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Unit 7 Gas Laws and Thermodynamics Chapters 11 & 16 Gases CHAPTER 11 Chapter 11 – Section 1: Gases and Pressure Pressure and Force • Pressure is the force per unit area on a surface. Pressure = Force Area Chapter 11 – Section 1: Gases and Pressure Gases in the Atmosphere • The atmosphere of Earth is a layer of gases surrounding the planet that is retained by Earth's gravity. • By volume, dry air is 78% nitrogen, 21% oxygen, 0.9% argon, 0.04% CO2, and small amounts of other gases. Chapter 11 – Section 1: Gases and Pressure Atmospheric Pressure • Atmospheric pressure is the force per unit area exerted on a surface by the weight of the gases that make up the atmosphere above it. Chapter 11 – Section 1: Gases and Pressure Measuring Pressure • A common unit of pressure is millimeters of mercury (mm Hg). • 1 mm Hg is also called 1 torr in honor of Evangelista Torricelli who invented the barometer (used to measure atmospheric pressure). • The average atmospheric pressure at sea level at 0°C is 760 mm Hg, so one atmosphere (atm) of pressure is 760 mm Hg. Chapter 11 – Section 1: Gases and Pressure Measuring Pressure (continued) •Pressure can also be measured in pascals (Pa): 1 Pa = 1 N/m2. •One pascal is very small, so usually kilopascals (kPa) are used instead. •One atm is equal to 101.3 kPa. 1 atm = 760 mm Hg (Torr) = 101.3 kPa Chapter 11 – Section 1: Gases and Pressure Units of Pressure Chapter 11 – Section 1: Gases and Pressure Converting Pressure Sample Problem The average atmospheric pressure in Denver, CO is 0.830 atm. Express this pressure in: a. millimeters of mercury (mm Hg) 0.830 atm x 760 mm Hg = 631 mm Hg 1 atm b. kilopascals (kPa) 0.830 atm x 101.3 kPa 1 atm = 84.1 kPa Chapter 11 – Section 1: Gases and Pressure Dalton’s Law of Partial Pressures • Dalton’s law of partial pressures - the total pressure of a gas mixture is the sum of the partial pressures of the component gases. PT = P1 + P2 + P3 … Chapter 11 – Section 1: Gases and Pressure Dalton’s Law of Partial Pressures Sample Problem A container holds a mixture of gases A, B & C. Gas A has a pressure of 0.5 atm, Gas B has a pressure of 0.7 atm, and Gas C has a pressure of 1.2 atm. a. What is the total pressure of this system? PT = P1 + P2 + P3 … PT = 0.5 atm + 0.7 atm + 1.2 atm = 2.4 atm b. What is the total pressure in mm Hg? 2.4 atm x 760 mm Hg = 1800 mm Hg 1 atm Chapter 11 – Section 2: The Gas Laws Gases and Pressure • Gas pressure is caused by collisions of the gas molecules with each other and with the walls of their container. • The greater the number of collisions, the higher the pressure will be. Chapter 11 – Section 2: The Gas Laws Pressure – Volume Relationship • When the volume of a gas is decreased, more collisions will occur. • Pressure is caused by collisions. • Therefore, pressure will increase. • This relationship between pressure and volume is inversely proportional. Chapter 11 – Section 2: The Gas Laws Boyle’s Law • Boyle’s Law – The volume of a fixed mass of gas varies inversely with the pressure at a constant temperature. P1V1 = P2V2 • P1 and V1 represent initial conditions, and P2 and V2 represent another set of conditions. Chapter 11 – Section 2: The Gas Laws Boyle’s Law Sample Problem A sample of oxygen gas has a volume of 150.0 mL when its pressure is 0.947 atm. What will the volume of the gas be at a pressure of 0.987 atm if the temperature remains constant? Solution: P1V1 = P2V2 (0.947 atm)(150.0 mL) = (0.987 atm)V2 (0.947 atm)(150.0 mL) = 144 mL V2 = (0.987 atm) Chapter 11 – Section 2: The Gas Laws Volume – Temperature Relationship • the pressure of gas inside and outside the balloon are the same. • at low temperatures, the gas molecules don’t move as much – therefore the volume is small. • at high temperatures, the gas molecules move more – causing the volume to become larger. Chapter 11 – Section 2: The Gas Laws Charles’s Law • Charles’s Law – The volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature. V1 V2 = T1 T2 • V1 and T1 represent initial conditions, and V2 and T2 represent another set of conditions. Chapter 11 – Section 2: The Gas Laws The Kelvin Temperature Scale • Absolute zero – The theoretical lowest possible temperature where all molecular motion stops. • The Kelvin temperature scale starts at absolute zero (-273oC.) • This gives the following relationship between the two temperature