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Aim: How is gas behavior explained?

Learn about standard temperature and pressure, gas behavior variables, and the Combined Gas Law with examples to understand how gases behave under different conditions.

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Aim: How is gas behavior explained?

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  1. DO NOW: What is standard temperature and pressure?Will atmospheric pressure be higher during a storm or on a sunny day?

  2. Aim: How is gas behavior explained?

  3. Gas Variables • 4 variables are used to explain gas behavior: • P=pressure (in atm or kPa) • V=volume (in L or mL) • T=temperature (in Kelvin) • n=moles

  4. Combined Gas Law • Relates P, V, and T while holding n constant. • P1V1/T1 = P2V2/T2

  5. Example #1 • A gas has a pressure of 110.0 kPa at 288K and fills a 1.5L container. What is the pressure exerted by the gas after the volume and temperature are increased to 2.5L and 303K, respectively? • P1=110.0 kPa V1=1.5 L T1=288 K • P2=? V2=2.5 L T2=303 K

  6. P1V1 = P2V2 T1 T2 (110.0kPa)(1.5L) = (P2)(2.5L) 288K 303K P2=69 kPa

  7. Example #2 • A sample of neon gas occupies 4.61 L at 2.83 atm and 277K. What will be the pressure of the gas in atm when 8.3 L are occupied at 293K? • P1=2.83 atm V1=4.61 L T1=277 K • P2=? V2=8.3 L T2=293 K

  8. P1V1 = P2V2 T1 T2 • (2.83 atm)(4.61 L) = (P2)(8.3 L) 277 K 293 K P2=1.7 atm

  9. Example #3 • A sample of gas has a volume of 500.0 L when its temperature is 25.0ºC and its pressure is 0.395 atm. What will the new volume be at STP? • P1=0.395 atm V1=500.0 L T1=298 K • P2=1.00 atm V2=? T2=273 K

  10. P1V1 = P2V2 T1 T2 • (0.395 atm)(500.0 L) = (1.00 atm)(V2) 298 K 273 K V2=181 L

  11. 1. At a temperature of 273 K, a 400.-milliliter gas sample has a pressure of one atm. If the pressure is changed to .5 atm, at which temperature will this gas sample have a volume of 551 milliliters? • 2. A gas occupies a volume of 444 mL at 273 K and 79.0 kPa. What is the final kelvin temperature when the volume of the gas is changed to 1880 mL and the pressure is changed to 38.7 kPa?

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