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KMT and Gas Laws States of Matter, Kinetic Molecular Theory, Diffusion, Properties of Gases, and Gas Laws.

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Presentation on theme: "KMT and Gas Laws States of Matter, Kinetic Molecular Theory, Diffusion, Properties of Gases, and Gas Laws."— Presentation transcript:

1 KMT and Gas Laws States of Matter, Kinetic Molecular Theory, Diffusion, Properties of Gases, and Gas Laws

2 Standards 4. The kinetic molecular theory describes the motion of atoms and molecules and explains the properties of gases. As a basis for understanding this concept: a. Students know the random motion of molecules and their collisions with a surface create the observable pressure on that surface. 4. b. Students know the random motion of molecules explains the diffusion of gases. 4. c. Students know how to apply the gas laws to relations between the pressure, temperature, and volume of any amount of an ideal gas or any mixture of ideal gases. 4. d. Students know the values and meanings of standard temperature and pressure (STP). 4. e. Students know how to convert between the Celsius and Kelvin temperature scales. 4. f. Students know there is no temperature lower than 0 Kelvin. 4. g.* Students know the kinetic theory of gases relates the absolute temperature of a gas to the average kinetic energy of its molecules or atoms. 4. h.* Students know how to solve problems by using the ideal gas law in the form PV = nRT. 4. i. * Students know how to apply Dalton’s law of partial pressures to describe the composition of gases and Graham’s law to predict diffusion of gases.

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5 States of Matter SolidLiquidGasPlasma Freezing Melting Deposition Sublimation Condensation Boiling Deionization Ionization

6 Condensation (Gas  Liquid)

7 Boiling (Liquid  Gas)

8 Sublimation (Solid  Gas)

9 Deposition (Gas  Solid)

10 Freezing (Liquid  Solid)

11 Melting (Solid  Liquid)

12 Molecular Motion of Gases

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14 KMT The path of any individual molecule could best be described as random. KMT – Kinetic Molecular Theory

15 Molecular Motion The state of matter depends on how much energy (motion) the molecules, atoms, or ions have. The state of matter also depends on how attracted the atoms, molecules, or ions are to each other.

16 Molecular Motion State of Matter O O Na + Cl – O H H + + – Gas Liquid Solid

17 Nonpolar molecules O O O O

18 Polar molecules O H H + + – O H H + + – O H H + + –

19 Ionic compounds Na + Cl – Na + Cl – Na + Cl – Na + Cl – – Na + + Cl –

20 Molecular Twist and Stretch

21 Diffusion Diffusion – when a substance spreads out in a gas or liquid. Examples: 1.Perfume eventually reaching the far side of a room. 2.Kool-Aid dissolving into water.

22 Diffusion of Gases

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24 Diffusion of Liquids

25 Temperature (T) Kinetic energy is the energy of motion. Temperature is defined as a measure of the average kinetic energy of the atoms or molecules.

26 Temperature (T) There are two scales and an absolute unit. (degrees Fahrenheit, degrees Celsius, and Kelvin)

27 Temperature Scales Water Boils Human Body Water Freezes 212°F 98.7°F 32°F 100°C 37°C 0°C 373K 310K 273K

28 Temperature Scales Surface of Sun Room Temp. Absolute Zero 9,941°F 70°F -460°F 5,505°C 21°C -273°C 5,778K 294K 0K

29 Converting Temperatures Fahrenheit  Celsius °C = (°F – 32)×(5/9) Celsius  Fahrenheit °F = °C ×(9/5) + 32 Celsius  Kelvin K = °C + 273.15

30 Absolute Zero At Zero Kelvin (0 K or –273.15 °C), atoms and molecules stop moving. There is no temperature lower than absolute zero (0 K).

31 Volume (V) How much space is occupied by a fluid. Liquid Gas Usually gases are measured in Liters (L)

32 Pressure (P) Defined as force divided by area. The force comes from atoms’ or molecules’ collisions with the wall of the container. The greater the number of collisions or the more energy with each collision, the greater the pressure.

33 Pressure (P) Defined as force divided by area. The force comes from atoms’ or molecules’ collisions with the wall of the container. The greater the number of collisions or the more energy with each collision, the greater the pressure.

34 Pressure Units Unit Unit Symbol1 atm = Atmospheresatm-- PascalsPa101,325 Pa KilopascalskPa101.3 kPa Pounds per square inch lbs. or psi14.7 psi Millimeters mercury mm Hg760 mm Hg in. 2

35 Pressure 0 atm Outer Space (a vacuum) 1,072 atm At the Bottom of Mariana Trench 1 atm Regular Atmosphere (at sea level) 0.33 atm Top of Mt. Everest

36 Gas Properties PropertySymbol Usual Unit Unit Symbol PressurePkilopascalkPa VolumeVlitersL TemperatureTKelvinK Molesnmolesmol

37 STP = Standard Temperature and Pressure Temperature is 0°C = 273.15 K and Pressure is 1 atm = 101.3 kPa

38 Gas Laws Most of the gas laws deal with taking a quantity of gas and changing one property (pressure, temperature, or volume) and predicting how the other properties will change in response.

