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Gas Laws

Gas Laws. The Gas Laws Kinetic Molecular Theory . Matter is composed of very tiny particles . Particles of matter are in continual motion . The total KE of colliding particles remains constant When individual particles collide, some lose energy while others gain energy.

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Gas Laws

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  1. Gas Laws • The Gas Laws • Kinetic Molecular Theory • .Matter is composed of very tiny particles • .Particles of matter are in continual motion • .The total KE of colliding particles remains constant • When individual particles collide, some lose energy while others gain energy. • No overall energy loss = elastic collisons

  2. Properties of a Gas • Expansion – no definite shape or volume • Low density – density of gas is about 1/1000 of that same substance in a liquid or a solid phase • Diffusion – process of spreading out spontaneously to occupy a space uniformly

  3. Ideal Gas • consists of very small independent particles. These particles move at random in space and experience elastic collisions • Extra – Approximate speed is 103 meters/sec, at 0 degrees C and 1 ATM of pressure • Undergo: 5 x 109 collisions/second

  4. Gases in Real life are not so ideal

  5. When would a gas not act “ideal”? • At low temperatures - the molecules may be moving slow enough to allow attractions to occur • At high pressures - the molecules may once again have attractive forces between them in effect

  6. Gases and Pressure Pressure –the total average force over a given area is called pressure. Although the collisions of any single molecule of gas within the walls of a container are intermittent, there are so many molecules in even a small container, that the tremendous number of collisions average out to a steady pressure.

  7. Gas and Temperature • Temperature – the average kinetic energy or motion energy of its molecules. • Avogadro’s hypothesis states that at any given temperature and pressure, equal volumes of gases, contain the same number of molecules.

  8. Avogadro • The following table supports the hypothesis by showing that one mole (6.02 x 1023 molecules) of 11 different gases occupies approximately the same volume. These volumes were derived experimentally at the same pressure and temperature.

  9. Table 1. Volumes of various gases at 0 C and 760 mm pressure 1 mole: Oxygen Occupies: 22.393 L Sulfur Dioxide Occupies: 21.888 L Hydrogen Occupies: 22.430 L Helium Occupies: 22.426 L Chlorine Occupies: 22.063 L Nitrogen Occupies: 22.403 L Ammonia Occupies: 22.094 L Hydrogen Chloride Occupies: 22.248 L Carbon Monoxide Occupies: 22.402 L Carbon Dioxide Occupies: 22.262 L

  10. Why should one mole of hydrogen occupy the same volume as one mole of oxygen? Since oxygen molecules are about 16 times as heavy as hydrogen molecules, it would seem more reasonable for a mole of oxygen molecules to occupy much more volume than a mole of hydrogen molecules at the same temperature and pressure.

  11. Remember the Demo? • The smaller balls make up in speed what they lack in size in “carving out” an area. This reasoning can be applied to the volume occupied by lighter gas molecules compared to the identical volume occupied by an equal number of heavier gas molecules at the same temperature and pressure.

  12. Van der Waal Forces • London Dispersion Forces - (Van der Waal’s Forces) – the attractive forces between gas molecules. • Even though the total electrical charge of the gas molecule is neutral, the positively charged nuclei of its atoms are not completely shielded by negatively charged electrons.

  13. Van der Waal’s In general, the magnitude of Van Der Waal’s forces increases with an increase in the number of electrons per atom and with an increase in size of the molecule. Few electrons = lame wave Many electrons = more powerful wave!!

  14. STP conditions • Standard temperature and pressure – useful when wanting to compare or measure gas volumes • ST = 0 C = 273K • SP = 760mm of Hg = 1atm = 760 Torr • STP = standard temperature and pressure

  15. Boyle’s Law • Boyle’s Law – the volume of a definite quantity of dry gas is inversely proportional to the pressure, provided temperature remains constant. (V  1/P) • Formulas or dimensional analysis? P1V1 = P2V2

  16. Boyle’s Law, Continues • Examples: • Heimlich maneuver (increase P in lungs by decreasing their V) • Syringe • Straw (Lungs are acting as the chamber that increases in V) • VP & VP

  17. Problem 1 Suppose a syringe plunger is pushed all the way in so that the syringe contains 0.10 mL of air at 1.00 ATM. The plunger is now pulled back quickly so that the total volume changes to 3.00 mL. Calculate the momentary pressure before any liquid comes into the syringe.

