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The Gas Laws Chemistry Dr. May

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Gaseous Matter Indefinite volume and no fixed shape Indefinite volume and no fixed shape Particles move independently of each other Particles move independently of each other Particles have gained enough energy to overcome the attractive forces that held them together as solids and liquids Particles have gained enough energy to overcome the attractive forces that held them together as solids and liquids

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Avogadros Number One mole of a gas contains Avogadros number of molecules One mole of a gas contains Avogadros number of molecules Avogadros number is Avogadros number is 602,000,000,000,000,000,000,000 6.02 x 10 23 or

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Diatomic Gas Elements Gas Hydrogen (H 2 ) Hydrogen (H 2 ) Nitrogen (N 2 ) Nitrogen (N 2 ) Oxygen (O 2 ) Oxygen (O 2 ) Fluorine (F 2 ) Fluorine (F 2 ) Chlorine (Cl 2 ) Chlorine (Cl 2 ) Molar Mass 2 grams/mole 28 grams/mole 32 grams/mole 38 grams/mole 70 grams/mole

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Inert Gas Elements Gas Helium Helium Neon Neon Argon Argon Krypton Krypton Xenon Xenon Radon Radon Molar Mass 4 grams/mole 20 grams/mole 40 grams/mole 84 grams/mole 131 grams/mole 222 grams/mole

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Other Important Gases Gas Carbon Dioxide Carbon Dioxide Carbon Monoxide Carbon Monoxide Sulfur Dioxide Sulfur Dioxide Methane Methane Ethane Ethane Freon 14 Freon 14 Formula Molar Mass CO 2 44 g/mole CO28 g/mole SO 2 64 g/mole CH 4 16 g/mole CH 3 CH 3 30 g/mole CF 4 88 g/mole

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One Mole of Oxygen Gas (O 2 ) Has a mass of 32 grams Has a mass of 32 grams Occupies 22.4 liters at STP Occupies 22.4 liters at STP 273 Kelvins (0 o C) 273 Kelvins (0 o C) One atmosphere (101.32 kPa)(760 mm) One atmosphere (101.32 kPa)(760 mm) Contains 6.02 x 10 23 molecules Contains 6.02 x 10 23 molecules (Avogadros Number)

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Mole of Carbon Dioxide (CO 2 ) Has a mass of 44 grams Has a mass of 44 grams Occupies 22.4 liters at STP Occupies 22.4 liters at STP Contains 6.02 x 10 23 molecules Contains 6.02 x 10 23 molecules

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One Mole of Nitrogen Gas (N 2 ) Has a mass of 28 grams Has a mass of 28 grams Occupies 22.4 liters at STP Occupies 22.4 liters at STP Contains 6.02 x 10 23 molecules Contains 6.02 x 10 23 molecules

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Mole of Hydrogen Gas (H 2 ) Mass Mass Volume at STP Volume at STP Molecules Molecules 2.0 grams 22.4 liters 6.02 x 10 23

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Standard Conditions (STP) Molar Volume Molar Volume Standard Standard Temperature Temperature Standard Pressure Standard Pressure 22.4 liters/mole 0 o C 273 Kelvins 1 atmosphere 101.32 kilopascals 760 mm Hg

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Gas Law Unit Conversions liters milliliters liters milliliters milliliters liters milliliters liters o C Kelvins o C Kelvins Kelvins o C Kelvins o C mm atm mm atm atm mm atm mm atm kPa atm kPa kPa atm kPa atm Multiply by 1000 Divide by 1000 Add 273 Subtract 273 Divide by 760 Multiply by 760 Multiply by 101.32 Divide by 101.32

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Charles Law At constant pressure, the volume of a gas is directly proportional to its temperature in Kelvins At constant pressure, the volume of a gas is directly proportional to its temperature in Kelvins V 1 = V 2 T 1 T 2 T 1 T 2 As the temperature goes up, the volume goes up As the temperature goes up, the volume goes up

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Boyles Law At constant temperature, the volume of a gas is inversely proportional to the pressure. At constant temperature, the volume of a gas is inversely proportional to the pressure. P 1 V 1 = P 2 V 2 As the pressure goes up, the volume goes down As the pressure goes up, the volume goes down

