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- 1. Gas Laws and Chemistry• Some Background • Standard Temperature and Pressure (STP) • Molar Volume of Gas @ STP • 22.4 liters
- 2. What is STP?• Standard Temperature oK= oC + 273 • Measured in Kelvin 20oC= 293oK not Celsius• Standard Pressure • 760 mm Lord Kelvin
- 3. Boyle’s Law• Volume varies inversely with pressure Robert Boyle
- 4. An Application of Boyle’s Law• What would be the volume of 15 liters of gas if the pressure changed from 700 mm to 850 mm?15 liters x 700 mm = 12.35 liters 850 mm
- 5. Charles’s Law• Volume varies directly with temperature Jacques Charles
- 6. An Application of Charles’s Law...• What would be the volume of 20 liters of gas at 25oC if the temperature were changed to 0oC? Convert oC to oK 25oC (25oC + 273 = 298oK) 0oC (0oC + 273 = 273oK) 20 liters x 273oK = 18.32 liters 298oK A balloon on dry ice
- 7. Gay-Lussac’s Law• Pressure varies directly with temperature Joseph Gay-Lussac
- 8. Application of Gay-Lussac’s Law• What would be the change in pressure of a ﬁxed volume of gas at 40oC and 600 mm pressure, if the temperature were increased to 80oC ? 600 mm x 353oK = 676.67 mm 313oK
- 9. Combined Gas Law Problem• What would be the volume of 75 liters of a gas at 30oC and 700 mm pressure if the temperature and pressure were changed to 70oC and 800 mm? 75 liters x 343oK x 700 mm = 74.29 liters 303oK 800 mm
- 10. Dalton’s Law of Partial Pressures• The total pressure of a mixture of gases is equal to the sum of the partial pressures of the gases
- 11. Combining the Gas Laws to Solve a Quantitative Problem Mg + 2 HCl MgCl2 + H2 • If 0.5 grams of Mg are completely reacted with HCl to form H2, how many liters of H2 will be collected over water at 30oC and a pressure of 700 mm?0.5 g Mg x 1 mole Mg x 1 mole H2 x 22.4 liters H2 = 0.46 liters H2 @ STP 24.31 g Mg 1 mole Mg 1 mole H2
- 12. The Final Step... • Take volume of H at STP and convert it to 2 lab conditions using Boyle’s, Charles’s, and Dalton’s LawsDalton’s Law correction700 mm (pressure at lab conditions) – 31.8 mm (water vapor pressure) = 668.2 mm0.46 liters H2 x 760 mm x 303oK = 0.58 liters of H2 at lab conditions 668.2 mm 273oK

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