  18. Solve using dimensional analysis • Initial P is 1.00 ATM so you have two possible conversion factors: • A. 0.10 mL OR B. 3.00 mL • ---------- ---------- • 3.00 mL 0.10 mL • Which do we choose? • Think: If we are increasing V we expect the gas P to decrease so we want the conversion factor that is less than one. 1.00 atm x 0.10 mL = 0.033 atm ---------- 3.00 mL

  19. The plug-and-chug (no thinking involved) method P1V1 = P2V2 1.00 atm x 0.10 mL = P2 x 3.00 mL P2 = 0.033 atm

  20. Problem 2 A small child carrying an inflated balloon with a volume of 2.00 liters gets on an airplane in N.Y. at a P of 760. Torr. What will the volume of his balloon be on arrival at Mexico City, which is 7347 feet above sea level and has an average P of 600. Torr?

  21. Answer to problem 2 2.00 L x 760. Torr ----------- = 2.53 Liters 600. Torr Solve, p&c method #2 = P1V1 = P2V2 760. Torr x 2.00 Liter = V2 x 600. Torr V2 = 2.53 Liters

  22. Charle’s Law • The volume of a gas is directly proportional to the Kelvin Temperature (Pressure remains a constant). V  T V1 = V2 T1 T2

  23. Problem 1 • The initial volume of a piston is 1.5 Liters at 300.K. What will be the final volume at 600. K? (P is constant) • Think: Since volume  temperature, as V increases then T must increase. So, use the conversion factor that is greater than 1

  24. Answer • 1.5 Liter x 600.K = 3.0L • 300.K Problem 2: Suppose a balloon had a volume of 2 x 105 liter when it was filled with hot air at a temp. of 500. C. What volume would it occupy if it were sealed and cooled at 27 C? • Or solve way #2 = V1/T1 = V2 /T2 • 1.5L / 300.K = V2 /600.K • V2 = 3.0L

  25. Answer to Problem 2 • ( 2 x 105 L ) x 300.K = 8 x 104 L • 773K

  26. Gay-Lussac’s Law The pressure inside a fixed volume is directly proportional to the Kelvin temperature. P  T Example: Spray can warning: “Do not incinerate!” P1 =P2 T1 T2

  27. Gay-Lussac’s, continued The contents of a spray can are all used up and the can thus contains gas at a pressure of 1.00 atm. The can is thrown into a fire where it’s T rises to 627 C. Why does the can explode? -- Started at a temperature of 27 C. • Think: If T increases, the P must increase if there is no change in volume. So, solve for pressure .

  28. Solving for this Law • The pressure inside the can, right before it blew up, would have been: • Remember if T , P • 1.00 atm x 900. K = 3.00 atm 300. K

  29. Problem 2 Problem: A pressure cooker is closed and is at 20. C. To what temperature must the gas be heated in order to create an internal pressure of 2.00 atm?

  30. Solving problem 2 Solve:P1 = 1.00 atm and it has to increase to 2.00 atm. So, P needs to increase and that means temperature must also increase. • 293 K x 2.00 atm • 1.00 atm = 586 K

  31. Combining the gas laws • P1V1 = P2V2 T1 T2 Whoops, wrong combining.

  32. Combination problem • Suppose you have a balloon that has a volume of 20.0 L at a pressure of 1.00 atm and a temperature of 32.0 C. If this balloon rose into the atmosphere where the temperature was only 12.0 C and the pressure only 0.82 atm, what would its new volume be?

