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Gas Laws

Gas Laws. Characteristics of Gases. highly compressible. occupy the full volume of their containers. exert a uniform pressure on all inner surfaces of a container diffuse (mix) easily and quickly have very low densities. Kinetic Molecular Theory.

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Gas Laws

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  1. Gas Laws

  2. Characteristics of Gases • highly compressible. • occupy the full volume of their containers. • exert a uniform pressure on all inner surfaces of a container • diffuse (mix) easily and quickly • have very low densities.

  3. Kinetic Molecular Theory • Gases consist of a large number of molecules in constant random motion. • Volume of individual molecules negligible compared to volume of container. • Intermolecular forces (forces between gas molecules) negligible. • Collision of gas particles are elastic so no kinetic energy is lost • As temperature increases the gas particles move faster, hence increased kinetic energy.

  4. Four Physical Quantities for Gases

  5. Temperature K = ºC + 273 ºF -459 32 212 ºC -273 0 100 K 0 273 373 Always use absolute temperature (Kelvin) when working with gases.

  6. Practise Absolute zero is –273C or 0 K Calculate the missing temperatures 0C = _______ K 100C = _______ K 100 K = _______ C –30C = _______ K 300 K = _______ C 403 K = _______ C 25C = _______ K 0 K = _______ C

  7. Kelvin Practice Absolute zero is –273C or 0 K What is the approximate temperature for absolute zero in degrees Celsius and kelvin? Calculate the missing temperatures 0C = _______ K 100C = _______ K 100 K = _______ C –30C = _______ K 300 K = _______ C 403 K = _______ C 25C = _______ K 0 K = _______ C 273 373 –173 243 27 130 298 –273

  8. Pressure Pressure (P ) is defined as the force exerted per unit area. The atmospheric pressure is measured using a barometer. Which shoes create the most pressure?

  9. Pressure • KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi • 1 atm = 760 mmHg = 760 torr = 101325 Pa.

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

  11. STP STP Standard Temperature & Pressure 273 K 101.325 kPa

  12. STP SLC Standard Laboratory Conditions 25°C or 298 K 101.325 kPa

  13. V T P The Gas Laws -BOYLES-CHARLE-GAY-LUSSAC

  14. Boyle’s Law P V The pressure and volume of a gas are inversely related at constant mass & temp. PV = k

  15. A. Boyle’s Law

  16. Practise A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL?

  17. 726 mmHg x 946 mL P1 x V1 = 154 mL V2 A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P2 = = 4460 mmHg

  18. Charles’ Law V T The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

  19. Charles’ Law

  20. Practise • A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? • If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be?

  21. V1 = 3.5 L, T1 = 300K, V2 = ?, T2 = 200K 3.5 L / 300 K = V2 / 200 K V2 = (3.5 L/300 K) x (200 K) = 2.3 L • A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? • If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be? V1 = 1 L, T1 = 22°C = 295 K V2 = ?, T2 = 100 °C = 373 K V1/T1 = V2/T2,1 L / 295 K = V2 / 373 K V2 = (1 L/295 K) x (373 K) = 1.26 L For more lessons, visit www.chalkbored.com

  22. 1.54 L x 398.15 K V2 x T1 = 3.20 L V1 A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1/T1 = V2/T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? T2 = = 192 K

  23. Gay-Lussac’s Law P T The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume

  24. Gay-Lussac’s Law

  25. Combined Gas Law P1V1 T1 P2V2 T2 = P1V1T2 =P2V2T1

  26. E. Gas Law Problems • A gas occupies 473 cm3 at 36°C. Find its volume at 94°C. CHARLES’ LAW GIVEN: V1 = 473 cm3 T1 = 36°C = 309K V2 = ? T2 = 94°C = 367K T V WORK: P1V1T2 = P2V2T1 (473 cm3)(367 K)=V2(309 K) V2 = 562 cm3

  27. E. Gas Law Problems • A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW GIVEN: V1 = 100. mL P1 = 150. kPa V2 = ? P2 = 200. kPa P V WORK: P1V1T2 = P2V2T1 (150.kPa)(100.mL)=(200.kPa)V2 V2 = 75.0 mL

  28. Practise • A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP.

  29. E. Gas Law Problems • A gas occupies 7.84 cm3 at 71.8 kPa & 25°C. Find its volume at STP. COMBINED GAS LAW P T V GIVEN: V1=7.84 cm3 P1=71.8 kPa T1=25°C = 298 K V2=? P2=101.325 kPa T2=273 K WORK: P1V1T2 = P2V2T1 (71.8 kPa)(7.84 cm3)(273 K) =(101.325 kPa)V2 (298 K) V2 = 5.09 cm3

  30. E. Gas Law Problems • A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW GIVEN: P1 = 765 torr T1 = 23°C = 296K P2 = 560. torr T2 = ? P T WORK: P1V1T2 = P2V2T1 (765 torr)T2 = (560. torr)(309K) T2 = 226 K = -47°C

  31. Avogadro’s Principle V n • Equal volumes of all gases contain equal numbers of moles at constant temp & pressure.

  32. The Ideal Gas Equation The gas laws can be combined into a general equation that describes the physical behavior of all gases. Charles’s law Avogadro’s law Boyle’s law 11.5 rearrangement PV = nRT R is the proportionality constant, called the gas constant.

