2. Gas Laws: Introduction At the conclusion of our time together, you should be able to: 1. List 5 properties of gases 2. Identify the various parts of the kinetic molecular theory 3. Define pressure 4. Convert pressure into 3 different units 5. Define temperature 6. Convert a temperature to Kelvin

3. 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 2 2 NaN 3 ---> 2 Na + 3 N 2

4. THREE STATES OF MATTER

5. 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. Gases can be expanded infinitely. Gases fill containers uniformly and completely. Gases fill containers uniformly and completely. Gases diffuse and mix rapidly. Gases diffuse and mix rapidly.

6. To Review  Gases expand to fill their containers  Gases are fluid – they flow  Gases have low density  1/1000 the density of the equivalent liquid or solid  Gases are compressible  Gases effuse and diffuse

7. Properties of Gases Gas properties can be modeled using math. This 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 unit MUST be in Kelvin!!! No Exceptions! n = amount (moles) n = amount (moles) P = pressure (atmospheres) P = pressure (atmospheres)

8. Ideal Gases Ideal gases are imaginary gases that perfectly fit all of the assumptions of the kinetic molecular theory.  Gases consist of tiny particles that are far apart relative to their size.  Collisions between gas particles and between particles and the walls of the container are elastic collisions  No kinetic energy is lost in elastic collisions

9. Ideal Gases Ideal Gases (continued)  Gas particles are in constant, rapid motion. They therefore possess kinetic energy, the energy of motion  There are no forces of attraction between gas particles  The average kinetic energy of gas particles depends on temperature, not on the identity of the particle.

10. Pressure  Is caused by the collisions of molecules with the walls of a container  Is equal to force/unit area  SI units = Newton/meter 2 = 1 Pascal (Pa)  1 atmosphere = 101,325 Pa  1 atmosphere = 1 atm = 760 mm Hg = 760 torr  1 atm = 29.92 in Hg = 14.7 psi = 0.987 bar = 10 m column of water.

11. Measuring Pressure The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17 th century. The device was called a “barometer” Baro = weight Meter = measure

12. An Early Barometer The normal pressure due to the atmosphere at sea level can support a column of mercury that is 760 mm high.

13. PressureColumn height measures Pressure of atmosphere 1 standard atmosphere (atm) * = 760 mm Hg (or torr) * = 29.92 inches Hg * = 14.7 pounds/in2 (psi) = 101.325 kPa (SI unit is PASCAL) = about 34 feet of water!

14. 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

15. Let’s Review: Standard Temperature and Pressure “STP” P = 1 atmosphere, 760 torr T =  C, 273 Kelvins The molar volume of an ideal gas is 22.42 liters at STP

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

17. 101.325 kPa 14.7 psi Pressure Conversions – Your Turn A. What is 2.00 atm expressed in torr? B. The pressure of a tire is measured as 32.0 psi. What is this pressure in kPa? = 1520 torr 2.00 atm x 760 torr 1 atm = 221 kPa 32.0 psi x

18. Converting Celsius to Kelvin Gas law problems involving temperature require that the temperature be in KELVINS! Kelvins =  C + 273 °C = Kelvins - 273

19. Gas Laws: Boyle’s and Charles’ Law At the conclusion of our time together, you should be able to: 1.Describe Boyle’s Law with a formula. 2.Use Boyle’s Law to determine either a pressure or volume 3.Describe Charles’ Law with a formula. 4.Use Charles’ Law to determine either a temperature or volume

20. Boyle’s Law P α 1/V This means Pressure and Volume are INVERSELY PROPORTIONAL if moles and temperature are constant (do not change). For example, P goes up as V goes down. Robert Boyle (1627-1691). Son of Earl of Cork, Ireland.

21. Boyle’s Law Summary Pressure is inversely proportional to volume when temperature is held constant.

22. Boyle’s Law Example 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.

23. Example P1P1 V1V1 P2P2 V2V2 (0.947 atm)(0.15 L)(0.987 atm)V2V2 0.14 L or 140 mL

24. Boyle’s Law Practice #1 P1P1 V1V1 P2P2 V2V2 (1 atm)(0.5 L)(0.5 atm)V2V2 1 L

25. Charles’ Law If n and P are constant, then V α T V and T are directly proportional. 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.

26. Charles’s Original Balloon Modern Long- Distance Balloon

27. Charles’ Law Summary The volume of a gas is directly proportional to temperature, and extrapolates to zero at zero Kelvin. (P = constant)

28. Charles’ Law

29. Example V1V1 T1T1 V2V2 T2T2 0.075 L 298 K V2V2 323 K 0.081 L

30. Charles’ Law Practice #1 V1V1 T1T1 V2V2 T2T2 2.75 L 293.0 K 2.46 L T2T2 262 K or -10.9 o C

31. Gas Laws: Gay-Lussac’s and Combined At the conclusion of our time together, you should be able to: 1.Describe Gay-Lussac’s Law with a formula 2.Use Gay-Lussac’s Law to determine either a temperature or volume 3.Combine all three laws into the Combined Gas Law 4.Use the Combined Gas Law to determine either temperature, volume or pressure

32. Gay-Lussac’s Law If n and V are constant, then P α T P and T are directly proportional. 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. Gay Lussac’s Law Summary The pressure and temperature of a gas are directly related, provided that the volume remains constant.

