A Molecular Approach, 4e - Notes for Chapter (5).doc

Chapter 5. Gases Chapter 5. Gases Chapter 5. Gases Student Objectives 5.1 Supersonic Skydiving and the Risk of Decompression Define pressure and understand how differences in pressure could have caused severe damage in a long free fall. 5.2 Pressure: The Result of Molecular Collisions Understand pressure from a molecular point of view. Understand examples of pressure: wind and pressure imbalance on the eardrum. Understand why pressure can be measured in mm of Hg, and convert between the different pressure units. Understand the origin of the two numbers for blood pressure. 5.3 The Simple Gas Laws: Boyle’s Law, Charles’s Law, and Avogadro’s Law Know and be able to rationalize Boyle’s law, the inverse relationship between volume and pressure. Use the inverse mathematical relationship between pressure and volume to solve initial and final states problems at constant temperature and amount. Know and be able to rationalize Charles’s law, the direct relationship between volume and temperature. Use the direct mathematical relationship between volume and temperature to solve initial and final states problems at constant pressure and amount. Know and be able to rationalize Avogadro’s law, the direct relationship between volume and amount (moles). Use the direct mathematical relationship between volume and amount to solve initial and final states problems at constant pressure and temperature. 5.4 The Ideal Gas Law Know and understand how the ideal gas law combines the three simple gas laws into one equation. Calculate using the ideal gas law and the gas constant, R, with the appropriate value and units. 5.5 Applications of the Ideal Gas Law: Molar Volume, Density, and Molar Mass of a Gas Define standard temperature and pressure and molar volume of an ideal gas. Know and understand the relationship between molar volume, molar mass, and density. Calculate using density, molar mass, and molar volume. 5.6 Mixtures of Gases and Partial Pressures Define and understand partial pressure of a gaseous component in a mixture. Define and determine mole fraction of a component in a mixture. Know and understand that Dalton’s law of partial pressures relates to mole fraction of a gas to the partial pressure of the gas. Understand how the total pressure affects the partial pressures of gases in blood, especially during deep-sea diving. Know and understand the technique of collecting gases over water. 5.7 Gases in Chemical Reactions: Stoichiometry Revisited Understand how stoichiometry applies to gases via the number of moles in the ideal gas law. Understand how stoichiometry relates to molar volume. 5.8 Kinetic Molecular Theory: A Model for Gases Define kinetic molecular theory for gases. Understand each of the three postulates/assumptions of the kinetic molecular theory. Understand how the kinetic molecular theory explains Boyle’s, Charles’s, Avogadro’s, and Dalton’s laws. Follow the derivation of the ideal gas law from the kinetic molecular theory. Understand that all gases at the same temperature have the same kinetic energy, and understand the relationship between speed and molar mass. Understand the graphical representation of the distribution of molecular speeds. 5.9 Mean Free Path, Diffusion, and Effusion of Gases Define and understand mean free path. Define diffusion and effusion and understand how they are related to the kinetic molecular theory. 5.10 Real Gases: The Effects of Size and Intermolecular Forces Understand that the ideal gas law is an approximation that works well under certain circumstances and not so well at low temperature and/or high pressure. Understand that nonideal behavior arises from the finite volume of gas particles and the intermolecular forces between particles. Recognize and identify the components of the van der Waals equation. Section Summaries Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples Teaching Tips Suggestions and Examples Misconceptions and Pitfalls Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 5.1 Supersonic Skydiving and the Risk of Decompression Pressure practical definition dangers of decompression Intro figure: