Unit 3: Phases of Matter

Transcription

1 Unit 3: Phases of Matter Student Name: Key Class Period: Page 1 of 63

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3 Unit 3 Vocabulary: 1. Avogadro s Hypothesis: Equal volumes of two ideal gases under the same conditions of temperature and pressure will contain equal number of molecules. 2. Boiling: the transition of a liquid into a gas at the vaporization point. 3. Condensing: the transition of a gas into a liquid at the vaporization point. 4. Equilibrium: the condition that exists when the rates of two opposing phase changes are equal. 5. Evaporating: the transition of the surface molecules of a liquid into a gas below the boiling point. 6. Freezing: the transition of a liquid into a solid at the solidification point. 7. Gas: the phase of matter with complete dissociation of particles of matter with a great average distance between the particles compared to the particle size. Negligible attractive forces between particles. 8. Heat of Fusion: energy required to liquefy a gram of solid at melting point. 9. Heat of vaporization: energy required to gasify a gram of liquid at its vaporization point. 10. Ideal Gas: a gas in which the molecules are infinitely small and far apart, the molecules travel with straight-line motion, all collisions are elastic (no net energy loss), no attractive forces between gas molecules and speed of molecules is directly proportional to Kelvin. Gasses act most ideal at high temperature and low pressures. 11. Liquid: a phase of matter characterized as loosely organized yet held together by intermolecular or ionic attractive forces. 12. Melting: the transition of a solid into a liquid at the liquefaction point. 13. Pressure: Force exerted over an area. Page 3 of 63

4 Unit 2 Vocabulary (Cont d) 14. Solid: a phase with matter arranged in regular geometric patterns (lattices) with only vibrational motion, and no relative motion. 15. Specific Heat: the energy required to heat one gram of a substance one Kelvin. 16. Sublimation: the transition of a solid directly into a gas, skipping liquefaction. 17. Vapor Pressure: the pressure exerted by vapor in a vapor-liquid mixture closed system at equilibrium. 18. Vapor-Liquid Equilibrium: a system where the rate of evaporation equals the rate of condensing. Page 4 of 63

5 Unit 3 Homework Assignments: Assignment: Date: Due: Page 5 of 63

6 Topic: Phases and Phase Change Objective: How does closeness of matter affect their properties? Positive and negative charges attract each other. Attractive forces between molecules are called intermolecular attractive forces (IMAF). The strength of these forces determines what phase of matter a substance is in at a given temperature. Substances having weak attractive forces (London dispersion) between their molecules tend to be gasses at room temperature, and substances with strong attractive forces (ionic) tend to be solids at room temperature. Phases of matter are simply stages of attraction. Gases are composed of molecules with little or no attractive forces, allowing the molecules to fly freely past each other. Liquids are made of molecules with stronger attractive forces, allowing molecules to flow past each other, but still stay together. Solids are made of molecules or ions with strong attractive forces, which lock the molecules into a crystal lattice where the particles are free to vibrate, but they cannot move relative to each other. Page 6 of 63

7 Topic: Properties of Solids Objective: How does closeness of solids affect their properties? Properties of Solids: 1. Molecules, atoms, or ions arranged into a regular, geometric pattern called a crystal lattice. 2. Molecules, atoms, or ions vibrate in place. They do NOT move relative to each other. 3. Solids have a definite shape and definite volume. Page 7 of 63

8 Topic: Properties of Liquids Objective: How does closeness of liquids affect their properties? Properties of Liquids: 1. Molecules, atoms, or ions may flow past each other. 2. Viscosity is a resistance to flow due to intermolecular attractive forces (IMAF). 3. Viscosity increases as temperature decreases and IMAF strength increases. 4. Liquid molecules near a gaseous surface may escape IMAF and enter the vapor phase at temperatures below the boiling point in a process called evaporation. (This is the source for vapor pressure) 5. Liquids take on the shape of the container they are in and have a definite volume. (In the absence of gravity or a container, liquids form a perfect sphere.) Page 8 of 63

9 Topic: Properties of Gases Objective: How does closeness of gases affect their properties? Properties of Gases: 1. Gas molecules are extremely far apart compared to the size of the molecules. 2. Gas molecules travel in a straight line until they collide with something. 3. Collisions are elastic, meaning they don t lose kinetic energy in the collision. 4. Gas molecules move faster when it is hotter (higher Kelvin). 5. The gas phase is the only phase that is affected by changes in pressure. 6. Gases spread out to take the shape and entire volume of whatever container they are placed in. Page 9 of 63

10 Topic: Phase Change Diagram Objective: How do the changes of phase relate to each other? Phase Change Diagram: Phase Symbols: Formula (s) = solid phase Formula (l) = liquid phase Formula (g) = gaseous phase Page 10 of 63

