Simple Gas Laws. To facilitate comparison of gases, the following standards are used: STP: O C (273 K) and kpa. SATP: 25 C (298 K) and 101.
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2 Simple Gas Laws To facilitate comparison of gases, the following standards are used: STP: O C (273 K) and kpa If assuming 1 mol, V = 22.4L SATP: 25 C (298 K) and kpa If assuming 1 mol, V = 24.5L Alexander Karen 2
3 Relationship between Pressure & Volume
4 Boyle s Law At a constant temperature, the volume occupied by a given quantity of gas is inversely proportional to the pressure of the gas. Alexander Karen 4
5 How can we investigate Boyle s Law? When investigating Boyles law, a given volume of gas is sucked into a cylinder and the end is sealed. (The temperature of the gas is kept constant.) Using several equal weights we can apply increasing pressure to the gas. We can calculate the pressure by dividing the force applied by the area of the top of the cylinder. The volume will be shown on the scale on the cylinder. Alexander Karen 5
6 Boyle s Law Apparatus Alexander Karen 6
7 Boyle s Law Alexander Karen 7
8 Boyle s Law Sample experiment results: Pressure P Volume V P x V Did you notice that if P is doubled, V is halved? If p increases to 3 times as much, V decreases to a 1/3 rd. Alexander Karen 8
9 Boyle s Law Alexander Karen 9
10 Boyle s Law Alexander Karen 10
11 Boyle s Law P1 = initial pressure [kpa] or [mmhg] V1 = initial volume [L] or [ml] P2 = final pressure [kpa] or [mmhg] V2 = final volume [L] or [ml] *On the condition that the number of moles (n) of gas and Alexander temperature Karen (T) are constant. 11
12 Boyle s Law Alexander Karen 12
13 Boyle s Law Example 1 A sample of helium gas is collected at room temperature in a 2.5 L balloon at normal atmospheric pressure. The balloon is then immersed in water, also at room temperature, so that the external pressure on it increases to kpa. What is the final volume of the balloon? P1 = kpa V1 = 2.5 L P2 = kpa V2 =? P1V1 = P2V2 V2 = (P1V1) = (101.3 kpa)(2.5 L) = 2.3 L P2 (110.6 kpa) The final volume of the balloon is 2.3 L. Alexander Karen 13
14 Boyle s Law Example 2 What happens to the volume if the pressure doubles? If the pressure exerted on a quantity of gas is doubled, the volume will decrease by half and vice versa. Alexander Karen 14
15 Boyle s Law Example 3 Which will go higher, a 3 L balloon half filled or a full 3 L balloon? Alexander Karen 15
16 Boyle s Law Example 4 a) b) c) Alexander Karen 16
17 Relationship between Temperature & Volume
18 Charles law Charles Law demo Note: Balloons keep a small amount of gas (air) at an approximately constant pressure. In this experiment, as a balloon is dipped into a beaker of liquid nitrogen (-196 C; -320 F), the air inside them quickly cools. The volume of the air inside the balloon decreases as the temperature of the balloon decreases. When heated with warm breathe, the air inside them heats up. The volume of the air inside the balloon increases as the temperature of the balloon increases. Alexander Karen 18
19 Charles Law At a constant pressure, the volume of a gas is directly proportional to the absolute temperature of the gas. Alexander Karen 19
20 Charles Law Alexander Karen 20
21 Charles Law Alexander Karen 21
22 Charles Law Alexander Karen 22
23 Charles Law Alexander Karen 23
24 Charles Law V1 = initial volume [L] or [ml] T1 = initial temperature [K] V2 = final volume [L] or [ml] T2 = final temperature [K] *On the condition that the number of moles (n) of gas and pressure (P) are constant. Alexander Karen 24
25 Charles Law Example 1 Calculate the volume that a sample of air would occupy at 40 0 C if it occupies 1.00 L at 20 0 C at a constant pressure. V1 = 1.00 L T1 = 20 C = 293K V2 =? T2 = 40 C = 313K V2 = (V1T2) = (1.00L)(313K) = 1.07 L T1 (293K) The final volume of the sample of air is 1.07 L. Alexander Karen 25
26 Charles Law Example 2 What temperature should a sample of gas be cooled from 25 0 C to reduce its volume to half its initial value at a constant P? V1 = 2.00 L T1 = 25 C = 298K V2 = 1.00 L T2 =? T2 = (V2T1) = (1.00L)(298K) = 149 K V1 (2.00L) What about in C? The final temperature of the sample of gas would be 149 K or -124 C. Alexander Karen 26
