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Intro to Gas Laws

Intro to Gas Laws. Chapter 13.1. Kinetic Molecular Theory. Gas Particles Are: NOT attracted or repelled by each other Smaller than the spaces between them In constant random motion No kinetic energy is lost when gas particles collide with each other or with the walls of their container

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Intro to Gas Laws

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  1. Intro to Gas Laws Chapter 13.1

  2. Kinetic Molecular Theory • Gas Particles Are: • NOT attracted or repelled by each other • Smaller than the spaces between them • In constant random motion • No kinetic energy is lost when gas particles collide with each other or with the walls of their container • All gases have the same kinetic energy at the same temperature

  3. Three factors that affect gases • Temperature • A measure of the kinetic energy of a gas • Measured in ºC or in K (but only Kelvin can be used in all calculations for gas laws) • Volume • A measure of the amount of space the gas occupies (equal to vol. of container) • Measured in L or mL • Pressure • A measure of the number of times the molecules collide (with each other and/or with the walls of the container) • Measured in mm Hg, kPa, or atm.

  4. Important Info • STP = Standard Temperature & Pressure • Pressure unit conversions: 1 atm = 760 mmHg = 101.3 kPa • Temperature unit conversions: Kelvin = °C + 273 • Temperature should always be in Kelvinunits for gas law problems

  5. Boyle’s Law • Pressure is inversely related to volume as long as temperature is constant. • In other words: as pressure increases, volume decreases • Boyle’s Law: • P1V1 = P2V2 • Just identify your variables and solve!

  6. Example 1 • A sample of compressed methane has a volume of 648 mL at a pressure of 503 kPa. To what pressure would the methane have to be compressed in order to have a volume of 216 mL? • Variables: P1= 503 kPa, V1= 649 mL, P2 = ?, V2 = 216 mL • 503 kPa x 649 mL = P2 x 216 mL • P2 = 1510 kPa

  7. Charles’ Law • Volume is directly proportional to temperature, as long as pressure is constant. • In other words: as temperature increases, volume increases • Charles’ Law • V1/T1 = V2/T2 • Just identify your variables and solve! Remember temp in K!

  8. Example 2 • A weather balloon contains 5.30 kL of He gas when the temperature is 12 ºC. At what temperature will the balloon’s volume have increased to 6.00 kL? • Convert ºC to K • 12 + 273 = 285 K • V1 = 5.30 kL, T1 = 285 K, V2 = 6.00 kL, T2 = ? • 5.30 kL/285 K= 6.00 kL/ T2 • T2 = 323 K

  9. Gay-Lussac’s Law • Pressure is directly proportional to temperature, as long as volume is constant • In other words: as temperature increases, pressure increases • Gay-Lussac’s Law • P1/T1 = P2/T2 • Just Identify your variables and solve! Remember temp in K!

  10. Example 3 • Pressure inside a jelly jar before it is sealed is 1.75 atm at 25 ºC. Volume is constant. After the jar is heated to 100 ºC, what is the new pressure? • Convert ºC to K (twice) • 25 + 273 = 298 K • 100 + 273 = 373 K • P1= 1.75 atm, T1= 298 K, P2= ?, T2= 373 K • 1.75 atm/ 298 K = P2/ 373 K • P2 = 2.19 atm

  11. Combined Gas Law • The combined laws of Boyle, Charles, and Gay-Lussac: • P1V1 /T1 = P2V2 /T2 • Nothing remains constant! Except the amount of gas! • Just identify your variables and solve. Temp in K!

  12. Example 4 • A student collects 285 mL of O2 gas at 15 ºC and a pressure of 99.3 kPa. The next day, the same sample occupies 292 mL at a temperature of 11 ºC. What’s the new pressure? • Convert ºC to K • 15 + 273 = 288 K • 11 + 273 = 284 K • P1= 99.3 kPa, V1= 285 mL, T1= 288 K, P2= ?, V2= 292 mL, T2= 284 K • (99.3 kPa x 285 mL)/288K=(P2 x 292 mL)/284 K • P2 = 95.6 kPa

  13. Avogadro’s Principle • The principle that equal volumes of all gases at the same conditions of temperature and pressure have the same number of molecules. • 1 mol = 22.4 L @ STP • Now you can interrelate the following for any gas: • Mass, moles, pressure, volume, and temperature

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