11 Topic: Phase Equilibrium Objective: What processes are at work during phase equilibrium? Phase equilibrium: Boiling and condensing both occur at the boiling point (100.0 C for pure water), freezing and melting both occur at the melting point (0.0 C for pure water). During the phase change, both phases exist at equilibrium. Equilibrium: a condition where the rates of opposing changes are equal. At equilibrium temperature a substance at the melting/freezing point is melting at the SAME rate that it is freezing. Equilibrium is the reason why water melts ABOVE 0 C and freezes BELOW 0 C. AT 0 C a mixture of water and ice will NOT change either direction. If you have a sealed flask containing 20.0 g of water ice and 20.0 g of liquid water and keep it at EXACTLY 0.0 C, there will still 20.0 g of water ice and 20.0 g of liquid water for as long as the temperature remains 0.0 C. Page 11 of 63

12 Topic: Sublimation Objective: What occurs during the solid to gaseous phase change? Sublimation: During sublimation the attractive forces between solid molecules are so weak that heating a solid causes it to go directly into the gas phase. There are two common substances that undergo this change; CO 2(s), known as dry ice, and I 2(s). Water also undergoes sublimation. Ice cubes shrink if left alone in a freezer. The opposite of sublimation is the process of DEPOSITION. This can be witnessed by the intricate patterns of ice found on window surfaces on those cold, dry upstate New York mornings when water vapor in the air contacts the cold glass and the gaseous water phase changes directly into solid water. Page 12 of 63

13 Notes page: Page 13 of 63

14 Topic: Heating Curve Diagrams Objective: What steps occur during constant heating of a substance? Page 14 of 63

15 Topic: Heating Curve Diagrams Objective: What steps occur during constant heating of a substance? Phase Change Diagrams (Heating/Cooling Curve) Page 14: When applying CONSTANT HEAT to a solid (point A), the temperature of the solid increases (segment AB) until the melting point (point B) is reached, for example of water ice at 0 C or 273 K. The solid then absorbs potential energy as it transitions from solid to liquid, or melts (segment BC). During the melting process, the temperatures (under CONSTANT HEAT) of both solid and liquid water BOTH remain constant until the ENTIRE solid has become liquid (point C). Once the substance is completely in the liquid phase, the temperature will increase (segment CD). Once the boiling temperature (point D) has been reached, for liquid water at 100 C or 373 K, the temperatures (under CONSTANT HEAT) of both liquid water and water vapor will remain constant in a process called boiling, or vaporization (segment DE). Once ALL the water is in the gas phase (point E), the temperature (under CONSTANT HEAT) of the water vapor will increase (segment EF). Both phase changes are ENDOTHERMIC here, because heat is a CONSTANT application at the same RATE. Page 15 of 63

16 Topic: Heating Curve for Water Objective: What steps occur during constant heating of water? Heating Curve for Water: The graph below shows a sample of water initially in the solid phase as it constantly heated from 200 K to 420 K Page 16 of 63

17 Topic: Heating Curve for Water Objective: What steps occur during constant heating of water? Heating Curve for Water Page 16: Note: The melting point of water is 0 C (273 K) and the boiling point of water is 100 C (373 K). 1. How many minutes pass from the first appearance of the liquid phase until the substance is entirely in the gas phases? a. The liquid first appears (point B) at 4 minutes, and the liquid has entirely transitioned to the gas phase (point E) at 34 minutes. The time from first liquid to last liquid is therefore: 34 mins - 4 mins = 30 mins 2. How many minutes will it take for this substance to completely undergo melting? a. The solid starts to melt (point B) at 4 minutes, and is entirely transitioned to the liquid phase (point C) at 8 minutes. The time from last solid to all liquid is therefore: 8 mins - 4 mins = 4 mins 3. For how many minutes is the water completely in a solid crystal lattice phase? a. H 2 0 is a crystal lattice in the solid phase (point A) from 0 minutes until the melting temperature (point B) is reached at 4 minutes. The time is therefore: 4 mins - 0 mins = 4 mins Page 17 of 63

18 4. Which line segment represents when H 2 0 is both in the liquid AND gas phases? a. H 2 0 is in both the liquid and the gas phase during boiling at 20 minutes (point D) to 34 minutes (point E). The time is therefore: 34 mins - 20 mins = 14 mins 5. For how many minutes is the water completely in a phase WITHOUT a definite shape or volume? a. The gas phase has no definite shape or volume. The H 2 O is completely in the gas phase from 34 minutes (point E) until the end of the heating curve (point F). The time therefore: 36 mins - 34 mins = 2 min 6. How many minutes will it take for the water to completely boil away, once the boiling point has been reached? a. H 2 O reaches the boiling point in 20 minutes (point D), and complete boiling takes until 34 minutes (point E). The time therefore: 34 mins - 20 mins = 14 mins Watch the Phase Change Diagram for water Page 18 of 63

