2. Gas Laws Gas Laws

3. 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. Characteristics of Gases Characteristics of Gases

4. 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.

5. Four Physical Quantities for Gases Phys. Qty.SymbolSI unitOther common units pressureP Pascal (Pa) atm, mm Hg, torr, psi volumeVm3m3 dm 3, L, mL, cm 3 temp.TK°C, °F molesnmol

6. TemperatureTemperature ºF ºC K -45932212 -2730100 0273373 K = ºC + 273 Always use absolute temperature (Kelvin) when working with gases.

7. Calculate the missing temperatures 0  C = _______ K100  C = _______ K 100 K = _______  C – 30  C= _______ K 300 K = _______  C 403 K = _______  C 25  C = _______ K0 K = _______  C Practise Absolute zero is – 273  C or 0 K

8. What is the approximate temperature for absolute zero in degrees Celsius and kelvin? Calculate the missing temperatures 0  C = _______ K100  C = _______ K 100 K = _______  C – 30  C= _______ K 300 K = _______  C 403 K = _______  C 25  C = _______ K0 K = _______  C Kelvin Practice 273373 – 173243 27130 298 – 273 Absolute zero is – 273  C or 0 K

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

10. PressurePressure b 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.

11. PressurePressure b Barometer measures atmospheric pressure Mercury Barometer Aneroid Barometer

12. STPSTP Standard Temperature & Pressure 273 K 101.325 kPa STP

13. STPSTP Standard Laboratory Conditions 25°C or 298 K 101.325 kPa SLC

14. -BOYLES -CHARLE -GAY-LUSSAC The Gas Laws

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

16. A. Boyle’s Law

17. PractisePractise 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?

18. P 1 x V 1 = P 2 x V 2 P 1 = 726 mmHg V 1 = 946 mL P 2 = ? V 2 = 154 mL P 2 = P 1 x V 1 V2V2 726 mmHg x 946 mL 154 mL = = 4460 mmHg

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

20. Charles’ Law

21. PractisePractise 2. A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? 3.If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be?

22. 2. A sample of gas occupies 3.5 L at 300 K. What volume will it occupy at 200 K? 3.If a 1 L balloon is heated from 22°C to 100°C, what will its new volume be? V 1 = 3.5 L, T 1 = 300K, V 2 = ?, T 2 = 200K 3.5 L / 300 K = V 2 / 200 K V 2 = (3.5 L/300 K) x (200 K) = 2.3 L V 1 = 1 L, T 1 = 22°C = 295 K V 2 = ?, T 2 = 100 °C = 373 K V 1 /T 1 = V 2 /T 2, 1 L / 295 K = V 2 / 373 K V 2 = (1 L/295 K) x (373 K) = 1.26 L For more lessons, visit www.chalkbored.com www.chalkbored.com

23. A sample of carbon monoxide gas occupies 3.20 L at 125 0 C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V 1 = 3.20 L T 1 = 398.15 K V 2 = 1.54 L T 2 = ? T 2 = V 2 x T 1 V1V1 1.54 L x 398.15 K 3.20 L = = 192 K V 1 /T 1 = V 2 /T 2

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

25. Gay-Lussac’s Law

26. Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1

27. GIVEN: V 1 = 473 cm 3 T 1 = 36°C = 309K V 2 = ? T 2 = 94°C = 367K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 E. Gas Law Problems b A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. CHARLES’ LAW TT VV (473 cm 3 )(367 K)=V 2 (309 K) V 2 = 562 cm 3

28. GIVEN: V 1 = 100. mL P 1 = 150. kPa V 2 = ? P 2 = 200. kPa WORK: P 1 V 1 T 2 = P 2 V 2 T 1 E. Gas Law Problems b A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. BOYLE’S LAW PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

29. PractisePractise b A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP.

30. GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25°C = 298 K V2 = ?V2 = ? P 2 = 101.325 kPa T 2 = 273 K WORK: P 1 V 1 T 2 = P 2 V 2 T 1 (71.8 kPa)(7.84 cm 3 )(273 K) =(101.325 kPa) V 2 (298 K) V 2 = 5.09 cm 3 E. Gas Law Problems b A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. P  T  VV COMBINED GAS LAW

31. GIVEN: P 1 = 765 torr T 1 = 23°C = 296K P 2 = 560. torr T 2 = ? WORK: P 1 V 1 T 2 = P 2 V 2 T 1 E. Gas Law Problems b A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? GAY-LUSSAC’S LAW PP TT (765 torr)T 2 = (560. torr)(309K) T 2 = 226 K = -47°C

