2. 1 Gases Properties of Gases Gas Pressure

3. 2 Gases What gases are important for each of the following: O 2, CO 2 and/or He? A. B. C. D.

4. 3 Gases What gases are important for each of the following: O 2, CO 2 and/or He? A. CO 2 B. O 2 /CO 2 C. O 2 D. He

5. 4 Some Gases in Our Lives Air: oxygen O 2 nitrogen N 2 ozone O 3 argon Ar carbon dioxide CO 2 water H 2 O Noble gases: helium He neon Ne krypton Kr xenon Xe Other gases: fluorine F 2 chlorine Cl 2 ammonia NH 3 methane CH 4 carbon monoxide CO nitrogen dioxide NO 2 sulfur dioxide SO 2

6. 5 The Nature of Gases Gases are compressible Why can you put more air in a tire, but can’t add more water to a glass full of water? Gases have low densities Dsolid or liquid = 2 g/mL Dgas 2 g/L

7. 6 Nature of Gases 1. Why does a round balloon become spherical when filled with air? 2. Suppose we filled this room halfway with water. Where would pressure be exerted?

8. 7 Nature of Gases Gases fill a container completely and uniformly Gases exert a uniform pressure on all inner surfaces of their containers

9. 8 Kinetic Theory of Gases The particles in gases Are very far apart Move very fast in straight lines until they collide Have no attraction (or repulsion) Move faster at higher temperatures

10. 9 Barometers 760 mmHg atm pressure

11. 10 Learning Check G1 A.The downward pressure of the Hg in a barometer is _____ than (as) the weight of the atmosphere. 1) greater2) less 3) the same B.A water barometer has to be 13.6 times taller than Hg barometer (D Hg = 13.6 g/mL) because 1) H 2 O is less dense 2) Hg is heavier 3) air is more dense than H 2 O

12. 11 Solution G1 A.The downward pressure of the Hg in a barometer is 3) the same (as) the weight of the atmosphere. B.A water barometer has to be 13.6 times taller than Hg barometer (D Hg = 13.6 g/mL) because 1) H 2 O is less dense

13. 12 Unit of Pressure One atmosphere (1 atm)  Is the average pressure of the atmosphere at sea level  Is the standard of pressure  P = Force Area 1.00 atm = 760 mm Hg = 760 torr

14. 13 Learning Check G2 When you drink through a straw you reduce the pressure in the straw. Why does the liquid go up the straw? 1) the weight of the atmosphere pushes it 2) the liquid is at a lower level 3) there is empty space in the straw Could you drink a soda this way on the moon? 1) yes 2) no 3) maybe Why or why not?

15. 14 Solution G2 When you drink through a straw you reduce the pressure in the straw. Why does the liquid go up the straw? 1) the weight of the atmosphere pushes it 3) there is empty space in the straw Could you drink a soda this way on the moon? 2) no Why or why not? Low atmospheric pressure

16. 15 Types of Pressure Units PressureUsed in 760 mm Hg or 760 torrChemistry 14.7 lb/in. 2 U.S. pressure gauges 29.9 in. HgU.S. weather reports 101.3 kPa (kilopascals)Weather in all countries except U.S. 1.013 barsPhysics and astronomy

17. 16 Learning Check G3 A. What is 475 mm Hg expressed in atm? 1) 475 atm 2) 0.625 atm 3) 3.61 x 10 5 atm B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 1) 2.00 mm Hg 2) 1520 mm Hg 3)22,300 mm Hg

18. 17 Solution G3 A. What is 475 mm Hg expressed in atm? 485 mm Hg x 1 atm = 0.625 atm (B) 760 mm Hg B. The pressure of a tire is measured as 29.4 psi. What is this pressure in mm Hg? 29.4 psi x 1.00 atm x 760 mmHg = 1.52 x 10 3 mmHg 14.7 psi 1.00 atm (B)

