2. 1 GASES Chemistry I – Chapter 11

3. 2 Importance of Gases Airbags fill with N 2 gas in an accident.Airbags fill with N 2 gas in an accident. Gas is generated by the decomposition of sodium azide, NaN 3.Gas is generated by the decomposition of sodium azide, NaN 3. 2 NaN 3 ---> 2 Na + 3 N 22 NaN 3 ---> 2 Na + 3 N 2

4. 3 THREE STATES OF MATTER

5. 4 General Properties of Gases There is a lot of “free” space in a gas.There is a lot of “free” space in a gas. Gases can be expanded infinitely. (they will fill whatever “container” they are in.)Gases can be expanded infinitely. (they will fill whatever “container” they are in.) Gases fill containers uniformly and completely.Gases fill containers uniformly and completely. Gases diffuse and mix rapidly.Gases diffuse and mix rapidly.

6. 5 Pressur e Pressure is the amount of force per unit area. 1.Gas Pressure is the pressure caused by particles of gas striking an object. 2.Air pressure is the pressure of the column of atmosphere above you,pressing down on you.

7. 6 Atmospheric Pressure Measured with a BAROMETER (developed by Torricelli in 1643)Measured with a BAROMETER (developed by Torricelli in 1643) Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). (Just like a straw in a soft drink)Hg rises in tube until force of Hg (down) balances the force of atmosphere (pushing up). (Just like a straw in a soft drink) Why is Hg so good for use in barometer?Why is Hg so good for use in barometer?

8. 7 Atm. Pressure About 34 feet of water = 1 atm pressure! Column height measures Pressure of atmosphere 1 standard atmosphere (atm) * = 760 mm Hg (or torr) * = 29.92 inches Hg * = 14.7 pounds/in2 (psi) * = 101.3 kPa (SI unit is PASCAL)

9. 8 Pressure Conversions Factor Label Method is how we will make conversions. Sample Problem, p 379 of text One standard atmosphere (1 atm) is equivalent to 760 mm of Hg. What is that height expressed in inches? (Hint: 1.00 in = 25.4 mm) Analysis: Identify 1.Your “given”: _____ 2.Your unknown (what you’re solving for):_____ 3.the relationship (formula, etc) between the given & the unknown: ___ Write an equation: 1.Your given is on the left, including the units. 2.The conversion factor is written next, but you must decide how to write it: 3.Write an “=“ 4.Write a blank for your answer, and the units your answer will be written in.

10. 9 Pressure Conversions A. What is 475 mm Hg expressed in atm? 1 atm 760 mm Hg B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 760 mm Hg 14.7 psi = 1.52 x 10 3 mm Hg = 0.625 atm 475 mm Hg x 29.4 psi x

11. 10 Pressure Conversions A. What is 2 atm expressed in torr? B. The pressure of a tire is measured as 32.0 psi. What is this pressure in kPa?

12. 11 Properties of Gases Gas properties can be modeled using math. Model depends on— V = volume of the gas (L)V = volume of the gas (L) T = temperature (K)T = temperature (K) –ALL temperatures in the entire chapter MUST be in Kelvin!!! No Exceptions! n = amount (moles)n = amount (moles) P = pressure (atmospheres)P = pressure (atmospheres)

13. Chapter 14 The Behavior of Gases (new page of your permanent Chem notebook) 12

14. Properties of Gases Compressibility-a measure of how much the volume of matter decreases under pressure. Gases are compressible b/c of the space btwn particles in a gas At room temp, distance btwn particles in an enclosed gas ~ 10x the diameter of the particles 13

15. Factors to Describe Gas Pressure Pressure in kPa Volume in L Temperature in K Number of moles, n 14

16. Number of Moles Pressure is caused by the particles striking the walls of the container. If the gas is in a rigid container, the volume is constant If you triple the number of gas particles (n), you triple the amount of pressure. (see p 415 of text) 15

17. Volume Reducing the volume in a container will raise the pressure (inverse relationship.) See p 416 If you press on a piston, no gas escapes, but the volume decreases. The same # of particles are striking a smaller area, so pressure increases. 16

18. Temperature As a gas is heated, the temperature increases If temperature increases, particle speed increases IF the container is rigid, the pressure will increase. If the container is not rigid, the volume will increase & pressure will remain the same. 17

19. Section 14.1 Assessment p 417 1. Why is a gas easy to compress? 2. List 3 factors that can affect gas pressure. 3. Why does a collision with an air bag cause less damage than a collision with a steering wheel? 4. How does a decrease in temp affect the pressure of a contained gas? 5. If the temp is constant, what change in volume would cause the pressure of an enclosed gas to be reduced to ¼ of its original value? 6. Assuming the gas in a container remains at a constant temp, how could you increase the gas pressure in a container a hundredfold? 18

20. 19 Properties of Gases, cont. We can study the relationship between 2 variables if we keep the others the same.

