UNIT #8: Low Density: Compression and Expansion: Diffusion:

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1 NAME: UNIT #8: Characteristics of Gases Gas Laws and Calculations Intermolecular Forces Phase Changes Energy Calculations Heating and Cooling Curves Vapor Pressure 1. GENERAL CHARACTERISTICS OF GASES a) Low Density: Gases have low densities due to the small number of gas particles in a relatively large space. The molecular particles of gas are much more distant from one another than they are in either a liquid or solid. That is why gases are transparent. Light can pass through the relatively large spaces between the molecules of gas without any distortion. Liquids distort light more so because the particles are closer and the light passing through a liquid can be deflected more easily. b) Compression and Expansion: Gases can be compressed by applying pressure and reducing their volume, causing the gas particles to come closer together and increasing their density. Likewise, gases will expand with the removal of pressure and increasing their volume, causing the gas particles to become farther apart and reducing their density. Ex. A propane tank used with gas grills contains propane molecules (C 3 H 8 ) that are under pressure. The pressure forces the gas molecules closer to on another, temporarily forcing them into a liquid state. When the tank s valve is opened, it releases pressure and allows the propane molecules to expand back into a gaseous state. The combustible gas then feeds into the grill and once it s ignited, provides the fuel for grilling. When the valve is closed, the remaining gas molecules stay under pressure until the grill is used again. c) Diffusion: Diffusion is the spreading out of one substance through another substance. The direction of diffusion is always from a region of high concentration to a region of low concentration. Ex. If a gas possessing an odor such as ammonia is allowed to escape an open bottle, the odor is quickly detected in the room as the gas vapors diffuse from the bottle (higher concentration) to the air (lower concentration). Smaller, lighter gas particles diffuse at a faster rate than larger, heavier gas particles. d) Effusion is the escape of a gas through a small hole. Ex. Helium balloons deflate as the gaseous helium molecules effuse through the pores in latex balloons over time. Smaller, lighter gas particles effuse at a faster rate than larger, heavier gas particles.

2 2. KINETIC MOLECULAR THEORY The experimental observation about the behavior of gases can be explained with a simple theoretical model known as the kinetic molecular theory. This theory is based on the following postulates or assumptions. a) Gases are composed of a large number of particles that behave like hard, spherical objects in a state of constant, random motion. b) These particles move in a straight line until they collide with another particle or the walls of the container. c) These particles are much smaller than the distance between particles. Most of the volume of a gas is therefore empty space. d) There is no force of attraction between gas particles or between the particles and the walls of the container. e) Collisions between gas particles or collisions with the walls of the container are perfectly elastic. None of the energy of a gas particle is lost when it collides with another particle or with the walls of the container. f) The average kinetic energy (energy of movement) of a collection of gas particles depends on the temperature of the gas. Temperature is a measure of the average kinetic energy of the particles in a sample of matter. At a given temperature, all gases have the same average kinetic energy. Gases at high temperatures have greater kinetic energy and move with greater velocity. Gases at low temperature have lower kinetic energy and move with slower velocity. Ex. If a balloon is inflated at room temperature, the balloon maintains its size because the gas molecules within the balloon have the same average kinetic energy; however, if the balloon is removed outside on a hot, summer day, it will quickly expand as the gas molecules are heated and expand with greater velocity. Likewise, if the balloon were taken into a refrigerated environment, the balloon would contract as the gas molecules cool and move at a lower velocity. 3. IDEAL VS. REAL GASES a) Ideal Gases Most discussions of gases assume that they behave ideally with the following characteristics: 1. Gas particles have no attraction for one another, regardless of the conditions; therefore, increasing pressure on a gas or cooling the temperature will never cause the gas particles to attract to the point of turning into a liquid. 2. Gas particles have no volume. 3. Gas laws apply to gases under all conditions of temperature and pressure. b) Real Gases Real gases exhibit the following behavior: 1. Under high pressure/low temperature, gas particles can be compressed enough to attract and liquefy.

