1. Pages 422-440 Notes One Unit Five Characteristics of Gases Pressure of fluids Standard Temperature and Pressure Converting Pressures Gas Laws

2. Pressure Versus Molecular Collision Pressure is caused by molecular collision A molecule colliding creates a force. Catching a ball creates a force. P=F/A pp 427

3. Pressure viewed as created in a fluid Created by the weight The deeper you go, the more weight. pp 427

4. Air is a fluid…just like water pp 427

5. Torricellian Barometer 760 torr 780 torr Air Pressure Mercury 740 torr pp 427

6. Pop can Demo

7. Standard Pressure, Temperature and Volume 1 atm = 14.7 psi 1 atm = 29.92 in Hg 1 atm = 760 mm Hg 1 atm = 760 torr 1 atm = 101,325 Pa 1 atm = 101.325 kPa 1 atm = 1.01325 bar 273K or 0 o C K= o C+273 22.4 Liter/mole for any gas at STP pp 427

8. Converting Pressures Convert 25 lb/in 2 to torr Convert 75 Kpa to in Hg

9. Starter The pressure of a gas is measured as 49 torr. Convert this pressure to atmospheres, kiloPascals, and mmHg.

10. Starter: Pressure Conversions The pressure of a gas is measured as 49 torr. Represent this pressure in atmospheres, Pascals, and mmHg.

11. Physical Properties of Gases Gas Laws: Relationships between volume, temperature, pressure, and amount of gas.

12. Boyle’s Law: P and V as one increases, the other decreases inversely proportional pressure is caused by moving molecules hitting container walls If V is decreased and the # of molecules stays constant, there will be more molecules hitting the walls per unit

13. Boyle’s Law: P and V Boyle’s Law: the V of fixed mass of gas varies inversely with P at a constant T. PV = k k is a constant for a certain sample of gas that depends on the mass of gas and T What kind of graph is V vs. P? If we have a set of new conditions for the same sample of gas, they will have same k so:

14. Boyle’s Law

15. Boyle’s Law: P and V Discovered by Irish chemist, Robert Boyle Used a J-shaped tube to experiment with varying pressures in multistory home and effects on volume of enclosed gas

16. Example: Boyle’s Law Consider a 1.53-L sample of gaseous SO 2 at a pressure of 5.6 x 10 3 Pa. If the pressure is changed to 1.5 x 10 4 Pa at constant temperature, what will be the new volume of the gas?

17. Charles’ Law: V and T if P is constant, gases expand when heated when T increases, gas molecules move faster and collide with the walls more often and with greater force to keep the P constant, the V must increase

18. Charles’ Law: V and T Problem: if we use Celsius, we could end up with negative values from calculations in gas laws for volumes we need a T system with no negative values: Kelvin Temperature Scale –starts at -273.15 ° C = absolute zero = 0 K –lowest possible temperature balloon going into liquid nitrogen

19. Charles’ Law: V and T Charles’ Law: the V of fixed mass of gas at constant P varies directly with Kelvin T. V = kT k is a constant for a certain sample of gas that depends on the mass of gas and P What kind of graph is V vs. T? If we have a set of new conditions for the same sample of gas, they will have same k so:

20. Charles’ Law discovered by French physicist, Jacques Charles in 1787 first person to fill balloon with hydrogen gas and make solo balloon flight

21. Example: Charles’ Law & Temp. A sample of gas at 15°C and 1 atm has a volume of 2.58 L. What volume will this gas occupy at 38°C and 1 atm?

22. Charles’ and Boyle’s Demo P1=P1=1.0atm V1=V1= 1.0L T1=T1=273K P2=P2= 2.0L T2=T2= 546K V2=V2= 1.0atm Pressure is constant. Moles are constant. V1V1 T1T1 = V2V2 T2T2 P1=P1= 1.0atm V1=V1=1.0L T1=T1= 273K P2=P2=2.0atm V2=V2=0.50L T2=T2=273K P 1 xV1V1 P 2 xV2V2 = Temperature is constant. Moles are constant. pp 433-440 1.0L 273K = 2.0L 546K 1.0a x1.0L2.0a x0.50L=

23. The Combined Gas Law

24. Combining the gas laws So far we have seen two gas laws: Jacques CharlesRobert Boyle P1V1P1V1 =P2V2P2V2 V1V1 T1T1 = V2V2 T2T2 These are all subsets of a more encompassing law: the combined gas law P1P1 T1T1 = P2P2 T2T2 Read pages 437, 438. Do Q 26 – 33 (skip 31) P 1 V 1 P 2 V 2 T 1 T 2 = Joseph Louis Gay-Lussac