19 Student name: Class Period: Please carefully remove this page from your packet to hand in. Phase and Phase Change homework Circle your answer for multiple choice questions. 1. Which sample below has molecules that flow past each other, but are still attracted to each other? a) C 6 H 12 O 6(s) b) C 6 H 12 O 6(l) c) C 6 H 12 O 6(g) 2. NaCl (l) is stated to be boiling when it undergoes a phase change to: a) Solid b) Liquid c) Gas d) Aqueous 3. What phase of matter has a definite volume, but not definite shape? a) Solid b) Liquid c) Gas 4. Motor oil viscosities are tested at different temperatures to see how effective the oil is at preventing friction. As temperature increases, what happens to the viscosity of the motor oil? a) Increases b) Decreases c) Stays the same 5. Explain your answer to question #4 in terms of molecular motion and average kinetic energy. As molecular motion becomes more rapid, kinetic energy increases, causing increased temperature 6. Which of the following phase changes are exothermic? a) LiF (s) LiF (l) c) LiF (s) LiF (g) b) LiF (l) LiF (s) d) LiF (l) LiF (g) Page 19 of 63

20 7. Explain why your choice for question #6 IS exothermic. Liquid has higher kinetic energy than solid, so energy is given off from the liquid to solid stage 8. Explain why the other three options for question #6 are NOT exothermic. Solid liquid absorbs (endo); liquid gas absorbs (endo) 9. Which phase of matter has the strongest attractive forces? a) Solid b) Liquid c) Gas 10. Explain why attractive forces between particles of a substance determine which phase a substance is in at a given temperature. Greater attractive forces in solids; decreases to liquid and decrease even more to gas phase A beaker containing g of substance X in the solid phase is heated over a hot plate at the CONSTANT rate of J/minute. The results were logged in the table below. Page 20 of 63

21 Student name: Class Period: Please carefully remove this page from your packet to hand in. Page 21 of 63

22 Answer the following questions based on the chart on pg Draw a graph (page 21) of the data on pg. 20, placing temperature on the Y-axis and time on the X-axis. Label ON THE GRAPH the three phases, and the positions where phase change occurs. 12. At what time does liquid substance X first appear? _4 mins_ 13. When is substance X initially completely in the liquid phase? _8 mins_ 14. At what time does gas substance X first appear? _14 mins_ 15. When is substance X initially completely in the gas phase? 21 mins 16. How long does it take for substance X to transition from the first liquid present to the total gas phase? _17 mins_ (21 mins - 4 mins) 17. How long does substance X exist completely in the liquid phase? _6 mins_ (14 mins - 8 mins) 18. How long does substance X exist completely as a crystal lattice? _4 mins_ (4 mins - 0 mins) 19. How long does substance X exist in a phase with no definite shape or volume? _2 mins_ (23 mins - 21 mins) Page 22 of 63

23 Student name: Class Period: Please carefully remove this page from your packet to hand in. Phase and Phase Change homework 20. Describe the changes, if any, of the VISCOSITY of substance X during heating in the liquid phase. As temperature increases, viscosity will decrease 21. Describe the changes, if any, in the KINETIC ENERGY of substance X between 4 minutes and 8 minutes. As there is NO increase in temperature there is no change in the average KE of the water 22. Describe the changes, if any, in the POTENTIAL ENERGY of substance X between 4 minutes and 8 minutes. As there is NO increase in temperature (avg KE), then the added energy increases the PE of the water 23. Water melts at 273 K and boils at 373 K. Circle and describe which substance, water or substance X, has the stronger intermolecular forces. X has stronger attractive forces as it has a higher boiling point than does water Page 23 of 63

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26 Topic: Energy for Phase Change Objective: How do we quantify the energy required to phase change? Melting and boiling are both endothermic because energy has to be constantly added in order for the change to be completed. Condensing and freezing are exothermic because energy has to be constantly removed in order for the change to be completed. Heat of Fusion: Heat of Fusion: the amount of heat energy needed to melt one gram of a solid at the substance s melting point. The heat of fusion for water (H 2 O) is 334 J/g as found in Reference Table B. To calculate the amount of heat energy (in joules) required to melt a sample, use the equation (Reference Table T) of: q = mh f (H f = Heat of Fusion for a substance) Heat of Fusion is the property used to defrost a car s windshield. Excess engine coolant heat is piped into the defroster to continually ADD heat (q) to the air that flows onto the inside of the windshield, adding heat to the ice on the windshield until all the ice has melted (reached the Heat of Fusion) for the ice. Page 26 of 63

27 Heat of Fusion may be used to determine the energy (J) required to freeze a liquid at the freezing point (same temperature as melting point). With freezing you REMOVE heat (q) from liquid. Water placed into ice cube trays in the freezer has the heat removed by the freezer components. This removed heat is transferred to the coils at the back of the refrigerator, which is why the back of a refrigerator is warmer, as heat is being removed from the inside of the refrigerator. i. How many joules of energy does take to completely melt grams of water at water s melting point? a. q = mh f g x 334 J/g = J or 3.34 x 10 4 J ii. How many joules of energy does it take to freeze 50.0 g of water at water s freezing point? a. q = mh f 50.0 g x 334 J/g = J or 1.67 x 10 4 J Watch Bozeman Science Endothermic and Exothermic video Page 27 of 63