27 Relationship of Boyle s & Charles Laws Alexander Karen 27
28 Relationship between Volume & Moles
29 Avogadro s Law Under the same temperature and pressure conditions, the volume of a gas is directly proportional to its quantity expressed in number of moles. Alexander Karen 29
30 Avogadro s Law Alexander Karen 30
31 Avogadro s Law Alexander Karen 31
32 Avogadro s Law Alexander Karen 32
33 Avogadro s Law V1 = initial volume [L] or [ml] n1 = initial quantity of gas [mol] V2 = final volume [L] or [ml] n2 = final quantity of gas [mol] *On the condition that temperature (T) and pressure (P) are constant. Alexander Karen 33
34 Avogadro s Law Example 1 A helium balloon occupies a volume of 15L and contains 0.50 mol of He at SATP. What will the new volume of the balloon be if 0.20 mol of He is added under the same conditions? V1 = 15 L n1 = 0.50 mol V2 =? n2 = mol = 0.70 mol V2 = (V1)(n2) = (15L)(0.70mol) = 21 L n1 (0.50 mol) The final volume of the helium balloon is 21 L. Alexander Karen 34
35 Avogadro s Law Alexander Karen 35
36 Avogadro s Law V1 = initial volume [L] or [ml] m1 = initial mass [g] V2 = final volume [L] or [ml] m2 = final mass [g] *On the condition that temperature (T) and pressure (P) are constant. Alexander Karen 36
37 Relationship between Pressure & Temperature
38 V is constant n is constant Gay-Lussac s Law At a constant volume, the pressure of a given quantity of gas is directly proportional to the absolute temperature of the gas. Alexander Karen 38
39 Gay-Lussac s Law Alexander Karen 39
40 Gay-Lussac s Law Alexander Karen 40
41 Gay-Lussac s Law P1 = initial pressure [kpa] or [mm Hg] T1 = initial temperature [K] P2 = final pressure [kpa] or [mm Hg] T2 = final temperature [K] *On the condition that the number of moles (n) of Alexander gas Karen and volume (V) are constant. 41
42 Gay-Lussac s Law Example 1 At 14.0 C, helium gas stored in a metal tank exerts a pressure of 507 kpa. What will the pressure be if the temperature increases to 40.0 C? P1 = 507 kpa T1 = 14.0 C = 287 K P2 =? T2 = 40.0 C = 313 K P2 = (P1)(T2) = (507 kpa)(313 K) = kpa T1 (287 K) The final pressure of the helium balloon will be 553 kpa. Alexander Karen 42
43 Simple Gas Laws REVIEW Alexander Karen 43
44 Relationship between P, V, T, n.
45 General Gas Law The General Gas Law can be used to predict the final conditions of a gas once its initial conditions have been modified. n1 n2 Alexander Karen 45
46 General Gas Law Example 1 A sample of O 2, occupies 5.0 L at 25 0 C and 500 kpa. Calculate the volume of the gas at STP. P1 = 500 kpa V1 = 5.0 L T1 = 25 C = 298 K P2 = kpa V2 =? T2 = 0.0 C = 273 K (500 kpa)(5.0 L) = (101.3 kpa)(v2) 298 K (273 K) The final volume of the gas will be 22.6 L. Alexander Karen 46
47 General Gas Law Example 2 A container with an initial volume of 1.0 L is occupied by a gas at a pressure of 150 kpa at 25 0 C. The pressure is increased to 600 kpa and the temperature is raised to C. Calculate the new volume. Alexander Karen 47
48 General Gas Law Example 2 A container with an initial volume of 1.0 L is occupied by a gas at a pressure of 150 kpa at 25 0 C. The pressure is increased to 600 kpa and the temperature is raised to C. Calculate the new volume. P1 = 150 kpa V1 = 1.0 L T1 = 25 C = 298 K P2 = 600 kpa V2 =? T2 = 100 C = 373 K (150 kpa)(1.0 L) = (600 kpa)(v2) 298 K (373 K) The final volume of the gas will be 0.31 L. Alexander Karen 48
49 General Gas Law - Experiment Alexander Karen 49
50 Relationship between Partial Pressures
51 Dalton s Law At a given temperature, the total pressure of a gas mixture equals the sum of the partial pressures of all the gases in the mixture. Alexander Karen 51
52 Dalton s Law Alexander Karen 52
53 Dalton s Law Alexander Karen 53
54 Dalton s Law Alexander Karen 54
55 Dalton s Law Example 1 A gaseous mixture contains three noble gases, He, Ar, and Kr. The total pressure exerted by the mixture is 151 kpa, the partial pressure of He and Ar are 33% and 40 %. Calculate the partial pressure of Kr. Alexander Karen 55