25. Combined Gas Law Equations P1P1 = P2T1V2P2T1V2 T2V1T2V1 V1V1 = P2T1V2P2T1V2 T2P1T2P1 T2T2 = P2T1V2P2T1V2 P1V1P1V1 T1T1 = P1T2V1P1T2V1 P2V2P2V2 P2P2 = P1T2V1P1T2V1 T1V2T1V2 V2V2 = P1T2V1P1T2V1 P2T1P2T1

26. Combined Gas Law Equation Is……..  pp 433-440

27. Combined gas Law Problem One A gas occupies 2.0 m 3 at 121.2 K, exerting a pressure of 100.0 kPa. What volume will the gas occupy at 410.0 K if the pressure is increased to 220.0 kPa? (2.0m 3 ) (100.0KPa) (121.2K ) (220.0KPa )(V 2 ) (410.0K ) = (2.0m 3 ) (100.0KPa)(410.0K ) (121.2K )(220.0KPa ) = V2V2 V2=V2=3.1m 3 pp 433-440 Assign variables and calculate V 2.

28. Combined gas Law Problem Two A 10.0 gram sample of ethane(C 2 H 6 ) gas is at STP. If the volume is changed to 26.0 liters, what is the new Kelvin temperature of the gas? 10.0g 30.0g/mol = 22.4L V 1 V1=V1=7.46L (101.325KPa) (7.46L) (273K) = (101.325KPa)(26.0L) (T 2 ) (273K)(101.325KPa)(26.0L) (101.325KPa) (7.46L) = T2T2 T2=T2=951K pp 433-440 C H 2x 6x 12.0 = 1.0 = 24.0 6.0 30.0g/m E # Mass 10.0 gram 1) Calculate Formula mass. 2) Calculate V 1. 3) Assign variables and calculate T 2.

29. Notes Two Unit Five Grahams’ Law Calculation Review Mass-Mass Calculation Mass-Volume Calculation @STP Volume-Mass Calculation @STP Pages 441-450

30. Graham’s Law Demo 17.0g/m36.5g/m pp 442

31. Graham’s Law Describes how speed compares between gas molecules with different masses. Two different gases: 1)Same Temperature 2)Different Masses Kinetic energy  ½ M 1 V 1 2 = ½ M 2 V 2 2 pp 442

32. V12V12 ½ M 2 V 2 2 = ½ M 1 V 1 2 Graham’s Equation pp 442 M 2 V 2 2 M 1 V 1 2 M1M1 M1M1 M2V22M2V22 M1M1 V12V12 = V22V22 V22V22 = ÷ by M 1 ÷ by V 2 2 Square root of both sides

33. Grahams’ Law Problem One At a high temperature molecules of chlorine gas travel 15.90cm. What is the mass of vaporized metal (gas) under the same conditions, if the metal travels 8.97cm? M2=M2=223g/m Cl2x35.5 = 71.0g/m E # Mass pp 442 (15.90cm) 8.97cm = = 14.9 2

34. Grahams’ Law Problem Two At a certain temperature molecules of chlorine gas travel at 0.450 km/s. What is the speed of sulfur dioxide gas under the same conditions? V2=V2=0.474Km/s pp 442 Cl2x35.5 = 71.0g/m E # Mass S O 1x 2x 32.1 = 16.0 = 32.1 32.0 64.1g/m E# Mass V2V2 =

35. Review Mass-Mass Calculation 1) grams H 2 to moles H 2 5.00g÷ 2) moles H 2 to moles O 2 2.5m H 2 x 3) moles O 2 to grams O 2 1.3mO 2 x 2.0gH 2 /m=2.5m H 2 32.0g/m =42gO 2 ( 1mO 2 ) ( 2mH 2 ) =1.3mO 2 212 H 2 x1.0 = 2.0g/m E # Mass O2x16.0 = 32.0g/m E # Mass How many grams of oxygen will react with 5.00 grams of hydrogen to make water? H 2 (g) + O 2 (g)  H 2 O(l) 42 5.0g X32.0g/m ___m ___g 2.5 ___m1.3 ÷2.0g/m

