1. I. Physical Properties I. Gases I. Gases

2. Nature of Gases b Gases have mass. b They can be compressed. b They completely fill their containers. b Representative particles can be atoms or molecules b Different gases can diffuse through one another. b They exert pressure. b The pressure of a gas increases with temperature.

3. A. Kinetic Molecular Theory b 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 avg. KE directly related to Kelvin temperature.

4. B. Real Gases b Particles in a REAL gas… have their own volume can become attracted to each other b Gas behavior is most ideal… at low pressures at high temperatures in nonpolar atoms/molecules

5. C. Characteristics of Gases b Gases expand to fill any container. random motion, no attraction b Gases are fluids (like liquids). no attraction b Gases have very low densities. no volume = lots of empty space

6. C. Characteristics of Gases b Gases can be compressed. no volume = lots of empty space b Gases undergo diffusion & effusion. random motion

7. Amount of a gas: n = moles molar mass = g/mol Volume: (V)The volume of a gas is the volume of the container. The metric unit of volume is the liter..

8. D. Temperature ºF ºC K -45932212 -2730100 0273373 K = ºC + 273 b Always use absolute temperature (Kelvin) when working with gases.

9. E. Pressure Which shoes create the most pressure?

10. E. Pressure b Barometer measures atmospheric pressure Mercury Barometer Aneroid Barometer

11. Pressure conversions 760mmHg = 1atm= 14.7 lb/in 2 =101.3kPa = 760Torr = 101,300Pa All of the above represent the force exerted by atmospheric pressure at sea level.

12. Pressure conversion problems Convert 750. mmHg to atm 750. mmHg 1 atm 760 mmHg = 0.987 atm Sig figs

13. Pressure is considered to be the most accurate measure of storm strength. One of the lowest recorded pressures in the Western Hemisphere is 88.86kPa. How high would a column of mercury be balanced by this atmospheric pressure? 88.86 kPa 760 mmHg 101.3kPa = 666.5mmHg

14. The air pressure inside the cabin of an airplane is 8.3 lb/in 2. What is this pressure in atm? 8.3 lb/in 2 1 atm 14.7 lb/in 2 = 0.56 atm

15. E. Pressure b KEY UNITS AT SEA LEVEL 101.325 kPa (kilopascal) 1 atm 760 mm Hg 760 torr 14.7 psi

16. F. STP Standard Temperature & Pressure 0°C 273 K 1 atm101.325 kPa -OR- STP

17. E. Pressure b Manometer measures contained gas pressure U-tube ManometerBourdon-tube gauge

18. PracticePractice b Find the manometer practice sheet that is in your packet. b Closed manometers are easy to read. Find the difference in the levels in the arms in the U-tube. b This difference is the pressure of the gas in mmHg. This amount can then be converted into any other unit you need.

19. Reading open ended manometers b In this manometer, one end is open to the atmosphere and the other end is connected to the gas. b When the valve is opened, the gas exerts pressure on the mercury that is within the U-tube. b Three possibilities can occur.

20. No difference in levels. b 1. The mercury level does not change once the valve that connects the gas to the manometer is opened. b This would mean that the atmospheric pressure is the same as the gas pressure.

21. Open end level is higher than level near the contained gas. b 2. If the mercury level is higher in the side of the U-tube that is connected to the atmosphere, the contained gas must be exerting a pressure that is equal to the atmospheric pressure PLUS the difference in the mercury levels. b Find the difference in the levels and then ADD this to the atmospheric pressure to find the gas pressure.

22. b Find the difference b in the two levels. b Add this difference b to the atmospheric b pressure. b Convert the b atmospheric pressure to mmHg b before you add. U-tube Manometer

23. Level is higher in the side connected to the contained gas. b 3. If the mercury is higher in the arm that is connected to the contained gas, the atmospheric pressure MUST be higher than the contained gas. b To find the pressure of the gas, determine the difference in the levels and then SUBTRACT this from the atmospheric pressure.

24. b Time to practice manometer problems.

25. II. The Gas Laws Gases Gases

26. A. Boyle’s Law P V PV = k

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

28. A. Boyle’s Law

29. V T B. Charles’ Law

30. V T b The volume and absolute temperature (K) of a gas are directly related at constant mass & pressure

31. B. Charles’ Law

32. P T C. Gay-Lussac’s Law

33. P T b The pressure and absolute temperature (K) of a gas are directly related at constant mass & volume

34. b http://www.mhhe.com/physsci/che mistry/essentialchemistry/flash/gas esv6.swf http://www.mhhe.com/physsci/che mistry/essentialchemistry/flash/gas esv6.swf

35. = kPV PTPT VTVT T D. Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1

36. GIVEN: V 1 = 473 cm 3 T 1 = 36.0°C = 309K V 2 = ? T 2 = 94.0°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 36.0°C. Find its volume at 94.0°C. CHARLES’ LAW TT VV (473 cm 3 )(367 K)=V 2 (309 K) V 2 = 562 cm 3

37. 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

38. GIVEN: V 1 = 7.84 cm 3 P 1 = 71.8 kPa T 1 = 25.0°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 & 25.0°C. Find its volume at STP. P  T  VV COMBINED GAS LAW

39. GIVEN: P 1 = 765 torr T 1 = 23.0°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 23.0°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

40. III. Ideal Gas Law Gases Gases

41. V n A. Avogadro’s Principle b Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas

42. PV T VnVn PV nT A. Ideal Gas Law = k UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K = R You don’t need to memorize these values! Merge the Combined Gas Law with Avogadro’s Principle:

43. A. Ideal Gas Law UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K PV=nRT You don’t need to memorize these values!