21. 20 Boyle’s Law 2 variables we will study: P & V All other variables kept the same (T, n, etc.) Ex: a rising balloon

22. 21 Boyle’s Law P α 1/V This means Pressure and Volume are INVERSELY PROPORTIONAL For example, P goes up as V goes down. For example, P goes up as V goes down. P 1 V 1 = P 2 V 2 Robert Boyle (1627-1691). Son of Early of Cork, Ireland.

23. 22 Boyle’s Law and Kinetic Molecular Theory P proportional to 1/V

24. 23 Boyle’s Law A bicycle pump is a good example of Boyle’s law. As the volume of the air trapped in the pump is reduced, its pressure goes up, and air is forced into the tire.

25. Sample Problem 14.1 (p 419) A balloon contains 30.0 L of helium gas at 103kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25.0 kPa? 24 Givens P1= 103 kPaP2 = 25.0kPa V1 = 30.0L Unknown V 2

26. 25 Charles’s Law 2 variables we will study: V & T All other variables kept the same (P, n, etc.)

27. 26 Charles’s Law If n and P are constant, then V α T V and T are directly proportional. V 1 V 2 = T 1 T 2 If one temperature goes up, the volume goes up!If one temperature goes up, the volume goes up! Jacques Charles (1746- 1823). Isolated boron and studied gases. Balloonist.

28. Sample Problem 14.2 Using Charles Law (p 421) A balloon inflated in a room at 24˚C has a volume of 4.00L. The balloon is then heated to a temperature of 48˚C. What is the new volume if the pressure remains constant? 27 Givens: T 1 = 24˚C + 273 = 297KT 2 = 48 + 273 =321K V 1 = 4.00 L Unknown = V 2 4.00L = V 2 297K 321K 4.00L*321K = V 2 297K

29. CW/HW Practice Problems, pp 419-423 # 7-12 28

30. 29 Charles’s original balloon Modern long-distance balloon

31. 30 Charles’s Law

32. 31 Gay-Lussac’s Law If n and V are constant, then P α T P and T are directly proportional. P 1 P 2 = T 1 T 2 If one temperature goes up, the pressure goes up!If one temperature goes up, the pressure goes up! Joseph Louis Gay- Lussac (1778-1850)

33. 32 Gas Pressure, Temperature, and Kinetic Molecular Theory P proportional to T

34. 33 Combined Gas Law Since all 3 gas laws are related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! P 1 V 1 = P 2 V 2 T 1 T 2 No, it’s not related to R2D2

35. 34 Combined Gas Law If you should only need one of the other gas laws, you can cover up the item that is constant and you will get that gas law! = P1P1 V1V1 T1T1 P2P2 V2V2 T2T2 Boyle’s Law Charles’ Law Gay-Lussac’s Law

36. 35 Combined Gas Law Problem A sample of helium gas has a volume of 180 mL, 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? Set up Data Table P 1 = 0.800 atm V 1 = 180 mL T 1 = 302 K P 2 = 3.20 atm V 2 = 90 mL T 2 = ??

37. 36 Calculation P 1 = 0.800 atm V 1 = 180 mLT 1 = 302K P 2 = 3.20 atm V 2 = 90 mL T 2 = ?? P 1 V 1 P 2 V 2 = P 1 V 1 T 2 = P 2 V 2 T 1 T 1 T 2 T 2 = P 2 V 2 T 1 P 1 V 1 T 2 = 3.20 atm x 90.0 mL x 302 K 0.800 atm x 180.0 mL T 2 = 604 K - 273 = 331 °C = 604 K

38. 37 Learning Check A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?

39. 38 One More Practice 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?

40. 39 And now, we pause for this commercial message from STP OK, so it’s really not THIS kind of STP… STP in chemistry stands for Standard Temperature and Pressure Standard Pressure = 1 atm (or an equivalent) Standard Temperature = 0 deg C (273 K) STP allows us to compare amounts of gases between different pressures and temperatures

41. 40 Try This One A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?

42. 41 Avogadro’s Hypothesis Equal volumes of gases at the same T and P have the same number of molecules. V and n are directly related. twice as many molecules

43. 42 Avogadro’s Hypothesis and Kinetic Molecular Theory P proportional to n The gases in this experiment are all measured at the same T and V.

44. 43 IDEAL GAS LAW Brings together gas properties. BE SURE YOU KNOW THIS EQUATION! P V = n R T

45. 44 Using PV = nRT P = Pressuren = number of moles V = VolumeT = Temperature R is a constant, called the Ideal Gas Constant= 8.31 L*kPa K*mol K*mol We must convert the units to match R.

