1. Unit 10 Gases

2. Lesson 1 Kinetic Molecular Theory and Gas Characteristics

3. Kinetic – Molecular Theory In order to understand gases you must understand the Kinetic Molecular Theory. The KMT assumes the following concepts about an ideal gas:

4. Kinetic- Molecular Theory The kinetic-molecular theory explains the behavior of particles of matter. It has these five points: 1.All matter is made of small particles (atoms and molecules) that do have mass. 2.For gases, distances between particles are very large. (1,000x solids and liquids)

5. Kinetic-Molecular Theory 3. These particles are in constant, rapid, random motion.

6. Kinetic-Molecular Theory 4. All collisions between particles are perfectly elastic. This means that there is no loss of energy OR in other words, the amount of total energy before the collision and after are equal. 5. The temperature of a substances is determined by the average kinetic energy of the particles. Why? Important to understanding the properties of gases later

7. Kinetic-Molecular Theory We also typically assume for gases that 1) particles have no attractive forces between one another and 2) the volume of the particles is insignificant compared to the volume of gas.

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

9. Pressure Applies to fluids i.e. liquids and gases Because of their nature they exert pressure in all directions Gas pressure is the result of gas particles colliding with the walls of their container

10. Measuring Gas Pressure Barometer – measures atmospheric pressure Mercury Barometer Aneroid Barometer

11. Measuring Gas Pressure Manometer measures contained gas pressure U-tube Manometer Bourdon-tube gauge

12. Units of Pressure Atmospheres (atm) Millimeters Mercury (mmHg) Torr (Torr) Note: Torr is the same as mmHg kiloPascals (kPa) 760. mmHg(Torr) = 1.00 atm = 101.3 kPa

13. Conversion Practice Convert each of the following into the other two pressure units. A.100. atm B. 50.kPa C. 200. mmHg

14. Temperature Temperature is the average kinetic energy of the particles in a material. If temperature is the average kinetic energy of particles and kinetic energy is due to the particle motion, then there should be a temperature where all motion of particles stops. The temperature where all particle motion has stopped is called absolute zero. Absolute zero is at –273 o C.

15. Temperature We use absolute zero for the basis of a new temperature scale called the Kelvin (K) scale. In this scale the size of each degree is the same as the size of a Celsius degree. So if we know -273 o C = 0 K and each degree is the same size, conversions are easy! K = o C + 273 o C = K - 273

16. So Why Use Kelvin? Note: We can never have a negative Kelvin temperature. (Ex. –5 K) Also, notice that we no longer use the degree symbol in the Kelvin scale. This is because " o " means the number is on a relative scale. The Kelvin is an absolute scale.

17. Conversion Practice Convert the following to the other temperature scale. a) 273 o C b) 273 K d) 173 o C e) 542. K g) 0 K

18. Lesson 2 Gas Laws

19. Standard temperature and pressure is abbreviated STP. Standard temperature is 0 o C (or 273 K) and standard pressure is 1.00 atm, 760. mmHg, or 101.3 kPa. We use cm 3, mL, or L to measure volume. Avogadro’s Law 1 mole of any gas at STP is 22.4L

20. Dalton’s Law of Partial Pressure Dalton's Law states that at constant volume and temperature, the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each gas. P total = P T = P 1 + P 2 + P 3..... Ex. What is the total pressure of a mixture of H 2, Ne, and Ar if the pressure of the H 2 is 41.1 kPa, the pressure of the argon is 37.2 kPa, and the pressure of the neon is 5.1 kPa? 83.4 kPa

21. Boyle’s Law – Pressure and Volume Boyle's Law states that at constant temperature, the pressure of a gas varies inversely with the volume. P 1 V 1 = P 2 V 2 So what’s mean? When you raise pressure, volume decreases and vice versa. When you lower pressure, volume increases and vice versa.

22. Boyle’s Law – How can I remember? When you boil (Boyle) something, you produce a VaPor So Boyle’s Law deals with V and P

23. Problem #5 If a child’s helium balloon has a volume of 2.45 L at 104 kPa and the child let’s go, when the balloon reaches an altitude where the pressure is 48.2 kPa what will be the volume of the balloon? P 1 V 1 = P 2 V 2 So (2.45 L)(104 kPa) = V 2 (48.2 kPa) 254.8 L*kPa = V 2 (48.2 kPa) V 2 = 5.2863 L After accounting for sig. fig. V 2 = 5.29 L

24. Charles’ Law – Volume and Temperature Charles' Law states that at constant pressure, the volume of a gas varies directly with Kelvin temperature. Charles helped to formulate the absolute or Kelvin temperature scale. V 1 = V 2 orT 1 V 2 = T 2 V 1 T 1 T 2 * When calculating any gas law problem using temperature, it must be in Kelvin.

25. Charles’ Law – How can I remember? Charlie Brown was on TV So Charles’ law deals with T and V

26. Gay – Lussac’s Law – Pressure and Temperature According to this law, at constant volume, the pressure of a gas varies directly with Kelvin temperature. P 1 = P 2 T 1 P 2 = T 2 P 1 T 1 T 2 How can I remember? It’s not the other two

27. 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 A gas’ pressure is 765 torr at 23°C. At what temperature will the pressure be 560. torr? PP TT (765 torr)T 2 = (560. torr)(296K) T 2 = 216.68 K = -56°C

28. Example: Gay-Lussac’s Law The gas in an aerosol can is at a pressure of 3.00 atm at 25°C. Directions on the can warn the user not to keep the can in a place where temperature exceeds 52°C. What would the gas pressure be in the can at 52°C?