44. GIVEN: P = ? atm n = 0.412 mol T = 16.0°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 C. Ideal Gas Law Problems b Calculate the pressure in atmospheres of 0.412 mol of He at 16.0°C & occupying 3.25 L.

45. GIVEN: V = ? n = 85 g T = 25.0°C = 298 K P = 104.5 kPa R = 8.315 dm 3  kPa/mol  K C. Ideal Gas Law Problems b Find the volume of 85 g of O 2 at 25.0°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

46. Gases Gases IV. Two More Laws

47. B. Dalton’s Law b The total pressure of a mixture of gases equals the sum of the partial pressures of the individual gases. P total = P 1 + P 2 +... When a H 2 gas is collected by water displacement, the gas in the collection bottle is actually a mixture of H 2 and water vapor.

48. GIVEN: P H2 = ? P total = 94.4 kPa P H2O = 2.72 kPa WORK: P total = P H2 + P H2O 94.4 kPa = P H2 + 2.72 kPa P H2 = 91.7 kPa B. Dalton’s Law b Hydrogen gas is collected over water at 22.5°C. Find the pressure of the dry gas if the atmospheric pressure is 94.4 kPa. Look up water-vapor pressure for 22.5°C. Sig Figs: Round to least number of decimal places. The total pressure in the collection bottle is equal to atmospheric pressure and is a mixture of H 2 and water vapor.

49. GIVEN: P gas = ? P total = 742.0 torr P H2O = 42.2 torr WORK: P total = P gas + P H2O 742.0 torr = P H2 + 42.2 torr P gas = 699.8 torr b A gas is collected over water at a temp of 35.0°C when the barometric pressure is 742.0 torr. What is the partial pressure of the dry gas? Look up water-vapor pressure for 35.0°C. Sig Figs: Round to least number of decimal places. B. Dalton’s Law The total pressure in the collection bottle is equal to barometric pressure and is a mixture of the “gas” and water vapor.

50. C. Graham’s Law b Diffusion Spreading of gas molecules throughout a container until evenly distributed. b Effusion Passing of gas molecules through a tiny opening in a container

51. C. Graham’s Law KE = ½mv 2 b Speed of diffusion/effusion Kinetic energy is determined by the temperature of the gas. At the same temp & KE, heavier molecules move more slowly.  Larger m  smaller v

52. C. Graham’s Law b Graham’s Law Rate of diffusion of a gas is inversely related to the square root of its molar mass. The equation shows the ratio of Gas A’s speed to Gas B’s speed.

53. b Determine the relative rate of diffusion for krypton and bromine. Kr diffuses 1.381 times faster than Br 2. C. Graham’s Law The first gas is “Gas A” and the second gas is “Gas B”. Relative rate mean find the ratio “v A /v B ”.

54. b A molecule of oxygen gas has an average speed of 12.3 m/s at a given temp and pressure. What is the average speed of hydrogen molecules at the same conditions? C. Graham’s Law Put the gas with the unknown speed as “Gas A”.

55. b An unknown gas diffuses 4.0 times faster than O 2. Find its molar mass. C. Graham’s Law The first gas is “Gas A” and the second gas is “Gas B”. The ratio “v A /v B ” is 4.0. Square both sides to get rid of the square root sign.

56. - Gases - Gases VI. Gas Stoichiometry at Non-STP Conditions

57. A. Gas Stoichiometry b Moles  Liters of a Gas: STP - use 22.4 L/mol Non-STP - use ideal gas law b Non- STP Given liters of gas?  start with ideal gas law Looking for liters of gas?  start with stoichiometry conv.

58. 1 mol CaCO 3 100.09g CaCO 3 B. Gas Stoichiometry Problem b What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? 5.25 g CaCO 3 = 1.26 mol CO 2 CaCO 3  CaO + CO 2 1 mol CO 2 1 mol CaCO 3 5.25 g? L non-STP Looking for liters: Start with stoich and calculate moles of CO 2. Plug this into the Ideal Gas Law to find liters.

59. WORK: PV = nRT (103 kPa)V =(1mol)(8.315 dm 3  kPa/mol  K )(298K) V = 1.26 dm 3 CO 2 B. Gas Stoichiometry Problem b What volume of CO 2 forms from 5.25 g of CaCO 3 at 103 kPa & 25ºC? GIVEN: P = 103 kPa V = ? n = 1.26 mol T = 25°C = 298 K R = 8.315 dm 3  kPa/mol  K

60. WORK: PV = nRT (97.3 kPa) (15.0 L) = n (8.315 dm 3  kPa/mol  K ) (294K) n = 0.597 mol O 2 B. Gas Stoichiometry Problem b How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? GIVEN: P = 97.3 kPa V = 15.0 L n = ? T = 21°C = 294 K R = 8.315 dm 3  kPa/mol  K 4 Al + 3 O 2  2 Al 2 O 3 15.0 L non-STP ? g Given liters: Start with Ideal Gas Law and calculate moles of O 2. NEXT 

61. 2 mol Al 2 O 3 3 mol O 2 B. Gas Stoichiometry Problem b How many grams of Al 2 O 3 are formed from 15.0 L of O 2 at 97.3 kPa & 21°C? 0.597 mol O 2 = 40.6 g Al 2 O 3 4 Al + 3 O 2  2 Al 2 O 3 101.96 g Al 2 O 3 1 mol Al 2 O 3 15.0L non-STP ? g Use stoich to convert moles of O 2 to grams Al 2 O 3.