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1 Chapter 7 Gases. 2 Particles of a gas Move rapidly in straight lines and are in constant motion. Move rapidly in straight lines and are in constant.

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Presentation on theme: "1 Chapter 7 Gases. 2 Particles of a gas Move rapidly in straight lines and are in constant motion. Move rapidly in straight lines and are in constant."— Presentation transcript:

1 1 Chapter 7 Gases

2 2 Particles of a gas Move rapidly in straight lines and are in constant motion. Move rapidly in straight lines and are in constant motion. Have kinetic energy that increases with an increase in temperature. Have kinetic energy that increases with an increase in temperature. Are very far apart. Are very far apart. Have essentially no attractive (or repulsive) forces. Have essentially no attractive (or repulsive) forces. Have very small volumes compared to the volume of the container they occupy. Have very small volumes compared to the volume of the container they occupy. Kinetic Theory of Gases

3 3 Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n). Gases are described in terms of four properties: pressure (P), volume (V), temperature (T), and amount (n). Properties of Gases

4 4 A barometer measures the pressure exerted by the gases in the atmosphere. A barometer measures the pressure exerted by the gases in the atmosphere. The atmospheric pressure is measured as the height in mm of the mercury column. The atmospheric pressure is measured as the height in mm of the mercury column. Barometer

5 5 A. The downward pressure of the Hg in a barometer is _____ than/as the weight of the atmosphere. 1) greater 2) less 3) the same 1) greater 2) less 3) the same B. A water barometer is 13.6 times taller than a Hg barometer (D Hg = 13.6 g/mL) because 1) H 2 O is less dense 2) H 2 O is heavier 3) air is more dense than H 2 O 3) air is more dense than H 2 O Learning Check

6 6 A.The downward pressure of the Hg in a barometer is 3) the same as the weight of the atmosphere. B. A water barometer is 13.6 times taller than a Hg barometer (D Hg = 13.6 g/mL) because 1) H 2 O is less dense Solution

7 7 A gas exerts pressure, which is defined as a force acting on a specific area. Pressure (P) = Force Area A gas exerts pressure, which is defined as a force acting on a specific area. Pressure (P) = Force Area One atmosphere (1 atm) is 760 mm Hg. One atmosphere (1 atm) is 760 mm Hg. 1 mm Hg = 1 torr 1.00 atm = 760 mm Hg = 760 torr 1 mm Hg = 1 torr 1.00 atm = 760 mm Hg = 760 torr Pressure

8 8 In science, pressure is stated in atmospheres (atm), millimeters of mercury (mm Hg), and Pascals (Pa). In science, pressure is stated in atmospheres (atm), millimeters of mercury (mm Hg), and Pascals (Pa). Units of Pressure

9 9 A. What is 475 mm Hg expressed in atm? 1) 475 atm 2) atm 3) 3.61 x 10 5 atm B. The pressure in a tire is 2.00 atm. What is this pressure in mm Hg? 1) 2.00 mm Hg 2) 1520 mm Hg 3)22,300 mm Hg Learning Check

10 10 A. What is 475 mm Hg expressed in atm? 2) atm 485 mm Hg x 1 atm = atm 760 mm Hg 760 mm Hg B. The pressure of a tire is measured as 2.00 atm. What is this pressure in mm Hg? 2) 1520 mm Hg 2.00 atm x 760 mm Hg = 1520 mm Hg 1 atm 1 atm Solution

11 11 The pressure of a gas is inversely related to its volume when T and n are constant. The pressure of a gas is inversely related to its volume when T and n are constant. If volume decreases, the pressure increases. If volume decreases, the pressure increases. Boyles Law

12 12 The product P x V remains constant as long as T and n do not change. 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 P 3 V 3 = 2.0 atm x 8.0 L = 16 atm L The product P x V remains constant as long as T and n do not change. 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 P 3 V 3 = 2.0 atm x 8.0 L = 16 atm L Boyles Law can be stated as P 1 V 1 = P 2 V 2 (T, n constant) Boyles Law can be stated as P 1 V 1 = P 2 V 2 (T, n constant) PV Constant in Boyles Law

13 13 Solving for a Gas Law Factor The equation for Boyles Law can be rearranged to solve for any factor. The equation for Boyles Law can be rearranged to solve for any factor. To solve for V 2, divide both sides by P 2. To solve for V 2, divide both sides by P 2. P 1 V 1 = P 2 V 2 Boyles Law P 2 P 2 P 2 P 2 P 1 V 1 =V 2 P 1 V 1 =V 2 P 2 P 2 Solving for P 2 Solving for P 2 P 2 = P 1 V 1 V 2 V 2

14 14 PV in Breathing Mechanics When the lungs expand, the pressure in the lungs decreases. When the lungs expand, the pressure in the lungs decreases. Inhalation occurs as air flows towards the lower pressure in the lungs. Inhalation occurs as air flows towards the lower pressure in the lungs.

