Unit 7 Gas Laws and Thermodynamics Chapters 11 & 16

CHAPTER 11 Gases

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% CO 2, 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/m 2. One pascal is very small, so usually kilopascals (kPa) are used instead. One atm is equal to kPa. 1 atm = 760 mm Hg (Torr) = 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 atm. Express this pressure in: a. millimeters of mercury (mm Hg) b. kilopascals (kPa) Chapter 11 – Section 1: Gases and Pressure atm atm mm Hg x = atm atm kPa x = 631 mm Hg 84.1 kPa

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. Chapter 11 – Section 1: Gases and Pressure P T = P 1 + P 2 + P 3 …

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? b. What is the total pressure in mm Hg? Chapter 11 – Section 1: Gases and Pressure P T = P 1 + P 2 + P 3 … P T = 0.5 atm atm atm = 2.4 atm atm mm Hg x = 1800 mm Hg

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. P 1 and V 1 represent initial conditions, and P 2 and V 2 represent another set of conditions. Chapter 11 – Section 2: The Gas Laws P 1 V 1 = P 2 V 2

Boyle’s Law Sample Problem A sample of oxygen gas has a volume of mL when its pressure is atm. What will the volume of the gas be at a pressure of atm if the temperature remains constant? Solution: Chapter 11 – Section 2: The Gas Laws P 1 V 1 = P 2 V 2 (0.947 atm) (150.0 mL) = (0.987 atm)V2V2 V2V2 = (0.947 atm) (150.0 mL) (0.987 atm) = 144 mL

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 – The volume of a fixed mass of gas at constant pressure varies directly with the Kelvin temperature. Charles’s Law Chapter 11 – Section 2: The Gas Laws V1V1 V2V2 = T1T1 T2T2 V 1 and T 1 represent initial conditions, and V 2 and T 2 represent another set of conditions.

The Kelvin Temperature Scale Absolute zero – The theoretical lowest possible temperature where all molecular motion stops. The Kelvin temperature scale starts at absolute zero (-273 o C.) This gives the following relationship between the two temperature scales: Chapter 11 – Section 2: The Gas Laws K = o C + 273

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: Chapter 11 – Section 2: The Gas Laws V1V1 V2V2 = T1T1 T2T2 752 mL V2V2 = 298 K 323 K K = o C T1T1 = = 298 T2T2 = = mL V2V2 = 298 K 323 K x= 815 mL

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. P 1 and T 1 represent initial conditions. P 2 and T 2 represent another set of conditions. P1P1 P2P2 = T1T1 T2T2 Chapter 11 – Section 2: The Gas Laws

The Combined Gas Law The combined gas law is written as follows: 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 P1P1 P2P2 = T1T1 T2T2 V1V1 V2V2

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 atm and 10.0°C? Solution: Chapter 11 – Section 2: The Gas Laws (1.08 atm) (0.855 atm) = 298 K 283 K K = o C T1T1 = = 298 T2T2 = = 283 (1.08 atm) V2V2 = (298 K) (283 K) = 60.0 L P1P1 P2P2 = T1T1 T2T2 V1V1 V2V2 (50.0 L) V2V2 (0.855 atm)

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: V1V1 V2V2 = n1n1 n2n2 Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law

Standard Molar Volume Standard Temperature and Pressure (STP) is 0 o C 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? b. What volume does moles of a gas occupy at STP? Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law 5.00 L L mol x = mol mol mol L x = 17.2 L

Volume Ratios You can use the volume ratios as conversion factors just like you would use mole ratios. 2CO(g) + O 2 (g) → 2CO 2 (g) 2 molecules 1 molecule2 molecules 2 mole 1 mole2 mol 2 volumes 1 volume2 volumes Example: What volume of O 2 is needed to react completely with L of CO to form CO 2 ? Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law L CO L CO L O x = L O 2

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 H 2 at STP reacts with excess CuO according to the following equation: CuO(s) + H 2 (g) → Cu(s) + H 2 O(g) a.How many moles of H 2 react? b.How many grams of Cu are produced? Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law 5.61 L H 2 L H 2 mol H x = mol H L H 2 L H 2 mol H x mol H 2 mol Cu 1 1 x g Cu x = 15.9 g Cu

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: R represents the ideal gas constant which has a value of (L atm)/(mol K). Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law PV = nRT

The Ideal Gas Law Sample Problem What is the pressure in atmospheres exerted by a mol sample of nitrogen gas in a 10.0 L container at 298 K? Solution: Chapter 11 – Section 3: Gas Volumes and the Ideal Gas Law PV = nRT P (10.0 L)= (0.500 mol)( L atm/mol K) P = (0.500 mol) (298 K) (10.0 L) = 1.22 atm (298 K) ( L atm/mol K)

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. Chapter 11 – Section 4: Diffusion and Effusion

CHAPTER 16 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: Where the variables stand for the following: Q = heat transferred (joules or calories) m = mass (g) c p = specific heat ∆T = change in temperature ( o C or K) Chapter 16 – Section 1: Thermochemistry Q = m x c p x ∆T

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? b. How much energy will the same glass sample gain when it is heated from 314 K to 344 K? Chapter 16 – Section 1: Thermochemistry Q = m x c p x ∆T 32 J =(4.0 g)(c p )(40. K) c p = (4.0 g)(40. K) 32 J = 0.20 J/(g K) Q = m x c p x ∆T Q =(4.0 g)(0.20 J/(g K))(30 K) = 24 J

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. Chapter 16 – Section 1: Thermochemistry ∆H = H products - H reactants

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 CH 4 : C(s) + 2H 2 (g) → CH 4 (g) ∆H f = ? The component reactions are: C(s) + O 2 (g) → CO 2 (g)∆H c = kJ H 2 (g) + ½O 2 (g) → H 2 O(l) ∆H c = kJ CH 4 (g) + 2O 2 (g) → CO 2 (g) + 2H 2 O(l) ∆H c = kJ CO 2 (g) + 2H 2 O(l) → CH 4 (g) + 2O 2 (g) ∆H c = kJ - 2 moles of H 2 are used to make CH 4, so multiply the 2 nd equation by 2 (including ∆H.) -CH 4 is on the products side, not the reactants side, so reverse the 3 rd reaction and change the sign on ∆H. -Cancel unwanted terms and add the ∆H’s. Chapter 16 – Section 1: Thermochemistry kJ kJ