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Gas Laws and Relationships between P, V, and T

Gas Laws and Relationships between P, V, and T. Boyle’s Law Charles’s Law Gay-Lusaac’s Law How to use each. Gases and their Variables. Four Variables when exploring gases: P , V , T , and n P : Pressure in atm’s V : Volume in L T : Temperature in K

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Gas Laws and Relationships between P, V, and T

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  1. Gas Laws and Relationships between P, V, and T Boyle’s Law Charles’s Law Gay-Lusaac’s Law How to use each

  2. Gases and their Variables • Four Variables when exploring gases: P, V, T, and n • P: Pressure in atm’s • V: Volume in L • T: Temperature in K • n: quantity of matter in moles How do they relate to one another? Let’s put on our PVT Cards and find out!

  3. Gas Properties can be modeled using MATH!And again, Scientists have done all of the work for us! http://www.youtube.com/watch?v=13WUqWd_Yk8&feature=related

  4. Boyle’s Law States: The volume of a sample of gas is inversely proportional to its pressure, if temperature remains constant. Translation: At constant temperature and n, 1 α P V Robert Boyle EX. Volume Decreases, Pressure Increases Volume Increases, Pressure Decreases In an inverse relationship, the product of the two quantities is a constant. P1 x V1 = P2 x V2 P1 V1 = P2 V2

  5. Boyle’s Law

  6. Example 1 - A sample of gas collected in a 350 cm3 container exerts a pressure of 103 kPa. What would be the volume of this gas at 150 kPa of pressure? (Assume that the temperature remains constant.) • Solving: If temperature remains constant - use Boyle’s Law. • Write the original formula: P1V1 = P2V2 • Then list what is given and what is unknown. • P1 = 103 kPa V1= 350 cm3 P2 = 150 kPa V2 = ? Then PLUG AND CHUG! Let’s work it together. 103 kPa (350 cm3) = 150 kPa V2 150 kPa 150 kPa 240 cm3=V2

  7. Charles’s Law States: The volume of a sample of gas is directly proportional to its Kelvin temperature, if pressure remains constant. Translation: At constant pressure and n, V α T **Temperature ALWAYS in Kelvin** Jacques Charles EX. Volume Increases, Temperature Increases Volume Decreases, Temperature Decreases In an direct relationship, the quotient of the two quantities is a constant. V1 / T1 = V2 / T2 V1 = V2 T1 T2

  8. Charles’s Law

  9. Ex. If a gas occupies 733 cm3 at 10.0 oC, at what temperature will it occupy 950 cm3? Assume that pressure remains constant. • Solving: If pressure remains constant - use Charles’s Law. • Write the original formula:V1 = V2 • T1 T2 • Then list what is given and what is unknown. • V1 = 733 cm3 T1= 10.0 oC V2 = 950 cm3 T2 = ? Then PLUG AND CHUG! Let’s work it together. First convert oC to Kelvin: K = 10.0 oC + 273 = 283 K 733 cm3 = 950 cm3 283 K T2 366.7 or 370 K= T2

  10. How does a hot air balloon work? TIP: Think about Charles’s Law.

  11. Gay-Lussac’s Law States: The Kelvin temperature of a sample of gas is directly proportional to pressure, if volume remains constant. Translation: At constant volume and n, T α P **Temperature ALWAYS in Kelvin** Joseph Gay-Lussac EX. Temperature Increases, Pressure Increases Temperature Decreases, Pressure Decreases In an direct relationship, the quotient of the two quantities is a constant. P1 / T1 = P2 / T2 P1 = P2 T1 T2

  12. Gay-Lussac’s Law

  13. EX. If a gas is cooled from 323.0 K to 273.15 K and the volume is kept constant what final pressure would result if the original pressure was 750.0 mm Hg? • Solving: If volume remains constant - use Gay-Lussac’s Law. • Write the original formula:P1 = P2 • T1 T2 • Then list what is given and what is unknown. • P1 = 750.0 mm Hg T1= 323.0 K P2 = ? T2 = 273.15 K Then PLUG AND CHUG! Let’s work it together. 750.0 mm Hg= P2 323.0 K 273.15 K 634.2 mm Hg= P2

  14. ONE MORE FUN FACT! Standard Temperature and Pressure (STP) • At STP: • Temperature = 273 K or 0 oC • Pressure = 1 atm = 760 mm Hg Gas Laws Song http://www.youtube.com/watch?v=13WUqWd_Yk8

  15. Let’s Practice! 1. A sample of neon has a volume of 239 cm3 at 2.00 atm of pressure. What would the pressure have to be in order for the gas to have a volume of 5.00 x 102 cm3? 2. A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 °C is placed inside an oven whose temperature is 50.0 °C. The pressure inside the container at 20.0 °C was at 3.00 atm. What is the pressure of the nitrogen after its temperature is increased? 3. A sample of gas at 3.00 x 103 mm Hg inside a steel tank is cooled from 500.0 °C to 0.00 °C. What is the final pressure of the gas in the steel tank? HOMEWORK = Practice problems with each law! FINISH LAB TO TURN IN AT END OF CLASS!

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