29. How to Remember which law goes with each pair of variables: P V T Boyle’s Charles’Gay-Lussac’s

30. Combined Gas Law Boyle’s, Charles’s and Gay-Lussac’s laws can be combined into a single law. The combined gas law states the relationship among pressure, volume, and temperature of a fixed amount of gas

31. Combined Gas Law Boyle's, Charles', and G - L Laws can be combined into one gas law called the Combined Gas Law. P 1 V 1 = P 2 V 2 T 1 T 2 orP 1 T 2 V 1 = P 2 T 1 V 2 How can I remember? It’s the only that combines P, V and T

32. = k PV T Combined Gas Law P1V1T1P1V1T1 = P2V2T2P2V2T2 P 1 V 1 T 2 = P 2 V 2 T 1

33. Example: Combined Gas Law A Helium-filled balloon has volume of 50.0 L at 25°C and 1.08 atm. What volume will it have at 0.855 atm and 10.°C?

34. 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 Gas Law Problems A gas occupies 7.84 cm 3 at 71.8 kPa & 25°C. Find its volume at STP. P  T  VV

35. 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 Gas Law Problems A gas occupies 473 cm 3 at 36°C. Find its volume at 94°C. TT VV (473 cm 3 )(367 K)=V 2 (309 K) V 2 = 562 cm 3

36. 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 Gas Law Problems A gas occupies 100. mL at 150. kPa. Find its volume at 200. kPa. PP VV (150.kPa)(100.mL)=(200.kPa)V 2 V 2 = 75.0 mL

37. UNIT 10 – GAS LAWS Lesson 3 Ideal and Graham’s Laws

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

39. PV T VnVn PV nT Ideal Gas Law = k UNIVERSAL GAS CONSTANT R=0.0821 L  atm/mol  K = R Merge the Combined Gas Law with Avogadro’s Principle:

40. Ideal Gas Law The Ideal Gas Law says that for an ideal gas pressure, volume, moles, and Kelvin temperature of particles are all related. P V = n R T P is the pressure in atmospheres V is the volume in Liters n is the number of moles T is the temperature in Kelvin R = 0.0821 L. atm mole. K

41. Molecular View of The Ideal Gas Law

43. 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=nRT V P=(3.25)=(0.412)(0.0821)(289) L mol L  atm/mol  K K _____________________ 3.25L P = 3.01 atm Ideal Gas Law Problems Calculate the pressure in atmospheres of 0.412 mol of He at 16°C & occupying 3.25 L.

44. GIVEN: V = ? n = 85 g T = 25°C = 298 K P = 104.5 KPa=1.03atm R = 0.0821atm-L /mol  K Ideal Gas Law Problems 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 (1.03)V=(2.7) (0.0821) (298) atm mol atm-L /mol  K K V = 64 L

45. Ideal Gas Law Remember: Something is ideal if it conforms perfectly to a set of conditions. Although ideal gases do not exist, the ideal gas law can be used to describe real gases under normal conditions.

46. When does the ideal gas law not apply? 1.Under very high pressures – volume becomes so small that the volume of gas particles become significant. 2.Under very low temperatures – gas no longer have enough kinetic energy to overcome attractive forces

47. Ideal Gas Law Variations We know the molar mass is grams of substance per mole of substance or MM = m/n So n = m / MM Substituting in we get PV = (m/MM)RT Bring MM up to the other side finally gives us: P V MM = m R T

48. Ideal Gas Law Variations We also know the density of a substance is its mass divided by its volume, or D = m/V Start with P V MM = m R T Divide V over and you get P V = (m/V) R T And you end up with: P MM = D R T

49. A Note About Kinetic Energy We’ve said that temperature is related to average kinetic energy and that gases at the same temperature have the same average kinetic energy The equation for kintetic energy is: KE = ½ m(v) 2 Bottom line: at the same temperature, gases have the SAME kinetic energy, but DIFFERENT SPEEDS

50. Diffusion Diffusion Diffusion – Spreading of gas molecules throughout a container until evenly distributed. – Movement toward lower concentration Effusion Effusion – Passing of gas molecules through a tiny opening in a container

51. Diffusion of gases Click in this box to enter notes. Copyright © Houghton Mifflin Company. All rights reserved. Go to Slide Show View (press F5) to play the video or animation. (To exit, press Esc.) This media requires PowerPoint® 2000 (or newer) and the Macromedia Flash Player (7 or higher). [To delete this message, click inside the box, click the border of the box, and then press delete.]

52. Effusion of a Gas Click in this box to enter notes. Copyright © Houghton Mifflin Company. All rights reserved. Go to Slide Show View (press F5) to play the video or animation. (To exit, press Esc.) This media requires PowerPoint® 2000 (or newer) and the Macromedia Flash Player (7 or higher). [To delete this message, click inside the box, click the border of the box, and then press delete.]

53. Graham’s Law Graham’s Law 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.

54. Graham’s Law – What does it really mean? The heavier a gas is, the slower it moves

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

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