19. Chapter 7 Gases Pressure and Volume (Boyle’s Law) Temperature and Volume (Charles’ Law) Temperature and Pressure (Gay-Lussac’s Law)

20. Pressure and Volume ExperimentPressureVolume P x V (atm) (L) (atm x L) 18.0 2.0 16 24.04.0_____ 32.08.0_____ 41.016_____ Boyle's Law P x V = k (constant) when T remains constant P1V1= 8.0 atm x 2.0 L = 16 atm L P2V2= 4.0 atm x 4.0 L = 16 atm L P1V1 = P2V2 = k Use this equation to calculate how a volume changes when pressure changes, or how pressure changes when volume changes. new vol. old vol. x Pfactor new P old P x Vfactor V2 = V1 x P1 P2 = P1 x V1 P2 V2

21. P and V Changes P1P1 P2P2 V1V1 V2V2

22. Boyle's Law The pressure of a gas is inversely related to the volume when T does not change Then the PV product remains constant P 1 V 1 = P 2 V 2 P 1 V 1 = 8.0 atm x 2.0 L = 16 atm L P 2 V 2 = 4.0 atm x 4.0 L = 16 atm L

23. PV Problem Freon-12, CCl 2 F 2, is used in refrigeration systems. What is the new volume (L) of a 1.6 L sample of Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg at constant T?

24. PV Calculation Prepare a data table DATA TABLE Initial conditionsFinal conditions P 1 = 50 mm HgP 2 = 200 mm Hg V 1 = 1.6 LV 2 = ? ?

25. Find New Volume (V 2 ) Solve for V 2 : P 1 V 2 = P 2 V 2 V 2 = V 1 x P 1 /P 2 V 2 = 1.6 L x 50 mm Hg = 0.4 L 200 mm Hg

26. Learning Check GL1 A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant) Explain. 1) 3.2 L 2) 6.4 L 3) 12.8 L

27. Solution GL1 A sample of nitrogen gas is 6.4 L at a pressure of 0.70 atm. What will the new volume be if the pressure is changed to 1.40 atm? (T constant) 6.4 L x 0.70 atm = 3.2 L (1) 1.40 atm Volume must decrease to cause an increase in the pressure

28. Learning Check GL2 A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain. 1) 200. mmHg 2) 400. mmHg 3) 1200 mmHg

29. Solution GL2 A sample of helium gas has a volume of 12.0 L at 600. mm Hg. What new pressure is needed to change the volume to 36.0 L? (T constant) Explain. 600. mm Hg x 12.0 L = 200. mmHg (1) 36.0 L Pressure decrease when volume increases.

30. Charles’ Law V = 125 mL V = 250 mL T = 273 K T = 546 K Observe the V and T of the balloons. How does volume change with temperature?

31. Charles’ Law: V and T At constant pressure, the volume of a gas is directly related to its absolute (K) temperature V 1 = V 2 T 1 T 2

32. Learning Check GL3 Use Charles’ Law to complete the statements below: 1. If final T is higher than initial T, final V is (greater, or less) than the initial V. 2. If final V is less than initial V, final T is (higher, or lower) than the initial T.

33. Solution GL3 V 1 = V 2 T 1 T 2 1. If final T is higher than initial T, final V is (greater) than the initial V. 2. If final V is less than initial V, final T is (lower) than the initial T.

34. V and T 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?

35. VT Calculation Complete the following setup: Initial conditionsFinal conditions V 1 = 785 mLV 2 = ? T 1 = 21°C = 294 KT 2 = 0°C = 273 K V 2 = _______ mL x __ K = _______ mL V 1 K Check your answer: If temperature decreases, V should decrease.

