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GASES.

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Presentation on theme: "GASES."— Presentation transcript:

1 GASES

2 The Kinetic-Molecular Theory for Ideal Gases
The kinetic-molecular theory describes the behavior of IDEAL gases in terms of particles in motion. Real gases do not obey the kinetic molecular theory.

3 The Kinetic-Molecular Theory for Ideal Gases
1. Gas particles are much smaller than the spaces between them. The gas particles themselves have virtually no (negligible) volume.

4 The Kinetic-Molecular Theory for Ideal Gases
2. Gas particles are in constant, random motion. Particles move in a straight line until they collide with other particles or with the walls of their container.

5 The Kinetic-Molecular Theory for Ideal Gases
3. Gas particles do not attract or repel each other. Therefore ideal gases would never condense to form liquids.

6 The Kinetic-Molecular Theory for Ideal Gases
4. No kinetic energy is lost when gas particles collide with each other or with the walls of their container. Collisions are perfectly elastic.

7 The Kinetic-Molecular Theory for Ideal Gases
5. All gases have the same average kinetic energy at a given temperature.

8 Gas Solubility Gases are less soluble in warm liquids than in cooler liquids.

9 Gas Solubility Gases are more soluble when under pressure.

10 Vapor Pressure The vapor pressure of water increases as the temperature increases.

11 STP STP stands for standard temperature and pressure.

12 Standard Pressure Standard pressure is 1 atm (atmosphere) which is equal to mm Hg, torr, or kPa (kilopascals).

13 Standard Temperature Standard temperature is 273 Kelvin.

14 Standard Atmospheric Pressure
1) Perform the following pressure conversions. a) 144 kPa = _____ atm (1.42) b) 795 mm Hg = _____ atm (1.05)

15 Standard Atmospheric Pressure
Perform the following pressure conversions. c) 669 torr = ______ kPa (89.2) d) 1.05 atm = ______ mm Hg (798)

16 Standard Atmospheric Pressure
Air pressure at higher altitudes, such as on a mountaintop, is slightly lower than air pressure at sea level.

17 Standard Atmospheric Pressure
Air pressure is measured using a barometer.

18 The Gas Laws

19 Boyle’s Law Boyle’s law states that the pressure and volume of a gas at constant temperature are inversely proportional. Inversely proportional means as one goes up the other goes down.

20 1 atm 4 Liters

21 As the pressure on a gas increases, the volume decreases.
2 atm 2 Liters

22 Boyle’s Law The P-V graph for Boyle’s law results in a hyperbola because pressure and volume are inversely proportional.

23 Boyle’s Law

24 Boyle’s Law P x V = K (K is some constant) P1 V1 = P2 V2

25 Example A balloon is filled with 25 L of air at 1.0 atm pressure. If the pressure is changed to 1.5 atm, what is the new volume?

26 Example First, make sure the pressure or volume units in the question match. A balloon is filled with 25 L of air at atm pressure. If the pressure is changed to 1.5 atm, what is the new volume? THEY DO!

27 Example A balloon is filled with 25 L of air at atm pressure. If the pressure is changed to 1.5 atm, what is the new volume? P1 V1 = P2 V2 V2 1.0 atm (25 L) 1.5 atm V2 = 17 L

28 Problem 2 A balloon is filled with 73 L of air at 1.3 atm pressure. What pressure is needed to change the volume to L? P2 = 2.2 atm

29 Problem 3 A gas is collected in a 242 cm3 container. The pressure of the gas in the container is measured and determined to be 87.6 kPa. What is the volume of this gas at standard pressure? V2 = 209 cm3

30 Problem 4 A gas is collected in a 24.2 L container. The pressure of the gas in the container is determined to be 756 mm Hg. What is the pressure of this gas if the volume increases to 30.0 L? P2 = 610. mm Hg

31 Charles’ Law Charles’ Law states that the volume of a gas is directly proportional to the Kelvin temperature if the pressure is held constant. Directly proportional means that as one goes up, the other goes up as well.

32 Charles’ Law The V-T graph for Charles’ law results in a straight line because volume and temperature are directly proportional.

33 Charles’ Law

34 Charles’ Law V / T = K (K is some constant) V1 V2 = T1 T2

35 Charles’ Law In any gas law problem involving temperature, temperature must be in Kelvin. K = °C

36 Example What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure?

37 Example First, make sure the volume units in the question match.
What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure? THEY DO!

