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- 1. CHEMISTRY - Chapter 12 Notes THE GAS LAWS I. Review of the Kinetic Theory - Gases: A. Definition of a molecule: B. Description of an ideal gas: C. Five main points of the kinetic theory-gases: 1. Gases are composed of very tiny particles with large distances between them. 2. Gas molecules are in constant motion with vibrational, rotational, and translational movements. The linear motion is interrupted by collisions with other molecules and/or the walls of the container. 3. The collisions between gas molecules and other molecules and the container are perfectly elastic. 4. Gas molecules exert no IMFs of attraction on each other. (This is also true of real gases when the molecules are far apart.) 5. The average kinetic energy [EK(ave)] of gas molecules increases directly with the kelvin temperature of the gas. II. Relationship Between Pressure and Volume at Constant Temperature A. Boyle’s Law: The volume of a definite quantity of dry gas is inversely proportional to the pressure on/by the gas, provided the temperature remains constant. 1. As P on/by a gas ↑, the V occupied by the gas . V 2. As P on/by a gas ↓, the V occupied by the gas . B. The volume is affected inversely by the pressure; PV = k C. If the pressure changes to P′, then the volume changes to V′ such that P′V′ = k P D. At constant temperature, the value of ‘k’ is constant. ∴ PV = P′V′ 1. P and P′ can be expressed in any pressure unit. 2. V and V′ can be expressed in any volume unit. E. Sample Boyle’s Law problems: 1. A sample of oxygen gas occupies a volume of 250.0 mL at 25°C and 750.0 torr. Calculate the volume of the gas at 25°C and 800.0 torr. 2. The pressure on the carbon dioxide gas in a 5.00 L tank is 3.50 atm. What is the pressure due to the carbon dioxide if it released into a 100.0 L tank?
- 2. CHEMISTRY - Chapter 12 Notes 2 III. Relationship Between Temperature and Volume at Constant Pressure A. As the temperature of a gas increases, the gas molecules move faster and exert a greater pressure. 1. 2. B. As the T↑, the V must to maintain constant pressure C. Conversely, as the T↓, the volume must to keep the pressure constant D. This shows a mathematical relationship between temperature and volume: V = kT or V/T = k V E. If the temperature changes to T′, the volume will change to V′ such that V′/T′ = k, provided the pressure is kept constant. F. At the same pressure, the value of ‘k’ is constant. ∴ V/T =V′/T′ T G. Experimental evidence indicates that the volume of a gas maintained at constant pressure changes by for each Celsius degree temperature change. + enough heat to raise the temperature of 1 V >1 V the gas by 1 C° @ @ 0°C and 1°C and 1.0 atm 1.0 atm H. Extrapolation of the T vs. V line to the point where the volume of the ideal gas is zero. V T I. The direct relationship between the temperature of a gas and the volume it occupies is valid provided the temperatures are expressed in kelvin (K). J. Charles’ Law: The volume of a definite quantity of dry gas is directly proportional to the kelvin temperature provided the pressure on the gas remains constant. V V′ = 1. V and V′ can be expressed in any volume unit. TK TK ′ 2. TK and TK must be expressed in kelvins (K). ′
- 3. CHEMISTRY - Chapter 12 Notes 3 K. Sample Charles’ Law problems: 1. A gas occupies a volume of 4500 cm3 at 40.0°C and standard pressure. Calculate the volume of the gas if the temperature drops to 10.0°C. 2. A balloon has a volume of 5.00 x 103 L at 17.0°C. Determine the temperature change needed to increase the volume to 5.20 x 103 L. IV. Combined Boyle’s and Charles’ Laws V V′ PV P ′V ′ A. Combining PV = P′V′ and = produces = known as the Combined Gas Law TK TK TK TK ′ ′ B. The Combined Gas Law also shows the relationship between the pressure exerted by a gas and its kelvin temperature provided the volume remains constant. This is called Gay-Lussac’s Law. C. Standardizing temperature and pressure (STP) 1. Standard temperature is defined as 0.00°C (273 K) 2. Standard pressure is defined as in Chapter 11 [1.0 atm = 760 torr = 101.3 kPa = 14.7 PSI] D. Sample Combined Gas Law problems: 1. A sample of nitrogen gas has a volume of 275 mL when measured at 10.0°C and 95.0 kPa pressure. Calculate the volume of the gas at STP. 2. The pressure required for a reaction is 3.00 atm.. The gases involved in the reaction currently occupy a volume of 2.00 L at 27.0°C and 1.05 atm pressure. If the gases are transferred to an 800.0 mL flask, what temperature must be reached before the reaction can begin? 3. One mole of an ideal gas should have a volume of 22.4 L at STP. What volume will the gas occupy when the conditions are 7.0°C and 745 torr? At -40.0°C and 1000.0 torr?
