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GAS LAWS

GAS LAWS. Kinetic Molecular Theory. Particles in an ideal gas… have no volume . have elastic collisions . are in constant , random, straight-line motion . don’t attract or repel each other. have an average KE directly related to Kelvin temperature. Real Gases.

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GAS LAWS

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  1. GAS LAWS

  2. Kinetic Molecular Theory • Particles in an ideal gas… • have no volume. • have elastic collisions. • are in constant, random, straight-line motion. • don’t attract or repel each other. • have an average KE directly related to Kelvintemperature.

  3. Real Gases • Particles in a REAL gas… • have their own volume • attract and repel each other • Gas behavior is most ideal… • at low pressures • at high temperatures ***Most real gases act like ideal gases except under high pressure and low temperature.

  4. Characteristics of Gases • Gases expand to fill any container. • Take the shape and volumeof their container. • Gases are fluids (like liquids). • Little to no attraction between the particles • Gases have very low densities. = lots of empty space between the particles

  5. Characteristics of Gases • Gases can be compressed. • lots of empty space between the particles • Indefinite density • Gases undergo diffusion. • random motion • scatter in all directions

  6. Pressure Which shoes create the most pressure?

  7. Pressure- how much a gas is pushing on a container. • Atmospheric pressure- atmospheric gases push on everything on Earth • UNITS AT SEA LEVEL 1 atm =101.3 kPa (kilopascal)= 760 mmHg =760 torr

  8. Aneroid Barometer Mercury Barometer Pressure • Barometer • measures atmospheric pressure

  9. U-tube Manometer Bourdon-tube gauge Pressure • Manometer • measures contained gas pressure C. Johannesson

  10. K = ºC + 273 ºF -459 32 212 ºC -273 0 100 K 0 273 373 Temperature=how fast the molecules are moving • Always use absolute temperature (Kelvin) when working with gases. C. Johannesson

  11. Standard Temperature & Pressure 0°C 273 K 1 atm 101.3 kPa 760 mm Hg -OR- STP -OR-

  12. Volume = how much space a gas occupies Units • L, mL, cm3 • 1000 mL = 1 L • 1 mL = 1 cm3

  13. V T P BASIC GAS LAWS

  14. V T Charles’ Law • T  V (temperature is directly proportional to volume) • T ↑ V↑ & T↓ V↓ • V1 = V2 T1 T2T is always in K • K = °C + 273 • P and n = constant

  15. V1V2 = T1 T2 Charles’ Law (Pressure is held constant) Timberlake, Chemistry 7th Edition, page 259

  16. Charles’ Law Timberlake, Chemistry 7th Edition, page 254

  17. Charles’ Law The egg out of the bottle Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  18. Charles’ Law Problem Mrs. Rodriguez inflates a balloon for a party. She is in an air-conditioned room at 27.0oC, and the balloon has a volume of 4.0 L. Because she is a curious and intrepid chemistry teacher, she heats the balloon to a temperature of 57.0oC. What is the new volume of the balloon if the pressure remains constant? • Given Unkown Equation • Substitute and Solve T1 = 27.0oC +273= 300 K V1 = 4.0 L T2 = 57.0oC +273= 330 K P1V1= P2V2 T1 V1T2 V2 = ? L 4.0 L = V2= 300 K 330K 4.4 L

  19. Charles’ Law Learning Check A 25 L balloon is released into the air on a warm afternoon (42º C). The next morning the balloon is recovered on the ground. It is a very cold morning and the balloon has shrunk to 22 L. What is the temperature in º C? • Given Unkown Equation • Substitute and Solve V1 = 25 L T1 = 42 oC +273= 315 K V2 = 22 L P1V1= P2V2 T1 V1T2 T2 = ? ºC 25 L = 22 L = 315 K T2 277.2 K -273 = 4.2 ºC

  20. P V Boyle’s Law • P↓ V ↑ & P↑ V ↓ • P  1/V (pressure is inversely proportional to volume) • P1V1 = P2V2 • T and n = constant

  21. Boyle’s Law P1V1 = P2V2 (Temperature is held constant) Timberlake, Chemistry 7th Edition, page 253

  22. Boyle’s Law • Marshmallows in a vacuum Timberlake, Chemistry 7th Edition, page 254

  23. Mechanics of Breathing Boyle’s Law Timberlake, Chemistry 7th Edition, page 254

  24. Boyle’s Law Problem P1V1= P2V2 T1 T2 V1 = 30 L P1 = 1 atm P2 = .25atm V2 = ? L • V2 0.25 atm= 30 L x 1.0 atm = 120 L A balloon is filled with 30.L of helium gas at 1.00 atm. What is the volume when the balloon rises to an altitude where the pressure is only 0.25 atm? Given UnkownEquation Substitute and Solve