38 Example Second, make sure to convert degrees Celsius to Kelvin. What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure? °C 25 K = K = 298 K

39 Example What is the temperature of a gas that is expanded from 2.5 L at 25 ºC to 4.1 L at constant pressure? V1 V2 2.5 L 4.1 L = T1 T2 298 K T2 = 489 K

40 Problem 5 What is the final volume of a gas that starts at 8.3 L and 17 ºC and is heated to 96 ºC? V2 = 11 L

41 Problem 6 A 225 cm3 volume of gas is collected at 57 ºC. What volume would this sample of gas occupy at standard temperature? V2 = 186 cm3

42 Problem 7 A 225 cm3 volume of gas is collected at 42 ºC. If the volume is decreased to 115 cm3, what is the new temperature? T2 = 161 K

43

44 Gay-Lusaac’s Law Gay-Lusaac’ Law states that the pressure of a gas is directly proportional to the Kelvin temperature if the volume is held constant.

45 Gay-Lusaac’s Law At higher temperatures, the particles in a gas have greater kinetic energy. They move faster and collide with the walls of the container more often and with greater force, so the pressure rises.

46 Gay-Lusaac’s Law The P-T graph for Gay-Lusaac’s law results in a straight line because pressure and temperature are directly proportional.

47 Gay-Lussac’s Law P / T = K (K is some constant) P1 P2 = T1 T2

48 Example What is the pressure inside a L can of deodorant that starts at 25 ºC and 1.2 atm if the temperature is raised to 100 ºC? Volume remains constant.

49 There is only one pressure unit!
Example First, make sure the pressure units in the question match. What is the pressure inside a L can of deodorant that starts at 25 ºC and 1.2 atm if the temperature is raised to 100 ºC? There is only one pressure unit!

50 Example Second, make sure to convert degrees Celsius to Kelvin. What is the pressure inside a L can of deodorant that starts at 25 ºC and 1.2 atm if the temperature is raised to 100 ºC? °C 25 K = K = 298 K

51 Example What is the pressure inside a L can of deodorant that starts at 25 ºC and 1.2 atm if the temperature is raised to 100 ºC? °C 100 K = K = 373 K

52 Example What is the pressure inside a L can of deodorant that starts at 25 ºC and 1.2 atm if the temperature is raised to 100 ºC? P1 P2 1.2 atm = T1 T2 298 K 373 K P2 = 1.5 atm

53 Problem 8 A can of deodorant starts at 43 ºC and 1.2 atm. If the volume remains constant, at what temperature will the can have a pressure of 2.2 atm? T2 = 579 K

54 Problem 9 A can of shaving cream starts at ºC and 1.30 atm. If the temperature increases to 37 ºC and the volume remains constant, what is the pressure of the can? P2 = atm

55 Problem 10 A 12 ounce can of a soft drink starts at STP. If the volume stays constant, at what temperature will the can have a pressure of 2.20 atm? T2 = 601 K

56 The Combined Gas Law The gas laws may be combined into a single law, called the combined gas law, which relates two sets of conditions of pressure, volume, and temperature by the following equation.

57 The Combined Gas Law P1 V1 P2 V2 = T1 T2

58 Example A 15 L cylinder of gas at 4.8 atm pressure at 25 ºC is heated to 75 ºC and compressed to 17 atm. What is the new volume?

59 There is only one volume unit!
Example First, make sure the volume units in the question match. A 15 L cylinder of gas at 4.8 atm pressure at 25 ºC is heated to 75 ºC and compressed to 17 atm. What is the new volume? There is only one volume unit!

60 Example Second, make sure the pressure units in the question match.
A 15 L cylinder of gas at 4.8 atm pressure at 25 ºC is heated to 75 ºC and compressed to 17 atm. What is the new volume? They do!

