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Pressure Force acting on a unit area of a surface

Pressure Force acting on a unit area of a surface. Standard Atmospheric Pressure Barometric Pressure. Air has weight and therefore exerts pressure. A standard Atmosphere (1atm) is the average pressure at sea level. What happens to air pressure as you increase your elevation?

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Pressure Force acting on a unit area of a surface

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  1. PressureForce acting on a unit area of a surface

  2. Standard Atmospheric PressureBarometric Pressure • Air has weight and therefore exerts pressure. • A standard Atmosphere (1atm) is the average pressure at sea level. • What happens to air pressure as you increase your elevation? • How do we measure pressure?

  3. Kinetic Molecular Theory • Particles in an ideal gas… • have elastic collisions. • are in constant, random, straight-line motion. • don’t attract or repel each other. • have an avg. KE directly related to Kelvin temperature. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  4. Characteristics of Ideal Gases Gases expand to fill any container. Gases are fluids (like liquids). Gases have very low densities. Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  5. Characteristics of Gases • Gases can be compressed. • Gases undergo diffusion. • random motion Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  6. Properties of Gases V = volume of the gas (liters, L) T = temperature (Kelvin, K) P = pressure (atmospheres, atm) n = amount (moles, mol) Gas properties can be modeled using math. Model depends on:

  7. PT V P T V PT V 1 V Boyle’s P a ___ a Charles V T a Gay-Lussac’s P T Pressure - Temperature - Volume Relationship

  8. Boyle’s Law P1V1 = P2V2 (Temperature is held constant)

  9. As the pressure on a gas increases - the volume decreases • Pressure and volume are inversely related 1 atm • As the pressure on a gas increases 2 atm 4 Liters 2 Liters

  10. Pressure vs. Volume for a Fixed Amount of Gas(Constant Temperature) Pressure Volume PV (Kpa) (mL) 100 500 50,000 150 333 49,950 200 250 50,000 250 200 50,000 300 166 49,800 350 143 50,500 400 125 50,000 450 110 49,500 600 500 400 300 Volume (mL) 200 100 0 100 200 300 400 500 Pressure (KPa)

  11. Boyle’s Law Illustrated Zumdahl, Zumdahl, DeCoste, World of Chemistry2002, page 404

  12. Pressure Units • Atmosphere • Feet of water • mm Hg • cm Hg • Torr • Barr • mbarr • kPa • Pa

  13. PV Calculation (Boyle’s Law) A quantity of gas has a volume of 120 dm3 when confined under a pressure of 93.3 kPa at a temperature of 20 oC. At what pressure will the volume of the gas be 30 dm3 at 20 oC? P1 x V1 = P2 x V2 (93.3 kPa) x (120 dm3) = (P2) x (30 dm3) P2 = 373.2 kPa

  14. One-liter flask Volume and Pressure Two-liter flask The average molecules hits the wall twice as often. The total number of impacts with the wall is doubled and the pressure is doubled. The molecules are closer together; the density is doubled. Bailar, Jr, Moeller, Kleinberg, Guss, Castellion, Metz, Chemistry, 1984, page 101

  15. V1 V2 = T1 T2 Charles' Law This means, for example, that Volumegoes up as Temperaturegoes up. V and T are directly related. A hot air balloon is a good example of Charles's law. Jacques Charles (1746 - 1823) Isolated boron and studied gases. Balloonist. (Pressure is held constant)

  16. Temperature • Raising the temperature of a gas increases the pressure if the volume is held constant. • The molecules hit the walls harder. • The only way to increase the temperature at constant pressure is to increase the volume.

  17. VT Calculation (Charles’ Law) At constant pressure, the volume of a gas is increased from 150 dm3 to 300 dm3 by heating it. If the original temperature of the gas was 20 oC, what will its final temperature be (oC)? 150 dm3 293 K 300 dm3 T2 T1 = 20 oC + 273 = 293 K T2 = X K V1 = 150 dm3 V2 = 300 dm3 = T2 = 586 K oC = 586 K - 273 T2 = 313 oC

  18. Temperature and the Pressure of a Gas High in the mountains, Richard checked the pressure of his car tires and observed that they has 202.5 kPa of pressure. That morning, the temperature was -19 oC. Richard then drove all day, traveling through the desert in the afternoon. The temperature of the tires increased to 75 oC because of the hot roads. What was the new tire pressure? Assume the volume remained constant. What is the percent increase in pressure? P1 = 202.5 kPa P2 = X kPa T1 = -19 oC + 273 = 254 K T2 = 75 oC + 273 = 348 K 202.5 kPa 254 K P2 348 K = P2 = 277 kPa % increase = 277 kPa - 202.5 kPa x 100 % 202.5 kPa or 37% increase

  19. P T Gay-Lussac’s Law The pressure and absolute temperature (K) of a gas are directly related • at constant mass & volume Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  20. The Combined Gas Law When measured at STP, a quantity of gas has a volume of 500 dm3. What volume will it occupy at 0 oC and 93.3 kPa? (101.3 kPa) x (500 dm3) = (93.3 kPa) x (V2) 273 K 273 K (101.3) x (500) = (93.3) x (V2) P1 = 101.3 kPa T1 = 273 K V1 = 500 dm3 P2 = 93.3 kPa T2 = 0 oC + 273 = 273 K V2 = X dm3 V2 = 542.9 dm3

  21. Standard Temperature and Pressure • 0o C • 1 atm = 760 mmHg = 760 torr = 101.3 kPa

  22. Collecting Gasses over a liquid (such as water) • Some water vapor will be mixed with the collected gas and will exert pressure. • The partial pressure of the water vapor for each temperature is constant. • So, take the pressure of the gas and subtract the partial pressure of water at the given temperature.

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