  25. Boyle’s Law Learning Check V1 = 100. mL = 0.100 L P1 = 150. kPa P2 = 200. kPa P1V1= P2V2 T1 T2 V2 = ? L • V2 x 200. kPa= 0.100 L x 150. kPa= 75.0 mL • 0.0750 L A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. Given UnkownEquation • Substitute and Solve

  26. AVOGADRO’S LAW • Vn Vn • V n (direct) • V1 = V2 n1 n2 • T & P Constant V n

  27. Avogadro’s Law Problem P1V1= P2V2 n1T1 n2T2 V1 = 3.0 L n1 = 7.0 mol V2 = 2.3 L n2 = ? mol 3.0 L = 2.3 L = 7.0 mol n2 mol 5.4 mol A 3.0 liter sample of gas contains 7.0 moles. How much gas will there be, in order for the sample to be 2.3 liters? P & T do not change Given UnkownEquation • Substitute and Solve

  28. P T Gay-Lussac’s Law • P1 = P2 T1 T2 • V & n constant • Direct relationship • PT PT 

  29. Gay-Lussac Law • Collapsing Barrel

  30. Gay-LussacLaw • Tank car implosion

  31. COMBINED IDEAL GAS LAW • P1V1 = P2V2 n1T1 n2T2 • If P, V, n, or T are constant then they cancel out of the equation. • n usually constant (unless you add or remove gas), so • P1V1 = P2V2 T1 T2

  32. Combined Gas Law Problem • Ms. Evans travels to work in a hot air balloon from the Rocky Mountains. At her launch site, the temperature is 5.00 °C, the atmospheric pressure is 0.801 atm, and the volume of the air in the balloon is 120.0 L. When she lands in Plano, the temperature is 28.0 °C and the atmospheric pressure is 101.3 kPa. What is the new volume of the air in the balloon? • Given Unkown Equation • Substitute and Solve T1 = 5.0oC +273= 278 K P1 = 0.801 atm V1 = 120.0 L T2 = 28.0oC +273= 301 K P2 = 101.3 kPa = 1 atm V2 = ? L V1 x P1= V 2 x P2 T 1 T 2 • V2 x 1 atm= 120.0 L x 0.801 atm= 104 L • 301K 278 K

  33. Combined Gas Law Learning Check • Nitrogen gas is in a 7.51 L container at 5.C and 0.58 atm. What is the new volume of the gas at STP? • Given Unkown Equation • Substitute and Solve T1 = 5.0oC +273= 278 K P1 = 0.58 atm V1 = 7.51 L T2 = 273 K P2 = 1 atm V2 = ? L V1 x P1= V 2 x P2 T 1 T 2 • V2 x 1.0 atm= 7.51L x 0.58atm= 4.3 L • 273 K 278 K

  34. Ideal Gas Law (“Pivnert”) PV=nRT R = The Ideal Gas Constant R = 0.0821 (L*atm) R = 62.4 (L*mm Hg) (mol*K) (mol*K) R = 8.31 (L*kPa) (mol*K) • V has to be in Liters, n in Moles, T in Kelvin, • P can be in atm, kPa or mmHg • * Choose which R to used based on the units of your pressure. P V = n R T (atm) (L) = (moles) (L*atm/mol*K) (K) (kPa) (L) = (moles) (L*kPa/mol*K) (K) mm Hg (L) = (moles) (L*mmHg/mol*K) (K)

  35. Ideal Gas Law Problem • A rigid steel cylinder with a volume of 20.0 L is filled with nitrogen gas to a final pressure of 200.0 atm at 27.0 oC. How many moles of gas does the cylinder hold? • Given Unkown Equation • Substitute and Solve V = 20.0 L P = 200.0 atm T =27.0oC +273= 300 K PV=nRT R= .0821 atm L/K Mole moles of nitrogen? • n0821 atm L/K Mole x 300 K = 200.0 atm x 20.0L= 162 moles