3 4. GAS LAWS 2. Gas particles do have a definite volume due to the size of their atoms and the length of the bonds between the atoms. a) Standard Temperature and Pressure (STP) Standard Temperature: 0 C/273 K K = C Standard Pressure: 1 atmosphere 1 atm. 760 mm Hg. 760 torr kilopascals kpa b) Conversions between pressure units are based on standard pressure equalities Ex. A gas is at 5.0 atm pressure. What is its pressure in mm Hg? 5.0 atm x 760 mm Hg = 3,800 mm Hg 1 atm A gas is at 350 torr pressure. What is its pressure in kilopascals? 350 torr x kpa = 47 kpa 760 torr c) Dalton s Law of Partial Pressures The total pressure of a mixture of gases is equal to the sum of the pressures of all the gases in the mixture. P TOTAL = P 1 + P 2 + P 3 etc. Ex. A mixture of gas consists of O 2 at 0.15 atm; N 2 at 0.12 atm; and CO 2 at 0.7 atm. What is the total pressure of the mixture? P TOTAL = 0.15 atm atm atm = 0.97 atm d) Boyle s Law Robert Boyle ( ) determined that the pressure of a gas (held at constant temperature) varies inversely with the volume of the gas. In other words, for a gas held at constant temperature, as pressure increases, volume decreases; likewise, as pressure decreases, volume increases. P 1 V 1 = P 2 V 2 where P is pressure and V is volume Ex. A sample of gas (at constant temperature) occupies 3.0 L at 1.5 atm. What is the volume of the gas if its pressure is decreased to 0.5 atm? This is a Boyle s law problem since volume and pressure are the units given. P 1 and V 1 represent the gas at its original condition P 1 = 1.5 atm V 1 = 3.0 L P 2 and V 2 represent the gas at its new condition P 2 = 0.5 atm V 2 =? (1.5 atm)(3.0 L) = (0.5 atm) V 2 (1.5 atm)(3.0 L) = 9 L V atm

4 e) Charles s Law Jacques Charles ( ) determined that at constant pressure, the volume of a gas is directly proportional to its temperature in Kelvin. In other words, for a gas held at constant pressure, as temperature increases, volume increases; likewise, as temperature decreases, volume decreases. V 1 = V 2 Where V is volume and T is temperature in K T 1 T 2 NOTE: When using gas equations involving temperature, temperature must always be converted to K before calculating. Ex. A sample of N 2 gas (at constant pressure) occupies of volume of 500 ml at 25 C. What is its volume if the temperature is increased to 50 C? This is a Charles s law problem since volume and temperature are the units given. V 1 and T 1 represent the gas at its original condition V 1 = 500 ml. T 1 = 25 C = 298 K V 2 and T 2 represent the gas at is new condition V 2 =? T 2 = 50 C = 323 K 500 ml = V K 323 K V 2 = (500 ml)(323 K) = 542 ml V K f) Combined Gas Law The combined gas law relates pressure, volume and temperature into one equation. P 1 V 1 = P 2 V 2 Where P is pressure, V is volume and T is temperature in K T 1 T 2 Ex. A gas at 110 kpa and 30 C fills a flexible container with an initial volume of 2.0 L. If the temperature is raised to 80 C and the pressure is increased to 440 kpa, what is the new volume? This is a combined gas law problem since the units given are pressure, temperature and volume. P 1, V 1 and T 1 represent the gas at its original conditions P 1 = 110 kpa V 1 = 2.0 L T 1 = 30 C = 303 K P 2, V 2 and T 2 represent the gas at its new condition (110kPa)(2.0L) = (440kPa)V K 353 K P 2 = 440 kpa V 2 =? T 2 = 80 C = 353 K

5 V 2 (440kPa)(303 K) = (110kPa)(2.0L)(353 K) V 2 = (110kPa)(2.0L)(353 K) V 2 = 0.58L (440kPa)(303 K) g) Ideal Gas Law The ideal gas law is the most comprehensive law because it relates pressure, volume, temperature and moles of gas into one equation. The ideal gas law includes the gas constant R which varies depending on the pressure unit given in the problem. Gas Constants with Corresponding Units: R = 8.31 L x kpa use this constant if pressure unit given is kpa K x moles R = L x atm use this constant if pressure unit given is atm K x moles R = 62.4 L x mmhg use this constant if pressure unit given is mmhg or torr K x moles NOTE: You are not expected to memorize the gas constants. R values will be given; however, you will need to know which constant to use for a given problem. Also, if volume units are given in ml, you will need to convert to Liters before solving since that is the volume unit in the gas constant. 1 Liter = 1,000 ml Liters ml mulnply by 1,000 (move decimal 3 places to right) ml Liters divide by 1,000 (move decimal 3 places to left) PV = nrt Where P is pressure, V is volume, n is moles of gas, R is the gas constant and T is temperature in K Ex. How many moles of hydrogen gas are there in 4.0L at a pressure of 120kPa and standard temperature? This is an ideal gas law problem since it asks for moles (the only gas equation that solves for moles) and volume, pressure and temperature are given. P = 120kPa V = 4.0L n =? R = 8.31 L x kpa K x moles T = 273 K (standard temperature) Rearranging the ideal gas equation to solve for n gives: n = PV n = (120kPa)(4.0L) RT (8.31 L kpa) x (273 K) K moles n = 0.21 moles