Space Helmet Figure 5.1 Gas Pressure 5.2 Pressure: The Result of Molecular Collisions Pressure molecular view units: millimeters of mercury, torr, atmosphere, pascal, pounds per square inch measurements mercury barometer manometer blood pressure unnumbered figure: regions of high and low atmospheric pressure Figure 5.2 Pressure and Particle Density Figure 5.3 Pressure Imbalance Figure 5.4 The Mercury Barometer Table 5.1 Common Units of Pressure Example 5.1 Converting between Pressure Units Figure 5.5 The Manometer Chemistry and Medicine: Blood Pressure 5.3 The Simple Gas Laws: Boyle’s Law, Charles’s Law, and Avogadro’s Law Simple gas laws Boyle’s law: volume and pressure V 1/P P1V1 P2V2 Charles’s law: volume and temperature V T V1/T1 V2/T2 Avogadro’s law: volume and amount V n V1/n1 V2/n2 Figure 5.6 The J-Tube Figure 5.7 Volume versus Pressure Figure 5.8 Molecular Interpretation of Boyle’s Law Figure 5.9 Increase in Pressure with Depth Example 5.2 Boyle’s Law Figure 5.10 Volume versus Temperature Chemistry in Your Day: Extra-long Snorkels Figure 5.11 Molecular Interpretation of Charles’s Law unnumbered figure: hot-air balloon unnumbered figure: balloon in liquid N2 Example 5.3 Charles’s Law Figure 5.12 Volume versus Number of Moles Example 5.4 Avogadro’s Law 5.4 The Ideal Gas Law Derived from simple gas laws: PV nRT units: Kelvin for T, moles for n gas constant R: function of P, V units unnumbered figure: relationship between ideal gas law and simple gas laws unnumbered figure: label on aerosol can Example 5.5 Ideal Gas Law I Example 5.6 Ideal Gas Law II Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 5.1 Supersonic Skydiving and the Risk of Decompression Some students may remember the jump. If not an internet search for news articles or videos may help. Some students do not realize that air is made of matter. 5.2 Pressure: The Result of Molecular Collisions Recognizing pressure as related to force will likely help most students. The difference can be explained by talking about snowshoes and ice skates. Large numbers of people have experienced their ears “popping” in an airplane or upon another ascent in altitude. Tornados, hurricanes, and cyclones are extreme examples of differential pressures in weather. Blood pressure is an interesting alternative since the previous examples deal with gases. Barometers and manometers combine liquids and gases. The end of the tube can be open, under a vacuum, or filled with air. The outcomes are different and can be confusing. 5.3 The Simple Gas Laws: Boyle’s Law, Charles’s Law, and Avogadro’s Law The easiest example of Boyle’s law is a cylinder with a sliding piston. A plastic syringe works well as long as one has a convenient connection to a gauge. Limited numbers of students will have personal experience with diving, though most students likely will have experience swimming underwater. Charles’s law can be demonstrated using balloons and dry ice/acetone or liquid nitrogen. Conceptual Connection 5.1 Boyle’s Law and Charles’s Law Avogadro’s law can be demonstrated by producing a gas over water to show displacement. An interesting challenge question comes from burning a candle within a sealed container connected to a pressure gauge. Oxygen is consumed, but carbon dioxide is produced. Breathing at increased depths of water becomes more difficult but is possible even to substantial depths. Hot-air and helium balloons rise because of decreased density compared with the surrounding air. Some students memorize the different equations for initial and final states problems, but all of them can be derived easily from the ideal gas law. In initial and final states problems with T, the ratio of the values must be in Kelvin and not in degrees Celsius. 5.4 The Ideal Gas Law A series of R values is possible with different units especially for pressure. Common problems include using mixtures of unit conversions: °C to K, mL to L, mmHg to atm. Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 5.5 Applications of the Ideal Gas Law: Molar Volume, Density, and Molar Mass of a Gas Standard temperature and pressure molar volume Density: molar volume/molar mass Finding the molar mass of a gas unnumbered figure: molar volumes of different gases at STP Example 5.7 Density Example 5.8 Molar Mass of a Gas 5.6 Mixtures of Gases and Partial Pressures Mixtures of gases partial pressure mole fraction Dalton’s law Deep-sea diving hypoxia oxygen toxicity nitrogen narcosis Collecting gases over water vapor pressure of water Table 5.3 Composition of Dry Air Example 5.9 Total Pressure and Partial Pressures Figure 5.12 Oxygen Partial Pressure Limits unnumbered figure: gas pressure in lungs Example 5.10 Partial Pressures and Mole Fractions Figure 5.14 Collecting a Gas over Water Table 5.4 Vapor Pressure of Water versus Temperature Example 5.11 Collecting Gases over Water 5.7 Gases in Chemical Reactions: Stoichiometry Revisited Calculations: moles from the ideal gas law P, V, T of gas A moles A moles B P, V, T of gas B Calculations: molar volume V of gas A moles A moles B mass of gas B Example 5.12 Gases in Chemical Reactions Example 5.13 Using Molar Volume in Gas Stoichiometric Calculations Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 5.5 Applications of the Ideal Gas Law: Molar Volume, Density, and Molar Mass of a Gas Conceptual Connection 5.2 Molar Volume Density can be calculated from molar volume and molar mass or from P, T and M. Conceptual Connection 5.3 Density of a Gas Ideal gases have a common molar volume but not a common mass. Standard temperature for gases is not the same standard as in thermochemistry and thermodynamics. 5.6 Mixtures of Gases and Partial Pressures The partial pressure of oxygen in air is important in physiology. Oxygen available in the lungs for transfer to the blood can be changed by a change in the total air pressure or by a change in the percent oxygen in air. Conceptual Connection 5.4 Partial pressures Partial pressure of oxygen plays a role in diving and in climbing at high altitude. Too little or too much can be dangerous. The vapor pressure of water must be considered when collecting gases over water, a common laboratory technique. The solubility of gases in water is generally small except for CO2. The vapor pressure of water is a function of temperature and is covered in more detail in Chapter 11. That gases behave ideally, independent of identity, gives rise to partial pressures. The sum of the mole fractions is 1. 5.7 Gases in Chemical Reactions: Stoichiometry Revisited The ideal gas law is used to obtain moles rather than using mass or volume and concentration of an aqueous solution. At the same pressure and temperature, the mole ratio from a balanced reaction is the same as the volume ratio for gaseous components. At the same volume and temperature, the mole ratio from a balanced reaction is the same as the pressure ratio for gaseous components. Conceptual Connection 5.5 Pressure and Number of Moles Stoichiometry problems involving gases are inherently the same as other stoichiometry problems. Gases simply present another means of obtaining moles of substances. Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 5.8 Kinetic Molecular Theory: A Model for Gases Kinetic molecular theory postulates/assumptions particles negligibly small average kinetic energy proportional to T elastic collisions Relevance to gas laws pressure as a force per unit area Boyle’s law Charles’s law Avogadro’s law Dalton’s law ideal gas law Temperature and molecular speeds Distribution of molecular speeds dependence on T dependence on M Figure 5.15 A Model for Gas Behavior Figure 5.16 Elastic versus Inelastic Collsions Figure 5.17 The Pressure on the Wall of a Container Figure 5.18 Velocity Distribution for Several Gases at 25 °C Figure 5.19 Velocity Distribution for Nitrogen at Several Temperatures Example 5.14 Root Mean Square Velocity 5.9 Mean Free Path, Diffusion, and Effusion of Gases Mean free path Diffusion Effusion Graham’s law: Figure 5.20 Mean Free Path Figure 5.21 Effusion Example 5.15 Graham’s Law of Effusion Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 5.8 Kinetic Molecular Theory: A Model for Gases The physics in the kinetic molecular theory is remarkably simple given all of the behaviors it explains. The ideal gas law can be derived from the definition