36. Mass-Volume Calculation @STP 1) grams H 2 to moles H 2 5.00g÷ 2) moles H 2 to moles O 2 2.5m H 2 x 3) moles O 2 to liters O 2 1.3mO 2 x 2.0gH 2 /m=2.5m H 2 22.4L/m =29LO 2 ( 1mO 2 ) ( 2mH 2 ) =1.3mO 2 212 H 2 x1.0 = 2.0g/m E # Mass O2x16.0 = 32.0g/m E # Mass How many liters of oxygen will react with 5.00 grams of hydrogen to make water? H 2 (g) + O 2 (g)  H 2 O(l) 29 5.0g X22.4g/m ___m ___L 2.5 ___m1.3 ÷2.0g/m pp 449

37. Volume-Mass Calculation @STP 1) Liters H 2 to moles H 2 56.0L÷ 2) moles H 2 to moles O 2 2.50m H 2 x 3) moles O 2 to grams O 2 1.25mO 2 x 22.4LH 2 /m=2.50m H 2 32.0g/m =40.0gO 2 ( 1mO 2 ) ( 2mH 2 ) =1.25mO 2 212 H 2 x1.0 = 2.0g/m E # Mass O2x16.0 = 32.0g/m E # Mass How many grams of oxygen will react with 56.0 liters of hydrogen to make water? H 2 (g) + O 2 (g)  H 2 O(l) 40.0 56.0L X32.0g/m ____m ____g 2.50 ____m1.25 ÷22.4L/m pp 449

38. Notes Three Unit Five Kinetic theory of gases Molar volume @ Non-STP Conditions R is Universal Gas Constant Pages 452-459

40. The Ideal Gas Law PV = nRT

41. Ideal Gases An “ideal” gas exhibits certain theoretical properties. Specifically, an ideal gas … Obeys all of the gas laws under all conditions. Does not condense into a liquid when cooled. Shows perfectly straight lines when its V and T & P and T relationships are plotted on a graph. In reality, there are no gases that fit this definition perfectly. We assume that gases are ideal to simplify our calculations. We have done calculations using several gas laws (Boyle’s Law, Charles’s Law, Combined Gas Law). There is one more to know…

42. THE KINETIC THEORY OF GASES Large number of particles 6.022x10 23 atoms/mole pp 426

43. Finding volumes @ Non-STP Conditions pp 446 PV=nRT What is… P? V? n? T? moles R?Universal Gas Constant P V n T R = (101.325kpa) (22.4L) (1 m)(273.15K) R = Ideal Gas Equation R =(8.314 ) Kpa -L m - K

44. The Ideal Gas Law PV = nRT P = Pressure (in kPa)V = Volume (in L) T = Temperature (in K) n = moles R = 8.31 kPa L K mol R is constant. If we are given three of P, V, n, or T, we can solve for the unknown value. Recall, From Boyle’s Law: P 1 V 1 = P 2 V 2 or PV = constant From combined gas law: P 1 V 1 /T 1 = P 2 V 2 /T 2 or PV/T = constant Or 0.0821 L-Atm/mol-K Or 62.4 L-Torr/mol-K

45. Developing the ideal gas law equation PV/T = constant. What is the constant? At STP: T= 273K, P= 101.3 kPa, V= 22.4 L/mol PV = constant T mol Mol is represented by n, constant by R: PV = R Tn Rearranging, we get: PV = nRT Because V depends on mol, we can change equation to: At STP: (101.3 kPa)(22.4 L) = (1 mol)(R)(273K) R = 8.31 kPa L K mol Note: always use kPa, L, K, and mol in ideal gas law questions (so units cancel)

46. Finding Volume at Non-STP Oxygen is made reacting 15.00g water at 209.0Kpa and 20.0 o C. 18.0g/m= 0.833m 22.4L/m =9.33LO 2 ( 1mO 2 ) (2m H 2 O) =0.417mO 2 a) What volume of oxygen would be made at STP? 1) grams H 2 O to moles H 2 O 2) moles H 2 O to moles O 2 3) moles O 2 to liters O 2 15.00g÷ 0.833m x 0.417mO 2 x 2H 2 O(l)  2H 2 (l) + 1O 2 (l) 15.00g X22.4L/m ______m ____L 0.833 0.417 ÷18.0g/m _____m 9.33 @STP E #Mass O 16.0 = 16.0 1 x 18.0g/m H 2 x1.0 = 2.0

47. Mass Volume @ non-STP How many Liters of Oxygen are made by decomposing12.13g sodium peroxide at 129.5Kpa and 22.1 o C. 2Na 2 O 2 (l)  2Na 2 O(l) + 1O 2 (g)

49. Molar Volume Lab Molar Volume Lab Data Molar Volume Lab Calculations

50. Molar Volume Lab Data 3.41 1.39 35.64 19.9 743.2 17.3 3.39 1.37 35.62 19.7 743.0 17.5 3.40 1.38 35.63 19.9 743.1 17.4

