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Entry Task: March 20-21 st Block #1

Entry Task: March 20-21 st Block #1. Question: State Boyles laws You have 5 minutes. Agenda:. Discuss Ch. 14 sec. 1-2 Go through how to do the math HW: Worksheet with these equations . I can…. Apply the 3 gas laws to problems involving pressure, volume and temperature of a gas.

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Entry Task: March 20-21 st Block #1

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  1. Entry Task: March 20-21st Block #1 Question: State Boyles laws You have 5 minutes

  2. Agenda: • Discuss Ch. 14 sec. 1-2 • Go through how to do the math • HW: Worksheet with these equations 

  3. I can… • Apply the 3 gas laws to problems involving pressure, volume and temperature of a gas. • Pressure v. Volume • Temperature v. Volume • Pressure v. Temperature

  4. WarningAll RED text slides are embedded notes

  5. The theory explains the behavior of particles that make up gas, liquids and solids- particles are in constant motion. What is the kinetic theory?

  6. Briefly summarize each assumption about gas behaviors (pg. 419-420) 1. Gas particles do not attract or repel each other. 2. Gas particles are much smaller than the distances between them. 3. Gas particles are in constant, random motion 4. No kinetic energy is lost when gas particles collide with each other or container. 5. All gases have the same average kinetic energy at a given temperature.

  7. 1. Number of gas particles. What are the four factors is the basis of the kinetic theory? 2. Temperature 3. Pressure 4. Volume

  8. Assuming that the number of particles and temperature is constant, what happens to the gas in a plastic balloon if you squeeze it? Provide the two factors that are affected? What is the relationship between the two? Pressure and volume are the two factors affected. As one increases the other decreases (inverse relationship)

  9. Its states that the volume of a given amount of gas held at a constant temperature varies inversely with the pressure. State Boyles Law

  10. P1 V1 = P2 V2 What is Boyles Law’s Formula

  11. 1. The volume of a gas at 99.0 kPa is 300 ml. If the pressure is increased to 188 kPa, what will be the new volume? P1 = 99.0 kPa V1 = 300 mL P2 = 188 kPa V2 = X (99.0 kPa) (300 mL) = (188 kPa) (X) 29700 mL = X 188 158 ml = X

  12. 2. The pressure of a sample of helium in a 1.99 –L container is 0.988 atm. What is the new pressure if the sample is placed in a 2.00 L container? P1 = 0.988 atm V1 = 1.99 L P2 = X V2 = 2.00 L (0.988 atm) (1.99 L) = (X) (2.00 L) 1.966 atm = X 2.00 0.983 atm = X

  13. 3. Air trapped in a cylinder fitted with a piston occupies 145.7 ml at 1.08 atm pressure. What is the new volume of air when the pressure is increased to 1.43 atm by applying force to the piston? P1 = 1.08 atm V1 = 145.7 mL P2 = 1.43 atm V2 = X (1.08 atm) (145.7 mL) = (1.43 atm) (X) 15735.6 mL = X 1.43 1.10 x102mL = X

  14. 4. If it takes 0.5000L of oxygen gas kept in a cylinder under pressure to fill an evacuated 4.00 L reaction vessel in which the pressure is 0.980 atm, what was the initial pressure of the gas in the cylinder? P1 = X atm V1 = 0.5000 L P2 = 0.980 atm V2 = 4.00 L (X) (0.5000 L) = (0.980 atm) (4.00L) X= 3.92 atm 0.5000 7.84 atm = X

  15. 5. A sample of neon gas occupies 0.220 L at 0.860 kPa. What will be its volume at 29.2 kPa pressure? P1 = 0.860 kPa V1 = 0.220 L P2 = 29.2 kPa V2 = X L (0.860 kPa) (0.220 L) = (29.2 kPa) (X L) 0.1892 L = X 29.2 6.00 x10-3L= X

  16. Its states that the volume of a given mass of a gas is directly proportional to its kelvin temperature at constant pressure. State Charles Law

  17. V1 V2T1 T2 What is Charles Law’s Formula =

  18. -273.15 ˚C After graphing temperature verses volume, and extrapolating temperature verses volume to zero volume, what is the temperature that would match this volume?

