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UNIT 4 NOTEPACK: States of Matter, Gas Laws, and Energy

UNIT 4 NOTEPACK: States of Matter, Gas Laws, and Energy. Unit 4 Objectives:. 1) Describe, at the molecular level, the difference between a gas, liquid, and solid phase. (CH 10) 2) Describe states of matter using the kinetic molecular theory. (CH 10)

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UNIT 4 NOTEPACK: States of Matter, Gas Laws, and Energy

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  1. UNIT 4 NOTEPACK:States of Matter, Gas Laws, and Energy

  2. Unit 4 Objectives: 1) Describe, at the molecular level, the difference between a gas, liquid, and solid phase. (CH 10) 2) Describe states of matter using the kinetic molecular theory. (CH 10) 3) Describe changes in states of matter (s, l, g) with respect to kinetic energy and temperature. (CH 10) 4) Summarize / explain the different variables (P, V, T) that describe a gas. (CH 11) 5) Understand the law of conservation of energy in terms of gaining and losing energy. (CH 20)

  3. OBJECTIVE 4: ● Summarize / explain the different variables that describe a gas.

  4. Kinetic Theory of GasesThe basic assumptions of the kinetic molecular theory: ● Gases are mostly empty space ● The molecules in a gas are separate, very small, and very far apart

  5. Kinetic Theory of GasesThe basic assumptions of the kinetic molecular theory: ● Gas molecules are in constant, chaotic motion ● Collisions between gas molecules are elastic (there is no energy gain or loss)

  6. Kinetic Theory of GasesThe basic assumptions of the kinetic molecular theory: ● As temperature increases, gas molecules move faster ● Gas pressure is caused by collisions of molecules with the walls of the container

  7. Behavior of Gases ● Gases have weight ● Gases take up space ● Gases exert pressure ● Gases fill their containers Gases doing all of these things!

  8. Variables that Describe a Gas ● Volume: measured in L, mL, cm3 (1 mL = 1 cm3) ● Amount: measured in grams (g) ● Temperature: measured in Kelvin (K)  K = ºC + 273 ● Pressure: measured in mm Hg, torr, atm, etc.  P = F / A (force per unit area)

  9. Moderate Force (about 100 lbs) Small Area (0.0625 in2) P = F /A Enormous Pressure (1600 psi)

  10. Bed of Nails Moderate Force Small Pressure P = F / A Large Surface Area (lots of nails)

  11. Units of Pressure ● 1 atm = 760 mm Hg ● 1 atm = 101.3 kPa

  12. BOYLE’S LAW: ● For a sample of gas at a constant temperature, the volume of the gas varies inversely with the pressure. ● As P, Vand vice versa…. ● Inverse relationship! Equation: P1V1 = P2V2

  13. Boyle’s Law and Kinetic Molecular Theory ● How does kinetic molecular theory explain Boyle’s Law? ● Gas molecules are in constant, random motion; ● Gas pressure is the result of molecules colliding with the walls of the container; ● As the volume of a container becomes smaller, the collisions over a particular area of container wall increase…the gas pressure increases!

  14. Pressure-Volume Calculations ● Example #1: Consider the syringe. Initially, the gas occupies a volume of 8 mL and exerts a pressure of 1 atm. ● What would the pressure of the gas become if its volume were increased to 10 mL?

  15. Equation for Boyle’s Law ● P1V1 = P2V2 ● where: P1 = initial pressure V1 = initial volume P2 = final pressure V2 = final volume

  16. P1V1 = P2V2 ● Using the same syringe example, just “plug in” the values & solve for the missing one: P1V1 = P2V2 (1 atm) (8 mL) = (P2) (10 mL)

  17. P1V1 = P2V2 (1 atm) (8 mL) = P2 (10 mL) P2 = 0.8 atm

  18. Example #2: A sample of gas occupies 12 L under a pressure of 1.2 atm. What would its volume be if the pressure were increased to 3.6 atm? (assume temp is constant) P1V1 = P2V2

  19. Example #2: A sample of gas occupies 12 L under a pressure of 1.2 atm. What would its volume be if the pressure were increased to 3.6 atm? (assume temp is constant) P1V1 = P2V2 (1.2 atm)(12 L) = (3.6 atm)V2

  20. Example #2: A sample of gas occupies 12 L under a pressure of 1.2 atm. What would its volume be if the pressure were increased to 3.6 atm? (assume temp is constant) P1V1 = P2V2 (1.2 atm)(12 L) = (3.6 atm)V2 (1.2 atm) (12 L) = V2 (3.6 atm)

  21. Example #2: A sample of gas occupies 12 L under a pressure of 1.2 atm. What would its volume be if the pressure were increased to 3.6 atm? (assume temp is constant) P1V1 = P2V2 V2 = 4.0 L

  22. Example #3: A sample of gas occupies 28 L under a pressure of 200 kPa. If the volume is decreased to 17 L, what is the new pressure? (assume temp is constant) P1V1 = P2V2

  23. Example #3: A sample of gas occupies 28 L under a pressure of 200 kPa. If the volume is decreased to 17 L, what is the new pressure? (assume temp is constant) P1V1 = P2V2 (200 kPa)(28 L) = (P2)(17 L)

  24. Example #3: A sample of gas occupies 28 L under a pressure of 200 kPa. If the volume is decreased to 17 L, what is the new pressure? (assume temp is constant) P1V1 = P2V2 (200 kPa)(28 L) = (P2)(17 L) (200 kPa) (28 L) = P2 (17 L)

  25. Example #3: A sample of gas occupies 28 L under a pressure of 200 kPa. If the volume is decreased to 17 L, what is the new pressure? (assume temp is constant) P1V1 = P2V2 P2 = 329 kPa

  26. Temperature-Volume Relationship: ● What happens to matter when it is heated? ● It EXPANDS! ● What happens to matter when it is cooled? ● It CONTRACTS! ● Gas samples expand and shrink to a much greater extent than either solids or liquids.