28 Topic: Energy for Phase Change Objective: How do we quantify the energy required to phase change? Heat of vaporization: Heat of Vaporization: the amount of heat energy needed to completely boil one gram of liquid at the substance s boiling point. The heat of vaporization for water (H 2 O) is 2260 J/g as found in Reference Table B. To calculate the amount of heat energy (in joules) required to melt a sample, use the equation (Reference Table T) of: q = mh v (H v = Heat of Vaporization for a substance) Heat of Vaporization is the property for drying a substance that was in solution. If you heat a solution to remove the water, you are left with the dry solute. Heat of Vaporization removes the water. Heat of vaporization may also be used for determining how many joules are released when a gas condenses back into a liquid. Use the same calculation. Which causes greater danger, having boiling water at 100 C splashed on you, or steam at 100 C? The water is hot (334 J/g) and your skin absorbs that heat. The steam was also hot (2260 J/g), and as the steam touches your skin, it releases more (in J/g) heat of vaporization, causing greater injury. Page 28 of 63

29 Heat of vaporization is absorbed from your skin as your sweat evaporates, removing heat from your skin (and blood), leaving your body cooler. i. How many joules of energy are required to completely boil away g of water at water s boiling point? a. q = mh v g x 2260 J/g = J or 2.26 x 10 5 J ii. How many joules of energy does it take to completely condense 50.0 g of water at water s boiling point? a. q = mh v 50.0 g x 2260 J/g = J or 1.13 x 10 5 J Page 29 of 63

31 Student name: Class Period: Please carefully remove this page from your packet to hand in. Energy Required for Phase Change homework Circle your answer for the multiple choice questions. 1. Which of the following phase changes requires only Heat of Fusion? a) H 2 O (s) H 2 O (g) b) H 2 O (g) H 2 O (l) c) H 2 O (l) H 2 O (g) d) H 2 O (s) H2O (l) 2. Which of the following phase change is endothermic? a) H 2 O (s) H 2 O (l) b) H 2 O (g) H 2 O (l) c) H 2 O (l) H 2 O (s) d) H 2 O (g) H 2 O (s) A mixture of 50.0 g of water ice (H 2 O (s) ) and 30.0 g water (H 2 O (l) ) is in a sealed flask at 0.0 C. 3. What will happen to the amount of ice in the flask if the mixture is left alone at a constant 0.0 C? a) Increase b) Decrease c) Stay the same 4. Explain your answer for question #3. At the EXACT melting point liquid and solid are in equilibrium Page 31 of 63

32 5. What will happen to the amount of ice in the flask if the temperature of the flask is lowered to -10 C? a) Increase b) Decrease c) Stay the same 6. Explain your answer for question #5. Below the melting (freezing) point the equilibrium shifts towards more solid than liquid Calculate the energy (in joules) required for the following. Show your setup and calculations! 7. Melt 20.0 g of H 2 O (s) at 0 C. 8. Boil 30.0 g H 2 O (l) at 100 C 9. Freeze g of H 2 O (l) at 0 C 10. Boil 50.0 g H 2 O (l) at 100 C Page 32 of 63

33 Topic: Gases and Pressure Objective: What assumptions are needed to set property standards? Kinetic-Molecular Theory (Ideal Gas Law): 1. Gases are made of molecules that are extremely small and very far apart. This allows gases to be easily compressed. 2. Gas molecules move in a straight-line motion until they encounter another molecule. 3. Any collision a gas molecule makes will be elastic with negligible loss of energy. Each time a gas molecule collides against an obstacle, it bounces off with the same energy it had before the collision. 4. There are negligible intermolecular attractive forces between ideal gas molecules. All matter has attractive forces, but distances between gas molecules are so great that attractive forces may be ignored. Gases with London dispersion forces behave the most like an ideal gas should, and gases with hydrogen bond attractions behave the least like an ideal gas should. 5. The average velocity of gas molecules is directly proportional to the Kelvin temperature. Velocity increases with increased temperature. Page 33 of 63

34 Topic: Ideal Gas Behavior Objective: What conditions are assumed for gas molecule behavior? Ideal Gas Behavior: Under conditions of HIGH temperature and LOW pressure, molecules behave more like an ideal gas. The smaller the gas molecule, the closer to ideal it will behave. Therefore hydrogen (H 2 ) and helium (He 2 ) are the gases that are the most ideal. Avogadro s Hypothesis: Avogadro s Hypothesis states if the conditions of temperature and pressure are the same, then equal numbers of gas molecules will occupy equal volumes. This is because gas molecules will spread out to an equal degree at the same temperature and pressure. We can apply these same math properties equally to all gases we treat as ideal. Page 34 of 63

