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Gas Laws Kinetic Theory

Gas Laws Kinetic Theory A physical model that explains the properties of gases, liquids, and solids in terms of: (A) forces between particles of matter (B) the en these particles possess. 3 Basic Assumptions of the Kinetic Theory (K.T.): 1) Matter is composed of tiny particles

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Gas Laws Kinetic Theory

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  1. Gas Laws Kinetic Theory A physical model that explains the properties of gases, liquids, and solids in terms of: (A) forces between particles of matter (B) the en these particles possess.

  2. 3 Basic Assumptions of the Kinetic Theory (K.T.): 1) Matter is composed of tiny particles Chemical properties of matter depend on their composition. Physical properties depend on the forces they exert on each other and the distance separating them.

  3. 2) Particles of Matter are in continual motion Their average kinetic en depends on temperature. 3) Total kinetic en of colliding particles remains constant When particles collide, some lose en while others gain en, but there is no overall en loss. Collisions of this

  4. type are said to be elastic. According to the K.T. a gas consists of very small independent particles which move randomly in space and experience elastic collisions. The size of gas molecules is insignificant when compared to the distance between molecules. Thus, we assume that the particles of a gas have

  5. no attraction for each other. They are called point masses since they are considered to have no real volume or diameter. A gas composed of point masses does not actually exist. This imaginary gas composed of molecules with mass, but not volume or mutual attraction is called an ideal gas.

  6. Gases have 4 main characteristics: 1) Expansion - no definite shape or volume. They expand to the shape and volume of their container. 2) Pressure - their pressure increases with an increase in temperature. 3) Low density - the density of a gas is approximately 1/1000 of the same density in the solid form.

  7. 4) Diffusion - the spontaneous (without help) spreading out of a material from areas of high concentration to areas of low concentration. Graham’s Law of Diffusion - the lighter gas diffuses faster At the same temperature two gases have the same ave K.E.

  8. K.E. = 1/2 mv2 (ex) Calculate the rate of diffusion of hydrogen gas to that of oxygen gas. (ex) Calculate the relative rate of diffusion of carbon monoxide gas to that of carbon dioxide gas.

  9. Calculate the velocity of sulfur dioxide gas if the velocity of bromine gas is .0892 m/s. • Calculate the mass of an unknown gas if the ratio of diffusion of oxygen gas to the unknown gas is 0.764.

  10. Variables we use to Measure a Gas: • (A) Volume (V) - the amount of space a material occupies. Units and conversions: 1 l = 1 dm3 = 1000 cm3 1 ml = 1cm3

  11. Avogadro’s Principle - at equal temps and pressures, equal volumes of gases contain the same number of molecules.

  12. (B) Pressure (P) - the force on a given amount of area caused by gas particles colliding with the walls of its container. Units of pressure: 1b/in2, mm Hg, torrs, pascals (pa), kilopascals (kpa), atmospheres (atm). Conversions for pressure: 1 atm = 760 mm Hg

  13. 1 mm Hg = 1 torr 1 mm Hg = 13.6 mm H2O 1 in Hg = 25.4 mm Hg 1 atm = 101.3 kpa 1 kpa = 7.5 mm Hg Standard pressure 1 atm = 101.3 kpa

  14. (C) moles (n) - the number of moles in a gas depends upon the pressure and the temp of the gas. (D) Temperature (T) - a measure of the ave K.E. Units: F, C,K Formulas (know these): F = (1.8XC) + 32 C = (F -32)/1.8

  15. K = C + 273 We need an equation that relates all 4 variables together (V, P, n, and T) = Ideal Gas Law! 3 of the variables will be measured experimentally and we will solve for the 4th.

  16. Ideal Gas Law Suppose we take a sample of a gas and measure its P, V, n, and T. If you calculate (Pa)(Va)/(na)(Ta) = # If you do the same for a different gas (Pb)(Vb)/(nb)(Tb) = #

  17. The numbers are the same = constant called the Universal Gas Constant = R The value of R depends on the units of the other 4 variables: If P = atm, V = l, n = moles, and T = K R = 0.082 l atm/K mol

  18. If P = kpa, V = dm3, n = moles, and T = K R = 8.31 kpa dm3/mol K Ideal Gas Law = PV/nT = R or PV = nRT T must be in K!

  19. (ex) What volume will 92 moles of oxygen gas occupy at 5.00 atm of pressure and 81.0 C? (534 = 530 l) (ex) A sample of gas is collected in a 78.02 l flask at 752 torrs and 22.0C. How many moles of gas are present? (0.287 moles)

  20. (ex) A sample of gas is collected at a pressure of 12.5 mm Hg. The gas has a volume of 17.2 ml and contains2.5 mol. What is the temp in F, C, and K? (0.0014 K, 1273 C, -459F) (ex) A sample of gas has a volume of 91.5 l and was collected at 72.4 F. What pressure will 36.2 mole of this gas exert on its container? (9.58 atm)

  21. Using the other value of R (ex) What pressure would be exerted by 0.622 mol of gas contained in a 9.22 dm3 vessel at 16.0 C? (162 kpa) (ex) At what temp is a gas if 0.0851 mol of it are found in a 604 cm3 vessel at 100.4 kpa? (85.8 K)

  22. Variations of the Ideal Gas Law: The ideal gas law can easily be altered to include mass. (g) (1 mole)/(fm or mw) = moles So if PV = nRT and n = mass/mw then PV = mRT/mw

  23. (ex) What is the mass of 21.5 l of hydrogen gas at 760 torrs and 0.0 C? (1.94 g) (ex) Find the molecular mass of a gas whose volume is 91.7 ml. It is collected at 37.2 atm and 71.4 C and has a mass of 0.20 g. (1.7 g)

  24. (ex) What is the temperature of 107.5 g of chlorine gas which occupies a volume of 95.2 l at a pressure of 205 torrs? (207 K) (ex) Suppose we measure the mass of the vapor of an unknown gas contained in a 273 cm3 bulb. We find that the bulb contains 0.750 g of gas

  25. at 97.2 kpa of pressure and 61.0 C. What is the molecular mass of the gas? (78.4 g/mol) (ex) What is the molecular weight of a gas if 150.0 cm3 has a mass of 0.922 g at 99.0 C and 107.0 kpa?

