Chemistry Lecture Slides Chapter 5: Properties of Gases PDF

Lecture Slides Chapter 5 Properties of Gases: The Air We Breathe Copyright © 2020 W. W. Norton & Company Chapter Outline 5.1 Air: An Invisible Necessity a 5.2 Atmospheric Pressure and Collisions 5.3 The Gas Laws 5.4 The Ideal Gas Law 5.5 Gases in Chemical Reactions 5.6 Gas Density...

Lecture Slides Chapter 5 Properties of Gases: The Air We Breathe Copyright © 2020 W. W. Norton & Company Chapter Outline 5.1 Air: An Invisible Necessity a 5.2 Atmospheric Pressure and Collisions 5.3 The Gas Laws 5.4 The Ideal Gas Law 5.5 Gases in Chemical Reactions 5.6 Gas Density 5.7 Dalton’s Law and Mixtures of Gases 5.8 The Kinetic Molecular Theory of Gases b 5.9 Real Gases a for interest only b Partially covered in CHEM 146 2 Learning Objectives Student will be able to: define, apply, and carry out calculations using ideal gas laws; carry out calculations related to gas reaction stoichiometry, density and molar mass; Dalton's Law of Partial Pressures, gas mixtures and the collection gases over water Understand the behaviour of real gases 3 Properties of a Gas 1. Neither definite shape nor definite volume 2. Uniformly fills any container 3. Exerts pressure on surroundings 4. Changes volume with temperature and pressure 5. Miscible (mixes together in any proportion) 6. Much less dense than solids or liquids Pressure Pressure (P): The ratio of a force (F) to the surface area (A) over which the force is applied Atmospheric pressure – force exerted by gases surrounding Earth on Earth’s surface Units of Pressure: SI units = Newton/meter2 = 1 Pascal (Pa) 1 standard atmosphere (1 atm) = 101,325 Pa 1 atm = 760 mmHg = 760 torr Converting Units of Pressure Elevation and Atmospheric Pressure Measurement of Pressure Barometer: instrument that measures atmospheric pressure Height of Hg column based on balance of forces: Gravity (pulls Hg down) Atmospheric pressure (pushes Hg up into evacuated tube) Measuring Pressure: Manometer  Ideal Gas Law Ideal Gas Law: PV = nRT Assumes that gas molecules: 1. Have tiny volumes compared with the R = universal gas constant = 0.08206 L atm/(K mol) collective volume they occupy P = pressure (in atm) 2. Move constantly and V = volume (in liters) randomly n = moles 3. Have average kinetic energy that is proportional to T = temperature (in kelvin) absolute temperature 4. Engage in elastic collisions with walls of container and other gas molecules Allows the calculation of gas properties when more than 5. Act independently of other one variable is changing gas molecules Practice: Ideal Gas Law Collect and Organize Calculate the pressure of 1.2 mol of methane gas in a 3.3 L container at 25C. We know the number of moles, volume, and temperature. We are asked to solve for pressure. n = 1.2 mol V = 3.3 L T = 25C = 298 K P=? Practice: Ideal Gas Law Summary Collect and Organize: We know the number of moles, volume, and temperature. We are asked to solve for pressure. Analyze: We know three of the four variables in the ideal gas law and the value of the gas constant R, which enables us to rearrange to solve for P. Solve: Think About It: The units of our calculated result are consistent with pressure, and the value appears to be reasonable. Combined Gas Law Boyle’s Law: PV = constant Charles’s Law: V/T = constant Avogadro’s Law: V/n = constant Combining the gas laws: If n is constant, then Reference Points for Gases Standard Temperature and Pressure (STP): P = 1 atm; T = 273 K (0.0C) Molar Volume: Volume occupied by one mole of an ideal gas at STP V = 22.4 L (calculated from the ideal gas law) Stoichiometry Calculations: Gases Stoichiometry calculations: Depend on mole/mole ratios of reactants and/or products Moles of gas can be calculated from ideal gas law if P, V, and T are known. Practice: Stoichiometry Calculations Oxygen generators in some airplanes are based on the chemical reaction between solid sodium chlorate (ℳ = 106.44 g/mol) and iron: NaClO3(s) + Fe(s) → O2(g) + NaCl(s) + FeO(s) The resultant O2 is blended with cabin air to provide 10–15 minutes of breathable air for passengers. How many grams of NaClO3 are needed in a typical generator to produce 125 L of O2 gas at 1.00 atm and 20.0°C? Practice: Stoichiometry Calculations Collect and Organize Oxygen generators in some airplanes are based on the chemical reaction between solid sodium chlorate (ℳ = 106.44 g/mol) and iron: NaClO3(s) + Fe(s) → O2(g) + NaCl(s) + FeO(s). The resultant O2 is blended with cabin air to provide 10– 15 minutes of breathable air for passengers. How many grams of NaClO3 are needed in a typical generator to produce 125 L of O2 gas at 1.00 atm and 20.0°C? We are given the volume of O2 to be generated at a particular pressure and temperature. Using this information, we can determine the mass of NaClO3 needed based on the stoichiometric relations in the balanced chemical equation. Practice: Stoichiometry Calculations Analyze Oxygen generators in some airplanes are based on the chemical reaction between solid sodium chlorate (ℳ = 106.44 g/mol) and iron: NaClO3(s) + Fe(s) → O2(g) + NaCl(s) + FeO(s). The resultant O2 is blended with cabin air to provide 10–15 minutes of breathable air for passengers. How many grams of NaClO 3 are needed in a typical generator to produce 125 L of O2 gas at 1.00 atm and 20.0°C? The solution requires two calculations. (1) According to the balanced chemical equation, 1 mole of NaClO3 is needed to produce 1 mole of O2. If we can determine how many moles of O2 occupy a volume of 125 L at 1.00 atm pressure and 20.0°C, we can determine the number