39 Boyle’s Law When given a certain amount of gas, if you increase the pressure, the volume decreases. If you decrease the pressure, the volume increases. Mathematically: P 1 V 1 = P 2 V 2 P 1 V 1 =P2P2 V2V2

40 Boyle’s Law When given a certain amount of gas, if you increase the pressure, the volume decreases. If you decrease the pressure, the volume increases. Mathematically: P 1 V 1 = P 2 V 2 P 1 V 1 =P2P2 V2V2 This assumes a constant temperature (T)

41 Boyle’s Law Example Your nephew is playing with a balloon in the car as your family drives over a mountain pass. The balloon initially had a volume of 1 L when the car was at the bottom of the mountain (and the air pressure was 100 kPa). Now that your family is at the top the air pressure is 70 kPa. What is the new volume of the balloon? P 1 V 1 =P2P2 V2V2

42 Boyle’s Law Example P 1 = 100 kPaP 2 = 70 kPa V 1 = 1 LV 2 = ? (100 kPa)(1 L) = (70 kPa) V 2 100 kPa L = 70 kPa V 2 1.43 L = V 2 P 1 V 1 = P 2 V 2 70 kPa 70 kPa

43 Boyle’s Law Example Normal P Low P

44 Boyle’s Law Example Normal PLow P V 1 = 1 L V 2 = 1.43 L

45 Charles’ Law When given a certain amount of gas, if you increase the temperature, the volume increases. If you decrease the temperature, the volume decreases. Mathematically: V 1 V 2 T 1 T 2 = V1T1V1T1 V2T2V2T2 = You must use Kelvin temperatures!

46 Charles’ Law When given a certain amount of gas, if you increase the temperature, the volume increases. If you decrease the temperature, the volume decreases. Mathematically: V 1 V 2 T 1 T 2 = V1T1V1T1 V2T2V2T2 = This assumes a constant pressure (P) You must use Kelvin temperatures!

47 Charles’ Law Example If 1.0 L of gas is contained within a piston at 27 ˚C (300 K), what will new volume be if the gas is cooled to -23 ˚C (250 K)? Assume that the pressure is constant. V1T1V1T1 V2T2V2T2 =

48 Charles’ Law Example V1T1V1T1 V2T2V2T2 = V 1 = 1.0 LV 2 = ? T 1 = 300 KT 2 = 250 K 1.0 L 300 K V 2 250 K = 1.0 300 V 2 250 = (250) V 2 = 0.83 L

49 Charles’ Law This assumes a constant pressure (P)

50 Charles’ Law This assumes a constant pressure (P)

51 Gay–Lussac’s Law When given a certain amount of gas, if you increase the temperature, the pressure increases. If you decrease the temperature, the pressure decreases. Mathematically: P 1 P 2 T 1 T 2 = P1T1P1T1 P2T2P2T2 =

52 Gay–Lussac’s Law When given a certain amount of gas, if you increase the temperature, the pressure increases. If you decrease the temperature, the pressure decreases. Mathematically: P 1 P 2 T 1 T 2 = P1T1P1T1 P2T2P2T2 = This assumes a constant volume (V)

53 Avogadro’s Law The volume of a gas at Standard Temperature and Pressure (STP) is directly proportional to the moles of the gas. V = k n At STP there are 22.4 L per mole of gas.

54 Avogadro’s Law Example How many liters will 3 moles of gas occupy at STP? V = k n V = 67.2 L V = (22.4 )(3 mol) L mol

55 Combined Gas Law This combines Boyle’s, Charles’, and Gay-Lussac’s Gas Laws. Mathematically: P 1 V 1 P 2 V 2 T 1 T 2 = Cancel out the properties that remain constant.

56 Combined Gas Law This combines Boyle’s, Charles’, and Gay-Lussac’s Gas Laws. Mathematically: P 1 V 1 P 2 V 2 T 1 T 2 = Cancel out the properties that remain constant.

57 Ideal Gas Law If we know 3 of the 4 gas properties (P, V, T, and n) we can solve for the missing one by using the formula: PV = nRT R is called the gas constant. R = 8.314 kPa L mol K

58 Ideal Gas Law A cylinder is filled with 0.2 moles of gas. The sealed cylinder has a volume 3.0 L and is heated with 3,000 J to a temperature of 300K. What is the pressure inside the cylinder? PV = nRT kPa L mol K P (3.0 L) = (0.2 mol)(8.314 )(300 K) P (3.0 L) = 499 kPa L P = 166 kPa 3.0 L 3.0 L

59 Gas Laws Summary Gas Law Formula Boyle’s LawP 1 V 1 = P 2 V 2 Charles’ Law V 1 V 2 T 1 T 2 Gay-Lussac’s Law P 1 P 2 T 1 T 2 Avogadro’s LawV = k n Ideal Gas LawPV = nRT = =

60 What is the change in volume? CH 4 (g) + H 2 O (g)  CO (g) + 3 H 2 (g) Methanewater carbon hydrogen monoxide + +

61 Graham’s Law of Diffusion From the simulations we saw, that lighter gas molecules move faster than heavier gas molecules. If we want to directly compare the speeds of gas molecules we can use: These are the molar masses vAMBvBMAvAMBvBMA = These are the average molecular speeds

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63 Ne F ONC B Be He Li H KrAr Cl Br Xe ISPSi Mg Al Ca Na K

64 Ne F ONC B Be He Li H KrAr Cl Br Xe ISPSi Mg Al Ca Na K

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66 4 e – in valence shell


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