51. Molar Volume Lab Calculations 3.40cm X ( 1.38g Mg) (100.00cm) = 0.0469g ÷24.3g/mol=0.00193moles 743.1 - 17.4= (725.7T) 1. Calculate the grams of magnesium reacted. 2. Calculate the moles of magnesium reacted. 3.Calculate the pressure of the dry hydrogen gas. 4. Calculate the volume (V 2 ) at STP from data. 5. What is the molar volume of hydrogen at STP. PH 2 = P air - P P P1P1 V1V1 T1T1 P2P2 V2V2 T2T2 = (725.7T)(35.63mL) (19.9 o C+273.15) (760T)V2V2 (273.15K) = V2=V2=31.70mL Volume of H 2 at STP Moles of H 2 31.70mL 0.00193m ==16400mL/m H2H2 = H2OH2O

52. Mass-Volume Calculation @STP 1) grams Al 2 (CO 3 ) 3 to moles Al 2 (CO 3 ) 3 6.00g÷ 2) moles Al 2 (CO 3 ) 3 to moles CO 2 0.0256mx 3) moles CO 2 to liters CO 2 0.0769mCO 2 x 234.0g/m=0.0256m 22.4L/m =1.72LCO 2 ( 3 m CO 2 ) ( 1 Al 2 (CO 3 ) 3 ) =0.0769m 1 1 3 Al 2 x27.0 = 54.0 E # Mass How many liters of CO 2 will form when 6.00 g of Al 2 (CO 3 ) 3 decomposes? Al 2 (CO 3 ) 3 (s)  Al 2 O 3 (s)+ CO 2 (g) 1.72 6.00g X22.4g/m ______m ____L 0.0256 ______m0.0769 ÷234.0g/m pp 449 C 3 x12.0 = 36.0 O 9 x16.0 = 144.0 234.0g/m

53. Unit Five Combined Quiz Review Unit Eight

54. Combined gas Law Quiz Example One A gas occupies 4.0 m 3 at 135.1 K, exerting a pressure of 101.3 kPa. What volume will the gas occupy at 390.0 K if the pressure is increased to 150.0 kPa? (4.0m 3 ) (101.3KPa) (135.1K ) (150.0KPa )(V 2 ) (390.0K ) = (4.0m 3 ) (101.3KPa)(390.0K ) (135.1K )(150.0KPa ) = V2V2 V2=V2=7.8M 3 Assign variables and calculate V 2.

55. A 53.0 gram sample of ethyne(C 2 H 2 ) gas is at STP. If the volume is changed to 41.0 liters, what is the new Kelvin temperature of the gas? 53.0g 26.0g/mol = 22.4L V 1 V1=V1=45.6L (101.325KPa) (45.6L) (273K) = (101.325KPa)(41.0L) (T 2 ) (273K)(101.325KPa)(41.0L) (101.325KPa) (45.6L) = T2T2 T2=T2=245K C H 2x 12.0 = 1.0 = 24.0 2.0 26.0g/m E # Mass 53.0 gram 1) Calculate Formula mass. 2) Calculate V 1. 3) Assign variables and calculate T 2. Combined gas Law Quiz Example Two

56. Eight Rows Seven Rows

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61. Eight Rows Seven Rows

62. Quiz Two Notes Mass-Volume Calculation @STP Grahams’ Law Problem

63. Mass-Volume Calculation @STP 1) grams N 2 to moles N 2 4.00g÷ 2) moles N 2 to moles O 3 0.143m N 2 x 3) moles O 3 to liters O 3 0.238mO 3 x 28.0gN 2 /m=0.143m N 2 22.4L/m =5.33L ( 5mO 3 ) ( 3mN 2 ) =0.238mO 3 353 N 2 x14.0 = 28.0g/m E # Mass O3x16.0 = 48.0g/m E # Mass How many liters of ozone will react with 4.00 grams of nitrogen to make dinitrogen pentoxide? N 2 (g) + O 3 (g)  N 2 O 5 (l) 5.33 4.00g X22.4g/m _____m ____L 0.143 ____m0.238 ÷28.0g/m pp 449