- 4. CHEMISTRY - Chapter 12 Notes 4 V. Gas Density and Changes in Pressure and/or Temperature. m A. Reminder: The formula for density is D = or m = DV V B. If the P on a gas ↑, the V of the gas ; therefore, the density of the gas must . If the P on a gas ↓, the V of the gas ; therefore, the density of the gas must . C. If the T of a gas ↑, the V of the gas ; therefore, the density of the gas must . If the T of a gas ↓, the V of the gas ; therefore, the density of the gas must . D. Combining the relationship between P and D with the relationship between T and D gives the following combined P, T, D formula: E. Sample gas density problems: 1. The density of a gas is 0.985 g/L at 25°C and 725 torr. What is its density at STP? 2. If the density of oxygen gas is 1.43 g/L at STP, determine the mass of 5.0 L of the gas at 27°C and 110.0 kPa pressure. 3. Calculate the density of sulfur trioxide gas at 50°C and 1200 torr. VI. Partial Pressure Problems. A. The pressure exerted by one gas molecule in a mixture of gases is the same as the pressure exerted by any other provided the conditions are the same for all molecules. 1. Larger, heavier molecules move more slowly, hitting less often and without the force of higher speed collisions. E K (l arge) = 1 m Heavyv 2 2 Slow 2. Smaller, lighter molecules move more rapidly, hitting more often and with the force of higher speed collisions. E K (small) = 1 m Light v2 2 Fast 3. The higher speed of the small molecules compensates for their smaller mass so the force of collisions per unit area is about the same as for large molecules. E K (l arge) ≈ E K (small) B. The partial pressure of a gas in a mixture is the pressure that the gas would exert in the volume if it was by itself. C. The partial pressure that each gas exerts in a mixture is in direct proportion to the fractional part of that gas in the mixture. PA = %A ⋅ PT D. Dalton’s Law of Partial Pressures: In a mixture of gases, the total pressure of the mixture is equal to the sum of the pressures that each gas would exert by itself in the same volume. PT = PA + PB + PC +K
- 5. CHEMISTRY - Chapter 12 Notes 5 E. Sample Dalton’s Law problems: 1. Air is about 21% oxygen by volume. Determine the partial pressure of oxygen in air at STP. If the partial pressure of nitrogen gas in the air at STP is 590 torr, what percentage of the air is nitrogen? 2. A mixture of three gases has the following partial pressures: PA = 25.0 kPa, PB = 35.0 kPa, and PC = 50.0 kPa. What is the composition of the mixture in terms of percentage of each gas? 3. Carbon dioxide gas was collected over water at 17°C and 100.0 kPa pressure. Look on page 275 in your textbook to find the PH 2 0 at 17°C. What is the PCO 2 ? If 500.0 mL of the CO2 /H2 0 vapor mixture was originally collected, determine the volume of the dry CO2 corrected to STP. VII. Rate of Diffusion Problems. A. Definition of diffusion: B. Since the average EK of a substance is determined by the temperature (EK α T), two different gases at the same temperature have the same average EK. C. Consider the diffusion of large, slow molecules vs. the diffusion of small, fast molecules, refer back to section VI., A., 1-3., where 1 m Lightv2 = 1 m Heavyv2 2 2 Fast Slow D. Derivation of Graham’s Law of Diffusion: E. Graham’s Law does not determine the actual average speed of the molecules but only the relative rate of the speed of one gas to another at the same temperature can be found using this law. F. Sample Graham’s Law problems: 1. Given CO2 and NO gases, which one will diffuse faster? How much faster? 2. Gas A (density = 1.25 g/L) and gas B are released from the same point at the same time. Gas B reaches the detection point in 45 seconds while it takes gas A 60 seconds to travel the same distance. Calculate the density of gas B.