6 5. INTERMOLECULAR FORCES a) Intermolecular forces are the force of attraction between molecules. The strength of this force depends on the type of molecules in a substance. b) Molecules with a weaker force do not require as much energy to loosen the attraction between the molecules and change their state. Thus, they have lower melting and boiling points. c) Molecules with a stronger force require more energy to loosen the attraction between the molecules and change their state. Thus, they have higher melting and boiling points. d) London/Dispersion Forces: Weakest intermolecular force between non-polar molecules. These molecules do not have δ + or δ - sides, so they are not attracted to one another as strongly. Any attraction is due to momentary shifts in the position of shared electrons. Ex. F-F Br-Br O=C=O CCl 4 These substances have the lowest melting and boiling points. e) Dipole Interactions: Medium strength intermolecular force between polar molecules. These molecules have δ + and δ - sides and the partial positive side of one molecule is attracted to the partial negative side of the neighboring molecule, so the molecules are more strongly attracted to one another. Ex. HCl PBr 3 CH 3 OH These substances have medium melting and boiling points. f) Hydrogen Bonds: Strongest intermolecular force between super polar molecules defined as those containing Hydrogen bound to the most electronegative atoms on the periodic table (Nitrogen, Oxygen and Fluorine). The partial charges in these bonds are more dramatic and molecules containing these bonds have stronger partial positive and partial negative sides, resulting in the greatest attraction between molecules. Ex. NH 3 H 2 O HF These substances have the highest melting and boiling points. 6. PHASE CHANGES a) Melting: changing a substance from a solid to a liquid. Melting is an endothermic process, meaning energy must be put into a substance in order to melt it. H fusion is the value used to calculate the energy required for melting a given substance and is always a positive value. Melting Point: temperature at which a given substance melts. MP WATER = 0 C b) Freezing: changing a substance from a liquid to a solid. Freezing (or solidification) is an exothermic process, meaning energy must be removed from a substance in order to freeze it. H solidification is the value used to calculate the energy that must be withdrawn from a given substance for freezing and is always a negative value. Freezing Point: temperature at which a given substance freezes. FP WATER = 0 C

7 c) Vaporization: changing a substance from a liquid to a gas (vapor). Vaporization is an endothermic process, meaning energy must be put into a substance in order to vaporize it. H vaporization is the value used to calculate the energy required for vaporizing a given substance and is always a positive value. Boiling Point: temperature at which a given substance turns from a liquid to a gas. BP WATER = 100 C d) Condensation: changing a substance from a gas to a liquid. Condensation is an exothermic process, meaning energy must be removed from a substance in order to condense it. H condensation is the value used to calculate the energy that must be withdrawn from a given substance for condensation and is always a negative value. Condensation Point: temperature at which a given substance turns from a gas to a liquid. CP WATER = 100 C e) Sublimation: changing a substance from a solid directly to a gas. Ex. Dry ice is CO 2 in solid form. When dry ice is exposed to warmer air, it changes directly to vapor instead of changing first to a liquid, then to a gas as most substances do. f) Deposition: changing a substance from a gas directly to a solid. Ex. Frost is the result of water vapor changing directly to solid ice crystals. 7. ENERGY CALCULATIONS a) Energy (Q) is most commonly expressed in units of calories, Joules(J) or kilojoules (kj). 1 calorie = 4.18 J = kj b) Phase Changes in a Substance 1. Phase changes involve the input or removal of energy, which is calculated using H values for a given substance. 2. H values are commonly expressed as energy unit/mole or energy unit/gram of substance and represent the amount of energy that must be put into or removed from that substance in order to change its phase. 3. Energy calculations for phase changes are determined as: Q = H x quantity of substance 4. H Values for Water: H fusion 6.01 kj/mole kj/gram 334 J/gram H solidification kj/mole kj/gram -334 J/gram H vaporization 40.7 kj/mole kj/gram 2261 J/gram H condensation kj/mole kj/gram J/gram 5. Example of phase change energy calculations a. How much energy is required to melt 5 moles (90 grams) of water? Q = H fusion x quantity Q = 6.01 kj/mole x 5 moles = kj Q = kj/gram x 90 grams = kj