15 15 PV in Breathing Mechanics When the lung volume decreases, pressure within the lungs increases. When the lung volume decreases, pressure within the lungs increases. Exhalation occurs as air flows from the higher pressure in the lungs to the outside. Exhalation occurs as air flows from the higher pressure in the lungs to the outside.

16 16 Freon-12, CCl 2 F 2, is used in refrigeration systems. What is the new volume (L) of a 8 L sample of Freon gas initially at 50 mm Hg after its pressure is changed to 200 mm Hg at constant T? 1. Set up a data table Conditions 1Conditions 2 P 1 = 50 mm HgP 2 = 200 mm Hg V 1 = 8 LV 2 = Calculation with Boyles Law ?

17 17 2. When pressure increases, volume decreases. Solve Boyles Law for V 2 : P 1 V 1 = P 2 V 2 V 2 = V 1 P 1 P 2 P 2 V 2 = 8 L x 50 mm Hg = 2 L 200 mm Hg pressure ratio decreases volume Calculation with Boyles Law (continued)

18 18 The helium in a cylinder has a volume of 120 mL and a pressure of 840 mm Hg. A change in the volume results in a lower pressure inside the cylinder. Does cylinder A or B represent the final volume? At a new pressure of 420 mm Hg, what is the new volume? 1) 60 mL 2) 120 mL3) 240 mL 1) 60 mL 2) 120 mL3) 240 mL Learning Check

19 19 The helium in a cylinder has a volume of 120 mL and a pressure of 840 mm Hg. A change in the volume results in a lower pressure inside the cylinder. Does cylinder A or B represent the final volume? B) If P decreases, V increases. At a new pressure of 420 mm Hg, what is the new volume of the cylinder? 3) 240 mL Solution

20 20 A sample of helium gas has a volume of 6.4 L at a pressure of 0.70 atm. What is the new volume when the pressure is increased to 1.40 atm (T constant)? A) 3.2 LB) 6.4 LC) 12.8 L Learning Check

21 21 A) 3.2 L Solve for V 2 : P 1 V 1 = P 2 V 2 V 2 = V 1 P 1 P 2 P 2 V 2 = 6.4 L x 0.70 atm = 3.2 L 1.40 atm 1.40 atm Volume decreases when there is an increase in the pressure (Temperature is constant). Solution

22 22 A sample of oxygen gas has a volume of 12.0 L at 600. mm Hg. What is the new pressure when the volume changes to 36.0 L? (T and n constant.) 1) 200. mm Hg 2) 400. mm Hg 3) 1200 mm Hg Learning Check

23 23 1) 200. mm Hg Data table Conditions 1Conditions 2 P 1 = 600. mm HgP 2 = ??? V 1 = 12.0 LV 2 = 36.0 L P 2 = P 1 V 1 V 2 V mm Hg x 12.0 L = 200. mm Hg 36.0 L Solution

24 24 The Kelvin temperature of a gas is directly related to the volume (P and n are constant). The Kelvin temperature of a gas is directly related to the volume (P and n are constant). When the temperature of a gas increases, its volume increases. When the temperature of a gas increases, its volume increases. Charles Law

25 25 For two conditions, Charles Law is written For two conditions, Charles Law is written V 1 = V 2 (P and n constant) T 1 T 2 T 1 T 2 Rearranging Charles Law to solve for V 2 Rearranging Charles Law to solve for V 2 V 2 = V 1 T 2 T 1 T 1 Charles Law V and T

26 26 Learning Check Solve Charles Law expression for T 2. V 1 = V 2 T 1 T 2

27 27 Solution V 1 = V 2 T 1 T 2 Cross multiply to give V 1 T 2 =V 2 T 1 Isolate T 2 by dividing through by V 1 V 1 T 2 =V 2 T 1 V 1 V 1 V 1 V 1 T 2 =V 2 T 1 V 1 V 1

28 28 A balloon has a volume of 785 mL at 21°C. If The temperature drops to 0°C, what is the new volume of the balloon (P constant)? 1.Set up data table: Conditions 1Conditions 2 V 1 = 785 mLV 2 = ? T 1 = 21°C = 294 KT 2 = 0°C = 273 K Be sure that you always use the Kelvin (K) temperature in gas calculations. Calculations Using Charles Law

