Simple Gas Laws To facilitate comparison of gases, the following standards are used: STP: O C (273 K) and 101.3 kpa If assuming 1 mol, V = 22.4L SATP: 25 C (298 K) and 101.3 kpa If assuming 1 mol, V = 24.5L Alexander Karen 2
Relationship between Pressure & Volume
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
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
Boyle s Law Apparatus Alexander Karen 6
Boyle s Law Alexander Karen 7
Boyle s Law Sample experiment results: Pressure P Volume V P x V 1.1 40 44 1.7 26 44 2.2 20 44 2.6 17 44 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
Boyle s Law Alexander Karen 9
Boyle s Law Alexander Karen 10
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
Boyle s Law Alexander Karen 12
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 110.6 kpa. What is the final volume of the balloon? P1 = 101.3 kpa V1 = 2.5 L P2 = 110.6 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
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
Boyle s Law Example 3 Which will go higher, a 3 L balloon half filled or a full 3 L balloon? Alexander Karen 15
Boyle s Law Example 4 a) b) c) Alexander Karen 16
Relationship between Temperature & Volume
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
Charles Law At a constant pressure, the volume of a gas is directly proportional to the absolute temperature of the gas. Alexander Karen 19
Charles Law Alexander Karen 20
Charles Law Alexander Karen 21
Charles Law Alexander Karen 22
Charles Law Alexander Karen 23
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
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
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
Relationship of Boyle s & Charles Laws Alexander Karen 27
Relationship between Volume & Moles
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
Avogadro s Law Alexander Karen 30
Avogadro s Law Alexander Karen 31
Avogadro s Law Alexander Karen 32
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
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 = 0.50 + 0.20 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
Avogadro s Law Alexander Karen 35
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
Relationship between Pressure & Temperature
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
Gay-Lussac s Law Alexander Karen 39
Gay-Lussac s Law Alexander Karen 40
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
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) = 552.9 kpa T1 (287 K) The final pressure of the helium balloon will be 553 kpa. Alexander Karen 42
Simple Gas Laws REVIEW Alexander Karen 43
Relationship between P, V, T, n.
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
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 = 101.3 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
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 100 0 C. Calculate the new volume. Alexander Karen 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 100 0 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
General Gas Law - Experiment Alexander Karen 49
Relationship between Partial Pressures
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
Dalton s Law Alexander Karen 52
Dalton s Law Alexander Karen 53
Dalton s Law Alexander Karen 54
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
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
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 = 200.0 kpa nt = 3.35+0.64+2.29 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
Dalton s Law Pressure of a DRY gas: Often a gas is collected by water displacement. Alexander Karen 58
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
Ideal Gas Law Calculate the constant for 1 mole of gas at STP. P = 101.3 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
Ideal Gas Law R is the ideal gas constant Alexander Karen 62
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 = 0.133322368 kilopascals P = 760 mm Hg = 101.3 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
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 20 000 kpa at 27 0 C. What is the mass of N 2 gas in the cylinder? P = 20 000 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 = (20 000 kpa)(20.0l) = 1333.3 mol N2 (1 mol)(300 K) m = (28 g/mol)(1333.3 mol) = 37 333 g The mass of N 2 gas in the cylinder is 37.3 x 10³ g. Alexander Karen 64
Graham s Law of Diffusion Alexander Karen 66
Graham s Law of Diffusion Graham measured the rate of effusion of gases Alexander Karen 67
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
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
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
Graham s Law of Diffusion So, which balloon will lose its gas first? Alexander Karen 71
Gas Stoichiometry REMEMBER: Alexander Karen 73
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 2 2 3 n KClO3 = (50 g)/ (122.5 g/mol) = 0.41 mol n O2: n O2 = 0.62 mol 2 mol KClO3 = 3 mol O2 0.41 mol KClO3 x KClO3 m = 50 g MM = 122.5 g/mol PV = nrt V = [(0.62 mol)(8.31)(298 K)] / (84 kpa) V = 18.2 L Alexander Karen 74
Molecular Motion There are three types of molecular motion: STATE MOTION Solid Vibration Liquid Vibration & Rotation Gas Vibration, Rotation, Translation Alexander Karen 76
Molecular Motion Bonds Vibrate: (solids, liquids and gases) Alexander Karen 77
Molecular Motion Bonds Rotate: (liquids and gas) Alexander Karen 78
Molecular Motion Bonds Translate: (gases) displacement of particles in a straight line Alexander Karen 79