34. Example P1P1 T1T1 P2P2 T2T2 3.00 atm 298 K P2P2 325 K 3.3 atm

35. Gay-Lussac’s Law Practice #1 P1P1 T1T1 P2P2 T2T2 1.8 atm 293 K 1.9 atm T2T2 310 K or 36 o C

36. Now let’s put all 3 laws together into one big law…….

37. Combined Gas Law The good news is that you don’t have to remember all three gas laws! Since they are all related to each other, we can combine them into a single equation. BE SURE YOU KNOW THIS EQUATION! No, it’s not related to R2D2

38. The Combined Gas Law The combined gas law expresses the relationship between pressure, volume and temperature of a fixed amount of gas. Boyle’s law, Gay-Lussac’s law, and Charles’ law are all derived from this by holding a variable constant.

39. 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

40. 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? 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 = ??

41. Calculation P 1 = 0.800 atm V 1 = 180 mL T 1 = 302 K 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 = 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

42. Gas Laws: Avogadro’s and Ideal At the conclusion of our time together, you should be able to: 1.Describe Law with a formula. 1.Describe Avogadro’s Law with a formula. 2.Use Law to determine either moles or volume 2.Use Avogadro’s Law to determine either moles or volume 3.Describe the Law with a formula. 3.Describe the Ideal Gas Law with a formula. 4.Use Law to determine either moles, pressure, temperature or volume 4.Use the Ideal Gas Law to determine either moles, pressure, temperature or volume 5.Explain the Kinetic Molecular Theory

43. Avogadro’s Law Equal volumes of gases at the same T and P have the same number of molecules. V = an V and n are directly related. twice as many molecules

44. Avogadro’s Law Summary  For a gas at constant temperature and pressure, the volume is directly proportional to the number of moles of gas (at low pressures). V = an a = proportionality constant V = volume of the gas n = number of moles of gas

45. Standard Molar Volume Equal volumes of all gases at the same temperature and pressure contain the same number of molecules. - Amedeo Avogadro

46. Avogadro’s Law Practice #1 V1V1 n1n1 V2V2 n2n2 4.00 L 0.21 mol 7.12 L n2n2 0.37 mol total 0.16 mol added

47. IDEAL GAS LAW Brings together gas properties. Can be derived from experiment and theory. BE SURE YOU KNOW THIS EQUATION! P V = n R T

48. Ideal Gas Law PV = nRT  P = pressure in atm  V = volume in liters  n = moles  R = proportionality constant  = 0.08206 L atm/ mol·   T = temperature in Kelvins Holds closely at P < 1 atm

49. Review of Kinetic Molecular Theory  Particles of matter are ALWAYS in motion  Volume of individual particles is  zero.  Collisions of particles with container walls cause pressure exerted by gas.  Particles exert no forces on each other.  Average kinetic energy  Kelvin temperature of a gas.

50. Deviations from Ideal Gas Law Real molecules have volume. The ideal gas consumes the entire amount of available volume. It does not account for the volume of the molecules themselves. There are intermolecular forces. An ideal gas assumes there are no attractions between molecules. Attractions slow down the molecules and reduce the amount of collisions. – Otherwise a gas could not condense to become a liquid.

51. R is a constant, called the Ideal Gas Constant Instead of learning a different value for R for all the possible unit combinations, we can just memorize one value and convert the units to match R. R = 0.08206 R = 0.08206 L atm mol K

52. Using PV = nRT How much N 2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to 745 mm Hg at 25 o C? Solution Solution 1. Get all data into proper units V = 27,000 L V = 27,000 L T = 25 o C + 273 = 298 K T = 25 o C + 273 = 298 K P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm P = 745 mm Hg (1 atm/760 mm Hg) = 0.98 atm And we always know R, 0.08206 L atm / mol K

53. How much N 2 is required to fill a small room with a volume of 960 cubic feet (27,000 L) to P = 745 mm Hg at 25 o C? Solution Solution 2. Now plug in those values and solve for the unknown. PV = nRT RT RT RT RT Using PV = nRT n = 1.1 x 10 3 mol (or about 30 kg of gas)

54. Ideal Gas Law Problems #1 (5.6 atm)(12 L)(0.08206 atm*L / mol*K )(T) 200 K (4 mol)

55. Gas Laws: Dalton’s Law At the conclusion of our time together, you should be able to: 1.Explain Dalton’s Law 2.Use Dalton’s Law to solve a problem

56. Dalton’s Law John Dalton 1766-1844

57. Dalton’s Law of Partial Pressures For a mixture of gases in a container, P Total = P 1 + P 2 + P 3 +... This is particularly useful in calculating the pressure of gases collected over water.