29 29 Calculations Using Charles Law (continued) 2. Solve Charles law for V 2 V 1 = V 2 T 1 T 2 V 2 = V 1 T 2 T 1 V 2 = 785 mL x 273 K = 729 mL 294 K

30 30 A sample of oxygen gas has a volume of 420 mL at a temperature of 18°C. At what temperature (in °C) will the volume of the oxygen be 640 mL (P and n constant)? 1) 443°C 2) 170°C 3) – 82°C Learning Check

31 31 170°C T 2 = T 1 V 2 V 1 V 1 T 2 = 291 K x 640 mL = 443 K 420 mL 420 mL = 443 K – 273 K = 170°C Solution

32 32 The pressure exerted by a gas is directly related to the Kelvin temperature of the gas at constant V and n. The pressure exerted by a gas is directly related to the Kelvin temperature of the gas at constant V and n. P 1 = P 2 P 1 = P 2 T 1 T 2 Gay-Lussacs Law: P and T

33 33 A gas has a pressure at 2.0 atm at 18°C. What is the new pressure when the temperature is 62°C? (V and n constant) 1. Set up a data table. Conditions 1Conditions 2 P 1 = 2.0 atmP 2 = T 1 = 18°C T 2 = 62°C = 291 K = 335 K = 291 K = 335 K Calculation with Gay-Lussacs Law ?

34 34 Calculation with Gay-Lussacs Law (continued) 2. Solve Gay-Lussacs Law for P 2 P 1 = P 2 P 1 = P 2 T 1 T 2 P 2 = P 1 T 2 T 1 T 1 P 2 = 2.0 atm x 335 K = 2.3 atm 291 K 291 K

35 35 Use the gas laws to complete with 1) Increases 2) Decreases 2) 2) A. Pressure _______ when V decreases. B. When T decreases, V _______. C. Pressure _______ when V changes from 12.0 L to 24.0 L. from 12.0 L to 24.0 L. D. Volume _______when T changes from 15.0 °C to 45.0°C °C to 45.0°C. Learning Check

36 36 Use the gas laws to complete with 1) Increases 2) Decreases 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 from 12.0 L to 24.0 L D. Volume 1) Increases when T changes from 15.0 °C to 45.0°C 15.0 °C to 45.0°C Solution

37 37 Next Time We complete Chapter 7 We complete Chapter 7

38 38 The combined gas law uses Boyles Law, Charles Law, and Gay-Lussacs Law (n is constant). The combined gas law uses Boyles Law, Charles Law, and Gay-Lussacs Law (n is constant). P 1 V 1 =P 2 V 2 T 1 T 2 T 1 T 2 Combined Gas Law

39 39 A sample of helium gas has a volume of L, a pressure of atm and a temperature of 29°C. At what temperature (°C) will the helium have a volume of 90.0 mL and a pressure of 3.20 atm (n constant)? 1. Set up Data Table Conditions 1Conditions 2 P 1 = atm P 2 = 3.20 atm V 1 = L (180 mL) V 2 = 90.0 mL T 1 = 29°C = 302 KT 2 = ?? Combined Gas Law Calculation

40 40 2. Solve for T 2 P 1 V 1 =P 2 V 2 T 1 T 2 T 1 T 2 T 2 = T 1 P 2 V 2 P 1 V 1 P 1 V 1 T 2 = 302 K x 3.20 atm x 90.0 mL = 604 K atm mL atm mL T 2 = 604 K – 273 = 331 °C Combined Gas Law Calculation (continued)

41 41 A gas has a volume of 675 mL at 35°C and atm pressure. What is the volume(mL) of the gas at –95°C and a pressure of 802 mm Hg (n constant)? A gas has a volume of 675 mL at 35°C and atm pressure. What is the volume(mL) of the gas at –95°C and a pressure of 802 mm Hg (n constant)? Learning Check

42 42 Data Table T 1 = 308 K T 2 = -95°C = 178K V 1 = 675 mL V 2 = ??? P 1 = 646 mm Hg P 2 = 802 mm Hg Solve for T 2 V 2 =V 1 P 1 T 2 P 2 T 1 P 2 T 1 V 2 = 675 mL x 646 mm Hg x 178K = 314 mL 802 mm Hg x 308K Solution

43 43 The volume of a gas is directly related to the number of moles of gas when T and P are constant. V 1 = V 2 n 1 n 2 The volume of a gas is directly related to the number of moles of gas when T and P are constant. V 1 = V 2 n 1 n 2 Avogadro's Law: Volume and Moles