36. Learning Check GL4 A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. What temperature (in °C) is needed to change the volume to 640 mL? 1) 443°C2) 170°C 3) - 82°C

37. Solution GL4 A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. What temperature (in °C) is needed to change the volume to 640 mL? 2) 170°C T 2 = 291 K x 640 mL = 443 K 420 mL = 443 K - 273 K = 170°C

38. Gay-Lussac’s Law: P and T The pressure exerted by a confined gas is directly related to the temperature (Kelvin) at constant volume. P (mm Hg)T (°C) 936100 761 25 691 0

39. Learning Check GL5 Use Gay-Lussac’s law to complete the statements below: 1. When temperature decreases, the pressure of a gas (decreases or increases). 2. When temperature increases, the pressure of a gas (decreases or increases).

40. Solution GL5 1. When temperature decreases, the pressure of a gas (decreases). 2. When temperature increases, the pressure of a gas (increases).

41. PT Problem A gas has a pressure at 2.0 atm at 18°C. What will be the new pressure if the temperature rises to 62°C? (V constant) T = 18°C T = 62°C

42. PT Calculation P 1 = 2.0 atmT 1 = 18°C + 273 = 291 K P 2 = ?T 2 = 62°C + 273 = 335 K What happens to P when T increases? P increases (directly related to T) P 2 = P 1 x T 2 T1T1 P 2 = 2.0 atm x K = atm K ?

43. Learning Check GL6 Complete with 1) Increases 2) Decreases 3) Does not change A. Pressure _____, when V decreases B. When T decreases, V _____. C. Pressure _____ when V changes from 12.0 L to 24.0 L (constant n and T) D. Volume _____when T changes from 15.0 °C to 45.0°C (constant P and n)

44. Solution GL6 A. Pressure 1) Increases, when V decreases B. When T decreases, V 2) Decreases C. Pressure 2) Decreases when V changes from 12.0 L to 24.0 L (constant n and T) D. Volume 1) Increases when T changes from 15.0 °C to 45.0°C (constant P and n)

45. LecturePLUS Timberlake44 Chapter 7 Gases The Combined Gas Law Volume and Moles (Avogadro’s Law) Partial Pressures

46. LecturePLUS Timberlake45 Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 Rearrange the combined gas law to solve for V 2 P 1 V 1 T 2 = P 2 V 2 T 1 V 2 = P 1 V 1 T 2 P 2 T 1

47. LecturePLUS Timberlake46 Combined Gas Law P 1 V 1 = P 2 V 2 T 1 T 2 Isolate V 2 P 1 V 1 T 2 = P 2 V 2 T 1 V 2 = P 1 V 1 T 2 P 2 T 1

48. LecturePLUS Timberlake47 Learning Check C1 Solve the combined gas laws for T 2.

49. LecturePLUS Timberlake48 Solution C1 Solve the combined gas law for T 2. (Hint: cross-multiply first.) P 1 V 1 = P 2 V 2 T 1 T 2 P 1 V 1 T 2 = P 2 V 2 T 1 T 2 = P 2 V 2 T 1 P 1 V 1

50. LecturePLUS Timberlake49 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?

51. LecturePLUS Timberlake50 Data Table Set up Data Table P 1 = 0.800 atm V 1 = 0.180 L T 1 = 302 K P 2 = 3.20 atm V 2 = 90.0 mL T 2 = ?? ??

52. LecturePLUS Timberlake51 Solution Solve for T 2 Enter data T 2 = 302 K x atm x mL = K atm mL T 2 = K - 273 = °C

53. LecturePLUS Timberlake52 Calculation Solve for T 2 T 2 = 302 K x 3.20 atm x 90.0 mL = 604 K 0.800 atm 180.0 mL T 2 = 604 K - 273 = 331 °C

54. LecturePLUS Timberlake53 Learning Check C2 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?

55. LecturePLUS Timberlake54 Solution G9 T 1 = 308 KT 2 = ? V 1 = 675 mLV 2 = 0.315 L = 315 mL P 1 = 0.850 atm P 2 = 802 mm Hg = 646 mm Hg T 2 = 308 K x 802 mm Hg x 315 mL 646 mm Hg 675 mL P inc, T inc V dec, T dec = 178 K - 273 = - 95°C

56. LecturePLUS Timberlake55 Volume and Moles How does adding more molecules of a gas change the volume of the air in a tire? If a tire has a leak, how does the loss of air (gas) molecules change the volume?