  36. Ideal Gas Law Learning Check • A balloon contains 2.00 mol of nitrogen at a pressure of 0.980 atm and a temperature of 37C. What is the volume of the balloon? • Given Unkown Equation • Substitute and Solve n = 2.00 mol P = 0.980 atm T =37.0oC +273= 310 K PV=nRT R= .0821 atm L/K Mole V in L? • 0.980 atm x V= 2.00 mol x .0821 atm L/K Mole x 310 K = 51.9 L

  37. Dalton’s Law of Partial Pressure • The total pressure of a mixture of gases is equal to the sum of the partial pressures of the component gases. • Ptotal = Pgas 1 + Pgas 2 + P­gas 3 + … A metal container holds a mixture of 2.00 atm of nitrogen, 1.50 atm of oxygen and 3.00 atm of helium. What is the total pressure in the canister? 6.5 atm

  38. 1 mol = 6.02 x 1023 particles Welcome to Mole Island

  39. Welcome to Mole Island 1 mol = molar mass

  40. Welcome to Mole Island 1 mole = 22.4 L @ STP

  41. Gas Stoichiometry Moles  Liters of a Gas: • 2C4H10 (g) + 13O2(g) ͢ 8CO2(g) + 10H2O(g) 2 mol + 13 mol ͢ 8 mol + 10 mol 2 L + 13 L ͢ 8 L + 10 L Recall:The coefficients in a chemical reaction represent molar amounts of substances taking part in the reaction. Avogadro’s principle states that one mole of any gas occupies 22.4 L at STP. Thus when gases are involved, the coefficients in a balanced chemical equation represent not only molar amounts but also relatives volumes Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  42. Gas Stoichiometry Problem In the following combustion reaction, what volume of methane (CH4) is needed to produce 26 L of water vapor? • CH4 (g) + 2O2(g) ͢ CO2(g) + 2H2O(g) x L ͢ 26 L 1 mol ͢ 2 mol 1 L ͢ 2 L x L = 26 L 1L 2L x= 13 L Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  43. Gas Stoichiometryuse ideal gas law PV=nRT • Looking for grams or moles of gas? • Step 1: start with ideal gas law to find moles of gas • Step 2: 1change to grams of gas Grams/mol? 1) Use Ideal Gas Law 2) Do stoichiometry calculations Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  44. Example 1 4 Al(s) + 3 O2(g)  2 Al2O3(s) How many grams of Al2O3 are formed from 15.0 L of O2 at 97.3 kPa & 21°C? PV=nRT • Given Unkown • VO2= 15.0 L O2 grams of Al2O3? R= .0821 atm L/K Mole • P O2 = 97.3 kPa= 0.9605 atm • T O2 =21oC +273= 294 K • Step 1: Calculate moles of O2 • n = PV = 0.9605 atm x 15.0 L = 0.5969 mol O2 • RT 0.0821 atm L/K Mole 294 K Given liters: Start with Ideal Gas Law and calculate moles of O2. Use stoich to convert moles of O2 to grams Al2O3. Step 2: Calculate mass of Al2O3 0.5969 molO2= X mol Al2O3= 0.3979 mol Al2O3 3moleO2 2 mole Al2O3 0.3979 mol Al2O3 x 101.96 g Al2O3= 1 mol Al2O3 41 g Al2O3 Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  45. Gas Stoichiometryuse ideal gas law PV=nRT • Looking for volume of gas? • Step 1: start with stoichiometry conversion to find moles of gas • Step 2: use ideal gas law to find the volume Liters ? 1) Do stoichiometry calculations 2) Use Ideal Gas Law Courtesy Christy Johannesson www.nisd.net/communicationsarts/pages/chem

  46. Example 2 What volume of CO2 forms from 5.25 g of CaCO3 at 101.3 kPa & 25ºC? CaCO3 CaO + CO2 • Given UnkownPV=nRT • m = 5.25 g CaCO3 volume of CO2? R= .0821 atm L/K Mole • P = 101.3 kPa = 1 atm T =25.0oC +273= 298 K • Step 1: Calculate moles of CO2 • 5.25 g CaCO3 x 1 mole CaCO3 = 0.0525 mol CaCO3 • 100 g CaCO3 • 1 mole CO2 = 1mole CaCO3 ; 0.0525 molCO2 Looking for liters: Start with stoich and calculate moles of CO2. Plug this into the Ideal Gas Law to find volume. Step 2: Calculate volume of CO2 V = nRT= 0.0525 mol CO2 x .0821 atm L/K Mole x298 K = 1.28 L P 1 atm

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