35 Topic: Ideal Gas Behavior Objective: What conditions are assumed for gas molecule behavior? Deviations from the Ideal Gas Law - Where and Why? The Ideal Gas law works well under STANDARD CONDITIONS of temperature and pressure (STP) as seen in Reference Table A. o Standard temperature is 0 C or 273 K. (K is more significant) o Standard pressure is atmosphere (atm) or kpa or mmhg (torr). Pressure: Pressure is force exerted over an area. The gases in an atmosphere exert pressure due to gravity s pull. On Earth the atmospheric pressure is about 14.7 pounds per square inch (psi). Therefore every square inch of surface area on Earth (including you) has about 14.7 pounds of force pressing on it when at sea level. Organisms born on Earth are born into the environment already pre-pressurized to withstand the atmospheric force. As you ascend (climb) in the atmosphere, there is less atmospheric mass above you, and therefore the atmospheric pressure drops. This causes body pressures to equalize some (ears pop, feet and hands swell slightly). Page 35 of 63

36 Topic: Vapor Pressure Objective: How does evaporation affect the liquid-gas environment? Pressure is measured in atmospheres (atm), kilopascals (kpa), or millimeters of mercury (mmhg). Pressure Conversions: 14.7 psi = 1.00 atm = kpa = mmhg Vapor Pressure: Vapor Pressure is the pressure exerted by a liquid s vapor in a sealed container in a vapor-liquid equilibrium at a given temperature. The vapor pressure of a liquid is NOT dependent on the mass or volume of the liquid. Vapor pressures are found on Reference Table H. The stronger the attractive intermolecular forces between liquid molecules, the lower the vapor pressure will be. Substances that have high vapor pressures evaporate quickly. Gasoline is a mixture of high vapor pressure liquids; it may be seen to evaporate even as it fills your tank. Alcohols and acetone have high vapor pressures, and therefore evaporate quickly. High vapor pressure substances are known as volatile, and many are also flammable, or easily burn or explode. Page 36 of 63

37 Topic: Boiling Point Objective: How do temperature and pressure affect boiling? Boiling Point: Boiling Point is the temperature at which a liquid s vapor pressure equals the pressure exerted on the liquid by outside forces. Use Reference Table H to determine a liquid s boiling point. Boiling point temperatures increase as system pressure increases. Normal Boiling Point: Normal boiling point is the boiling point of a liquid under a standard pressure of atmosphere (atm). Substances with higher boiling points have stronger attractive intermolecular forces holding their molecules together in the liquid phase, requiring more energy input to overcome the attractions and permit boiling. The normal boiling point of water at sea level is 100 C. At sea level, the atmosphere is exerting atm of pressure. As you ascend into the atmosphere, the mass of air above decreases, as well as pressure. Lower atmospheric pressure cannot keep water in a liquid state at higher temperatures, and the boiling point will be below 100 C. There are special cooking directions for high-altitude cooking due to the lower heat in boiling water at high altitudes. Page 37 of 63

38 Topic: Boiling Point Objective: How do temperature and pressure affect boiling? How to use Table H: 1) What is the vapor pressure of at C? a. What is the vapor pressure of ethanol at 40 C? Start at the 40 C point on the X axis, trace a line up to the ethanol curve, then move a cross and read the vapor pressure off of the y axis. The vapor pressure is 17 kpa. 2) What is the boiling point of at a pressure of kpa? a. What is the boiling point of water at a pressure of 30 kpa? Start at the 30 kpa point on the Y axis, trace a line to the water curve, then go down and read the temperature off the X axis. The boiling point is70 C. 3) What is the normal boiling point of? a. The normal boiling point is the temperature at which s liquid boils under standard pressure, kpa. There is a dashed line going across the table at kpa and is labeled kpa. Follow the kpa line to the curve you want, and go down to read the boiling point temperature off the X axis. Page 38 of 63

39 How to read Reference Table H; Vapor Pressure of Four Liquids Notice that the scales for the X axis are very different than the scales for the Y axis. Each interval on the Y axis is 10 kpa and each interval on the X axis is 5 C. Page 39 of 63

41 Student name: Class Period: Please carefully remove this page from your packet to hand in. Gases and Pressure homework Circle your answer for the multiple choice questions. 1. In which way do real gases deviate from ideal gas behavior? a) Real gas molecules are extremely far apart b) Real gas molecules have attractive forces c) Real gas molecules move faster at higher temperatures d) Real gas molecules travel in a straight line 2. Under which conditions will O 2 behave MOST ideally? a) 100 K and 1 atm c) 300 K and 1 atm b) 200 K and 1 atm d) 400 K and 1 atm 3. Which gas behaves most like an ideal gas at STP? a) O 2 b) N 2 c) H 2 d) F 2 4. Describe the quality of an ideal gas that your choice above best matches. Smaller molecules behave most ideally 5. Water boils at 100 C at sea level. At the highest point in New York State, Mt. Marcy, water will boil: a) At 100 C b) Below 100 C c) Above 100 C 6. Explain your answer for question 5 in terms of vapor pressure and boiling point. Higher altitudes have less atmospheric mass above, reducing the vapor pressure, lowering boiling point Page 41 of 63