64. Volume-Mass Calculation @STP 1) Liters N 2 to moles N 2 56.0L÷ 2) moles N 2 to moles O 3 2.50m N 2 x 3) moles O 3 to grams O 3 4.17mO 3 x 22.4LN 2 /m=2.50m N 2 48.0g/m =200.g ( 5mO 3 ) ( 3mN 2 ) =4.17mO 3 353 N 2 x14.0 = 28.0g/m E # Mass O3x16.0 = 48.0g/m E # Mass How many grams of ozone will react with 56.0 liters of nitrogen to make dinitrogen pentoxide? N 2 (g) + O 3 (g)  N 2 O 5 (l) 200. 56.0L X48.0g/m ____m ____g 2.50 ____m4.17 ÷22.4L/m pp 449

65. Grahams’ Law Problem Three =770m/s pp 442 He1x4.0 = 4.0g/m E # Mass V2V2 = S F 1x 6x 32.1 = 19.0 = 32.1 114.0 146.1g/m E# Mass M2M2 M1M1 = V1V1 V2V2 = V2V2 146.1g/m 4.0g/m 128m/s 146.1g/mx(128m/s) 4.0g/m If a sulfur hexafluoride molecule has a speed of 128m/s, what would helium’s speed be at the same temperature?

66. Seven/Eight Rows

68. Final Quiz Examples Fill in: Word Bank Combined gas Law Problems:11 and 12 Volume at Non-STP: Items 13.1 and 13.2 Grahams’ Law: Item 14 Written Response: Item 15

69. Word Bank: Items 1 through 10

70. Combined Gas Law: Problems 11 and 12 A gas occupies 7.1 m 3 at 162.3 K, exerting a pressure of 121.5 kPa. What volume will the gas occupy at 250.0 K if the pressure is increased to 172.1 kPa? (7.1m 3 ) (121.5KPa) (162.3K ) (172.1KPa )(V 2 ) (250.0K ) = (7.1m 3 ) (121.5KPa) (250.0K ) (162.3K )(172.1KPa ) = V2V2 V2=V2=7.7M 3 Assign variables and calculate V 2.

71. Volume at Non-STP: Items 13.1 and 13.2 PV=nRT V= (0.0778m)(8.314 LKpam -1 K -1 )(295.3K) (129.5KPa) V= 1.47 L b) What is the volume of gas at reaction conditions? P V nTR = Oxygen is made decomposing12.13g sodium peroxide at 129.5Kpa and 22.1 o C.

72. Grahams’ Law: Item 14 V2=V2=670m/s pp 442 H 2 x1.0 = 2.0g/m E # Mass O2x16.0 = 32.0g/m E # Mass V2V2 = If oxygen molecules have a speed of 168m/s, what would the speed of hydrogen molecules be at the same temperature?

73. Written Response: Item 15 Why is the molar volume quantity, 22.4L/m, different at non-standard conditions? Molar volume is different under non-standard conditions due to gas volume being dependent on pressure and temperature. According to Boyle’s law, if the volume is decrease, diagrams 1, the pressure will increase. This increase is due to the number of collisions occurring at a smaller volume. More collisions mean higher pressure. diagrams 1 Before After diagrams 2 Before After According to Charles’ Law, if the temperature is increased at constant temperature, the volume of a gas will increase. The increase occurs due the molecules striking the container walls with more force and causing the volume to expand. See diagrams 2.

74. end

75. Calculate the mean free path of a nitrogen molecule at 1 Torr, 298 K. (Nitrogen, which makes up 80% of the atmosphere, has a molecular diameter of 3.7Å)

77. Finding Volume at Non-STP Oxygen is made decomposing 26.31g iron(III) sulfate at 79.0Kpa and 21.1 o C. Fe 2 (SO 4 ) 3 (s)  FeO (s) + SO 3 (g) + O 2 (g) a) What volume of oxygen would be made at STP? 1) grams Fe 2 (SO 4 ) 3 to moles Fe 2 (SO 4 ) 3 26.31g÷ 2) moles Fe 2 (SO 4 ) 3 to moles O 2 0.06579m Fe 2 (SO 4 ) 3 x 3) moles O 2 to liters O 2 0.03290mO 2 x b) What is the volume of gas at reaction conditions? c) How many moles of the gas is produced? d) How many grams O 2 gas are produced? 399.9gFe 2 (SO 4 ) 3 /m=0.06579m Fe 2 (SO 4 ) 3 22.4L/m = 0.737LO 2 ( 1mO 2 ) (2m Fe 2 (SO 4 ) 3 =0.03290mO 2 1 23 0.5 PV=nRT V= (0. 03290m)(8.314 LKpam -1 K -1 )(294.3K) (79.0KPa) V=1.02L n=0.03290mO 2 n x g/m= g 0. 03290m x32.0g/m=1.05gO 2