of pressure as force per unit area. Two of the three assumptions in the kinetic molecular theory give rise to nonideal behavior. Conceptual Connection 5.6 Kinetic Molecular Theory I Conceptual Connection 5.7 Root Mean Square Velocity Conceptual Connection 5.8 Kinetic Molecular Theory II One of the main uses of going through a derivation is pointing out that knowing where an equation comes from and the assumptions required to derive it gives clues concerning its range of applicability. The analogy of billiard balls to atoms in an elastic collision is partially flawed since the balls lose some energy to friction with the table. Students are likely to have difficulty with distributions of velocities within a collection of molecules. 5.9 Mean Free Path, Diffusion, and Effusion of Gases Diffusion is easy for students to understand, as they have experience with odors (especially bad ones) spreading throughout a room. An analogy for diffusion is thermal equilibration, as they both are illustrations of the second law of thermodynamics. Most students are surprised by the high speeds of atoms and molecules at room temperature. Lecture Outline Terms, Concepts, Relationships, Skills Figures, Tables, and Solved Examples 5.10 Real Gases: The Effects of Size and Intermolecular Forces Real gases high pressures: finite particle volume low temperatures: intermolecular forces van der Waals equation Figure 5.22 Molar Volumes of Real Gases Figure 5.23 Particle Volume and Ideal Behavior Figure 5.24 The Effect of Particle Volume Table 5.5 Van der Waals Constants for Common Gases Figure 5.25 The Effect of Intermolecular Forces Figure 5.26 Real versus Ideal Behavior Teaching Tips Suggestions and Examples Misconceptions and Pitfalls 5.10 Real Gases: The Effects of Size and Intermolecular Forces Two of the three assumptions from the kinetic molecular theory break down and give rise to nonideal behavior. Each of the effects can be demonstrated independently. Conceptual Connection 5.9 Real Gases Why do we use the ideal gas law if we know it is “wrong”? For some gases under some conditions, the corrections are unnecessary. Additional Problem for Converting between Pressure Units (Example 5.1) A normal systolic blood pressure value is 110. mmHg. What is the pressure in pascal? Sort The problem gives a pressure in mmHg and asks you to find the equivalent in pascal. Given 110 mmHg Find Pa Strategize Since Table 5.1 does not have a direct conversion factor between mmHg and Pa, but does provide relationships between both of these units and atmospheres, you can convert to atm as an intermediate step. Conceptual Plan mmHg atm Pa Relationships Used 1 atm 760 mmHg 101,325 Pa 1 atm (both from Table 5.1) Solve Follow the conceptual plan to solve the problem. Begin with 110 mmHg and use the conversion factors to arrive at the pressure in Pa. Solution Check The units (Pa) are correct. The magnitude of the answer (14,700) makes physical sense since the blood pressure is a fraction of one atm. Additional Problem for Ideal Gas Law (Example 5.5) Calculate the volume in L occupied by 1.43 mol of oxygen gas at a pressure of 3.25 atm and a temperature of 298 K. Sort The problem gives you the number of moles of oxygen gas, the pressure, and the temperature. You are asked to find the volume in L. Given n 1.43 mol P 3.25 atm T 298 K Find V in L Strategize You are given three of the four variables (n, P, T) in the ideal gas law and asked to find the fourth (V). The conceptual plan shows how the ideal gas law provides the relationship between the given quantities and the quantity to be found. Conceptual Plan n, P, T V Relationships Used PV nRT (ideal gas law) Solve To solve the problem, first solve the ideal gas law for V. Then substitute the given quantities to compute V. Solution Check The units of the answer are correct and the magnitude makes sense. Additional Problem for Density (Example 5.7) Calculate the density of oxygen gas at 37.4 °C and a pressure of 720 mmHg. Sort The problem gives you the temperature and pressure of a gas and asks you to find its density. The problem also states that the gas is oxygen. Given T 37.4 °C P 720 mmHg Find d