44 44 Learning Check If 0.75 mole of helium gas occupies a volume of 1.5 L, what volume will 1.2 moles of helium occupy at the same temperature and pressure? 1) 0.94 L 2)1.8 L 3) 2.4 L

45 45 Solution 3) 2.4 L Conditions 1Conditions 2 V 1 = 1.5 LV 2 = ??? n 1 = 0.75 mole Hen 2 = 1.2 moles He V 2 = V 1 n 2 n 1 V 2 = 1.5 L x 1.2 moles He = 2.4 L 0.75 mole He 0.75 mole He

46 46 The volumes of gases can be compared when they have the same conditions of temperature and pressure (STP). The volumes of gases can be compared when they have the same conditions of temperature and pressure (STP). Standard temperature (T) 0°C or 273 K Standard pressure (P) 1 atm (760 mm Hg) STP

47 47 At STP, 1 mole of a gas occupies a volume of 22.4 L. At STP, 1 mole of a gas occupies a volume of 22.4 L. The volume of one mole of a gas is called the molar volume. The volume of one mole of a gas is called the molar volume. Molar Volume

48 48 The molar volume at STP can be used to form conversion factors L and 1 mole 1 mole 22.4 L The molar volume at STP can be used to form conversion factors L and 1 mole 1 mole 22.4 L Molar Volume as a Conversion Factor

49 49 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 8.00 L of gas at STP? 1) 25.6 g2) g3) 1.43 g Learning Check

50 50 A. 1) 5.60 L 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 g CH 4 1 mole CH 4 B. 3) 1.43 g 8.00 L x 1 mole He x 4.00 g He = 1.43 g He 22.4 L 1 mole He 22.4 L 1 mole He Solution

51 51 The relationship between the four properties (P, V, n, and T) of gases can be written equal to a constant R. The relationship between the four properties (P, V, n, and T) of gases can be written equal to a constant R. PV = R nT Rearranging this expression gives the expression called the ideal gas law. Rearranging this expression gives the expression called the ideal gas law. PV = nRT Ideal Gas Law

52 52 The universal gas constant, R, can be calculated using the molar volume of a gas at STP. The universal gas constant, R, can be calculated using the molar volume of a gas at STP. At STP (273 K and 1.00 atm), 1 mole of a gas occupies 22.4 L. At STP (273 K and 1.00 atm), 1 mole of a gas occupies 22.4 L. P V P V R = PV = (1.00 atm)(22.4 L) nT (1 mole) (273K) n T n T = L atm mole K mole K Note there are four units associated with R. Note there are four units associated with R. Universal Gas Constant, R

53 53 Another value for the universal gas constant is obtained using mm Hg for the STP pressure. What is the value of R when a pressure of 760 mm Hg is placed in the R value expression? Learning Check

54 54 What is the value of R when the STP value for P is 760 mmHg? R = PV = (760 mm Hg) (22.4 L) R = PV = (760 mm Hg) (22.4 L) nT (1 mole) (273K) nT (1 mole) (273K) = 62.4 L mm Hg mole K Solution

55 55 Dinitrogen oxide (N 2 O), laughing gas, is used by dentists as an anesthetic. If a 20.0 L tank of laughing gas contains 2.8 moles N 2 O at 23°C, what is the pressure (mm Hg) in the tank? Learning Check

56 56 1. Adjust the units of the given properties to match the units of R. V = 20.0 L, T = 296 K, n = 2.8 moles, P = ? V = 20.0 L, T = 296 K, n = 2.8 moles, P = ? 2. Rearrange the ideal gas law for P. P = nRT V P = (2.8 moles)(62.4 L mm Hg)(296 K) (20.0 L) (mole K) (20.0 L) (mole K) = 2.6 x 10 3 mm Hg = 2.6 x 10 3 mm Hg Solution

57 57 A cylinder contains 5.0 L of O 2 at 20.0°C and 0.85 atm. How many grams of oxygen are in the cylinder? Learning Check

58 58 1. Determine the given properties. P = 0.85 atm, V = 5.0 L, T = 293 K, n (or g =?) 2. Rearrange the ideal gas law for n (moles). n = PV RT RT = (0.85 atm)(5.0 L)(mole K) = 0.18 mole O 2 = (0.85 atm)(5.0 L)(mole K) = 0.18 mole O 2 (0.0821atm L)(293 K) (0.0821atm L)(293 K) 3. Convert moles to grams using molar mass. = mole O 2 x 32.0 g O 2 = 5.8 g O 2 1 mole O 2 1 mole O 2 Solution

