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The Behavior of Gases Unit 9

The Behavior of Gases Unit 9. GENERAL CHEMISTRY SPRING 2010 Mr. Hoffman Mrs. Paustian. Air Pressure. Pressure equation. Newtons (N). Square meters (m 2 ). Unit for Air Pressure. f P a. Pascals 1 N/m 2 = 1 Pascal (Pa) Standard pressure is 101.325 kPa on Earth.

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The Behavior of Gases Unit 9

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  1. The Behavior of GasesUnit 9 GENERAL CHEMISTRYSPRING 2010 Mr. HoffmanMrs. Paustian

  2. Air Pressure • Pressure equation Newtons (N) Square meters (m2)

  3. Unit for Air Pressure f P a • Pascals • 1 N/m2 = 1 Pascal (Pa) • Standard pressure is 101.325 kPa on Earth 1 kPa = 1000 Pa! Blaise Pascal (French philosopher and mathematician) Pascal’s Triangle

  4. Air Pressure f P a • Without changing the mass (force), how can you increase pressure? • Without changing the mass (force), how can you decrease pressure? Decrease the area Increase area

  5. Practice Problem f P a • The mass of a brick is 2.0 kg (F=19.6N), the sides are 0.05m and 0.03m. What is the pressure exerted?

  6. Pressure Units Torr is named after Evangelista Torricelli • Ways to represent pressure

  7. Etymology of “barometer” In Greek, “baros”= weight Meter= measure Literally means “measure the weight of air” or air pressure. The Barometer Bariatric surgeory is weight loss surgeory $5 Footlong

  8. How a barometer works Air presses down on an open tray of Hg This downward pressure pushes the Hg up into a tube Higher air pressure causes the mercury to go higher up in the tube (measured as height, mm Hg) The Barometer

  9. Compressibility A gas will expand to fill its container Compressibility A measure of how much the volume of matter decreases under pressure. Gases easily compress because of the space between the particles http://www.garagelibrary.com/images/airbag.jpg

  10. Kinetic Molecular Theory • Postulates the theory is based on… • Gases consist of tiny particles (of negligible mass) with great distances separating them • Gases are in constant random motion • Molecular collisions are elastic • Average kinetic energy is dependent only ontemperature

  11. The Kelvin Scale • As T increases, so does kinetic energy • Theoretically, kinetic energy can be zero, but it hasn’t been achieved and probably won’t ever be achieved • Absolute zero- The temperature at which a substance would have zero kinetic energy • The Kelvin Scale- a temperature scale directly related to kinetic energy • Zero on the Kelvin scale corresponds to zero kinetic energy

  12. The Kelvin Scale • Units are Kelvins (K), with no degree (o) sign Kelvin relates temperature to kinetic energy!

  13. Easy to convert between Celsius and Kelvin How do you think? oC  K? Add 273 K  oC? Substract 273 25oC  K? (25+273)= 298 K 310 K  oC? (310-273) = 37oC Fahrenheit Celsius? (oF – 32oF) x 5/9 = oC (oC x 9/5) + 32oF = oF Temperature Conversions

  14. Factors Affecting Gas Pressure Amount of gas Volume Temperature http://www.bmumford.com/photo/highspeed/Ted1.jpg

  15. Boyle’s Law • “Boyles inverse” • States that the volume of a gas varies inversely (opposite) with pressure if temperature and amount are held constant • Written: • P1V1 = P2V2 Robert Boyle

  16. Boyle’s Law in motion http://www.grc.nasa.gov/WWW/K-12/airplane/Animation/gaslab/Images/chprmt.gif

  17. Boyle’s Practice 1 • A tank contains a volume of 3 L and a pressure of 4 atmospheres. What volume would the gas from this tank fill up at a pressure of 1 atmosphere? • P1 = 4atm • V1 = 3L • P2 = 1 atm • V2 = ? Substitute into Boyle’s Law equation and solve for V2 P1V1 = P2V2 (4 atm)(3L) = (1atm)(V2) 12 atm*L = V2 1 atm V2= 12 L

  18. Boyle’s Practice 2 • Find the volume of a cylinder needed if you want to put 50 atmospheres of pressure with a volume of 3 L into a cylinder that can hold a pressure of no greater than 20 atmospheres. • P1= 50 atm • V1= 3 L • P2= 20 atm • V2= ? Substitute into Boyle’s Law equation and solve for V2 P1V1 = P2V2 (50 atm)(3 L) = (20 atm)(V2) 7.5 L = V2