42 7. Which substance listed on Reference Table H has the strongest attractive forces holding its molecules together? a) Water c) Propanone b) Ethanol d) Ethanoic acid 8. Explain your answer for question 7 above in terms of normal boiling point. Ethanoic acid reaches vapor pressure at 1 atm between 115C and 120C, which is a higher normal boiling point than the other three choices 9. Which sample of gas contains the same number of molecules as a 2.0 L sample of O 2(g) at 300 K and 100 kpa? a) 1.0 L sample of H 2(g) at 300 K and 100 kpa b) 2.0 L sample of N 2(g) at 300 K and 150 kpa c) 2.0 L sample of F 2(g) at 350 K and 150 kpa d) 2.0 L sample of Cl 2(g) at 300 K and 100 kpa 10. Explain your answer for question 9 above in terms of Avogadro s Hypothesis. d) has the same P and T and V as the question; therefore the ame number of moles as stated by Avogadro s hypothesis Perform the following conversions, showing your work atm = 2.0 atm x kpa 1 atm = kpa = 2Ō0 kpa kpa = kpa x 1 atm kpa = atm = atm Page 42 of 63

43 Student name: Class Period: Please carefully remove this page from your packet to hand in. Gases and Pressure homework 13. Why is the atmospheric pressure lower at the top of Mount Everest (highest point on Earth s surface) than it is at the surface of the Dead Sea (lowest point on Earth s surface)? Explain using the quantity of air above each location. Higher altitudes have less atmospheric mass above, reducing the vapor pressure, lowering boiling point 14. In which location in question 13 above will water boil at a higher temperature? Explain using your answer above. Water will boil at a higher temperature at the Dead Sea surface as the atmospheric (and vapor pressure) will be higher there due to more atmospheric mass above the Dead Sea than on Mt. Everest Based on Reference Table H, what is the vapor pressure of: C? kpa C? kpa 17. Ethanoic 110 C? kpa Page 43 of 63

44 Based on Reference Table H, what is the boiling point of: 18. Ethanol under 70 kpa pressure? _ _ C 19. Water under 10 kpa pressure? _ _ C 20. Ethanoic acid under 120 kpa pressure? _ _ C 21. Based on Reference Table H, what is the normal boiling point of propanone? _ _ C 22. Why do cooking directions require you to boil your food longer at higher elevations than the time required at lower elevations? Higher altitudes have less atmospheric mass above, reducing the vapor pressure, lowering boiling point 23. A flask containing only air is filled partway with water then sealed. What will happen to the pressure inside the flask over time? Explain in terms of vapor pressure. If the flask is sealed, vapor pressure will reach equilibrium as equal #s of molecules will be entering the gas phase as are returning to the liquid phase Page 44 of 63

45 Topic: The Gas Laws Objective: How can models of particle behavior be used to predict? The Gas Laws are relationships between temperature, pressure, and volume of a gas. Gas law equations are used to determine what effect changing one of those variables will have on any of the others. What are the units for pressure, volume, and temperature? Pressure: atm or kpa Volume: ml or L Temperature: K The Gas Laws are based on one fundamental truth known as Avogadro s Hypothesis which states: i. Equal volumes of two samples of ideal gases contain equal numbers of particles under the same conditions of temperature and pressure. Watch Bozeman Science Gas Laws video Page 45 of 63

46 Topic: The Gas Laws Objective: How are models of particle behavior used for predictions? Consider two 4.00 L containers, both at 298 K and 1.00 atm. Container A holds nitrogen (N 2 ) gas, while container B holds carbon dioxide (CO 2 ) gas. If container A holds 2.00 moles of N 2 gas, how many moles of CO 2 gas must be present in container B? i. As both containers hold equal volumes of gases under the same conditions of temperature and pressure, Avogadro s Hypothesis states that both containers will hold equal numbers of molecules, and then hold equal numbers of moles of molecules. Therefore, if container A holds 2.00 moles of gas, so container B MUST also hold 2.00 moles of gas. This means that all gases behave more or less equally to changes in temperature, pressure, and volume, so one equation may be used to describe these changes and also equally applied to all gases, if the assumption is made that they exhibit IDEAL gas behavior. Do equal volumes of gases under the same conditions of temperature and pressure have the same MASS? NO! Each element has its own unique atomic mass. Page 46 of 63

47 Topic: Solving Gas Law Problems Objective: How can we use Gas Laws to solve gas reaction problems? Solving Gas Law problems: 1. Get rid of the words. Create a data table to organize the numbers. a) If the units are volume (ml or L), pick out the V 1 and V 2. b) If the units are pressure (atm or kpa), pick out P 1 and P 2. c) If the units are temperature (K), pick out T 1 and T 2. NOTE: If temperature is given in C, convert it to K by adding C to get K, as Kelvin MUST be used in gas laws. 2. Write the gas law equation down. Write it on your homework page. 3. Circle the variable you are solving for. Then use basic algebra to rearrange the equation. 4. Substitute the numbers into the newly rearranged equation. Ensure you don t lose numbers and units in the process. Enter the numbers into your calculator. 5. Round your answer. Gas laws are multiplication and division, so round to the fewest number of sig figs in the original values. Make sure you have cancelled all units you can and place the proper units on your answer. Page 47 of 63