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Combined Gas Law P 1 V 1 = P 2 V 2 P 1 V 1 = P 2 V 2 T 1 T 2 T 1 T 2 Standard Pressure (P) = 101.32 kPa, 1 atm, or 760 mm Hg Standard Pressure (P) = 101.32 kPa, 1 atm, or 760 mm Hg Standard Temperature (T) is 273 K Standard Temperature (T) is 273 K Volume (V) is in liters, ml or cm 3 Volume (V) is in liters, ml or cm 3

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Charles Law Problem A balloon with a volume of 2 liters and a temperature of 25 o C is heated to 38 o C. What is the new volume? 1. Convert o C to Kelvins 25 + 273 = 298 K 38 + 273 = 311 K 2. Insert into formula

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Charles Law Solution V 1 = V 2 T 1 T 2 T 1 T 2 V 1 = 2 litersV 2 = Unknown V 1 = 2 litersV 2 = Unknown T 1 = 298 KT 2 = 311 K T 1 = 298 KT 2 = 311 K 2 = V 2 298 311

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Charles Law Solution 2 = V 2 2 = V 2 298 311 298 311 298 V 2 = (2) 311 V 2 = 622 298 298 V 2 = 2.09 liters

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Charles Law Problem Answer A balloon with a volume of 2 liters and a temperature of 25 o C is heated to 38 o C. What is the new volume? V 2 = 2.09 liters

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Boyles Law Problem A balloon has a volume of 2.0 liters at 743 mm. The pressure is increased to 2.5 atmospheres (atm). What is the new volume? 1. Convert pressure to the same units 743 760 =.98 atm 2. Insert into formula

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Boyles Law Solution P 1 V 1 = P 2 V 2 P 1 = 0.98 atmP 2 = 2.5 atm V 1 = 2.0 litersV 2 = unknown 0.98 (2.0) = 2.5 V 2

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Boyles Law Solution P 1 V 1 = P 2 V 2 0.98 (2.0) = 2.5 V 2 V 2 = 0.98 (2.0) 2.5 2.5 V 2 = 0.78 liters

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Boyles Law Problem Answer A balloon has a volume of 2.0 liters at 743 mm. The pressure is increased to 2.5 atmospheres (atm). What is the new volume? V 2 = 0.78 liters

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Combined Gas Law Problem A balloon has a volume of 2.0 liters at a pressure of 98 kPa and a temperature of 25 o C. What is the volume under standard conditions? 1. Convert 25 o C to Kelvins 25 + 273 = 298 K 2. Standard pressure is 101.32 kPa 3. Standard temperature is 273 K 4. Insert into formula

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Combined Gas Law Solution P 1 V 1 = P 2 V 2 T 1 T 2 T 1 T 2 P 1 = 98 kPaP 2 = 101.32 kPa V 1 = 2.0 litersV 2 = unknown T 1 = 298 KT 2 =273 K

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Combined Gas Law Solution P 1 V 1 = P 2 V 2 T 1 T 2 T 1 T 2 98 (2.0) = 101.32 V 2 298 273 298 273 (298) (101.32) V 2 = (273) (98) (2.0)

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Combined Gas Law Solution P 1 V 1 = P 2 V 2 T 1 T 2 T 1 T 2 (298) (101.32) V 2 = (273) (98) (2.0) V 2 = 273 (98) (2.0) (298) (101.32) (298) (101.32)

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Combined Gas Law Solution P 1 V 1 = P 2 V 2 T 1 T 2 T 1 T 2 V 2 = 273 (98) (2.0) (298) (101.32) (298) (101.32) V 2 = 53508 = 1.77 liters 30193 30193

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Combined Gas Law Problem Answer A balloon has a volume of 2.0 liters at a pressure of 98 kPa and a temperature of 25 o C. What is the volume under standard conditions? V 2 = 1.77 liters

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Combined Gas Law – V 2 P 1 V 1 = P 2 V 2 T 1 T 2 T 1 T 2 P 1 V 1 T 2 = P 2 V 2 T 1 P 1 V 1 T 2 = V 2 P 2 T 1 P 2 T 1