58. 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

59. 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% Ar7.1 mm Hg 0.03% CO 2 0.2 mm Hg P AIR = P N + P O + P Ar + P CO = 760 mm Hg 2 2 2 Total Pressure = 760 mm Hg

60. Collecting a gas “over water” Gases, since they mix with other gases readily, must be collected in an environment where mixing can not occur. The easiest way to do this is under water because water displaces the air. So when a gas is collected “over water”, that means the container is filled with water and the gas is bubbled through the water into the container. Thus, the pressure inside the container is from the gas AND the water vapor. This is where Dalton’s Law of Partial Pressures becomes useful.

61. Table of Vapor Pressures for Water

62. Solve This! A student collects some hydrogen gas over water at 20 degrees C and 768 torr. What is the pressure of the H 2 gas? 768 torr – 17.5 torr = 750.5 torr

63. Dalton’s Law of Partial Pressures Also, for a mixture of gases in a container, because P, V and n are directly proportional if the other gas law variables are kept constant: n Total = n 1 + n 2 + n 3 +... V Total = V 1 + V 2 + V 3 +... This is useful in solving problems with differing numbers of moles or volumes of the gases that are mixed together.

64. Gas Laws: Gas Stoichiometry At the conclusion of our time together, you should be able to: 1.Use the Ideal Gas Law to solve a gas stoichiometry problem.

65. 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.

66. 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

67. Gas Stoichiometry: Practice! How many grams of He are present in 8.0 L of gas at STP? = 1.4 g He 8.0 L He x 1 mol He 22.4 L He x 4.00 g He 1 mol He

68. Gas Stoichiometry Trick If reactants and products are at the same conditions of temperature and pressure, then mole ratios of gases are also volume ratios. 3 H 2 (g) + N 2 (g)  2NH 3 (g) 3 moles H 2 + 1 mole N 2  2 moles NH 3 67.2 liters H 2 + 22.4 liter N 2  44.8 liters NH 3

69. Gas Stoichiometry Trick Example How many liters of ammonia can be produced when 12 liters of hydrogen react with an excess of nitrogen in a closed container at constant temperature? 3 H 2 (g) + N 2 (g)  2NH 3 (g) 12 L H 2 L H 2 = L NH 3 L NH 3 3 2 8.0

70. What if the problem is NOT at STP? 1. You will need to use PV = nRT

71. Gas Stoichiometry Example on HO How many liters of oxygen gas, at 1.00 atm and 25 o C, can be collected from the complete decomposition of 10.5 grams of potassium chlorate? 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) 10.5 g KClO 3 1 mol KClO 3 122.55 g KClO 3 3 mol O 2 2 mol KClO 3 0.13 mol O 2

72. Gas Stoichiometry Example on HO How many liters of oxygen gas, at 1.00 atm and 25 o C, can be collected from the complete decomposition of 10.5 grams of potassium chlorate? 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) 3.2 L O 2 (1.0 atm)(V)(0.08206 atm*L / mol*K )(298 K)(0.13 mol)

73. Gas Laws: Dalton, Density and Gas Stoichiometry At the conclusion of our time together, you should be able to: 1.Explain Dalton’s Law and use it to solve a problem. 2.Use the Ideal Gas Law to solve a gas density problem. 3.Use the Ideal Gas Law to solve a gas stoichiometry problem.

74. GAS DENSITY Highdensity Lowerdensity 22.4 L of ANY gas AT STP = 1 mole

75. Gas Density … so at STP…

76. Density and the Ideal Gas Law Combining the formula for density with the Ideal Gas law, substituting and rearranging algebraically: M = Molar Mass P = Pressure R = Gas Constant T = Temperature in Kelvins

77. Gas Stoichiometry #4 How many liters of oxygen gas, at 37.0  C and 0.930 atmospheres, can be collected from the complete decomposition of 50.0 grams of potassium chlorate? 2 KClO 3 (s)  2 KCl(s) + 3 O 2 (g) = “n” mol O 2 50.0 g KClO 3 1 mol KClO 3 122.55 g KClO 3 3 mol O 2 2 mol KClO 3 = 0.612 mol O 2 = 16.7 L

78. 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)

79. Gas Laws: Finding “R” At the conclusion of our time together, you should be able to: 1.Use the Ideal Gas Law to find the value of “R”. See the problem on p. 50

80. Partial Pressure of CO 2 Carbon Dioxide gas over water at 17.0 degrees C and 97.932 kPa. What is the pressure of the CO 2 gas? 97.932 kPa – 1.90 kPa = 96.032 kPa = 0.948 atm 1.90 kPa

81. Determine the Moles of CO 2 Used: Mass of canister before-mass of canister after 17.099 g – 16.524 g= 0.575 g 0.575 g CO 2 1 mol CO 2 44.01 g CO 2 0.0131 mol CO 2

82. (0.948 atm)(0.347 L) (R)(290 K) 0.0868 atm*L / mol*K (0.0131 mol)

83. 0.08206 atm*L / mol*K (standard) 0.0868 atm*L / mol*K (experimental results) - 0.0048 atm*L / mol*K 0.08206 atm*L / mol*K x 100 - 5.85 % error - 0.00474 atm*L / mol*K (experimental error)