PRESSUREPRESSURE

19. Combined Gas Law Shows the relationship between pressure, temperature, and volume Combines Boyles, Charles’s, and Gay-Lussac’s laws into one equation

20. Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 Useful when dealing with many variables at one time

21. Combined Gas Law A gas at 2 atm and 289 o K fills a balloon with an initial volume of.05L. If the temperature is raised to 346 o K and the pressure increases to 5.6 atm, what is the new volume of the balloon?

22. Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 (2atm)(.05L) = (5.6atm)(V 2 ) 289oK 346oK V 2 =.214 L

23. Assignment P. 422 1-5 P. 425 6-8 P. 427 9-13

24. Avagadro’s Principle Equal volumes of gases that are at the same temperature and pressure have the same number of particles –1 L of He at 303 o K and 1 atm and 1 L of Ne at 303 o K and 1 atm contain the same number of particles

25. Molar Volume The volume that a gas occupies at 0 o C and at 1 atm of pressure –These conditions are known as STP (standard temperature and pressure) –Every gas occupies 22.4 L at STP

26. Molar Volume –So at STP 1 mole of any gas occupies 22.4 L –This can be used as a conversion factor when solving stoichiometry problems 22.4L 1 mol

27. Problem How many particles are in a sample of gas that has a volume of 3.73L at STP? 3.73L x 1 mol x 6.02 x 10 23 part= 22.4 L 1 mol 9.99 x 10 22 particles

28. Ideal Gas Law PV = nRT Since we now know how to determine the number of moles of a gas, we can use the above formula to relate four variables: Temp, Pressure, volume, and number of moles

29. Ideal Gas Law The R in the formula is the ideal gas constant. –Varies with pressure units –See chart on p 435

30. How do ideal gases differ from real gases? At low temp or high pressure gases behave differently because of intermolecular force interaction. Large molecules and polar molecules also deviate from ideal behavior

31. Ideal gas problem How many moles of a gas are contained in a 4.5L container at 1235K with a pressure of 3.50 atm?

32. Answer PV = nRT (3.5atm)(4.5L) = n(.0821L. atm/mol. K)(1235K) n=.15 mol

33. Gas Stoichiometry Same as stoichiometry for mass/mass or mass/volume problems. –Write what you are given first –Cancel units until you get to the unit you need

34. Sample problem What volume of H 2 will be needed to react with 2.3 L of O 2 in order for complete reaction of hydrogen with oxygen?

35. Answer 2.3L O 2 x 2 volumes H 2 = 4.6 L H 2 1 volumes O 2

36. Assignment P. 432 25-27 odd P. 433 29-33 odd P. 437 41-45 odd P. 438 45-49 odd P. 441 57-59 odd P. 443 61-65 odd