46. 45 Using PV = nRT Ex: p 439, Q#55Ex: p 439, Q#55 1.24 moles of gas at 35 C and 96.2 kPa pressure. What is the volume the gas occupies?1.24 moles of gas at 35 C and 96.2 kPa pressure. What is the volume the gas occupies? V=?V=? n= 1.24 moln= 1.24 mol T = 35 + 273 = 308KT = 35 + 273 = 308K P = 96.2kPaP = 96.2kPa R=8.31L*kPa/K*molR=8.31L*kPa/K*mol (96.2kPa)V = (1.24 mol ) (8.21L*kPa/K*mol) (308K)(96.2kPa)V = (1.24 mol ) (8.21L*kPa/K*mol) (308K) V=33LV=33L

47. 46 Learning Check Dinitrogen monoxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mm Hg) in the tank in the dentist office?

48. 47 Learning Check A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?

49. 48 Gases in the Air The % of gases in air Partial pressure (STP) 78.08% N 2 593.4 mm Hg 20.95% O 2 159.2 mm Hg 0.94% Ar 7.1 mm Hg 0.03% CO 2 0.2 mm Hg P AIR = P N 2 + P O 2 + P Ar + P CO 2 = 760 mm Hg (Total Pressure = 760mm Hg)

50. 49 Dalton’s Law of Partial Pressures What is the total pressure in the flask? P total in gas mixture = P A + P B +... Therefore, P total = P H 2 O + P O 2 = 0.48 atm Dalton’s Law: total P is sum of PARTIAL pressures. 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) 0.32 atm 0.16 atm 0.32 atm 0.16 atm

51. 50 Dalton’s Law John Dalton 1766-1844

52. 51 Health Note When a scuba diver is several hundred feet under water, the high pressures cause N 2 from the tank air to dissolve in the blood. If the diver rises too fast, the dissolved N 2 will form bubbles in the blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O 2 in scuba tanks used for deep descents.

53. 52 GAS DENSITY Highdensity Lowdensity 22.4 L of ANY gas AT STP = 1 mole

54. 53 Gases and Stoichiometry 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the volume of O 2 at STP? Bombardier beetle uses decomposition of hydrogen peroxide to defend itself.

55. 54 Gases and Stoichiometry 2 H 2 O 2 (l) ---> 2 H 2 O (g) + O 2 (g) Decompose 1.1 g of H 2 O 2 in a flask with a volume of 2.50 L. What is the volume of O 2 at STP? Solution 1.1 g H 2 O 2 1 mol H 2 O 2 1 mol O 2 22.4 L O 2 34 g H 2 O 2 2 mol H 2 O 2 1 mol O 2 34 g H 2 O 2 2 mol H 2 O 2 1 mol O 2 = 0.36 L O 2 at STP

56. 55 Gas Stoichiometry: Practice! A. What is the volume at STP of 4.00 g of CH 4 ? B. How many grams of He are present in 8.0 L of gas at STP?

57. 56 What if it’s NOT at STP? 1. Do the problem like it was at STP. (V 1 ) 2. Convert from STP (V 1, P 1, T 1 ) to the stated conditions (P 2, T 2 )

58. 57 Try this one! How many L of O 2 are needed to react 28.0 g NH 3 at 24°C and 0.950 atm? 4 NH 3 (g) + 5 O 2 (g) 4 NO(g) + 6 H 2 O(g)

59. 58 GAS DIFFUSION AND EFFUSION diffusion is the gradual mixing of molecules of different gases.diffusion is the gradual mixing of molecules of different gases. effusion is the movement of molecules through a small hole into an empty container.effusion is the movement of molecules through a small hole into an empty container. HONORS only

60. 59 GAS DIFFUSION AND EFFUSION Graham’s law governs effusion and diffusion of gas molecules. Thomas Graham, 1805-1869. Professor in Glasgow and London. Rate of effusion is inversely proportional to its molar mass. HONORS only

61. 60 GAS DIFFUSION AND EFFUSION Molecules effuse thru holes in a rubber balloon, for example, at a rate (= moles/time) that is proportional to Tproportional to T inversely proportional to M.inversely proportional to M. Therefore, He effuses more rapidly than O 2 at same T. He HONORS only

62. 61 Gas Diffusion relation of mass to rate of diffusion HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl HCl heavier than NH 3 Therefore, NH 4 Cl forms closer to HCl end of tube. HCl and NH 3 diffuse from opposite ends of tube. Gases meet to form NH 4 Cl HCl heavier than NH 3 Therefore, NH 4 Cl forms closer to HCl end of tube. HONORS only