48 Topic: Gas Property Relationships Objective: What relationships exist between P, V, & T in gases? Relationships between variables of Pressure (P), Volume (V), and Temperature (T): 1. Pressure vs. Volume (Constant Temperature)-as gas pressure is increased, gas volume is decreased (indirect relationship). Boyle s Law (Pressure vs Volume at constant temperature) P x V = k: P 1 V 1 = k, P 2 V 2 = k, therefore P 1 V 1 = P 2 V 2 Note: k here means a constant relationship, NOT Kelvin A 10.0 L (V 1 ) sample of gas (at constant temperature) under a pressure of 1.0 atm (P 1 ) is trapped in a cylinder with a moveable piston. The cylinder is moved until the pressure doubles to 2.0 atm (P 2 ). The new volume of the gas in the cylinder is now 5.0 L (V 2 ). If pressure is doubled, volume is halved. Page 48 of 63

49 Constant Temperature Example: A sample of gas occupies a volume of 2.00 L at Standard Temperature and Pressure (STP). If the pressure is increased to 2.00 atm at constant temperature, what will be the new gas volume? a) Set up a table of givens: (P 1 is given as STP, look at Ref. Table A = 1.00 atm) P 1 = 1.00 atm; V 1 = 2.00 L; P 2 = 2.00 atm; V 2 = X b) Rearrange the Combined Gas Law Equation (Ref. Table T) to solve for the desired variable, omitting the variable that is held constant (temperature in this case): has temperature (T) omitted to become: P 1 V 1 = P 2 V 2 c) Algebraically solve for V 2 to give the equation of: P 1 V 1 / P 2 = V 2 d) Substitute the given numbers and units into the new equation and solve: 1.00 atm 2.00 L = 1.00 L 2.00 atm Page 49 of 63

50 Topic: Gas Property Relationships Objective: What relationships exist between P, V, & T in gases? Relationships between variables of Pressure (P), Volume (V), and Temperature (T): 2. Volume vs. Temperature (Constant Pressure)-as gas temperature is increased, gas volume is increased (direct relationship). Charles Law (Volume vs. Temperature at constant pressure) V / T = k: V 1 / T 1 = k, V 2 / T 2 = k therefore V 1 / T 1 = V 2 / T 2 Note: k here means a constant relationship, NOT Kelvin A 1.0 L (V 1 ) sample of gas (at constant pressure) at a temperature of K (T 1 ) is trapped in a cylinder with a moveable piston. The cylinder is heated until the temperature doubles to K (T 2 ). The new gas volume will be 2.0 L (V 2 ). If the temperature is doubled, then volume doubles as well. Page 50 of 63

51 Constant Pressure example: A sample of gas at K occupies a volume of 5.00 L. If the temperature is doubled under constant pressure, what will be the new volume? a) Set up a table of givens: V 1 = 5.00 L; T 1 = K; V 2 = X; T 2 = K (T 1 doubled) b) Rearrange the Combined Gas Law Equation (Ref. Table T) to solve for the desired variable, omitting the variable that is held constant (pressure in this case): has pressure (P) omitted to become: V 1 / T 1 = V 2 / T 2 c) Algebraically solve for V 2 to give the equation of: V 1 T 2 / T 1 = V 2 d) Substitute the given numbers and units into the new equation and solve: 5.00 L K = 10.0 L K Page 51 of 63

52 Topic: Gas Property Relationships Objective: What relationships exist between P, V, & T in gases? Relationships between variables of Pressure (P), Volume (V), and Temperature (T): 1. Temperature vs. Pressure (Constant Volume)-as gas temperature is increased, gas pressure is increased (direct relationship). Gay-Lussac s Law (Pressure vs. temperature at constant pressure) P/ T = k: P 1 / T 1 = k, P 2 / T 2 = k, therefore P 1 /T 1 = P 2 /T 2 Note: k here means a constant relationship, NOT Kelvin A sample of gas at K (T 1 ) and 2.0 atm (P 1 ) is trapped in a rigid cylinder (constant volume). When the cylinder is heated until the temperature doubles to K (T 2 ), the new pressure of the gas will be 4.0 atm (P 2 ). If temperature is doubled, then pressure doubles as well. Page 52 of 63

53 Constant Volume example: A 10.0 L sample of gas in a rigid container at 1.00 atm and K is heated to K. Under constant volume, what will the new pressure of the gas be? a) Set up a table of givens: P 1 = 1.00 atm T 1 = K P 2 = X T 2 = K b) Rearrange the Combined Gas Law Equation (Ref. Table T) to solve for the desired variable, omitting the variable that is held constant (volume in this case): has volume (V) removed to become: P 1 / T 1 = P 2 / T 2 c) Algebraically solve for P 2 to give the equation of: P 1 T 2 / T 1 = P 2 d) Substitute the given numbers and units into the new equation and solve: 1.00 atm K = 4.00 atm K Page 53 of 63