of moles of NaClO3 we need. To determine the moles of O2 , we can use the ideal gas law. (2) Then we use our calculated value of moles of O2 and the balanced chemical equation to determine the number of moles and number of grams of NaClO3(s) required. To estimate the answer, notice that the volume of gas we wish to make (125 L) is about five times the molar volume of an ideal gas at STP (22.4 L). The difference between the temperature at STP (0°C = 273 K) and the temperature in this problem (20°C = 293 K) is relatively small, so we need about 5 moles of O2, which requires about 5 moles or about 500 g NaClO3. Practice: Stoichiometry Calculations Summary Collect and Organize: We are given the volume of O2 to be generated at a pressure and temperature. We can determine the mass of NaClO3 needed based on the stoichiometric relations in the balanced chemical equation. Analyze: The difference between the temperature at STP and the temperature in this problem is relatively small, so we need about 5 moles of O2, which requires about 5 moles or about 500 g NaClO3. Solve: Use the rearranged ideal gas equation to solve for the moles of O2: Then, use the calculated value of moles of O2 and the balanced chemical equation to determine the number of moles and number of grams of NaClO3(s) required: 1 mol NaClO3 106.44 g NaClO3 5.20 mol O2 ´ ´ = 553 g NaClO3 1 mol O2 1 mol NaClO3 Think About It: We predicted that about 500 g NaClO3 would be needed, which is within about 10% of the calculated value. Gas Density Density can be calculated from molar mass (ℳ) and molar volume (V/n). From the ideal gas law: Density: When P in atm and T in kelvin, d = g/L Buoyancy: Gas Densities Buoyancy depends on differences in gas densities. He(g) = 0.169 g/L* N2(g) = 1.19 g/L* CO2(g) = 1.86 g/L* Depends on: Molar masses Temperature Charles’s Law: Density  as temp *At 15°C and 1 atm Practice: Gas Density Calculate the density of air at 1.00 atm and 302 K and compare your answer with the density of air at STP (1.29 g/L). Assume that air has an average molar mass of 28.8 g/mol. Practice: Gas Density Collect and Organize Calculate the density of air at 1.00 atm and 302 K and compare your answer with the density of air at STP (1.29 g/L). Assume that air has an average molar mass of 28.8 g/mol. We are given the average molar mass, temperature, and atmospheric pressure of air, which we can use to calculate density. Practice: Gas Density Solve Calculate the density of air at 1.00 atm and 302 K and compare your answer with the density of air at STP (1.29 g/L). Assume that air has an average molar mass of 28.8 g/mol. Insert the values of pressure, temperature, and molar mass and solve for density.  Practice: Gas Density Summary Calculate the density of air at 1.00 atm and 302 K and compare your answer with the density of air at STP (1.29 g/L). Assume that air has an average molar mass of 28.8 g/mol. Collect and Organize: We are given the average molar mass, temperature, and atmospheric pressure of air, which we can use to calculate density. Analyze: The density of a gas is inversely proportional to temperature, and the temperature in this exercise (302 K) is higher than the temperature at STP, so the density of the air should be less than 1.29 g/L. Solve: Think About It: The calculated density of air at 302 K is less than the value at 273 K (1.29 g/L) as we expected. The density of air (and most gases) is considerably less than that of liquids. For example, the density of liquid water is 1.00 g/mL or 1.00 kg/L, about 1000 times greater than the density of air. Dalton’s Law of Partial Pressures For a mixture of gases in a container: PTotal = P1 + P2 + P3 +... Pressure depends on ntotal, not the identity of the gas. Mole Fraction and Partial Pressure Mole fraction: Ratio of the number of moles of a component in a mixture to the total number of moles in a mixture Mole fraction in terms of pressure: When V, T are constant, P1  n1. Partial Pressure By substitution: so, and… Collecting a Gas over Water 2 KClO3(s) → 2 KCl(s) + 3 O2(g) Gases collected: O2(g) and H2O(g) Kinetic Molecular Theory (KMT) Assumes that gas molecules: 1. Have tiny volumes compared with the collective volume they occupy 2. Move constantly and randomly 3. Have average kinetic energy that is proportional to absolute temperature 4. Engage in elastic collisions with walls of container and other gas molecules 5. Act independently of other gas molecules KMT and Gas Laws Gas laws involving pressure, volume: Boyle’s Law: P  1/V Decreasing volume increases number of collisions per area; P increases. Dalton’s Law: Ptotal = P1 + P2 + P3 +… Total pressure depends only on total number of moles of gas, not on their identities. Avogadro’s Law: V  n Increasing n increases the number of collisions; gas expands to keep pressure constant. Average Kinetic Energy Average kinetic energy: KEavg = ½m(urms)2 urms = the root-mean-square speed of the molecules Real vs. Ideal Gases Assumptions of kinetic molecular theory: #1: Vgas is negligible compared to Vcontainer. #5: Gas molecules act independently. Valid at STP, but not at higher pressures: At high P, low V: Vgas not negligible. Attractive forces between gas molecules are significant. Deviations from Ideal Behavior Vgas molecules  Vcontainer The curve for real gases drops below the horizontal line because of attraction between gas molecules. Real Gases Corrections to ideal gas law: van der Waals equation Corrections to the Ideal Gas Law a – corrects for attractive forces that affect pressure b – corrects for volume occupied by molecules 37 38 39 40