59 59 What is the molar mass of a gas if g of the gas occupy 215 mL at atm and 30.0°C? 1. Solve for the moles (n) of gas. n = PV = (0.813 atm) (0.215 L) RT ( L atm/mole K)(303K) n = PV = (0.813 atm) (0.215 L) RT ( L atm/mole K)(303K) = mole = mole 2. Set up the molar mass relationship. Molar mass = g = g = 35.6g/mole mole mole Molar mass = g = g = 35.6g/mole mole mole Molar Mass of a Gas

60 60 Gases in Equations The amounts of gases reacted or produced in a chemical reaction can be calculated using the ideal gas law and mole factors. The amounts of gases reacted or produced in a chemical reaction can be calculated using the ideal gas law and mole factors.Problem: What volume (L) of Cl 2 gas at 1.2 atm and 27°C is needed to completely react with 1.5 g of aluminum? 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s)

61 61 Gases in Equations (continued) 2Al(s) + 3Cl 2 (g) 2AlCl 3 (s) 1.5 g ? L 1.2 atm, 300K 1.5 g ? L 1.2 atm, 300K 1. Calculate the moles of Cl 2 needed. 1.5 g Al x 1 mole Al x 3 moles Cl 2 = mole Cl g Al 2 moles Al 27.0 g Al 2 moles Al 2. Place the moles Cl 2 in the ideal gas equation. V = nRT = (0.083 mole Cl 2 )( Latm/moleK)(300K) P 1.2 atm P 1.2 atm = 1.7 L Cl 2

62 62 What volume (L) of O 2 at 24°C and atm are needed to react with 28.0 g NH 3 ? 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) Learning Check

63 63 1. Calculate the moles of O 2 needed g NH 3 x 1 mole NH 3 x 5 mole O g NH 3 4 mole NH g NH 3 4 mole NH 3 = 2.06 mole O 2 = 2.06 mole O 2 2. Place the moles O 2 in the ideal gas equation. V = nRT = (2.06 moles)( L atm/moleK)(297K) P atm = 52.9 L O 2 Solution

64 64 In a mixture of gases, the partial pressure of each gas is the pressure that gas would exert if it were by itself in the container. In a mixture of gases, the partial pressure of each gas is the pressure that gas would exert if it were by itself in the container. Partial Pressure

65 65 The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture. 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 Daltons Law of Partial Pressures

66 66 The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles. The total pressure of a gas mixture depends on the total number of gas particles, not on the types of particles. Partial Pressures

67 67 For example, at STP, one mole of gas particles in a volume of 22.4 L will exert the same pressure as one mole of a mixture of gas particles in 22.4 L. For example, at STP, one mole of gas particles in a volume of 22.4 L will exert the same pressure as one mole of a mixture of gas particles in 22.4 L. V = 22.4 L V = 22.4 L Total Pressure 0.5 mole O mole He 0.2 mole Ar 1.0 mole 1.0 mole N mole O mole He 1.0 mole 1.0 atm

68 68 A scuba tank contains O 2 with a pressure of atm and He at 855 mm Hg. What is the total pressure in mm Hg in the tank? Learning Check

69 69 1. Convert the pressure in atm to mm Hg atm x 760 mm Hg = 342 mm Hg = P O 1 atm 2 2. Calculate the sum of the partial pressures. P total = P O + P He 2 P total = 342 mm Hg mm Hg = 1197 mm Hg Solution

70 70 Gases We Breathe

71 71 When a scuba diver makes a deep dive, the increased pressure causes more N 2 (g) to dissolve in the blood. When a scuba diver makes a deep dive, the increased pressure causes more N 2 (g) to dissolve in the blood. If a diver rises too fast, the dissolved N 2 will form bubbles in the blood, a dangerous and painful condition called "the bends." If a diver rises too fast, the dissolved N 2 will form bubbles in the blood, a dangerous and painful condition called "the bends." Helium, which does not dissolve in the blood, is mixed with O 2 to prepare breathing mixtures for deep descents. Helium, which does not dissolve in the blood, is mixed with O 2 to prepare breathing mixtures for deep descents. Health Note: Scuba Diving

72 72 For a deep dive, some scuba divers are using a mixture of helium and oxygen gases with a pressure of 8.00 atm. If the oxygen has a partial pressure of 1280 mm Hg, what is the partial pressure of the helium? 1) 520 mm Hg 2) 2040 mm Hg 3) 4800 mm Hg Learning Check

73 73 3) 4800 mm Hg P Total = 8.00 atm x 760 mm Hg = 6080 mm Hg 1 atm 1 atm P Total = P O + P He 2 P He = P Total - P O 2 2 P He = 6080 mm Hg mm Hg = 4800 mm Hg = 4800 mm Hg Solution


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