54 Topic: Combined Gas Law Objective: How do we combine the three laws into a combined law? The three given Gas Laws, Boyle s Law, Charles Law, and Gay-Lussac s Law, are combined to give the Combined Gas Law found on Reference Table T as shown. You MUST use KELVIN temps! NOTE: Ref Table T does NOT state temperature in K. You MUST convert C to K by adding 273 to C! If a variable is given as a constant, it may be ignored using this equation. Set up a table of the given variables and solve for the unknown. There MAY NOT be any CONSTANT. If no constant is stated, use all the given variables. Page 54 of 63

55 Combined Gas Law examples: 1. A 2.00 L sample of a gas at STP is heated to K and compressed to kpa. What will the new volume of the gas be? a) Set up a table of givens: P 1 = kpa; V 1 = 2.00 L; T 1 = 273 K; P 2 = kpa; V 2 = X; T 2 = K b) Rearrange the Combined Gas Law equation to solve for the desired new volume: solved for V 2 becomes the equation of: (P 1 V 1 T 2 ) / (P 2 T 1 ) = V 2 c) Substitute the given numbers and units into the new equation and solve: kpa 2.00 L K = 1.86 L kpa 273 K Page 55 of 63

56 Combined Gas Law examples: 2. A 2.00 L sample of a gas at 1.00 atm and K is heated to K and compressed to a volume of 1.00 L. What will the new pressure of the gas be? a) Set up a table of givens: P 1 = 1.00 atm; V 1 = 2.00 L; T 1 = K; P 2 = X; V 2 = 1.00 L; T 2 = K b) Rearrange the Combined Gas Law equation to solve for the desired new volume: solved for P 2 becomes the equation of: (P 1 V 1 T 2 )/ (V 2 T 1 ) = P 2 c) Substitute the given numbers and units into the new equation and solve: 1.00 atm 2.00 L K = 3.33 atm 1.00 L K Page 56 of 63

57 Combined Gas Law examples: 3. A 2.00 L sample of a gas at K and a pressure of 80.0 kpa is placed into a 1.00 L container and pressurized to kpa. What will the new temperature of the gas be? a) Set up a table of givens: P 1 = 80.0 kpa; V 1 = 2.00 L; T 1 = K; P 2 = kpa; V 2 = 1.00 L; T 2 = X b) Rearrange the Combined Gas Law equation to solve for the desired new volume: solved for T 2 becomes the equation of: (P 2 V 2 T 1 )/ (P 1 V 1 ) = T 2 c) Substitute the given numbers and units into the new equation and solve: (240.0 kpa 1.00 L K) = 450. K 80.0 kpa 2.00 L Page 57 of 63

59 Student name: Class Period: Please carefully remove this page from your packet to hand in. The Gas Laws homework Answer the following questions, showing ALL work, including conversions, units, and proper significant figures. Circle your answer. 1. A sample of hydrogen gas has a volume of 1.00 L at a pressure of kpa. If the temperature is kept constant and the pressure is raised to kpa, what is the new volume of the hydrogen gas sample? 2. A gas sample occupies 10.0 ml at 1.00 atm of pressure. If the gas sample is placed into a sealed 20.0 ml container and the temperature remains constant, what is the new pressure of the gas sample? Page 59 of 63

60 3. When ml of hydrogen gas is heated from 30.0 C to 60.0 C at a constant pressure, what will be the volume of the gas at 60.0 C? 4. Temperatures in New York State can normally range from near 100 F (38 C) to -20 F (-29 C) across the state from season to season. Assuming no air is lost in a tire (constant volume of L) that started with a pressure of 35 psi (241 kpa); find the change in tire pressure from the hottest temperature to coldest temperature given above. 241 kpa kpa = 52 kpa ( P) which is 7.5 psi loss Page 60 of 63

61 5. A sample of gas at K occupies a volume of 10.0 L at a pressure of 2.00 atm. To what temperature must the gas be raised to double both the gas volume and the gas pressure? 6. A 2.00 L sample of gaseous oxygen exerts a pressure of 2.00 atm at 273 K. If the temperature of the oxygen is raised to 546 K and the volume decreased to 1.00 L, what will be the final pressure of the gas? 7. A gas sample occupies 10.0 L of volume at STP. How much volume will it occupy at a pressure of 2.00 atm and temperature of K? Page 61 of 63

62 A CO 2 fire extinguisher contains 15.5 L of CO 2(g) under a pressure of atm at 22.0 C. When activated, the pressure of the CO 2 drops to 95.0 kpa and the volume of CO 2 released is 458 L. 8. What is the temperature of the released CO 2? 9. CO 2 undergoes sublimation at a temperature of C. What phase does the CO 2 in the scenario above exit the fire extinguisher? Solid CO 2 - the temperature that the CO 2 exits at (16.3 K) is C, which is FAR below the sublimation point of CO 2 Page 62 of 63