Strategize Equation 5.6 provides the relationship between the density of a gas and its temperature, pressure, and molar mass. The temperature and pressure are given, and you can compute the molar mass from the formula of the gas, which we know is O2. Conceptual Plan P, T, M d Relationships Used molar mass O2 32.00 Solve To solve the problem, you must gather each of the required quantities in the correct units. Convert the temperature to kelvins and the pressure to atmospheres. Now substitute the quantities into the equation to compute density. Solution T(K) 37.4 °C 273 310.4 K Check The units of the answer (g/L) are correct. The magnitude of the answer (1.19) makes physical sense compared to the density of gases calculated earlier. Additional Problem for Molar Mass of a Gas (Example 5.8) A sample of gas has a mass of 0.205 g. Its volume is 0.112 L at a temperature of 25 °C and a pressure of 740 mmHg. Find its molar mass. Sort The problem gives you the mass of a gas sample, along with its volume, temperature, and pressure. You are asked to find the molar mass. Given m 0.205 g V 0.112 L T(°C) 25 °C P 740 mmHg Find molar mass (g/mol) Strategize The conceptual plan has two parts: 1) Use the ideal gas law to find the number of moles of gas. 2) Use the definition of molar mass to find the molar mass. Conceptual Plan P, V, T n n, m molar mass Relationships Used PV nRT Solve Convert the temperature to K and pressure to atm. Then find the number of moles, solving the ideal gas law for n. Use the number of moles (n) and the given mass (m) to find the molar mass. Solution T(K) 25 °C 273 298 K Check The units (g/mol) are correct. The magnitude of the answer (46.0) seems to make sense for a gas. For example, it could be NO2. Additional Problem for Total Pressure and Partial Pressures (Example 5.9) A 1.50-L mixture of helium, neon, and argon has a total pressure of 754 mmHg at 310 K. If the partial pressure of helium is 431 mmHg and partial pressure of neon is 211 mmHg, what mass of argon is present in the mixture? Sort The problem gives you the partial pressures of two of the three components in a gas mixture, along with the total pressure, the volume, and the temperature, and asks you to find the mass of the third component. Given PHe 431 mmHg PNe 211 mmHg Ptotal 754 mmHg V 1.50 L T 310 K Find mAr Strategize You can find the mass of argon from the number of moles of argon, which you can calculate from the partial pressure of argon and the ideal gas law. Begin by finding the partial pressure of argon from Dalton’s law of partial pressures. Then use the partial pressure of argon together with the volume of the sample and the temperature to find the number of moles of argon. Finally, use the molar mass of argon to compute the mass of argon from the number of moles of argon. Conceptual Plan Ptotal, PHe, PNe PAr PAr, V, T nAr nAr mAr Relationships Used Ptotal PHe PNe PAr (Dalton’s law) PV nRT (ideal gas law) molar mass Ar 39.95 g/mol Solve Follow the conceptual plan. To find the partial pressure of argon, solve the equation for PAr, and substitute the values of the other partial pressures. Convert the partial pressure from mmHg to atm and use it in the ideal gas law to compute the amount of argon in moles. Use the molar mass of argon to convert from amount of argon in moles to mass of argon. Solution Ptotal PHe PNe PAr PAr Ptotal PHe PNe 754 mmHg 431 mmHg 211 mmHg 112 mmHg Check The units of the answer are correct and the magnitude makes sense. Additional Problem for Gases in Chemical Reactions (Example 5.12) Methanol (CH3OH) can be synthesized by the reaction: CO(g) 2 H2(g) CH3OH(g) What volume (in liters) of methanol gas, measured at a temperature of 473 K and a pressure of 820 mmHg, is produced from 100.0 g of carbon monoxide CO? Sort You are given the mass of carbon monoxide, a reactant of a chemical reaction. You are asked to find the volume of the product produced (methanol) at a specified temperature and pressure. Given 100.0 g CO T 473 K P 820 mmHg Find Vmethanol Strategize The volume of methanol produced can be calculated from the number of moles of methanol, which can be obtained from the number of moles of carbon