  33. B. Ideal Gas Law PV=nRT UNIVERSAL GAS CONSTANT R = 8.3145 J/mol·KR=0.0821 Latm/molK

  34. R = 0.0821 liter·atm/mol·K R = 8.3145 J/mol·KR = 8.2057 m3·atm/mol·KR = 62.3637 L·Torr/mol·K or L·mmHg/mol·K

  35. B. Ideal Gas Law • Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821Latm/molK WORK: PV = nRT P(3.25)=(0.412)(0.0821)(289) L mol Latm/molK K P = 3.01 atm

  36. B. Ideal Gas Law WORK: 85 g 1 mol = 2.7 mol 32.00 g • Find the volume of 85 g of O2 at 25°C and 104.5 kPa. IDEAL GAS LAW GIVEN: V=? n=85 g T=25°C = 298 K P=104.5 kPa R=8.315dm3kPa/molK = 2.7 mol PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm3kPa/molKK V = 64 dm3

  37. Under Pressure Dalton's Law of Partial Pressures

  38. 50 kPa 100 kPa 150 kPa 1 L oxygen 1 L nitrogen 1 L mixed gas Summary Dalton found that the total pressure of mixed gases is equal to the sum of their individual pressures (provided the gases do not react). Note: all of these volumes are the same + =

  39. Gas Mixtures Each component of a gas mixture exerts a pressure independent of the other components. The total pressure is the sum of the partial pressures.

  40. Practise • Calculating partial pressures

  41. Vapour Pressure Defined • Vapour pressure is the pressure exerted by a vapour. E.g. the H2O(g) in a sealed container. Eventually the air above the water is filled with vapour pushing down. As temperature , more molecules fill the air, and vapour pressure .

  42. Collecting gases over water • Many times gases are collected over H2O • Often we want to know the volume of dry gas at STP (useful for stoichiometry). For this we must make 3 corrections: • The level of water inside and outside the tube must be level (so pressure inside is equal to the pressure outside). • The water vapour pressure must be subtracted from the total pressure (to get the pressure of the dry gas). • Finally, values are converted to STP using the combined gas law.

  43. (P1)(V1)(T2) (100.4 kPa)(325 mL)(273 K) (P2)(T1) (101.325 kPa)(294 K) Sample calculation A gas was collected over 21°C H2O. After equal-izing water levels, the volume was 325 mL. Give the volume of dry gas at STP (Patm=102.9 kPa). Step 1: Determine vapour pressure (pg. 464) At 21°C vapour pressure is 2.49 kPa Step 2: Calculate the pressure of dry gas Pgas = Patm - PH2O = 102.9 - 2.49 = 100.41 kPa Step 3: List all of the data T1 = 294 K, V1 = 325 mL, P1 = 100.41 kPa Step 4: Convert to STP = 299 mL V2= =

  44. Assignment • 37.8 mL of O2 is collected by the downward displacement of water at 24°C and an atmospheric pressure of 102.4 kPa. What is the volume of dry oxygen measured at STP? • Try questions 8 – 10 on page 465. • 236 mL of H2 is collected over water at 22°C and at an atmospheric pressure of 99.8 kPa. What is the volume of dry H2 at STP? • If H2 is collected over water at 22°C and an atmospheric pressure of 100.8 kPa, what is the partial pressure of the H2 when the water level inside the gas bottle is equal to the water level outside the bottle?

  45. P1V1 P2V2 = T1 T2 (99.42 kPa)(37.8 mL) (101.3 kPa)(V2) = (297 K) (273 K) (99.42 kPa)(37.8 mL)(273 K) = 34.1 mL (V2) = (297 K)(101.3 kPa) Vapor pressure at 24C = 2.98 kPa Pgas = Patm - Pvapor = 102.4 kPa - 2.98 kPa = 99.42 kPa = P1 1) V1 = 37.8 mL, P1 = 99.42 kPa, T1 = 297 K V2 = ?, P2 = 101.3 kPa, T2 = 273 K

  46. P1V1 P2V2 = T1 T2 (97.16 kPa)(236 mL) (101.3 kPa)(V2) = (295 K) (273 K) (97.16 kPa)(236 mL)(273 K) = 209 mL (V2) = (295 K)(101.3 kPa) Vapor pressure at 22C = 2.64 kPa Pgas = Patm - Pvapor = 99.8 kPa - 2.64 kPa = 97.16 kPa = P1 3) V1 = 236 mL, P1 = 97.16 kPa, T1 = 295 K V2 = ?, P2 = 101.3 kPa, T2 = 273 K

  47. Answers 4 - Total pressure = PH2 + PH2O 100.8 kPa = PH2 + 2.64 kPa 100.8 kPa - 2.64 kPa = PH2 = 98.16 kPa For more lessons, visit www.chalkbored.com

  48. Ideal and real gas • Ideal gas: • Ideal gas molecule have - zero volume -zero attraction between molecules • Real gases behave ideally at - high temperature - low pressure

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