61 Example Third, make sure to convert degrees Celsius to Kelvin. A 15 L cylinder of gas at 4.8 atm pressure at 25 ºC is heated to 75 ºC and compressed to 17 atm. What is the new volume? °C 25 K = K = 298 K

62 Example A 15 L cylinder of gas at 4.8 atm pressure at 25 ºC is heated to 75 ºC and compressed to 17 atm. What is the new volume? °C 75 K = K = 348 K

63 Example P1 V1 P2 V2 = T1 T2 V2 = 4.9 L 4.8 atm (15 L) 17 atm 298 K
A 15 L cylinder of gas at 4.8 atm pressure at 25 ºC is heated to 75 ºC and compressed to 17 atm. What is the new volume? P1 V1 P2 V2 4.8 atm (15 L) 17 atm = T1 T2 298 K 348 K V2 = 4.9 L

64 Problem 11 If 6.2 L of gas at 723 mm Hg at 21 ºC is compressed to 2.2 L at mm Hg, what is the temperature of the gas? T2 = 594 K

65 Problem 12 A sample of nitrogen monoxide has a volume of 72.6 mL at a temperature of 16 °C and a pressure of kPa. What volume will the sample occupy at 24 °C and 99.3 kPa? V2 = 78.2 mL

66 Problem 13 A balloon is filled to 1.00 L at sea level and a temperature of 27 °C. At an altitude of 7000 m, atmospheric pressure drops to 300. mm Hg and the temperature cools to - 33 °C. What would its volume be when it reached the height of 7000 m? V2 = 2.03 L

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68

69 Avogadro’s Law Avogadro’s law states equal volumes of all gases, at the same temperature and pressure, have the same number of molecules. For a given mass of an ideal gas, the volume and amount (moles) of the gas are directly proportional if the temperature and pressure are constant.

70 Avogadro’s Law The molar volume for a gas is the volume that one mole occupies at 0.00ºC and 1.00 atm. 1 mole = 22.4 L at STP (standard temperature and pressure).

71 Example How many moles are in 63.2 L of a gas at STP? 1 mol 63.2 L = 2.82 moles 22.4 L

72 Problem 14 How many moles are in 45.0 L of a gas at STP? 2.01 moles

73 Problem 15 How many liters are in moles of a gas at STP? 14.2 L

74 Avogadro’s Law V / n = K (K is some constant) V1 V2 = n1 n2

75 Example Consider two samples of nitrogen gas. Sample 1 contains 1.5 mol and has a volume of 36.7 L. Sample 2 has a volume of 16.5 L at the same temperature and pressure. Calculate the number of moles of nitrogen in sample 2.

76 Example Sample 1 contains 1.5 mol and has a volume of 36.7 L. Sample 2 has a volume of 16.5 L. Calculate the number of moles of nitrogen in sample 2. V1 V2 36.7 L 16.5 L = n1 n2 1.5 mol n2 = 0.67 mol

77 Problem 16 If mol of argon gas occupies a volume of 652 mL at a particular temperature and pressure, what volume would mol of argon occupy under the same conditions? V2 = 1140 mL

78 Problem 17 If 46.2 g of oxygen gas (O2) occupies a volume of 100. L at a particular temperature and pressure, if the final volume is 10.8 L, how many moles of oxygen gas occupies this volume under the same conditions? n2 = mol

79 Problem 18 A 6.0 L sample at 25 °C and 2.00 atm of pressure contains 0.50 moles of a gas. If an additional 0.25 moles of gas at the same pressure and temperature are added, what is the final total volume of the gas? V2 = 9.0 L

80 Daltons’ Law of Partial Pressures
Dalton’s law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture, as shown below. PTotal = P1 + P2 + P3 + …

81 Example Determine the pressure in the fourth container if all of the gas molecules from the 1st three containers are placed in the 4th container. ?? 6 atm 2 atm 1 atm 3 atm

82 Problem 19 What is the total pressure in a balloon filled with air if the pressure of the oxygen is 170 mm Hg and the pressure of nitrogen is 620 mm Hg? 790 mm Hg

83 Problem 20 In a second balloon the total pressure is 1.30 atm. What is the pressure of oxygen (in mm Hg) if the pressure of nitrogen is 720. mm Hg?

84 Problem 20 1.30 atm 760 mm Hg 988 mm Hg = 1 atm
In a second balloon the total pressure is 1.30 atm. What is the pressure of oxygen (in mm Hg) if the pressure of nitrogen is 720. mm Hg? The two gas units do not match. We must convert the 1.30 atm into mm Hg. 1.30 atm 760 mm Hg 988 mm Hg = 1 atm

85 Problem 20 Pt = P1 + P2 + P3 + … 988 mm Hg = 720 mm Hg + Poxygen
Poxygen = 268 mm Hg

86 Problem 21 A container has a total pressure of 846 torr and contains carbon dioxide gas and nitrogen gas. What is the pressure of carbon dioxide (in kPa) if the pressure of nitrogen is 50. kPa? 63 kPa