  33. First convert the temperatures to Kelvin: 12.0 + 273 = 285 K 32.0 + 273 = 305 K Second: either plug in values into your combination formula or do the THINKING method P so V & T so V • 20.0 L x 1.00 atm x 285 K = 23 L 0.82 atm 305 K

  34. Dalton’s Law of Partial Pressure • the total pressure of the mixture of gases is the sum of their partial pressures • PTOTAL = PA + PB + PC + ….ETC

  35. Problem 1 A mixture of gases in a 3.00 L flask consists of Nitrogen at a partial pressure of 100 Torr and Oxygen at a PP of 300 Torr. What is the volume of each gas? Calculate the total pressure in the flask. Since the presence of another gas has no effect on the volume available to the first gas, both Nitrogen and Oxygen have a volume of 3.00L

  36. Ans. PTOTAL = 100 Torr + 300 Torr = 400 Torr Problem 2 Air has a relative constant proportion of N2 (78.10%), O2 (21.00%), Ar (0.90%), and CO2 (0.03%). The partial pressure of each gas in normal dry air at sea level is proportional to the number of molecules. What is the Total atmospheric pressure?

  37. N2 = 0.7810 atm. + O2 = 0.2100 atm. + • Ar = 0.0090 atm. + CO2 = 0.0003 atm. = answer. PTOTAL = 1.0003 atm

  38. Graham’s Law of Diffusion • Diffusion rate depends on 3 factors: • Speed of the molecules (the higher the temperature, the higher the average KE) • Diameter of the molecule (the larger the molecule, the slower the diffusion rate) • Potential Attractive forces (the greater the potential attractive forces between the molecules, the slower the diffusion rate)

  39. Graham’s Law • For an average molecule - A and an average molecule - B, use the following formula to determine velocity: VA = MB VB M A

  40. Graham’s Law Sample Problem • At room temperature, an average hydrogen molecule travels at a speed of 1700. meters/second (about 3000 miles/hr). What is the speed of an average oxygen molecule under the same conditions?

  41. Solving for Sample Problem 1 1700. m/sec = 32.0 amu VB 2.02 amu So, VB = 425 m/sec

  42. Another awesome formula • Problem: A steam autoclave is used to sterilize hospital instruments. Suppose an autoclave with a volume of 15.0 liters contains pure steam (water) at a temperature of 121 C and a P of 1550 Torr. How many grams of water does it contain? • Solve: PV = nRT • R = 62.4 L * Torr / Mole * K

  43. Solving using the Ideal Gas Law • 1550 Torr x 15.0 L = n x 62.4 L * Torr x 394 K Mol * K • n = 0.946 moles of H2O • 0946 moles of H2O x 18.0g H2O = 17.0 g H2O 1 mole H2O • NOTE: You could solve using your knowledge of combination gas laws and molar volume (at STP)

  44. Another Sample Problem • Example 2: A hospital uses an oxygen gas cylinder containing 3.50 kg of O2 gas in a volume of 20.0 L and at a T of 24 C. What is the P in atm in the cylinder?

  45. Problem Solved PV = nRT n = 3500 g O2 x 1 mole/32.0 g = 109 moles P( 20.0 L) = (109 moles)(0.0821 L x atm/mole x K)(297 K) P = 133 atm

  46. Eudiometer Tubes(mercury or water displacement) • Eudiometer – a tube used to collect a gas. It is closed at one end and is graduated. • Reaction = 2HCl (aq) + Ca (s)  H2 (g) + CaCl2 (aq)

  47. Eudiometer with mercurySituation #1 • If there is just enough gas to make the mercury level inside the tube the same as the level of the mercury in the bowl then the pressure of the hydrogen gas is the same as that of the atmosphere!

  48. Eudiometer, Situation #2 • Suppose enough hydrogen gas is added to make the level inside the tube lower than the level of the mercury in the bowl. The P of the hydrogen gas is greater than the atmospheric pressure. (To determine the hydrogen gas pressure one must add the level difference to the barometric reading.) • Problem situation 2: The volume of oxygen in a eudiometer is 37.0 mL. The mercury level inside the tube is 25.00mm lower than the outside. The barometric reading is 742.0 mm Hg.

  49. Solving for Problem 2 Solution 2: 742.0mm + 25.00mm = 767.0mm

  50. Eudiometer with mercury, situation #3 Suppose there was not enough gas to make the mercury level the same. Then, the pressure of hydrogen gas would not be the same as the pressure of the air outside the tube. (So, we must subtract the level difference from the barometric reading.) • Problem situation 3: What is the P of the gas in an eudiometer, when the mercury level in the tube is 14mm higher, than that outside? That barometer reads 735mm Hg.

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