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Chapter 13: Gases

Chapter 13: Gases. Nature of gases. Assumptions of Kinetic-Molecular theory are based on four factors: Number of particles present Temperature Pressure Volume When one variable changes, it affects the other three. Boyle’s Law.

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Chapter 13: Gases

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  1. Chapter 13: Gases

  2. Nature of gases • Assumptions of Kinetic-Molecular theory are based on four factors: • Number of particles present • Temperature • Pressure • Volume • When one variable changes, it affects the other three

  3. Boyle’s Law • Boyle’s Law: volume of a given amount of gas held at constant temperature varies inversely with pressure. • Increase volume = decrease pressure (less collisions) • Decrease volume = increase pressure (more collisions)

  4. Boyle’s Law P1V1 = P2V2 initial final *** You MUST memorize this equation!!!

  5. Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L?

  6. Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? What equation do we use?

  7. Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? What equation do we use? P1V1 = P2V2

  8. Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? Known: Unknown:

  9. Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? Known: V1 = 4.0 L V2 = 2.5 L P1 = 210 kPa Unknown: P2 = ?

  10. Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? P1V1 = P2V2

  11. Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? (210 kPa) (4.0 L) = (P2) (2.5 L) P1V1 = P2V2

  12. Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? (210 kPa) (4.0 L) = (P2) (2.5 L) P1V1 = P2V2

  13. Using Boyle’s Law A sample of helium gas in a balloon is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0-L volume is 210 kPa, what will the pressure be at 2.5 L? (210 kPa) (4.0 L) = (P2) (2.5 L) 340 kPa= (P2) P1V1 = P2V2

  14. Practice Problem: Boyle’s Law Example: The pressure of a sample of helium in a 1.00-L container is 0.988 atm. What is the new pressure if the sample is placed in a 2.00-L container?

  15. Practice Problem: Boyle’s Law Example: The pressure of a sample of helium in a 1.00-L container is 0.988 atm. What is the new pressure if the sample is placed in a 2.00-L container? P1V1 = P2V2

  16. Charles’s Law • Charles’s Law: volume of a given mass of gas is directly proportional to its kelvin temperature at constant pressure. • Increase temperature = Increase volume (faster particles) • Decrease temperature = Decrease volume (slower particles)

  17. Charles’s Law V1 T1 V2 T2 = initial final *** You MUST memorize this equation!!!

  18. Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant?

  19. Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant? What equation do we use? V1 T1 V2 T2 =

  20. Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant? Known: Unknown:

  21. Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant? Known: T1 = 40.0 °C V1 = 2.32 L T2 = 75.0 °C Unknown: V2 = ?

  22. Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant? V1 T1 V2 T2 =

  23. Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant? 2.32 L 40.0 °C V2 75.0 °C =

  24. Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant? 2.32 L 40.0 °C V2 75.0 °C x =

  25. Using Charles’s Law A gas sample at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C, what will the volume be, assuming the pressure remains constant? 2.58 L V2 =

  26. Practice Problem: Charles’s Law The celsius temperature of a 3.00-L sample of gas is lowered from 80.0 °C to 30.0 °C. What will be the resulting volume of this gas?

  27. Gay-Lussac’s Law • Gay-Lussac’s Law: pressure of a given mass of gas varies directly with kelvin temperature when the volume remains constant. • Increase temperature = Increase pressure • Decrease temperature = Decrease pressure

  28. Gay-Lussac’s Law P1 T1 P2 T2 = *** You MUST memorize this equation!!!

  29. Using Gay-Lussac’s Law The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank?

  30. Using Gay-Lussac’s Law The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank? What equation do we use?

  31. Using Gay-Lussac’s Law The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank? P1 T1 P2 T2 =

  32. Using Gay-Lussac’s Law Known: Unknown: The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank?

  33. Using Gay-Lussac’s Law Known: T1 = 22.0 °C P1 = 3.20 T2 = 60.0 °C Unknown: P2 = ? The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank?

  34. Using Gay-Lussac’s Law The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank? P1 T1 P2 T2 =

  35. Using Gay-Lussac’s Law The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank? 3.20 atm 22.0 °C P2 60.0 °C =

  36. Using Gay-Lussac’s Law The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank? 3.20 atm 22.0 °C 60.0 °C X P2 =

  37. Using Gay-Lussac’s Law The pressure of a gas in a tank in 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will be the pressure in the tank? 3.61 atm P2 =

  38. Overview of Gas Laws

  39. 14.2 Combined Gas Law • We can combine Boyle’s Law, Charles’s Law and Gay-Lussac’s Law into one law (“Combined Gas Law”). • States the relationship between pressure, volume, and temperature of a fixed amount of gas. P1V1 T1 P2V2 T2 =

  40. Converting to Kelvin (K) • For the rest of the chapter, we NEED to convert temperature to Kelvin (K) first, BEFORE we use the combined gas law. • To convert to Kelvin temperature (K), use the following conversion: • TK = 273 + TC

  41. Converting to Kelvin (K) Convert the following temperatures to Kelvin. 1) -25.0 °C 2) 0 °C 3) 23 °C 4) 80.0 °C

  42. Solving for a Variable Rearrange the Combined Gas Law to isolate the appropriate variable. 1) Solve for P1 2) Solve for V1 3) Solve for T1 4) Solve for T2 5) Solve for V2 P1V1 T1 P2V2 T2 =

  43. Using the Combined Gas Law A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume?

  44. Using the Combined Gas Law A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? What equation do we use?

  45. Using the Combined Gas Law A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? What equation do we use? P1V1 T1 P2V2 T2 =

  46. Using the Combined Gas Law A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? FIRST, what do we do to the temperature values?

  47. Using the Combined Gas Law A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? FIRST, what do we do to the temperature values? CONVERT TO KELVIN!

  48. Using the Combined Gas Law A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °Cand the pressure increased to 440 kPa, what is the new volume? Convert to Kelvin: T1 = 30.0 °C + 273 = 303 K T2 = 80.0 °C + 273 = 353 K

  49. Using the Combined Gas Law A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °Cand the pressure increased to 440 kPa, what is the new volume? Known Values: Unknown:

  50. Using the Combined Gas Law A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °Cand the pressure increased to 440 kPa, what is the new volume? Known Values: T1 = (30.0 °C + 273) 303 K P1 = 110 kPa V1 = 2.00 L T2 = (80.0 °C + 273) 353 K P2 = 440 kPa Unknown: V1 = ?

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