57. LecturePLUS Timberlake56 Learning Check C3 True (1) or False(2) 1.___The P exerted by a gas at constant V is not affected by the T of the gas. 2.___ At constant P, the V of a gas is directly proportional to the absolute T 3.___ At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.

58. LecturePLUS Timberlake57 Solution C3 True (1) or False(2) 1. (2)The P exerted by a gas at constant V is not affected by the T of the gas. 2. (1) At constant P, the V of a gas is directly proportional to the absolute T 3. (1) At constant T, doubling the P will cause the V of the gas sample to decrease to one-half its original V.

59. LecturePLUS Timberlake58 Avogadro’s Law When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas V 1 = V 2 n 1 n 2 initial final

60. LecturePLUS Timberlake59 STP The volumes of gases can be compared when they have the same temperature and pressure (STP). Standard temperature 0°C or 273 K Standard pressure 1 atm (760 mm Hg)

61. LecturePLUS Timberlake60 Learning Check C4 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? P 1 = V 1 = T 1 = K P 2 = V 2 = ?? T 2 = K V 2 = 15 L x atm x K = 6.8 L atm K

62. LecturePLUS Timberlake61 Solution C4 P 1 = 1.0 atm V 1 = 15 L T 1 = 273 K P 2 = 2.0 atm V 2 = ?? T 2 = 248 K V 2 = 15 L x 1.0 atm x 248 K = 6.8 L 2.0 atm 273 K

63. LecturePLUS Timberlake62 Molar Volume At STP 4.0 g He 16.0 g CH 4 44.0 g CO 2 1 mole 1 mole1mole (STP) (STP)(STP) V = 22.4 L V = 22.4 L V = 22.4 L

64. LecturePLUS Timberlake63 Molar Volume Factor 1 mole of a gas at STP = 22.4 L 22.4 L and 1 mole 1 mole 22.4 L

65. LecturePLUS Timberlake64 Learning Check C5 A.What is the volume at STP of 4.00 g of CH 4 ? 1) 5.60 L2) 11.2 L3) 44.8 L B. How many grams of He are present in 8.0 L of gas at STP? 1) 25.6 g2) 0.357 g3) 1.43 g

66. LecturePLUS Timberlake65 Solution C5 A.What is the volume at STP of 4.00 g of CH 4 ? 4.00 g CH 4 x 1 mole CH 4 x 22.4 L (STP) = 5.60 L 16.0 g CH 4 1 mole CH 4 B. How many grams of He are present in 8.0 L of gas at STP? 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 He 1 mole He

67. LecturePLUS Timberlake66 Daltons’ Law of Partial Pressures Partial Pressure Pressure each gas in a mixture would exert if it were the only gas in the container Dalton's Law of Partial Pressures The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture. P T = P 1 + P 2 + P 3 +.....

68. LecturePLUS Timberlake67 Gases in the Air The % of gases in air Partial pressure (STP) 78.08% N 2 593.4 mmHg 20.95% O 2 159.2 mmHg 0.94% Ar 7.1 mmHg 0.03% CO 2 0.2 mmHg P AIR = P N + P O + P Ar + P CO = 760 mmHg 2 2 2 Total Pressure760 mm Hg

69. LecturePLUS Timberlake68 Learning Check C6 A. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 1) 35.6 2) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557 2) 9.143) 0.109

70. LecturePLUS Timberlake69 Solution C6 A. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 2) 156 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557

71. LecturePLUS Timberlake70 Partial Pressures The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles. P = 1.00 atm 0.5 mole O 2 + 0.3 mole He + 0.2 mole Ar 1 mole H 2

72. LecturePLUS Timberlake71 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.

73. LecturePLUS Timberlake72 Learning Check C7 A 5.00 L scuba tank contains 1.05 mole of O 2 and 0.418 mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?