8 b. How much energy is required to vaporize 5 moles (90 grams) of water? Q = H vaporization x quantity Q = 40.7 kj/mole x 5 moles = kj Q = kj/gram x 90 grams = kj c. How much energy is released when 5 moles (90 grams) of water freezes? Q = H solidification x quantity Q = kj/mole x 5 moles = kj Q = kj/gram x 90 grams = kj d. How much energy is released when 5 moles (90 grams) of water condenses? Q = H condensation x quantity Q = kj/mole x 5 moles = kj Q = kj/gram x 90 grams = kj c) Temperature Changes in a Substance 1. Heating is an endothermic process (heat in) and energy is required to raise the temperature of a substance. 2. Cooling is an exothermic process (heat out) and energy must be released to lower the temperature of a substance. 3. Specific heat(c) is defined as the amount of heat needed to increase the temperature of 1 gram of a substance 1 C. Energy units for specific heat values are most often in Joules or calories. Specific heat of frozen water (ice): 2.1 J/g C or 0.5 cal/g C Specific heat of liquid water: 4.18 J/g C or 1.00 cal/g C Specific heat of gaseous water (steam): 1.9 J/g C or 0.45 cal/g C 4. T is the change in temperature of a substance. It can be calculated as: T = T FINAL - T INITIAL In the case of heating, T will be positive since the final temperature will be higher than the starting/initial temperature. In the case of cooling, T will be negative since the final temperature will be lower than the starting/initial temperature. 5. Energy calculations for temperature changes are determined as: Q = mc T Where Q is energy, m is mass, c is specific heat and T is change in temperature 6. Example of temperature change calculations a. How much energy is required to heat 100 grams of iron from 25 C to 300 C? The specific heat of iron is 0.46 J/g C Q = (100g.)(0.46 J/g C)(300 C 25 C) Q = 12,650 J b. How much energy is released in cooling 100 grams of aluminum from 250 C to 25 C? The specific heat of aluminum is 0.90 J/g C Q = (100g.)(0.90 J/g C)(25 C C) Q = -20,250 J

9 d) Calculating Total Energy 1. Total energy calculations must account for the energy required to change the temperature of a substance and the energy involved in any phase changes that the substance undergoes in the process. 2. Example How much energy is required in heating a 50 gram sample of water from -15 C to 150 C? First, you must determine if the substance will undergo phase changes. Since water melts at 0 C and boils at 100 C, it will undergo 2 phase changes in the heating process. a. Melting: Q = H fusion x mass Q = 334 J/gram x 50g. = 16,700 J b. Boiling: Q = H vaporization x mass Q = 2261 J/gram x 50g. = 113,050 J Next, you must calculate the amount of energy required to raise the temperature of water. Water is more difficult because its specific heat(c) varies with the phase. a. -15 C 0 C solid phase (ice) where c is 2.1 J/g C T= 0 C-(-15 C) = 15 C Q = mc T Q = (50g.)(2.1 J/g C)(15 C) = 1,575 J b. 0 C 100 C liquid phase where c is 4.18 J/g C T= 100 C- 0 C = 100 C Q = mc T Q = (50g.)(4.18 J/g C)(100 C) = 20,900 J c. 100 C 150 C gaseous phase (steam) where c is 1.9 J/g C T= 150 C- 100 C = 50 C Q = mc T Q = (50g.)(1.9 J/g C)(50 C) = 4,750 J d. Total Energy = 16,700 J + 113,050 J + 1,575 J + 20,900 J + 4,750 J = 156,975 J An easier example: How much energy is required in heating 100 grams of ethanol from -130 C to 90 C? Melting Point: -114 C H Fusion : 107 J/gram Boiling Point: 78 C H Vaporization: 839 J/gram Specific Heat: 2.4 J/g C a. Phase Changes: ethanol will melt and boil in this temperature range Q = 107 J/gram x 100g. = 10,700 J Energy required to melt Q = 839 J/gram x 100g. = 83,900 J Energy required to vaporize b. Temperature Change: ethanol has 1 specific heat, so one calculation can be performed for the energy required to change the temperature from -130 C to 90 C T = 90 C (-130 C) = 220 C Q = (100g.)(2.4 J/g C)(220 C) = 52,800 J Energy required to raise temp. c. Total Energy = 10,700 J + 83,900 J + 52,800 J = 147,400 J