  27. CHARLES’ LAW: ● The volume of a sample of gas is directly proportional to the Kelvin temperature. ● As T , V and vice versa…. ● Direct relationship! ● EQUATION:

  28. Temperature-Volume Relationship: ● Doubling the Kelvin temperature of a gas doubles its volume; ● Reducing the Kelvin temperature by one half causes the gas volume to decrease by one half… ● WHY KELVIN? The Kelvin scale never reaches “zero” or has negative values

  29. Converting Kelvin ● To convert from Celsius to Kelvin: add 273 Example: What is 110 ºC in Kelvin? SOLVE: + =

  30. Converting Kelvin ● To convert from Celsius to Kelvin: add 273 Example: What is 110 ºC in Kelvin? SOLVE: 110 ºC + 273 =

  31. Converting Kelvin ● To convert from Celsius to Kelvin: add 273 Example: What is 110 ºC in Kelvin? SOLVE: 110 ºC + 273 = 383 K

  32. Converting Kelvin ● To convert from Kelvin to Celsius: subtract 273 Example: 555 K in Celsius? SOLVE: - =

  33. Converting Kelvin ● To convert from Kelvin to Celsius: subtract 273 Example: 555 K in Celsius? SOLVE: 555 K - 273 =

  34. Converting Kelvin ● To convert from Kelvin to Celsius: subtract 273 Example: 555 K in Celsius? SOLVE: 555 K - 273 = 282 ºC

  35. Example #1: A sample of nitrogen gas occupies 117 mL at 100.°C. At what temperature would it occupy 234 mL if the pressure does not change? V1 = ; T1 = V2 = ; T2 = ???

  36. Example #1: A sample of nitrogen gas occupies 117 mL at 100.°C. At what temperature would it occupy 234 mL if the pressure does not change? V1 = 117 mL; T1 = 100 + 273 = 373 K V2 = 234 mL; T2 = ???

  37. Example #1: A sample of nitrogen gas occupies 117 mL at 100.°C. At what temperature would it occupy 234 mL if the pressure does not change? V1 = 117 mL; T1 = 100 + 273 = 373 K V2 = 234 mL; T2 = ???

  38. Example #1: A sample of nitrogen gas occupies 117 mL at 100.°C. At what temperature would it occupy 234 mL if the pressure does not change? V1 = 117 mL; T1 = 100 + 273 = 373 K V2 = 234 mL; T2 = ???

  39. Example #1: A sample of nitrogen gas occupies 117 mL at 100.°C. At what temperature would it occupy 234 mL if the pressure does not change? V1 = 117 mL; T1 = 100 + 273 = 373 K V2 = 234 mL; T2 = ???

  40. Example #1: A sample of nitrogen gas occupies 117 mL at 100.°C. At what temperature would it occupy 234 mL if the pressure does not change? T2 = 746 K = 473°C

  41. Example #2: A sample of oxygen gas occupies 65 mL at 28.8°C. If the temperature is raised to 72.2°C, what is the new volume of the gas? V1 = ; T1 = V2 = ??? ; T2 =

  42. Example #2: A sample of oxygen gas occupies 65 mL at 28.8°C. If the temperature is raised to 72.2°C, what is the new volume of the gas? V1 = 65 mL; T1 = 28.8 + 273 = 301.8 K V2 = ??? mL; T2 = 72.2 + 273 = 345.2 K

  43. Example #2: A sample of oxygen gas occupies 65 mL at 28.8°C. If the temperature is raised to 72.2°C, what is the new volume of the gas? V1 = 65 mL; T1 = 28.8 + 273 = 301.8 K V2 = ??? mL; T2 = 72.2 + 273 = 345.2 K

  44. Example #2: A sample of oxygen gas occupies 65 mL at 28.8°C. If the temperature is raised to 72.2°C, what is the new volume of the gas? V1 = 65 mL; T1 = 28.8 + 273 = 301.8 K V2 = ??? mL; T2 = 72.2 + 273 = 345.2 K

  45. Example #2: A sample of oxygen gas occupies 65 mL at 28.8°C. If the temperature is raised to 72.2°C, what is the new volume of the gas? V1 = 65 mL; T1 = 28.8 + 273 = 301.8 K V2 = ??? mL; T2 = 72.2 + 273 = 345.2 K

  46. Example #2: A sample of oxygen gas occupies 65 mL at 28.8°C. If the temperature is raised to 72.2°C, what is the new volume of the gas? V2 = 74.3 mL

  47. Temperature – Pressure Relationships ● Picture a closed, rigid container of gas (such as a scuba tank) – the volume is CONSTANT. ● So, what would happen to the kinetic energy of the gas molecules in the container if you were to heat it up? ● How would this affect pressure?

  48. Temperature – Pressure Relationships ● Raising the Kelvin temperature of the gas will cause an INCREASE in the gas pressure. ● WHY? ● With increasing temperature, the K.E. of the gas particles increases – they move faster! ● The molecules collide more often and with more energy with the walls of the container…increase in pressure!

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