scales: K = oC + 273 Chapter 11 – Section 2: The Gas Laws Charles’s Law Sample Problem A sample of neon gas occupies a volume of 752 mL at 25°C. What volume will the gas occupy at 50°C if the pressure remains constant? Solution: K = oC + 273 V1 V2 T1 = 25 + 273 = 298 = T1 T2 T2 = 50 + 273 = 323 752 mL V2 = 298 K 323 K 752 mL V2 = x 323 K = 815 mL 298 K Chapter 11 – Section 2: The Gas Laws Pressure – Temperature Relationship • Increasing temperature means increasing kinetic energy of the particles. • The energy and frequency of collisions depend on the average kinetic energy of the molecules. • Therefore, if volume is kept constant, the pressure of a gas increases with increasing temperature. Chapter 11 – Section 2: The Gas Laws Gay-Lussac’s Law • Gay-Lussac’s Law – The pressure of a fixed mass of gas varies directly with the Kelvin temperature. P1 P2 = T1 T2 • P1 and T1 represent initial conditions. P2 and T2 represent another set of conditions. Chapter 11 – Section 2: The Gas Laws The Combined Gas Law • The combined gas law is written as follows: P1 V1 P2 V2 = T1 T2 • Each of the other gas laws can be obtained from the combined gas law when the proper variable is kept constant. Chapter 11 – Section 2: The Gas Laws The Combined Gas Law Sample Problem A helium-filled balloon has a volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.0°C? K = oC + 273 Solution: T1 = 25 + 273 = 298 P1V1 P2V2 T2 = 10 + 273 = 283 = T1 T2 (1.08 atm)(50.0 L) (0.855 atm) V2 = 298 K 283 K (1.08 atm)(50.0 L) (283 K) V2 = = 60.0 L (298 K) (0.855 atm) Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law Avogadro’s Law • In 1811, Amedeo Avogadro discovered that the volume of a gas is proportional to the number of molecules (or number of moles.) • Avogadro’s Law - equal volumes of gases at the same temperature and pressure contain equal numbers of molecules, or: V1 V2 = n1 n2 Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law Standard Molar Volume • Standard Temperature and Pressure (STP) is 0oC and 1 atm. • The Standard Molar Volume of a gas is the volume occupied by one mole of a gas at STP. It has been found to be 22.4 L. Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law Molar Volume Conversion Factor • Standard Molar Volume can be used as a conversion factor to convert from the number of moles of a gas at STP to volume (L), or vice versa. Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law Molar Volume Conversion Sample Problem a. What quantity of gas, in moles, is contained in 5.00 L at STP? 5.00 L x 1 mol 22.4 L = 0.223 mol b. What volume does 0.768 moles of a gas occupy at STP? 0.768 mol x 22.4 L 1 mol = 17.2 L Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law Volume Ratios • You can use the volume ratios as conversion factors just like you would use mole ratios. 2CO(g) + O2(g) → 2 molecules 1 molecule 2 mole 1 mole 2 volumes 1 volume 2CO2(g) 2 molecules 2 mol 2 volumes • Example: What volume of O2 is needed to react completely with 0.626 L of CO to form CO2? 0.626 L CO x 1 L O2 2 L CO = 0.313 L O2 Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law The Mole Map • You can now convert between number of particles, mass (g), and volume (L) by going through moles. Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law Gas Stoichiometry Sample Problem Assume that 5.61 L H2 at STP reacts with excess CuO according to the following equation: CuO(s) + H2(g) → Cu(s) + H2O(g) a. How many moles of H2 react? 5.61 L H2 x 1 mol H2 = 0.250 mol H2 22.4 L H2 b. How many grams of Cu are produced? 5.61 L H2 x 1 mol H2 x 1 mol Cu x 63.5 g Cu = 15.9 g 22.4 L H2 1 mol H2 1 mol Cu Cu Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law The Ideal Gas Law • All of the gas laws you have learned so far can be combined into a single equation, the ideal gas law: PV = nRT • R represents the ideal gas constant which has a value of 0.0821 (L•atm)/(mol•K). Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law The Ideal Gas Law Sample Problem What is the pressure in atmospheres exerted by a 0.500 mol sample of nitrogen gas in a 10.0 L container at 298 K? Solution: PV = nRT P (10.0 L) = (0.500 mol)(0.0821 L•atm/mol•K) (298 K) (0.500 mol) (0.0821 L•atm/mol•K) (298 K) P= (10.0 L) 1.22 = atm Chapter 11 – Section 4: Diffusion and Effusion Diffusion and Effusion • Diffusion is the gradual mixing of two or more gases due to their spontaneous, random motion. • Effusion is the process whereby the molecules of a gas confined in a container randomly pass through a tiny opening in the container. Chapter 11 – Section 4: Diffusion and Effusion Graham’s Law of Effusion • Light molecules move faster than heavy ones. • Graham’s law of effusion says the greater the molar mass of a gas, the slower it will effuse. Thermochemistry CHAPTER 16 Chapter 16 – Section 1: Thermochemistry Heat and Temperature • Temperature – a measure of the average kinetic energy of the particles in a sample of matter. • The greater the kinetic energy of the particles in a sample, the hotter it feels. • Heat – energy transferred between samples of matter due to a difference in their temperatures. • Heat always moves spontaneously from matter at a higher temperature to matter at a lower temperature. Chapter 16 – Section 1: Thermochemistry Measuring Heat • Heat energy is measured in joules (or calories – food only) • Chemical reactions usually either absorb or release energy as heat. • The energy absorbed or released as heat in a chemical or physical change is measured in a calorimeter. Chapter 16 – Section 1: Thermochemistry Specific Heat • A quantity called specific heat can be used to compare heat absorption capacities for different materials. • Specific heat – the amount of energy required to raise the temperature of one gram of a substance by 1°C or 1 K. • Specific heat can be measured in units of J/(g•°C), J/(g•K), cal/(g•°C), or cal/(g•K). Chapter 16 – Section 1: Thermochemistry Heat Transfer Equation • Specific heat can be used to find the quantity of heat energy gained or lost with a change in temperature according to the following equation: Q = m x cp x ∆T • Where the variables stand for the following: Q = heat transferred (joules or calories) m = mass (g) cp = specific heat ∆T = change in temperature (oC or K) Chapter 16 – Section 1: Thermochemistry Heat Transfer Equation Sample Problem A 4.0 g sample of glass was heated from 274 K to 314 K, a temperature increase of 40. K, and was found to have absorbed 32 J of energy as heat. a. What is the specific heat of this type of glass? Q = m x cp x ∆T 32 J = (4.0 g)(cp) (40. K) 32 J = 0.20 cp = (4.0 g)(40. K) J/(g•K) b. How much energy will the same glass sample gain when it is heated from 314 K to 344 K? Q = m x cp x ∆T Q = (4.0 g)(0.20 J/(g•K))(30 K) = 24 J Chapter 16 – Section 1: Thermochemistry Enthalpy of Reaction • The enthalpy of reaction (H) is the quantity of energy transferred as heat during a chemical reaction. • The change in enthalpy (∆H) of a reaction is always the difference between the enthalpies of the products and the reactants. ∆H = Hproducts - Hreactants Chapter 16 – Section 1: Thermochemistry Exothermic Reactions • In an exothermic reaction, energy is released. Therefore, the energy of the products must be less than the energy of the reactants, and ∆H is negative. • The great majority of chemical reactions in nature are exothermic. Chapter 16 – Section 1: Thermochemistry Endothermic Reactions • In an endothermic reaction, energy is absorbed. Therefore, the energy of the products must be greater than the energy of the reactants, and ∆H is positive. Chapter 16 – Section 1: Thermochemistry Entropy and Reaction Tendency • Entropy – a measure of the degree of randomness in a system. • Processes in nature are driven in two directions: towards decreasing enthalpy and towards increasing entropy. • The combined enthalpy-entropy function is called the Gibbs free energy (G). Chapter 16 – Section 1: Thermochemistry Hess’s Law • Hess’s Law – the overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process. • Possible steps in Hess’s Law: 1. Reverse equation/ change sign on ΔH. 2. Multiply or Divide coefficients/multiply or divide ΔH. Chapter 16 – Section 1: Thermochemistry Hess’s Law Sample Problem Calculate the enthalpy of formation for CH4: C(s) + 2H2(g) → CH4(g) ∆Hf = ?-74.3 kJ The component reactions are: C(s) + O2(g) → CO2(g) ∆Hc = -393.5 kJ -571.6 kJ kJ 2H2(g) + ½O2(g) →2 H2O(l) ∆Hc = -285.8 CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆Hc = -890.8 kJ CO2(g) + 2H2O(l) → CH4(g) + 2O2(g) ∆Hc = +890.8 kJ -74.3 kJ - 2 moles of H2 are used to make CH4, so multiply the 2nd equation by 2 (including ∆H.) - CH4 is on the products side, not the reactants side, so reverse the 3rd reaction and change the sign on ∆H. - Cancel unwanted terms and add the ∆H’s.