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Gas Laws

Understand the basic principles of gas laws - Boyle's Law, Charles' Law, and Gay Lussac's Law - and how they relate to pressure, volume, and temperature. Learn the mathematical equations and apply them to solve numerical problems.

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Gas Laws

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  1. Gas Laws

  2. Gas Pressure • Pressureis defined as force per unit area. • Gas particles exert pressure when they collide with the walls of their container. • The SI unit of pressure is the pascal (Pa). • However, there are several units of pressure • Pascal (Pa) • Kilopascal (KPa) • Atmosphere (atm) • mmHg • Torr

  3. Boyle’s Law: Pressure and Volume • Boyle was an Irish chemist who studied the relationship between volume and pressure • Boyle’s law states that the pressure and volume of a gas at constant temperature are inversely proportional.

  4. Boyle’s Law: Pressure and Volume • At a constant temperature, the pressure exerted by a gas depends on the frequency of collisions between gas particles and the container. • If the same number of particles is squeezed into a smaller space, the frequency of collisions increases, thereby increasing the pressure.

  5. Boyle’s Law: Pressure and Volume • In mathematical terms, this law is expressed as follows. • P1 = initial pressure • V1 = initial volume • P2 = final pressure • V2 = final volume • P1 & P2 can be in anything as long as they are the same • V1 & V2 can be in anything as long as they are the same

  6. Example • A sample of Helium gas is compressed from 4.0 L to 2.5 L at a constant temperature. If the pressure of the gas in the 4.0 L volume is 210 KPa, what will the pressure be at 2.5 L?

  7. Example • P1 = 210 KPa • P2 = ? • V1 = 4.0 L • V2 = 2.5 L • P1V1 = P2V2 • (210 KPa)(4.0L) = (P2)(2.5 L) • P2 = 340 KPa

  8. Charles’ Law: Volume & Temperature • Charles was a French physicist who looked at the relationship between temperature and volume • He noted that as temperature went up, so did volume when pressure was held constant

  9. Charles’ Law: Volume & Temperature • This observation is Charles’s law, which can be stated mathematically as follows.

  10. Charles’ Law: Volume & Temperature • V1 = V2 T1 T2 • V1 = initial volume • V2 = final volume • T1 = initial temperature • T2 = final temperature • V1 & V2 can be in any unit as long as they are the same • T1 & T2 MUST be in Kelvin

  11. Temperature conversions K = 273 + °C °C = 0.56 (°F – 32) °F = 1.8 °C + 32

  12. Example • A sample of gas at 40.0 °C occupies a volume of 2.32 L. If the temperature is raised to 75.0 °C what will the new volume be?

  13. Example • V1 = 2.32 L • V2 = ? • T1 = 40.0 °C = 313 K • T2 = 75.0 °C = 348 K • V1 = V2 T1 T2 • 2.32 L = V2 313K 348 K • V2 = 2.58 L

  14. Gay Lussac’s Law: Pressure & Temperature • Gay Lussac studied the relationship between pressure and temperature • He noticed that at a constant volume a direct relationship existed between the Kelvin temperature and volume • Giving the equation: • P1 = P2 T1 T2

  15. Gay Lussac’s Law: Pressure & Temperature • P1 = P2 T1 T2 • P1 = initial pressure • P2 = final pressure • T1 = initial temperature • T2 = final temperature • P1 & P2 can be in any unit as long as they are the same • T1 & T2 MUST be in Kelvin

  16. Example • The pressure of a gas in a tank is 3.20 atm at 22.0 °C. If the temperature rises to 60.0 °C, what will the new pressure in the tank be?

  17. Example • P1 = 3.20 atm • P2 = ? • T1 = 22.0 °C = 295 K • T2 = 60.0 °C = 333 K • P1 = P2 T1 T2 • 3.20 atm = P2 295K 333K • P2 = 3.61 atm

  18. Combined Gas Law P1V1 = P2V2 T1 T2 • Instead of memorizing all three equations, you can simply memorize this one • Just delete what you don’ t need

  19. Example • A gas at 110.0 kPa and 30.0°C fills a flexible container to a volume of 2.00 L. If the temperature was raised to 80.0°C and the pressure was increased to 440.0 kPa, what is the new volume?

  20. Example • P1V1 = P2V2 T1 T2 • P1 = 110.0 kPa • V1 = 2.00 L • T1 = 30.0 °C = 303 K • P2 = 440.0 kPa • V2 = ? • T2 = 80.0 °C = 353 K

  21. Example • P1V1 = P2V2 T1 T2 • (110.0)(2.00L) = (440.0kPa)(V2) 303K 353K • V2 = 0.583 L

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