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Behavior of GASES

Behavior of GASES. Gases are made up of atoms and molecules just like all other compounds, but because they are in the form of a gas we can learn a great deal more about these molecules and compounds. It might seem a bit confusing because we can’t see most gases, but we know they exist.

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Behavior of GASES

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  1. Behavior of GASES

  2. Gases are made up of atoms and molecules just like all other compounds, but because they are in the form of a gas we can learn a great deal more about these molecules and compounds. It might seem a bit confusing because we can’t see most gases, but we know they exist.

  3. Elements that exist as gases at 250C and 1 atmosphere

  4. I. Let’s look at some of the Nature of Gases: • 1. Expansion– gases do NOT have a definite shape or volume. • 2.Fluidity– gas particles glide past one another, called fluid just like a liquid.

  5. Nature of Gases cont. • 3.Compressibility– can be compressed because gases take up mostly empty space. • 4.Diffusion– gases spread out and mix without stirring and without a current. Gases mix completely unless they react with each other.

  6. The word KINETIC refers to motion • Kinetic energy= energy an object has because of its motion Collisions of Gas Particles

  7. Collisions of Gas Particles

  8. II. Kinetic Molecular Theory of Gases Particles of matter (any type) are in constant motion! Because we know this we have a few assumptions that we make about gases, called the Molecular Theory of Gases:

  9. Kinetic theory: 1.Particles of a gas are in constant, straight-line motion, until they collide. • They move independently from each other Molecular Motion

  10. Kinetic theory: 2. Gases consist of a large number of tiny particles (moleculesor atoms) ; these particles are very far apart, therefore gas is mostly empty space. • There are no forces of attraction or repulsion between particles of gases.

  11. elastic collisions inelastic collisions Kinetic theory: • 3.Collisions between particles of a gas and the container wall are elastic. Which means there is no loss of energy. • Total Kinetic energy remains constant.

  12. Kinetic theory: • The average kinetic energy of gas particles depends on the temperature of the gas. (It is directly proportional) • KE=1/2 mv2 • (m=mass in kg and v=velocity is m/sec) • Calories (cal) ..Joules (j) measure Enegy • 1 cal= 4.18 J • 1Cal = 1000 cal

  13. III. Volume, Pressure, Temperature, Number of Moles (Descriptions of Gases) 1.Volume – refers to the space matter (gas) occupies. • Measured in liters (L). 1.00 dm3 = 1.00L = 1000 cm3 = 1000mL Volume (V)

  14. Pressure (P) • 2. Pressure(P) – the number of times particles collide with each other and the walls of the container (force exerted on a given area). • A vacuum is empty space= It has no pressure Pressure Simulation

  15. Atmospheric Pressure • The gases in the air are exerting a pressure called atmospheric pressure • Atmospheric pressure is a result of the fact that air has mass and is colliding with everything under the sun with a force.

  16. Atmospheric Pressure

  17. Atmospheric Pressure • Atmospheric pressure varies with altitude • The lower the altitude, the longer and heavier the column of air above an area of the earth. • Check the back of a box of cake mix for the difference in baking times based on the atmospheric pressure in your region.

  18. Atmospheric Pressure • Knowing atmospheric pressure is how forecasters predict the weather. • Low pressure or dropping pressure = a change of weather from fair to rain. • High pressure = clear skies & sun.

  19. Pressure is measured with a device called a barometer.(They operate on the change of pressure due to the weather)

  20. Barometer • At 1 atm (one atmospheric pressure) a column of mercury 760 mm high. 1 atm Pressure Column of Mercury Dish of Mercury

  21. Barometer • At 1 atm a column of mercury 760 mm high. • A 2nd unit of pressure is mm Hg (mercury) 1 atm = 760 mm Hg • A 3rd unit & the SI unit is the Pascal (Pa) 1 atm = 101.3 kPa 1 atm Pressure 760 mm

  22. Measured in atmospheres (atm). • 1atm = 760 millimeters Hg (Barometers use Hg) • 1atm = 760 torr (Named after Torricelli for the invention of the barometer) • 1atm = 101.3 kPa – kilopascals • 1 atm = 760 mm Hg = 101.3 kPa

  23. Practice: Convert 4.40 atm to mmHg. Convert 212.4kPa to mmHg.

  24. Temperature (T) • 3. Temperature (T) – as temperature increases gas particles move faster, as temperature decreases gas particles move slower. • measured with a thermometerin Celsius. • calculations involving gases are made after converting the Celsius to Kelvin temperature. • Measured in Kelvin (K). Kelvin = 273 + C Celsius = K - 273

  25. Practice: Convert 32.0°C to K. Convert 400. K to °C.

  26. Amount (n) 4. Number of Moles – tells you how much of a certain gas you have 1 mole = number of grams of the compound or element (molar mass) 6.02 x 1023 molecules per mole of the gas.