  19. Charles’ Law • “Charles direct” • The volume of a gas varies directly with temperature if the pressure and amount remain constant • Mathematically:

  20. Charles’ Law in Motion

  21. Charles’ Law Problem 1 Always convert temperature to Kelvin (K) A tank contains a volume of 3 L and a temperature of 100oC. What volume would the gas from this tank fill up at a temperature of 200oC? V1= 3 L T1= 100oC = 373K V2= ? T2= 200oC= 473K Substitute into equation and solve for V2 3L V2 = 373K 473K (3L)(473K) = (V2)(373K) 4L = V2

  22. Charles’ Law Problem 2 A 275 L helium balloon is heated from 20oC to 40oC. Calculate the final volume assuming pressure remains constant. V1= 275 L T1= 20oC (+273K)= 293K V2= ? T2= 40+273= 313K 275L = V2 293K 313K (293)(v2) = (275L)(313K) V2= 294 L

  23. Temperature-Pressure Relationships Gay-Lussac’s Law The pressure of a gas varies directly w/ temperature if volume and amount remain constant. Mathematically: Tire inflation…. Summer vs. winter…hmmm how do these two differ? Joseph Louis Gay-Lussac 1778-1850

  24. Gay-Lussac Problem 1 Temperature in Kelvin (K) A tank contains a pressure of 3 atm and a temperature of 100oC. What pressure would the gas from this tank be at a temperature of 200oC? P1= 3 atm T1= 100oC (+273K)= 373K P2= ? T2= 200oC= 473K 3 atm = P2 373 K 473 K P2= 4 atm

  25. Gay-Lussac Problem 2 Find the pressure needed if you wanted to put gas at 50oC and 75 atm into a vessel that is at 65oC. P1= 75 atm T1= 50oC= 323 K P2= ? T2= 65oC= 338 K 75 atm = P2 323 K 338 K P2 = 78 atm

  26. Combined Gas Law Puts together several scientists’ work on how gases behave when conditions are changes. You can change three things about a gas Amount of gas- this stays the same for now Temperature Volume Pressure changes in response to these STP??Temperature= 273KPressure= 1 atm

  27. CBL Problem 1 A 50 mL sample of hydrogen gas is collected at 772 mm Hg and 21oC. Calculate the volume of hydrogen at STP. P1= 772 mm Hg  atm V1= 50 mL  L T1= 21oC= 294 K P2= 1 atm V2= X T2= 273K Since your final conditions are at STP, you need to convert initial conditions to atm and L

  28. CBL Problem 1 solution P1= 772 mm Hg x 1 atm/760 mm Hg= 1.016 atm V1= 50 mL / 1000 mL= 0.05 L (1.016 atm)(0.05L) = (1atm)(V2) 294 K 273 K Cross multiply, then solve for V2 (1.016 atm)(0.05L)(273 K) = (1atm)(294K)(X) 0.05 L = V2

  29. Ideal Gases • Ideal gases are… • Gases that behave under all conditions as predicted by the kinetic molecular theory • Gases are not ideal… • When they don’t behave as predicted by kinetic molecular theory • Kinetic energy is… • Energy associated with motion

  30. The Ideal Gas Law “piv-nert” PV=nRT P= Pressure (atm) V= Volume (L) n= # moles (mol) R= Ideal gas constant= 0.0821 (L*atm)/(mol*K) T= Temperature (ALWAYS Kelvin)

  31. IGL Problem 1 If a container has a volume of 3 L and is at a temperature of 60oC and a pressure of 6 atm, how many moles of gas are in the container? V= 3L T= 60oC= 333K P= 6 atm R= 0.0821 L*atm/mol*K n= ?

  32. IGL Problem 1 Solution Substitute in PV=nRT and solve for n (6 atm)(3 L) = (n)(0.0821)(333 K) n= 0.7 mol The unit for R is L*atm mol*K Be sure units match this, and units will cancel

  33. IGL Problem 2 If a container has 50 moles of O2 and is at a temperature of 40oC and a pressure of 3 atm, how many liters are in this container? n= 50 mol O2 T= 40oC = 313 K P= 3 atm R= 0.0821 L*atm/mol*K V= ?

  34. IGL Problem 2 Solution Substitute in PV=nRT and solve for V (3atm)(V) = (50 mol)(0.0821)(313 K) V = 428 L

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