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Fundamentals of Gas Laws: The Kinetic Molecular Theory

Understand gas behavior based on the Kinetic Molecular Theory, covering Boyle's Law, Charles' Law, Gay-Lussac's Law, and Avogadro's Principle. Learn how temperature, volume, and pressure affect gases with real-life examples and equations. Enhance your knowledge of gas laws and their applications.

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Fundamentals of Gas Laws: The Kinetic Molecular Theory

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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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