87 Daltons’ Law of Partial Pressures
It is common to synthesize gases and collect them by displacing a volume of water.

88 Problem 22 Hydrogen was collected over water at 21°C on a day when the atmospheric pressure is 748 torr. The volume of the gas sample collected was 300. mL. The vapor pressure of water at 21°C is torr. Determine the partial pressure of the dry gas. torr

89 Problem 23 A sample of oxygen gas is saturated with water vapor at 27ºC. The total pressure of the mixture is 772 mm Hg and the vapor pressure of water is 26.7 mm Hg at 27ºC. What is the partial pressure of the oxygen gas? 745.3 mm Hg

90 Ideal Gases Ideal gases do not really exist, assuming that all gases are ideal makes the math easier and is a close approximation. Real gases behave more ideally at high temperature and low pressure.

91 Ideal Gases At high temperature, the gas molecules move more quickly, so attractive forces are negligible. At low pressure, the molecules are farther apart so attractive forces are negligible.

92 P V = n R T The Ideal Gas Law
Pressure times volume equals the number of moles (n) times the ideal gas constant (R) times the temperature in Kelvin.

93 The Ideal Gas Law Volume must be in liters.
If given milliliters divide the number by 1000 to convert it to liters.

94 The Ideal Gas Law R = 0.0821 (L atm)/(mol K) R = 8.314 (L kPa)/(mol K)
R = 62.4 (L mm Hg)/(mol K) The one you choose depends on the unit for pressure!

95 Example How many moles of air are there in a 2.0 L bottle at 19 ºC and 747 mm Hg? Choose the value of R based on the pressure unit. Since mm Hg are use, R = 62.4.

96 Example Second, make sure to convert degrees Celsius to Kelvin. How many moles of air are there in a 2.0 L bottle at 19 ºC and 747 mm Hg? °C 19 K = K = 292 K

97 Example How many moles of air are there in a L bottle at 19 ºC and 747 mm Hg? 292 K 747 P (2.0) V = n 62.4 R (292) T 1494 = n n = mol

98 Example What is the pressure in atm exerted by 1.8 g of H2 gas in a 4.3 L balloon at 27 ºC? Choose the value of R based on the pressure unit. Since atm is requested, R =

99 Example Second, make sure to convert degrees Celsius to Kelvin. What is the pressure in atm exerted by 1.8 g of H2 gas in a 4.3 L balloon at 27 ºC? °C 27 K = K = 300. K

100 Example Next, convert grams to moles.
What is the pressure in atm exerted by 1.8 g of H2 gas in a 4.3 L balloon at 300. K? 1.8 g H2 1 mol H2 = 0.90 mol H2 2.0 g H2

101 Example What is the pressure in atm exerted by 1.8 g of H2 gas in a 4.3 L balloon at 27 ºC? 300. K P V = n R T (4.3) 0.90 (0.0821) (300.) P = 5.2 atm

102 Problem 24 Sulfur hexafluoride (SF6) is a colorless, odorless and very unreactive gas. Calculate the pressure (in atm) exerted by moles of the gas in a steel vessel of volume 5.43 L at 69.5 ºC. P = 9.42 atm

103 Problem 25 Calculate the volume (in liters) occupied by 7.40 g of CO2 at STP. V = 3.77 L

104 Problem 26 A sample of nitrogen gas kept in a container of volume 2.30 L and at a temperature of 32 ºC exerts a pressure of 476 kPa. Calculate the number of moles of gas present. n = mol

105 Problem 27 A 1.30 L sample of a gas has a mass of 1.82 g at STP. What is the molar mass of the gas? 31.4 g/mol

106 Problem 28 Calculate the mass of nitrogen gas that can occupy 1.00 L at STP. 28.0 g

107 Proportionality P V = n R T
Variables on the same side of the equals sign are inversely proportional. This means as one goes up the other must go down.

108 Proportionality P V = n R T
Variables on opposite sides of the equals sign are directly proportional. This means as one goes up the other must go up.

109 Proportionality directly proportional volume decreases
29) How are pressure and temperature related? 30) If pressure increases, what happens to volume if temperature and number of moles stay constant ? directly proportional volume decreases

110 Proportionality volume decreases inversely proportional
31) If number of moles decreases, what happens to volume if temperature and pressure stay constant? 32) How are moles and temperature related? volume decreases inversely proportional


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