10 8. HEATING AND COOLING CURVES a) Graphical representations of the temperature and phase changes associated with heating or cooling a given substance. b) Heating Curve for Water c) Cooling Curve for Water d) Kinetic and Potential Energy 1. Kinetic energy is the energy atoms or molecules possess due to their motion. Temperature is a measure of the average kinetic energy of a substance. Molecules of a substance at high temperatures move faster than the molecules of a substance at low temperatures. 2. Chemical potential energy is energy that is stored in atoms and the bonds between atoms and is released during chemical reactions in which bonds are broken and then reformed. Potential energy can be converted to kinetic energy in the form of heat in an exothermic reaction. a. When gasoline is burned in a car engine, the potential energy stored in the bonds of the gasoline molecules is released mainly as heat. The heat

11 drives the engine to turn the wheels and consequently, converts the potential energy to kinetic energy in the car s movement. b. The human body is fueled by food and potential energy is stored in the bonds between atoms in carbohydrate molecules. When someone runs, combustion of carbohydrate molecules (cellular respiration) results in the release of energy which drives muscle movement and ultimately, converts the potential energy to kinetic energy in the person s movement. 3. Potential and Kinetic Energy Changes During Heating and Cooling a. Heating 1) Kinetic energy increases only when the temperature of the substance increases. On a heating curve, KE increases correspond to segments in which the line is rising. 2) There is no change in kinetic energy during a phase change, thus no temperature change occurs in the substance. 3) Potential energy increases only during phase changes as the substance melts or vaporizes. On a heating curve, PE increases correspond to the segments in which the line is flat. b. Cooling 1) Kinetic energy decreases only when the temperature of the substance decreases. On a cooling curve, KE decreases correspond to segments in which the line is dropping. 2) There is no change in kinetic energy during a phase change, thus no temperature change occurs in the substance. 3) Potential energy decreases only during phase changes as the substance freezes or condenses. On a cooling curve, PE decreases correspond to the segments in which the line is flat. c. Summary Temp. Rising PE same/ke Temp. Dropping PE same/ke Melting PE /KE same Freezing PE /KE same Vaporization PE /KE same Condensing PE /KE same 9. VAPOR PRESSURE a) As heat is applied to a liquid substance, molecules at the surface will begin to evaporate as the input of energy loosens the intermolecular forces between molecules. Evaporating gaseous molecules will begin to exert an upward force. b) Vapor pressure is the pressure exerted by a gas over a liquid. c) A substance boils when its vapor pressure is equal to or greater than the external pressure on the substance. d) Water in an open container boils at 100 C at standard air pressure. 100 C is the temperature water much reach before the pressure exerted by evaporating molecules equals or exceeds standard air pressure.

12 e) Water in a vacuum (lower air pressure) boils at a temperature much below 100 C since its external pressure is lower. Vapor pressure exceeds external pressure at a lower temperature. f) Phase diagrams are graphs showing the conditions at which a given substance exists as a solid, liquid or vapor at varying temperatures and pressures. 1. Critical Point: temperature and pressure at which a substance can no longer exist as a liquid. The critical point for water is 218 atm pressure and 374 C. 2. Triple Point: temperature and pressure at which a substance can exist as a solid, liquid and gas. The triple point for water is atm pressure and 0.01 C. g) Phase Diagram for Water:

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43 Unit 8 Note Quiz Questions Unit 8.1: States of Matter and IMFs 1. a a 4. a 6. a 5. a 7. a 8. a

44 9. 10 Unit 8.2: Thermodynamics & Thermochemistry 1. a 4. How many kj is 3.9 x 10 3 cal? a kj b kj c kj d. 933 kj a 3. a

45 6. Use the following diagram to answer the next four (4) questions: 7. Which shows the activation energy of the reaction? 8. Which letter is the enthalpy of the reaction? 9. Which letter shows the potential energy of CH 4 and O 2? 10. Is the reaction endo/exothermic

46 Unit 8.3: Energy Changes 1. A 2. A 3. A 4. A 5. A 6. A

47 7. A 8. A 9. A 10. A

48 Unit 8.4 : Phase Change 1. A tire has an air pressure of 109kPa, this is equal to how many atm? a atm b atm c atm d. 109 atm 2. A bag of potato chips is sealed at sea level, when the atmospheric pressure is 761.3mmHg. What is the pressure in the bag in Pa? a. 1 x 10 3 Pa b x 10 5 Pa c x 10 5 Pa d Pa 3. a 4. a

49 5. a 7. a 6. a

50 8. a Use the following diagram for the next two (2) questions: 9. What is the name of point 2? a. Triple point b. Critical point c. Melting point d. Boiling point 10.

51 Unit 8.5: HAS NO NOTE QUIZ Unit 8.6: Gas Laws (pt 1) 1. A 8. A 2. A 9. A 3. A 4. A 5. A 6. A 10. A 7. A Unit 8.7: HAS NO NOTE QUIZ