74. LecturePLUS Timberlake73 Solution C7 P = nRT P T = P O + P He V 2 P T = 1.47 mol x 0.0821 L-atm x 298 K 5.00 L(K mol) =7.19 atm

75. Lecture PLUS Timberlake 200074 Ideal Gas Law The equality for the four variables involved in Boyle’s Law, Charles’ Law, Gay-Lussac’s Law and Avogadro’s law can be written PV = nRT R = ideal gas constant

76. Lecture PLUS Timberlake 200075 Ideal Gases Behave as described by the ideal gas equation; no real gas is actually ideal Within a few %, ideal gas equation describes most real gases at room temperature and pressures of 1 atm or less In real gases, particles attract each other reducing the pressure Real gases behave more like ideal gases as pressure approaches zero.

77. Lecture PLUS Timberlake 200076 PV = nRT R is known as the universal gas constant Using STP conditions P V R = PV = (1.00 atm)(22.4 L) nT (1mol) (273K) n T = 0.0821 L-atm mol-K

78. Lecture PLUS Timberlake 200077 Learning Check G15 What is the value of R when the STP value for P is 760 mmHg?

79. Lecture PLUS Timberlake 200078 Solution G15 What is the value of R when the STP value for P is 760 mmHg? R = PV = (760 mm Hg) (22.4 L) nT (1mol) (273K) = 62.4 L-mm Hg mol-K

80. Lecture PLUS Timberlake 200079 Learning Check G16 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 (mmHg) in the tank in the dentist office?

81. Lecture PLUS Timberlake 200080 Solution G16 Set up data for 3 of the 4 gas variables Adjust to match the units of R V = 20.0 L20.0 L T = 23°C + 273 296 K n = 2.86 mol2.86 mol P = ? ?

82. Lecture PLUS Timberlake 200081 Rearrange ideal gas law for unknown P P = nRT V Substitute values of n, R, T and V and solve for P P = (2.86 mol)(62.4L-mmHg)(296 K) (20.0 L) (K-mol) = 2.64 x 10 3 mm Hg

83. Lecture PLUS Timberlake 200082 Learning Check G17 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?

84. Lecture PLUS Timberlake 200083 Solution G17 Solve ideal gas equation for n (moles) n = PV RT = (735 mmHg)(5.0 L)(mol K) (62.4 mmHg L)(293 K) = 0. 20 mol O 2 x 32.0 g O 2 = 6.4 g O 2 1 mol O 2

85. Lecture PLUS Timberlake 200084 Molar Mass of a gas What is the molar mass of a gas if 0.250 g of the gas occupy 215 mL at 0.813 atm and 30.0°C? n = PV = (0.813 atm) (0.215 L) = 0.00703 mol RT (0.0821 L-atm/molK) (303K) Molar mass = g = 0.250 g = 35.6 g/mol mol 0.00703 mol

86. Lecture PLUS Timberlake 200085 Density of a Gas Calculate the density in g/L of O 2 gas at STP. From STP, we know the P and T. P = 1.00 atm T = 273 K Rearrange the ideal gas equation for moles/L PV = nRTPV = nRT P = n RTV RTV RT V

87. Lecture PLUS Timberlake 200086 Substitute (1.00 atm ) mol-K = 0.0446 mol O 2 /L (0.0821 L-atm) (273 K) Change moles/L to g/L 0.0446 mol O 2 x 32.0 g O 2 = 1.43 g/L 1 L 1 mol O 2 Therefore the density of O 2 gas at STP is 1.43 grams per liter

88. Lecture PLUS Timberlake 200087 Formulas of Gases A gas has a % composition by mass of 85.7% carbon and 14.3% hydrogen. At STP the density of the gas is 2.50 g/L. What is the molecular formula of the gas?

