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Gas Laws

Gas Laws. Kinetic Theory. Particles of matter are ALWAYS in motion . The volume of individual particles is approximately zero. Collision of particles with container wall causes pressure. Particles exert no forces on each other.

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Gas Laws

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  1. Gas Laws

  2. Kinetic Theory Particles of matter are ALWAYS in motion. The volume of individual particles is approximately zero. Collision of particles with container wall causes pressure. Particles exert no forces on each other. The average kinetic energy is approximately equal to the Kelvin temperature of gases.

  3. Measuring Pressure • The first device for measuring atmospheric pressure was developed by Evangelista Torricelli during the 17th century • The device was called a “barometer” • baro = weight • meter = measure

  4. The Early Barometer • The normal pressure due to the atmosphere at sea level can support a column of mercury that is 760 mm high

  5. Modern Day Barometers

  6. Caused by the collisions of molecules colliding with the wall of the container equal to force/unit area SI unit = Newton/meter2 = 1 pascal (Pa) 1 standard atmosphere = 101.31 kPa 1 standard atmosphere = 1 atm = 760 mmHg = 760 torr Pressure Units

  7. Pressure Units(KNOW THIS CHART!!!!!)

  8. Converting Celsius to Kelvin For every 1 °C you cool a gas, the volume decreases by 1/273. –273 °C is absolute zero (0 K) – the coldest temperature possible; all motion ceases. K = °C + 273 °C = K - 273

  9. Converting Celsius to Fahrenheit •  F = 9/5  C +32 •  C = 5/9( F-32)

  10. Standard Temperature and Pressure (STP) Pressure: 1 atm or 760 mmHg or 101.3 kPa Temperature: 0° C or 273 K MEMORIZETHIS!!!!

  11. The Nature of Gases Gases: expand to fill their container are fluid – they flow have low density are compressible effuse and diffuse

  12. Boyle’s Law Robert Boyle (1627-1691), Irish chemist The volume of a given amount of gas held at a constant temperature varies inversely with the pressure. As pressure increases, volume decreases. P1V1 = P2V2

  13. Boyle’s Law Solve the following: 825 Torr = _____ kPa Solve the following: The volume of oxygen at 120 kPa is 3.20 L. What is the volume of oxygen at 101.3 kPa? 101.3 kPa / 760 Torr = x kPa / 825 Torr 109.96 kPa = x P1V1 = P2V2 (120 kPa)(3.20L) = (101.3 kPa)V2 3.79 L = V2

  14. Charles’ Law Jacques Charles (1746-1823), French physicist The volume of a gas varies directly with its absolute temperature. As temperature increases, volume increases. V1 = V2 T1 T2

  15. Charles’ Law Cont. Solve the following: The volume of a gas at 40.0 °C is 4.50 L. Find the volume of the gas at 80 °C . 40°C + 273 = 313 K 80°C + 273 = 353 K V1/T1 = V2/T2 4.5L / 313 K = V2 / 353 K 9 L = V2

  16. Gay-Lussac’s Law Joseph Gay-Lussac The pressure of a given mass of gas varies directly with the Kelvin temperature when the volume remains constant. P1 = P2 T1 T2 *** Remember – temperature must be in Kelvin

  17. Solve the following: 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 be the gas pressure in the tank? Gay-Lussac’s Law Temp must be in Kelvin, so convert °C to K T1 = 22.0 °C + 273 = 295K T2 = 60°C + 273 = 333K P1/T1 = P2/T2 3.2 atm / 295K = x atm / 333K 3.61 atm = P2

  18. Combined Gas Law Boyle’s, Charles’s, and Gay Lussac’s laws can be combined into a single law. The combined gas law states the relationship among pressure, volume, and temperature of a fixed amount of gas. P1V1 = P2V2 T1 T2 *** Remember – temperature must be in Kelvin

  19. Combined Gas Law Solve the following: A gas at 110 kPa and 30.0 °C fills a flexible container with an initial volume of 2.00 L. If the temperature is raised to 80.0 °C and the pressure increased to 440 kPa, what is the new volume? Convert temp to Kelvin: T1 = 30°C + 273 = 303K T2 = 80°C + 273 = 353K P1V1/T1= P2V2/T2 (110 kPa)(2.00L) / 303K = (440 kPa)V2 / 353K 0.582 L = V2

  20. Ideal Gas Law In all the other gas laws, the relationships hold true for a “fixed mass” or “given amount” of a gas sample. Because pressure, volume, temperature, and the number of moles present are all interrelated, one equation is used to describe their relationship. PV = nRT n = moles R = ideal gas law constant V must be in liter T must be in Kelvin

  21. The value of R depends on pressure unit given. If pressure unit is mm Hg orTorr: R = 62.4L • mm Hg or Torr mol • K If pressure is atm: R = 0.0821 L • atm mol • K If pressure iskPa: R = 8.31 L • kPa mol • K

  22. Ideal Gas Law Solve the following: 44.01 g of CO2 occupies a certain volume at STP. Find that volume. n = sample mass molar mass n = 44.01 g 44.0098g/mol n = 1 mol (1atm)(V) = (1mol)(0.0821) (273 K) V = 22.4 L

  23. Solve the following: Find the molar mass of a gas that weights 0.7155 g and occupies a volume of 250. mL at STP. (1 atm)(0.250 L) = (n)(0.0821)(273 K) n = 0.011 mol 0.011 mol = 0.755 g X X = 68.64 g/mol

  24. Dalton’s Law of Partial Pressure The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone. Pt is the total pressure of a sample which contains a mixture of gases P1, P2, P3, etc. are the partial pressures of the gases in the mixture Pt = P1 + P2 + P3 + ...

  25. The pressure of a mixture of nitrogen, carbon dioxide, and oxygen is 150 kPa. What is the partial pressure of oxygen if the partial pressures of the nitrogen and carbon dioxide are 100 kPA and 24 kPa, respectively? Pt = P1 + P2 + P3 + ... P = PN2 + PCO2 + PO2 150 kPa = 100 kPa + 24 kPa + PO2 PO2 = 150 kPa - 100 kPa - 24 kPa PO2 = 26 kPa

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