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

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

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

35. b R = 0.0821 liter·atm/mol·K R = 8.3145 J/mol·K R = 8.2057 m 3 ·atm/mol·K R = 62.3637 L·Torr/mol·K or L·mmHg/mol·K

36. GIVEN: P = ? atm n = 0.412 mol T = 16°C = 289 K V = 3.25 L R = 0.0821 L  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 B. Ideal Gas Law b Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L. IDEAL GAS LAW

37. GIVEN: V = ?V = ? n = 85 g T = 25°C = 298 K P = 104.5 kPa R = 8.315 dm 3  kPa/mol  K B. Ideal Gas Law b Find the volume of 85 g of O 2 at 25°C and 104.5 kPa. = 2.7 mol WORK: 85 g 1 mol = 2.7 mol 32.00 g PV = nRT (104.5)V=(2.7) (8.315) (298) kPa mol dm 3  kPa/mol  K K V = 64 dm 3 IDEAL GAS LAW

39. 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). + 1 L oxygen 50 kPa 1 L nitrogen 100 kPa 1 L mixed gas 150 kPa = Note: all of these volumes are the same

40. 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.

41. Practise Calculating partial pressures

42. Vapour Pressure Defined Vapour pressure is the pressure exerted by a vapour. E.g. the H 2 O(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 .

43. Collecting gases over water Many times gases are collected over H 2 O Often we want to know the volume of dry gas at STP (useful for stoichiometry). For this we must make 3 corrections: 1.The level of water inside and outside the tube must be level (so pressure inside is equal to the pressure outside). 2.The water vapour pressure must be subtracted from the total pressure (to get the pressure of the dry gas). 3.Finally, values are converted to STP using the combined gas law.

44. Sample calculation A gas was collected over 21°C H 2 O. After equal- izing water levels, the volume was 325 mL. Give the volume of dry gas at STP (P atm =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 P gas = P atm - P H2O = 102.9 - 2.49 = 100.41 kPa Step 3: List all of the data T 1 = 294 K, V 1 = 325 mL, P 1 = 100.41 kPa Step 4: Convert to STP (P 1 )(V 1 )(T 2 ) (P 2 )(T 1 ) V2=V2= (100.4 kPa)(325 mL)(273 K) (101.325 kPa)(294 K) = = 299 mL

45. Assignment 1.37.8 mL of O 2 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? 2.Try questions 8 – 10 on page 465. 3.236 mL of H 2 is collected over water at 22°C and at an atmospheric pressure of 99.8 kPa. What is the volume of dry H 2 at STP? 4.If H 2 is collected over water at 22°C and an atmospheric pressure of 100.8 kPa, what is the partial pressure of the H 2 when the water level inside the gas bottle is equal to the water level outside the bottle?

46. 1) V 1 = 37.8 mL, P 1 = 99.42 kPa, T 1 = 297 K V 2 = ?, P 2 = 101.3 kPa, T 2 = 273 K P1V1P1V1 T1T1 = P2V2P2V2 T2T2 (99.42 kPa)(37.8 mL) (297 K) = (101.3 kPa)(V 2 ) (273 K) (99.42 kPa)(37.8 mL)(273 K) (297 K)(101.3 kPa) =(V 2 )=34.1 mL Vapor pressure at 24  C = 2.98 kPa Pgas = Patm - Pvapor = 102.4 kPa - 2.98 kPa = 99.42 kPa = P 1

47. 3) V 1 = 236 mL, P 1 = 97.16 kPa, T 1 = 295 K V 2 = ?, P 2 = 101.3 kPa, T 2 = 273 K P1V1P1V1 T1T1 = P2V2P2V2 T2T2 (97.16 kPa)(236 mL) (295 K) = (101.3 kPa)(V 2 ) (273 K) (97.16 kPa)(236 mL)(273 K) (295 K)(101.3 kPa) =(V 2 )=209 mL Vapor pressure at 22  C = 2.64 kPa Pgas = Patm - Pvapor = 99.8 kPa - 2.64 kPa = 97.16 kPa = P 1

48. Answers 4 - Total pressure = P H 2 + P H 2 O 100.8 kPa = P H 2 + 2.64 kPa 100.8 kPa - 2.64 kPa = P H 2 = 98.16 kPa For more lessons, visit www.chalkbored.com www.chalkbored.com

49. 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