1. Gas Laws: Physical Properties Part 1

2. Kinetic Molecular Theory b The tiny particles in matter are in constant motion due to kinetic energy. b 1. A gas is composed of particles (atoms or molecules).  Gases have mass and take up space. b 2. The particles in a gas move in constant, random, rapid motion.  (kinetic energy (motion)  As Temp  ) b 3. All collisions are perfectly elastic. (elastic = energy is not lost)

3. Kinetic Molecular Theory Animation b http://www.tutorvista.com/physics/a nimations/kinetic-molecular-theory- animation http://www.tutorvista.com/physics/a nimations/kinetic-molecular-theory- animation b http://www.mpcfaculty.net/mark_bis hop/KMT.htm http://www.mpcfaculty.net/mark_bis hop/KMT.htm b https://www.youtube.com/watch?v= _rsqBNhFG1Y https://www.youtube.com/watch?v= _rsqBNhFG1Y

4. Gas Characteristics 3 Characteristics of ALL Gases: 1. Take up space 2. Exert pressure 3. Have energy

5. Gases take up space b Measure volume, or amount of space occupied by a gas. 1L = 1000 mL = 1000 cm 3 b A gas takes the shape and volume of its container.

6. Gases exert pressure b Pressure = billions of air molecules colliding with objects/ surfaces. Units used to measure pressure b 1 atm = 760 mmHg = 101.3 kPa = 760 torr = 14.7 psi atmospheres (atm) millimeters of mercury (mmHg) kiloPascals (kPa) 760 torr Pounds per square inch (psi)

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

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

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

10. Gases and energy b Temperature = measure of the average kinetic energy (motion) of gas particles: Measured in Kelvin (K) or Celcius (°C)  K = °C + 273 Increase temperature = Increase kinetic energy b At 0 K ( - 273 °C ) = 0 (Zero) kinetic energy No movement

11. Fahrenheit, Celsius, or Kelvin? b Three types of scales Three types of scales

12. D. Temperature ºF ºC K -45932212 -2730100 0273373 K = ºC + 273

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

14. b 1. A gas occupies a volume of 57.0 cm 3 at a temperature of 23.0 °C and pressure of 97.5 kPa. Convert to K, L, atm: a. temperature to K (K = °C + 273) K = 23.0 + 273 = 296 K b. volume to L (1000 cm 3 = 1 L) 57.0 cm 3 X ___1L__ 1000 cm 3 = 0.057 L c. pressure to atm (1 atm = 101.3 kPa) 97.5 kPa X __1 atm_ 101.3 kPa = 0.96 atm Converting Example:

15. 2.Which has more kinetic energy: oxygen gas particles at 75.0 o C or oxygen gas particles at 288.6 K? 1 st StepK = o C + 273 K = 75.0 + 273 = 348 K  HINT: The higher the temperature, the higher the energy.  Answer: Oxygen particles at 75.0 o C have more kinetic energy. Which would you expect to move at a faster speed?  Answer: Oxygen at 75 o C because it is a higher temperature and the gas particles would have more kinetic energy and move at a faster speed.

16. Converting Example: 3. A gas occupies 20.0 L at 100 o C and 780 mmHg. Convert to mL, K, atm. ANSWER: a. L to mL (1 L = 1000 mL) 20.0 L X __ 1000 mL__ 1 L = 20000 mL b. o C to K (K = o C + 273) K = 100 + 273 = 373 K c. mm Hg to atm (760 mm Hg = 1 atm) 780 mmHg X __ 1atm__ 760 mmHg = 1.026 atm

17. G. Boyle’s Law b At a constant mass & temperature, the pressure and volume of a gas are inversely related. If volume (V) increases, pressure (P) decreases (and vice versa) P V PV = k

18. G. Boyle’s Law P V PV = k

19. Boyle’s Law Video Clip Boyle's Law

20. Equation for Boyle’s Law b P1 x V1 = P2 x V2 b P1 = Initial Pressure b V1 = Initial Volume b P2 = Final Pressure b V2 = Final Volume b Use unit atm for pressure b Use units L for volume

21. Boyle’s Law Example #1 The air inside a sealed 10.0 L container exerts a pressure of 1.0 atm. If the container is compressed to a volume of 5.0 L, what is the pressure inside? b ANSWER: b P1 = 1.0 atmP1 V1 = P2 V2 (1.0 atm) (10.0 L) =P2(5.0 L) b P2 = ? b V1 = 10.0 L b V2 = 5.0 L b ANSWER: 2 atm

22. Boyle’s Law Example #2 If the pressure of a gas is doubled then the volume of the same gas is ________________ at constant temperature. b ANSWER: Reduced by half!!!

23. Boyle’s Law: P1V1 = P2V2

25. The volume of a gas varies directly with the absolute temperature at constant pressure.  V = KT  V 1 / T 1 = V 2 / T 2 Charles’ Law Jacques-Alexandre Charles Mathematician, Physicist, Inventor Beaugency, France November 12, 1746 – April 7, 1823

26. V T B. Charles’ Law

27. Charles’s Law b At a constant mass (n) and pressure (P), the volume (V) of a gas changes directly (same) with the temperature (T) b Direct Relationship If the Volume (V) of a gas increases then the Temperature (T) of the gas increases (and vice versa).

