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Unit 5: Gas and Atmospheric Chemistry

Unit 5: Gas and Atmospheric Chemistry. 11.2 & 11.3 Gas Laws. Context. In a gas, the particles are very far apart This means that there is a lot of empty space between them This makes gases compressible. Consider this….

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Unit 5: Gas and Atmospheric Chemistry

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  1. Unit 5: Gas and Atmospheric Chemistry 11.2 & 11.3 Gas Laws

  2. Context • In a gas, the particles are very far apart • This means that there is a lot of empty space between them • This makes gases compressible

  3. Consider this… • What happens to the volume of a gas when you increase the pressure? Ex: press a syringe that is stoppered) • Why? • Gases are compressible

  4. Animation • http://group.chem.iastate.edu/Greenbowe/sections/projectfolder/flashfiles/gaslaw/boyles_law_graph.html

  5. Boyle’s Law • In the 17th, Robert Boyle described this as “the spring in the air” • As pressure on a gas increases, the volume of the gas decreases proportionally, if temperature and amount of gas (moles) remain constant. • P α 1/V • PV = k P1V1 = P2V2

  6. Example Problem • A 550 L weather balloon at 98 kPa is released from the ground and rises into the atmosphere. It is caught a later height and instruments indicate the air pressure is 75 kPa. What is the volume of the captured balloon? P1= V1= P2= V2= 98 kPa 550 L 75kPa ? P1V1=P2V2 (98 kPa)(550L) = (75 kPa)(V2) V2= 719 L

  7. Real Life Applications • Scuba Diving • As they dive down underwater, the pressure increases  volume? • What happens to volume as you ascend in water? • Ears popping!

  8. Charles’ Law Temperature and Volume

  9. Absolute Zero • This temperature is absolute zero, the lowest possible temperature. • A theoretical temperature at which matter has no kinetic energy (volume of 0) and therefore transmit no thermal energy. • -273.15 ˚C

  10. Temperature Scales Kelvin devised a new scale that would have no negative values • 1˚C = 1 K • ˚C = K - 273.15 • K = ˚C + 273.15 In this unit, we will ALWAYS use K

  11. Charles’ Law As the temperature of a gas increases, the volume increases proportionally, when pressure and amount of gas (moles) remains constant. Temperature (in Kelvins) is directly proportional to volume A direct relationship V α T V = kT Vi = Vf Ti Tf

  12. V1 = V2 T1 T2 Charles’ Law Ex. A balloon with a volume of 1.0L at 25°C is cooled to -190°C. The new volume is… V1 = V2V1= 1.0L T1 = 25 + 273 = 298K T1 T2 T2 = -190 + 273 = 83K V2 = 1.0 L x 83 K 298K V2 = 0.28L = 280mL

  13. Pressure and Temperature: Gay-Lussacs Law Pressure increases proportionally as temperature increases, provided volume and amount remain constant. Pressure is directly proportional to temperature (in Kelvins) P  T A direct relationship P = kT P = k T P1 = P2 T1 T2

  14. Example Ex. A gas cylinder with a pressure of 1000kPa at 25°C is placed in a boiling water bath. What will the new pressure reading be? P1 = P2 P1 = 1000kPa T1 = 25 + 273 = 298K T1 T2 P2 = ? T2 = 100 + 273 = 373K P2 = 1000kPa x 373 K 298 K P2 = 1252 kPa

  15. Video!

  16. Practice!! • Popsicle stick analogy • P. 435 # 2,5 • P. 451 # 2-5 • Worksheet – “Problems – Gas Laws”

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