monoxide and the stoichiometry of the reaction. Find the number of moles of carbon monoxide from its mass by using the molar mass. Use the stoichiometric relationship from the balanced equation to find the number of moles of methanol formed from that quantity of carbon monoxide. Substitute the moles of methanol, together with the pressure and temperature into the ideal gas law to find the volume of hydrogen. Conceptual Plan g CO mol CO mol CH3OH n (mol CH3OH), P, T Vmethanol Relationships Used PV nRT (ideal gas law) 1 mol CO : 1 mol CH3OH (from balanced equation) molar mass CO 28.01 g Solve Follow the conceptual plan to solve the problem. Begin by using the mass of carbon monoxide to get the number of moles of CO. Convert the number of moles of carbon monoxide to moles of methanol. Convert the pressure to atmospheres and use the ideal gas law to find the volume of methanol. Solution 3.570 mol CO 3.570 mol CH3OH Check The units of the answer are correct. The magnitude makes sense. You start with about 3.5 moles of reactant. If the reaction was conducted at STP, you would expect about 80 L. Since the temperature is much higher than 298 K and the pressure is somewhat higher, the result should be greater. Additional Problem for Molar Volume and Stoichiometry (Example 5.13) How many grams of water form when 2.41 L of oxygen gas at STP completely react with an appropriate amount of H2? 2 H2(g) O2(g) 2 H2O(g) Sort You are given the volume of oxygen gas (a reactant) at STP and asked to determine the mass of water that forms upon complete reaction. Given 2.41 L O2 Find g H2O Strategize Since the reaction occurs under standard conditions, you can convert directly from the volume (in L) of hydrogen gas to the amount in moles. Then use the stoichiometric relationship from the balanced equation to find the number of moles of water formed. Finally, use the molar mass of water to obtain the mass of water formed. Conceptual Plan L O2 mol O2 mol H2O g H2O Relationships Used 1 mol 22.4 L (at STP) 1 mol O2 : 2 mol H2O (from balanced equation) molar mass H2O 18.02 g/mol Solve Follow the conceptual plan to solve the problem. Solution Check The units of the answer are correct. The magnitude makes sense since you start with about one-tenth of a molar volume (22.4 10 2.2) and make twice as many moles of water as oxygen. The mass of water is about one-fifth of a molar mass (18 5 3.6). Additional Problem for Root Mean Square Velocity (Example 5.14) Calculate the root mean square velocity of oxygen molecules at 100 °C. Sort You are given the kind of molecule and the temperature and asked to find the root mean square velocity. Given O2 T 100 °C Find urms Strategize The conceptual plan for this problem simply shows how the molar mass of oxygen and the temperature (in kelvins) can be used with the equation that defines the root mean square velocity. Conceptual Plan M, T urms Relationships Used 1 J 1 kg m2 / s2 Solve The quantities must be in the correct units—temperature in K and molar mass in kg/mol. Substitute the quantities into the equation. Solution T 100 °C 273 373 K Check The units of the answer (m/s) are correct, and the magnitude seems reasonable compared with the velocity previously calculated for 25 °C. Additional Problem for Graham’s Law of Effusion (Example 5.15) An unknown gas effuses at a rate that is 0.578 times that of nitrogen gas at the same temperature. Calculate the molar mass of the unknown gas in g/mol. Sort You are given the ratio of effusion rates for the unknown gas and nitrogen and asked to find the molar mass of the unknown gas. Given Find Munk Strategize The conceptual plan uses Graham’s law of effusion. You are given the ratio of rates and you know the molar mass of nitrogen. You can use Graham’s law to find the molar mass of the unknown gas. Conceptual Plan rate ratio, Mnitrogen Munk Relationships Used (Graham’s law) Solve Solve the equation for Munk and substitute the correct values to compute it. Solution Check The units of the answer (g/mol) are correct, and the magnitude makes sense. The unknown gas appears to be krypton. 66 Copyright © 2017 by Education, Inc. 67 Copyright © 2017 by Education, Inc.