28. STP (Standard Temp & Pressure) STP (Standard Temp & Pressure) b ST (Standard Temperature) 273 K Conversion Equation: K = o C + 273 b SP (Standard Pressure) 1 atm

29. Charles Law Video Clip b Charles law Charles law

30. Daily Life Examples of Charles’s Law b Soda cans and bottles have a label on them stating “Store in a cool, dry place”. b It can be observed that when you use a pressure deodorant can of any sort and spray for a few seconds, the can tends to become cooler. b When a ping pong ball gets dented without being punctured the best solution is to dip it for a while in warm water. b Inflated football deflates in winter. b During outdoor parties a common nuisance is having to replace popped party balloons.

31. Charles’ Law Demonstration b Egg Trick Egg Trick

32. B. Charles’ Law

33. Charles’ Law: V 1 /T 1 = V 2 /T 2

35. Charles’s Law Equation b V 1 = V 2 T 1 T 2 b V 1 = Initial Volume b T 1 = Initial Temperature b V 2 = Final Volume b T 2 = Final Temperature  Temperature needs to be in Kelvin (K)  Volume needs to be in L

36. Charles’s Law Examples b If a balloon has a volume of 0.50 L at 298 K, what is its new volume at 310 K ? V 1 = 0.50 L T 1 = 298 K V 2 = ? T 2 = 310 K V 1 = V 2 T 1 T 2 (0.50L) = V 2 (298K) (310K) V 2 = 0.52 L

37. Charles’s Law Examples b If the same initial balloon from #1 is cooled to 15 o C, what is the final volume of the balloon? V 1 = 0.50 L T 1 = 298 K V 2 = ? T 2 = K= (15 + 273) =288 K V 1 = V 2 T 1 T 2 (0.50L) = V 2 (298K) (288K) V 2 = 0.48 L

38. P T C. Gay-Lussac’s Law b At a constant mass and volume, the pressure and absolute temperature (K) of a gas are directly related. b If the pressure of a gas increases, the temperature of the gas increases (and vice versa).

39. P T C. Gay-Lussac’s Law

40. Gay-Lussac’s Law

41. 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. Gas Law Problems b The pressure of a gas 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

42. J. Combined Gas Law b Combines all relationships between pressure, temperature, and volume into one equation.

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

44. Combined Gas Law Equation b Equation: P 1 x V 1 = P 2 x V 2 T 1 T 2 b P = pressure b V = volume b T = temperature b Any of the 3 variables may change.  Pressure needs to be in atm  Volume needs to be L  Temperature needs to be in Kelvin (K)

45. Combined Gas Law Examples b 1. If a gas has a volume of 12 L at STP what will its volume be when temperature is 25 o C and pressure is 1.2 atm? P 1 = 1 atm V 1 = 12 L T 1 = 273 K P 2 = 1.2 atm V 2 = ? T 2 = (25 + 273) =298 K

46. Example 1 Continued b P 1 V 1 = P 2 V 2 T 1 T 2 b (1) (12) = (1.2) V 2 (273 K) (298 K) b V 2 = 10.92 L

47. 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

48. 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

49. 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

50. 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

51. III. Ideal Gas Law (p. 334-335, 340-346) Ch. 10 & 11 - Gases

52. V n A. Avogadro’s Principle

53. V n b Equal volumes of gases contain equal numbers of moles at constant temp & pressure true for any gas

54. PV T VnVn PV nT B. Ideal Gas Law = k UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K = R

55. B. Ideal Gas Law UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K R=8.315 dm 3  kPa/mol  K PV=nRT

56. 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

57. 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

58. Ch. 10 & 11 - Gases IV. Gas Stoichiometry at Non-STP Conditions (p. 347-350)

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

60. 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.

61. 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

62. 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 = ?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 

63. 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.

64. Ch. 10 & 11 - Gases V. Two More Laws (p. 322-325, 351-355)

65. A. 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 +... P atm = P H2 + P H2O

66. 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 A. 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 on p.899 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.

67. 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? DALTON’S LAW Look up water-vapor pressure on p.899 for 35.0°C. Sig Figs: Round to least number of decimal places. A. 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.

68. B. 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

69. B. 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 because…

70. B. Graham’s Law b Graham’s Law Rate of diffusion of a gas is inversely related to the square root of its molar mass. Ratio of gas A’s speed to gas B’s speed

71. b Determine the relative rate of diffusion for krypton and bromine. Kr diffuses 1.381 times faster than Br 2. B. 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 ”.

72. 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? B. Graham’s Law Put the gas with the unknown speed as “Gas A”.

73. b An unknown gas diffuses 4.0 times faster than O 2. Find its molar mass. B. 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.

74. TEAM PRACTICE! b Work the following problems in your book. Check your work using the answers provided in the margin. p. 324  SAMPLE PROBLEM 10-6  PRACTICE 1 & 2 p. 355  SAMPLE PROBLEM 11-10  PRACTICE 1, 2, & 3

75. Common Gases Lab b The following gases are going to be produced during this lab: H 2, O 2, CO 2 and NH 3. b Determine which gas is produced in each of the test tubes. H 2 : creates a popping sound O 2 : reignites / burns CO 2 : extinguishes a flame NH 3 : distinctive odor

76. Composition of the Atmosphere