56 Dalton s Law To determine the partial pressure of one specific gas within a mixture, the molar proportion of that gas is multiplied by the total pressure of the mixture. PA = Partial pressure of gas A in [kpa] or [mm Hg] na = Quantity of gas A in [moles] nt = Total quantity of gas in [moles] PT = Total pressure of the gas mixture in [kpa] or [mm Hg] Alexander Karen 56
57 Dalton s Law Example 2 At a given temperature, a mixture of gas contains 3.35 mol of Ne, o.64 mol of Ar and 2.19 mol of Xe. What is the partial pressure of xenon if the total pressure of the mixture is 200.0kPa? nne = 3.35 mol nar = 0.64 mol nxe = 2.19 mol PXe =? PT = kpa nt = mol = 6.18 mol PXe = nxe * PT = 2.19mol * 200kPa = 70.9kPa nt 6.18mol The partial pressure of xenon is 70.9 kpa. Alexander Karen 57
58 Dalton s Law Pressure of a DRY gas: Often a gas is collected by water displacement. Alexander Karen 58
59
60 Ideal Gas Law An ideal gas is defined as one in which all collisions between atoms or molecules are perfectly elastic and in which there are no intermolecular attractive forces. Alexander Karen 60
61 Ideal Gas Law Calculate the constant for 1 mole of gas at STP. P = kpa V = 22.4 L n = 1 mol R =? T = O C = 273 K R = (101.3 kpa)(22.4l) = 8.31 kpa L (1 mol)(273 K) mol K The constant for 1 mole of gas at STP is 8.31 (kpa)(l)/(mol)(k). Alexander Karen 61
62 Ideal Gas Law R is the ideal gas constant Alexander Karen 62
63 Ideal Gas Law Example 1 Calculate the constant for 1 mol of gas at a pressure of 760 mm of Hg, 273 K and 22.4 L NOTE: 1 mmhg = kilopascals P = 760 mm Hg = kpa V = 22.4 L n = 1 mol R =? T = 273 K R = (101.3 kpa)(22.4l) = 8.31 kpa L (1 mol)(273 K) mol K The constant for 1 mole of gas at STP is 8.31 (kpa)(l)/(mol)(k). Alexander Karen 63
64 Ideal Gas Law Example 2 A rigid steel cylinder with a volume of 20.0 L is filled with nitrogen gas to a final pressure of kpa at 27 0 C. What is the mass of N 2 gas in the cylinder? P = kpa V = 20.0 L n =? R = 8.31 (kpa)(l)/(mol)(k) T = 27 C = 300 K MM = 28 g/mol N2 m =? n = ( kpa)(20.0l) = mol N2 (1 mol)(300 K) m = (28 g/mol)( mol) = g The mass of N 2 gas in the cylinder is 37.3 x 10³ g. Alexander Karen 64
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66 Graham s Law of Diffusion Alexander Karen 66
67 Graham s Law of Diffusion Graham measured the rate of effusion of gases Alexander Karen 67
68 Graham s Law of Diffusion The rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Alexander Karen 68
69 Graham s Law of Diffusion The rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Alexander Karen 69
70 Graham s Law of Diffusion The rate of diffusion of an unknown gas is four times faster than oxygen gas. Calculate the molar mass of the unknown gas. Rate Unknown = 4m/s Rate O2 = 1m/s MM Unknown =? MM O2 = 32 g/mol (4 m/s)² = (32 g/mol) (1 m/s) ² MM MM Unknown = 2.0 g/mol Therefore the unknown gas is H2 (hydrogen gas.) Alexander Karen 70
71 Graham s Law of Diffusion So, which balloon will lose its gas first? Alexander Karen 71
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73 Gas Stoichiometry REMEMBER: Alexander Karen 73
74 Gas Stoichiometry Calculate the volume of oxygen produced at 25 and 84 kpa when 50 g of KClO3 is heated according to the following equation: P = 84 kpa V =? n = 0.62 mol R = 8.31 (kpa)(l)/(mol)(k) T = 25 = 298 K n KClO3 = (50 g)/ (122.5 g/mol) = 0.41 mol n O2: n O2 = 0.62 mol 2 mol KClO3 = 3 mol O mol KClO3 x KClO3 m = 50 g MM = g/mol PV = nrt V = [(0.62 mol)(8.31)(298 K)] / (84 kpa) V = 18.2 L Alexander Karen 74
75
76 Molecular Motion There are three types of molecular motion: STATE MOTION Solid Vibration Liquid Vibration & Rotation Gas Vibration, Rotation, Translation Alexander Karen 76
77 Molecular Motion Bonds Vibrate: (solids, liquids and gases) Alexander Karen 77
78 Molecular Motion Bonds Rotate: (liquids and gas) Alexander Karen 78
79 Molecular Motion Bonds Translate: (gases) displacement of particles in a straight line Alexander Karen 79
Chapter 13 Gases and Pressure. Pressure and Force. Pressure is the force per unit area on a surface. Force Area. Pressure =
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