  27. STP – “standard temperature and pressure”.( Measured at Sea Level) • Standard Temperature0C= 273 K • Standard Pressure 1.00atm= 760 torr = 760 mmHg = 101.325 kPa

  28. Gas Laws - How do all of pressure, temperature, volume, and amount of a gas relate to each other? Combined GAS Law(Initial) (Final) Peas x Vegetables P1 x V1 = P2 x V2 Table T1 T2

  29. Rules for solving gas law problems: • 1st write down what is given and what is unknown, • 2nd identify the gas law you want to use, and • 3rd rearrange the formula to solve for the unknown and • 4th solve the problem. (If temperature is involved, it MUST be converted to Kelvin! K = 273 + C)

  30. A. Boyle’s Law - Pressure and Volume (when temperature remains constant) V1 = initial or old volume V1P1 = V2P2 V2 = final or new volume P1 = initial or old pressure P2 = final or new pressure

  31. Inverse Relationship(As pressure increases, volume decreases and as pressure decreases, volume increases.) P1 x V1 = P2 x V2 T1 T2

  32. #2 from Boyles Law Problem Sheet A sample of carbon dioxide occupies a volume of 3.50 liters at 125 kPa pressure. What pressure would the gas exert if the volume was decreased to 2.00 liters? * P1 x V1 = P2 x V2 P1 = 125 kPa P2 = X V1 = 3.50 L V2 = 2.00 L 125 kPa x 3.50 L = P2 X 2.00L 2.00L 2.00L 219 kPa = P2

  33. B. Charles’ Law -Volume and Temperature (when pressure is constant) V1 = V2 T1 = initial or old temperature T1 T2 T2 = final or new temperature • Direct Relationship (As temperature increases, volume increases and as temperature decreases, volume decreases.)

  34. P1 = P2 = V1 = .3 L V2 = 6.5 L T1 = 40oC + 273 = 3l3 K T2 = X • #2 From Charles Law Problem Sheet • Oxygen gas is at a temperature of 40O C when it occupies a volume of 2.3 liters. To what temperature should it be raised to occupy a volume of 6.5 liters? V1 = V2 T1 T2 2.3L = 6.5L 3l3 K T2 12.3 L x T2 = 6.5L X 313K 2.3L 2.3L T2 = 880K

  35. Boyle’s Law Review P1V1 = P2V2 1621-1691

  36. Boyle’s Law BOYLE's Law in Action

  37. Volume Pressure How does Pressure and Volume of gases relate graphically? PV = k Temperature, # of particles remain constant

  38. Boyle’s Mathematical Law: If we have a given amount of a gas at a starting pressure and volume, what would happen to the pressure if we changed the volume? Or to the volume if we changed the pressure? since PV equals a constant P1V1 = P2V2 Ex: A gas has a volume of 3.0 L at 2 atm. What will its volume be at 4 atm?

  39. Boyle’s Mathematical Law: • List the variables or clues given: • P1 = 2 atm • V1 = 3.0 L • P2 = 4 atm • V2 = ? • determine which law is being represented: P1V1 = P2V2 P1V1 = V2 P2 3) Plug in the variables & calculate: (2 atm) (3.0 L) = (4 atm) (V2) 1.5 L

  40. Charles’s Law Review 1746-1823 Any relation to Bernet??

  41. Charles’ Law CHARLES' Law in Action

  42. Volume Temp How does Temperature and Volume of gases relate graphically? V/T = k Pressure, # of particles remain constant

  43. V1 V2 = T1 T2 Charles’s Mathematical Law: If we have a given amount of a gas at a starting volume and temperature, what would happen to the volume if we changed the temperature? Or to the temperature if we changed the volume? since V/T = k Ex: A gas has a volume of 3.0 L at 400K. What is its volume at 500K?

  44. V1 V2 = T1 T2 X L = 400K 500K Charles’s Mathematical Law: • List the variables or clues given: • T1 = 400K • V1 = 3.0 L • T2 = 500K • V2 = ? • determine which law is being represented: 3.0L 3) Plug in the variables & calculate: 3.8 L

  45. C. Gay-Lussac’s Law - Pressure and Temperature (when volume is constant) • P1 = P2 T1 T2 • Direct Relationship (As temperature increases, pressure increases and as temperature decreases, pressure decreases.) P1 x T2 = P2 xT1

  46. #2 From Gay-Lussac’s Problem Sheet • A gas has a pressure of 0.370 atm at 50.0 °C. What is the pressure at standard temperature? (STP =Remember O oC or 273 K) (Change 50.0 °C to Kelvin) P1 = P2 T1 T2 • P1=0.370 atm • P2 = 0.820 atm • V1 = • V2 = • T1 =40 oC + 273 = 323 K • T2 = O oC + 273 = 273 K • 0.370 atm = P2 • 323 K 273 K P2 = 0.820 atm

  47. D. Combined Gas Law - Pressure, Temperature, and Volume (None of the variables are constant) Combined GAS Law(Initial) (Final) Peas x Vegetables P1 x V1 = P2 x V2 Table T1 T2 P1 x V1 x T2 = P2 x V2 x T1

  48. Ex: Find the final volume of 25.0 ml of a Gas at STP. If the conditions change to 14 oC and 740 mmHg P1 x V1 = P2 x V2 T1 T2 P1=760 mmHg P2 = 740 mmHg V1 = 25.0mL V2 = ? T1 =0 oC + 273 = 273 K T2 = 14 oC + 273 = 287 K (760 mmHg) (25.0 ml) = (740 mmHg) (V2) 273 K 287 K (287K) (69.6 mmHg . ml) = (740 mmHg) (V2) (287K) K 287 K STP Temperature = 273K Pressure =760 mmHg 19974 mmHg.mL = (740 mmHg) (V2) 740 mmHg 740 mmHg 27ml = V2

  49. P1 P2 = T1 T2 Gay-Lussac’s Law 1778-1850

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