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The Gas Laws Chemistry Dr. May Gaseous Matter Indefinite volume and no fixed shape Indefinite volume and no fixed shape Particles move independently.

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Presentation on theme: "The Gas Laws Chemistry Dr. May Gaseous Matter Indefinite volume and no fixed shape Indefinite volume and no fixed shape Particles move independently."— Presentation transcript:

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2 The Gas Laws Chemistry Dr. May

3 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

4 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, x or

5 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

6 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

7 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

8 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 ( kPa)(760 mm) One atmosphere ( kPa)(760 mm) Contains 6.02 x molecules Contains 6.02 x molecules (Avogadros Number)

9 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 molecules Contains 6.02 x molecules

10 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 molecules Contains 6.02 x molecules

11 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

12 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 kilopascals 760 mm Hg

13 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 Divide by

14 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

15 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

16 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) = kPa, 1 atm, or 760 mm Hg Standard Pressure (P) = 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

17 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 = 298 K = 311 K 2. Insert into formula

18 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

19 Charles Law Solution 2 = V 2 2 = V V 2 = (2) 311 V 2 = V 2 = 2.09 liters

20 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

21 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 =.98 atm 2. Insert into formula

22 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

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

24 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

25 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 = 298 K 2. Standard pressure is kPa 3. Standard temperature is 273 K 4. Insert into formula

26 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 = kPa V 1 = 2.0 litersV 2 = unknown T 1 = 298 KT 2 =273 K

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

28 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)

29 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 = = 1.77 liters

30 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

31 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

32 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 Please send suggestions and comments to

33 The Ideal Gas Law Chemistry Dr. May

34 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

35 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

36 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 ( P in atm) or 8.3 (P in kPa)

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

38 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

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

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

41 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

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

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

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

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

46 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 = since P is in atms. R = since P is in atms. Find moles (n), then grams Find moles (n), then grams

47 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)

48 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

49 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

50 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 Please send suggestions and comments to


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