89. Lecture PLUS Timberlake 200088 Formulas of Gases Calculate Empirical formula 85.7 g C x 1 mol C = 7.14 mol C/7.14 = 1 C 12.0 g C 14.3 g H x 1 mol H = 14.3 mol H/ 7.14 = 2 H 1.0 g H Empirical formula = CH 2 EF mass = 12.0 + 2(1.0) = 14.0 g/EF

90. Lecture PLUS Timberlake 200089 Using STP and density ( 1 L = 2.50 g) 2.50 g x 22.4 L = 56.0 g/mol 1 L 1 mol n = EF/ mol = 56.0 g/mol = 4 14.0 g/EF molecular formula CH 2 x 4 = C 4 H 8

91. Lecture PLUS Timberlake 200090 Gases in Chemical Equations On December 1, 1783, Charles used 1.00 x 10 3 lb of iron filings to make the first ascent in a balloon filled with hydrogen Fe(s) + H 2 SO 4 (aq)  FeSO 4 (aq) + H 2 (g) At STP, how many liters of hydrogen gas were generated?

92. Lecture PLUS Timberlake 200091 Solution lb Fe  g Fe  mol Fe  mol H 2  L H 2 1.00 x 10 3 lb x 453.6 g x 1 mol Fe x 1 mol H 2 1 lb 55.9 g 1 mol Fe x 22.4 L H 2 = 1.82 x 10 5 L H 2 1 mol H 2 Charles generated 182,000 L of hydrogen to fill his air balloon.

93. Lecture PLUS Timberlake 200092 Learning Check G18 How many L of O 2 are need 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)

94. Lecture PLUS Timberlake 200093 Solution G18 Find mole of O 2 28.0 g NH 3 x 1 mol NH 3 x 5 mol O 2 17.0 g NH 3 4 mol NH 3 = 2.06 mol O 2 V = nRT = (2.06 mol)(0.0821)(297K) = 52.9 L P 0.950 atm

95. Lecture PLUS Timberlake 200094 Summary of Conversions with Gases Volume A Volume B Grams A Moles A Moles B Grams B Atoms or molecules A molecules B

96. Lecture PLUS Timberlake 200095 Daltons’ Law of Partial Pressures The % of gases in air Partial pressure (STP) 78.08% N 2 593.4 mmHg 20.95% O 2 159.2 mmHg 0.94% Ar 7.1 mmHg 0.03% CO 2 0.2 mmHg P AIR = P N + P O + P Ar + P CO = 760 mmHg 2 2 2 Total Pressure760 mm Hg

97. Lecture PLUS Timberlake 200096 Learning Check G19 A. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 1) 35.6 2) 156 3) 760 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557 2) 9.143) 0.109

98. Lecture PLUS Timberlake 200097 Solution G19 A. If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O 2 in the air? 2) 156 B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N 2 in the air? 1) 557

99. Lecture PLUS Timberlake 200098 Partial Pressure Pressure each gas in a mixture would exert if it were the only gas in the container Dalton's Law of Partial Pressures The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture. P T = P 1 + P 2 + P 3 +.....

100. Lecture PLUS Timberlake 200099 Partial Pressures The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles. STP P = 1.00 atm P = 1.00 atm 1.0 mol He 0.50 mol O 2 + 0.20 mol He + 0.30 mol N 2

101. Lecture PLUS Timberlake 2000100 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.

102. Lecture PLUS Timberlake 2000101 Learning Check G20 A 5.00 L scuba tank contains 1.05 mole of O 2 and 0.418 mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?

103. Lecture PLUS Timberlake 2000102 Solution G20 P = nRT P T = P O + P He V 2 P T = 1.47 mol x 0.0821 L-atm x 298 K 5.00 L(K mol) =7.19 atm