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The End This presentation was created for the benefit of our students by the Science Department at Howard High School of Technology This presentation was created for the benefit of our students by the Science Department at Howard High School of Technology Please send suggestions and comments to rmay@nccvt.k12.de.us Please send suggestions and comments to rmay@nccvt.k12.de.us

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The Ideal Gas Law Chemistry Dr. May

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Kinetic Molecular Theory Molecules of an ideal gas Are dimensionless points Are dimensionless points Are in constant, straight-line motion Are in constant, straight-line motion Have kinetic energy proportional to their absolute temperature Have kinetic energy proportional to their absolute temperature Have elastic collisions Have elastic collisions Exert no attractive or repulsive forces on each other Exert no attractive or repulsive forces on each other

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Ideal Gas Law PV = nRT P = pressure in kilopascals (kPa) or atmospheres (atm) V = volume in liters n = moles T = temperature in Kelvins R = universal gas constant

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Ideal Gas Law: PV = nRT Pressure (P) Pressure (P) Volume (V) Volume (V) Moles (n) Moles (n) Temperature (T) Temperature (T) The universal gas constant (R) The universal gas constant (R) Atm or kPa Always liters Moles Kelvins 0.0821 ( P in atm) or 8.3 (P in kPa)

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Universal Gas Constant R = 0.0821 if P = atmospheres R = 8.3 if P = kilopascals R = PV nT nT

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Deriving R for P in Atmospheres R = PV nT nT Assume n = 1 mole of gas Standard P = 1 atmosphere Standard V = molar volume = 22.4 liters Standard T = 273 Kelvins

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R Value When P Is In Atmospheres R = PV nT nT R = (1) (22.4) (1) 273 (1) 273 R = 0.0821 atm Liters mole Kelvins mole Kelvins

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Deriving R For P In Kilopascals R = PV nT nT Assume n = 1 mole of gas Standard P = 101.32 kilopascals Standard V = molar volume = 22.4 liters Standard T = 273 Kelvins

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R Value When P Is In Kilopascals R = PV nT nT R = (101.32) (22.4) (1) 273 (1) 273 R = 8.3 kPa Liters mole Kelvins mole Kelvins

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Ideal Gas Law - Pressure PV = nRT P = nRT V V Solves for pressure when moles, temperature, and volume are known

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Ideal Gas Law - Volume PV = nRT V = nRT P P Solves for volume when moles, temperature, and pressure are known

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Ideal Gas Law - Temperature PV = nRT T = PV nR nR Solves for temperature when moles, pressure, and volume are known

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Ideal Gas Law - Moles PV = nRT n = PV RT RT Solves for moles when pressure, temperature, and volume are known

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Ideal Gas Law Problem What is the mass of nitrogen in a 2.3 liter container at 1.2 atmospheres, and 25 o C ? V = 2.3 liters V = 2.3 liters P = 1.2 atmospheres P = 1.2 atmospheres T = 25 o C = 298 Kelvins T = 25 o C = 298 Kelvins R = 0.0821 since P is in atms. R = 0.0821 since P is in atms. Find moles (n), then grams Find moles (n), then grams

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Ideal Gas Law Solution (moles) PV = nRT 1.2 (2.3) = n (0.0821) (298) n = 1.2 ( 2.3) = 0.11 moles n = 1.2 ( 2.3) = 0.11 moles (0.0821) (298) (0.0821) (298)

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Ideal Gas Law Solution (Grams) Grams = moles x molecular weight (MW) Moles = 0.11 Moles = 0.11 Molecular Weight of N 2 = 28 g/mole Molecular Weight of N 2 = 28 g/mole Grams = 0.11 x 28 = 3.1 grams Grams = 0.11 x 28 = 3.1 grams

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Ideal Gas Law Answer What is the mass of nitrogen in a 2.3 liter container at 1.2 atmospheres, and 25 o C ? The answer is 0.11 moles and 3.1 grams

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The End This presentation was created for the benefit of our students by the Science Department at Howard High School of Technology This presentation was created for the benefit of our students by the Science Department at Howard High School of Technology Please send suggestions and comments to rmay@nccvt.k12.de.us Please send suggestions and comments to rmay@nccvt.k12.de.us

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