- 6. CHEMISTRY - Chapter 12 Notes 6 VIII. Avogadro’s Hypothesis A. From previous theory, all gas molecules, regardless of molecular mass, collide with the same force and exert the same pressure at the same temperature. B. It follows that if two different gases are contained in identical volumes at the same temperature, the only variable that could produce different pressures is the . C. ∴ If the volumes, temperatures, and pressures of two different gases are the same, the number of molecules in each volume must be the same. Avogadro’s Hypothesis D. Stated differently, equal numbers of different gas molecules under the same conditions of temperature and pressure must occupy equal volumes. E. Avogadro’s Hypothesis led to the concept of the mole (n) where a mole represents a fixed number of particles. (NA = 6.02 x 102 3) NA is called . F. The volume occupied by one mole of an ideal gas (called the molar volume) at STP is . G. The mass of a molar volume of a substance is called the . H. Reminder: 1 mole = 6.02 x 102 3 particles = MM grams = 22.4 L at STP IX. Ideal Gas Law 1 A. From Boyle’s Law: V α P B. From Charles’ Law: V α TK C. From Avogadro’s Law: V α n D. Combine all three laws to give: E. Determine the value(s) of the gas constant: Consider one mole of an ideal gas at STP. F. Sample Ideal Gas Law problems: 1. Calculate the volume occupied by 3.5 moles of carbon dioxide gas at 25°C and 720 torr. 2. A 12.5 gram sample of dinitrogen tetraoxide gas is confined to a 250.0 mL bottle at 40.0°C. What is the partial pressure of N2 O4 in kPa? 3. An acetylene gas (C2 H2 ) filled tank has a volume of 10.0 L with a pressure of 8.50 atm at 30.0°C. How many grams of C2 H2 (g) are in the tank?
- 7. CHEMISTRY - Chapter 12 Notes 7 X. Stoichiometry Problems Involving Gases Under Non-STP Conditions. A. Gases under “dry” conditions (ie. not collected over water) 1. Calculate the volume of CO2 gas produced by the complete decomposition of 20.0 g of calcium carbonate at 120.0°C and 0.85 atm pressure. 2. How many grams of sulfur will be burned to sulfur dioxide if the only oxygen available to the reaction is contained in 2500 L of air at 27.0°C and 800.0 torr pressure? Air is 21% oxygen by volume. B. Gases under “wet” conditions (ie. collected over water or by water displacement) (PH 2 O table on page 275 ) 1. When 5.00 grams of zinc metal are reacted with excess hydrochloric acid, the products of the reaction are zinc chloride and hydrogen gas. Determine the volume of hydrogen collected by water displacement at 22.0°C and 105.0 kPa pressure. 2. A 500.0 mL sample of nitrogen gas was collected over water at 50.0°C and 1.02 atm pressure. How many grams of N2 are available for a reaction with hydrogen gas to make ammonia? How many grams of NH3 will be produced?
- 8. CHEMISTRY - Chapter 12 Notes 8 XI. Deviations From Ideal Gas Behavior. A. Real gases differ from an ideal gas in two ways: 1. 2. B. When the pressure on a gas increases to the point where the molecules are 3-4 molecular diameters apart, the IMFs of attraction between real gas molecules must be considered. 1. Real gas molecules exert IMFs of attraction when the molecules are close enough to react to the attractive forces. Strong IMF gases do not have to be pushed as close for the IMFs to act. 2. The IMFs cause a greater-than-expected reduction in volume resulting in the product of P and V being smaller than predicted by Boyle’s Law. C. When the pressure on a gas becomes extreme, the molecules are pushed close enough together such that the volume occupied by the molecules themselves can no longer be ignored. 1. Real gas molecules have a finite size and when high pressure is applied, they resist moving too close together. 2. An increase in pressure does not produce the reduction in volume that Boyle’s Law predicts and the product of P and V is more than expected. D. Graphical illustration: PV Product PV product for an ideal gas Pressure

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