  19. -273.15 ˚C or 0 Kelvin, is absolute zero What is the temperature of absolute zero? ___________ Define what IS the theory behind absolute zero? The kinetic energy is 0, no motion!

  20. 6. A gas at 89 C ˚ occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.12 L? T1 = 89 + 273 = 362 K V1 = 0.67 L T2 = X V2 = 1.12 L 362K X = 0.67 L 1.12 L 405.44 = (X) (0.67 L) 405.44 K = X 0.67 605 – 273 = 332 ˚C

  21. 7. 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 gas? T1 = 80 + 273 = 353 K V1 = 3.00 L T2 = 30 + 273 = 303 K V2 = X L 353 K 303 K = 3.00 L X 909 = (X) (353 K) 909 L = X 353 2.58 L

  22. 8. What is the volume of the air in a balloon that occupies 0.620 L at 25˚C if the temperature is lowered to 0.00˚C? T1 = 25 + 273 = 298 K V1 = 0.620 L T2 = 0.0 + 273 = 273 K V2 = X L 298 K 273 K = 0.620 L X 169 = (X) (298 K) 169 L = X 298 0.568 L

  23. Its states that the pressure of a given mass of gas varies directly with the kelvin temperature when the volume remains constant. State Gay-Lussac’s Law

  24. P1 P2T1 T2 What is Gay-Lussac’s Formula =

  25. 9. A gas in a sealed container has a pressure of 125 kPa at a temperature of 30.0 ˚C. If the pressure in the container is increased to 201 kPa, what is the new temperature- in Celsius? P1 = 125 kPa T1 = 30 + 273 = 303K P2 = 201 kPa T2 = X 125 kPa 201 kPa = 303 K X 60903 = (X) (125 kPa) 60903 K = X 125 487 – 273 = 214 ˚C

  26. 10. The pressure in an automobile tire is 1.88 atm at 25 ˚C. What will be the pressure if the temperature warms up to 37.0 ˚C? P1 = 1.88 atm T1 = 25 + 273 = 298K P2 = X T2 = 37 + 273 = 310K 1.88 atm X = 298 K 310 K 583 = (X) (298K) 583 atm = X 298 1.96 atm or 2.0 atm

  27. 11. Helium gas in a 2.00 L cylinder is under 1.12 atm pressure. At 36.5 ˚C, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas cylinder? P1 = 1.12 atm T1 = X P2 = 2.56 atm T2 =36.5 + 273 = 309.5K 1.12 atm 2.56 atm = X K 309.5 K 347 = (X) (2.56 atm) 347 K = X 2.56 135 K- 273 = -138 ˚C

  28. 12. A gas sample has a pressure of 30.7 kPa at 0.00˚C, by how much does the temperature have to decrease to lower the pressure to 28.4 kPa? P1 = 30.7 kPa T1 = 0.0 + 273 = 273 K P2 = 28.4 kPa T2 = X 30.7 kPa 28.4 kPa = 273 K X 7753.2 = (X) (30.7 kPa) 7753 K = X 30.7 253 K- 273 = -20 ˚C

  29. 13. A ridged plastic container holds 1.00 L methane gas at 660 torr pressure when the temperature is 22.0˚C. How much more pressure will the gas exert if the temperature is raised to 44.6˚C? P1 = 660 torr T1 = 22 + 273 = 295K P2 = X T2 = 44.6 + 273 = 317.6K 660 torr X torr = 295 K 317.6 K 209616 = (X) (295 torr) 209616 K = X 295 711 torr- 660 = 51 torr

  30. BOYLES CHARLES & GAY-LUSSACWorksheet

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