  26. (ex) What will be the density of oxygen gas at 100.5 kpa and 23.0 C? (1.31 g/cm3) Other Gas Laws: What happens if we change the conditions? Squeeze it? Heat it?

  27. Add more moles to it? What will happen to the other variables? Conditions 1Conditions 2 n and T stay the same! For state 1For state 2 P1V1/n1T1 = R P2V2/n2T2 =R

  28. Found that if n and T are constant, then as P increases, V decreases - inversely related. (ex) Squeezing a ball Scuba diver Exploding teeth

  29. To get Boyle’s Law from P1V1/n1T1 = R P2V2/n2T2 =R P1V1/n1T1 = P2V2/n2T2 Let n1 = n2 and T1 = T2 since they are constant. Then P1V1= P2V2 Boyle’s Law

  30. Note: R is gone, so the only requirements for units is that all the pressure units must be the same and all the volume units must be the same. (ex) 12.5 l of a gas at 5.43 atm of pressure contract to a new volume as the pressure increases to 9.21 atm. What is the new volume? (7.37 l)

  31. (ex) If a gas at 920. mm Hg occupies a volume of 21. Ml, what will be the new pressure if the volume is cut in half? (2.41 atm) (ex) A gas is collected in a 242 cm3 container. The pressure of the gas in the container is measured and determined to be 87.6 kpa. What is the new of this gas at standard pressure? (209 cm3)

  32. (ex) Correct the following volume of a gas from the indicated pressures to the standard pressures: 1) 273 cm3 at 59.4 kpa (160. cm3) 2) 50.0 m3 at 55.1 kpa (27.2 m3)

  33. Demo: soda bottle and soda can. Pressure is constant as well as the number of moles! T increases, and volume increases T decreases, V decreases T and V are directly related. (ex) tires (winter vs. summer)

  34. To get Charles’ Law: P1V1/n1T1 = R P2V2/n2T2 =R P1V1/n1T1 = P2V2/n2T2 Let P1 = P2 and n1 = n2 since they are constant. Then V1/T1= V2/T2 Charles' Law Note: Volume can be in any units, but temperature must be in Kelvin!

  35. (ex) 19.21 l of a gas is at 27.2 C. Find the volume if the temperature rises to 39.2 C. (20.0 l) (ex) A gas expands to 25.7 ml and is at 27.5 F. Its original volume was 10.2 ml. What was its original temperature? (117 K)

  36. (ex) 225 cm3 of a gas is collected at 58.0 C. What volume would this sample of gas occupy at standard temp? Assume P is constant. (186 cm3) (ex) Correct the following volumes of gases for a change from the temperature indicated to standard temp. Assume P is constant. (1) 617 cm3 at 9 C

  37. 2) 617 cm3 at 282 K (597 cm3) Combined Gas Law - This equation allows one to calculate a change in P, V, or T, assuming only n stays the same. P1V1/n1T1 = R P2V2/n2T2 =R P1V1/n1T1 = P2V2/n2T2 Let n1 = n2 since they are constant.

  38. Then P1V1/T1= P2V2/T2 Combined Gas Law Combination of Charles’ and Boyle’s Law. Note: Volume and Pressure can be in any units, but they must match. Temperature must be in Kelvin!

  39. (ex) 200.0 ml of a gas at 299 torrs and 25.2 C. Find the new volume if the pressure increases to 760. torrs and the temp decreases to -5.02 C (70.7 ml) (ex) 905 ml of a gas are at 25.0 atm and 44.04 C. “Correct to standard conditions” means find the new volume at standard temp and pressure.

  40. (19.5 l) (ex) The volume of a gas measured at 75.6 kpa of pressure and 60.0 C is corrected to correspond to the volume it would occupy at STP. The measured volume of the gas is 10.0 cm3. (6.12 cm3)

  41. John Dalton’s Law of Partial Pressures - the total pressure in a container is the sum of the partial pressures of the gases in the container. In other words, each gas would exert the same pressure it would if it were present alone. PT = P1 + P2 + P3 + ...

  42. Therefore, if you collect a gas over water the water vapor pressure exerts a pressure and must be subtracted to find the pressure of a dry gas. Pg = PT - Pa

  43. Collecting Gases To measure the pressure on a gas we use a eudiometer. Over Mercury (3 cases) 1) levels are equal Pg = Pa 2) level inside is higher than the level outside Pa>Pg

  44. Pa = Pg + Pc Pg = Pa - Pc 3) level inside is lower than the level outside Pg > Pa Pg = Pa + Pc

  45. Examples (1) Calculate the pressure of a gas inside a eudiometer collected over Hg if the barometric pressure is 750. mm Hg and (a) the levels are equal. (750. mm Hg)

  46. (B) the inside level is 60. mm higher than the outside level. (690 mm Hg) (C) the inside level is 40. mm Hg lower than the outside level. (790 mm Hg)

  47. Collecting Gases Over Water Hg is expensive and toxic! Water is 1/13.6 times as dense as Hg, therefore, a given gas pressure will support a column of water 13.6 X as great as an equivalent column of Hg.

  48. (ex) Oxygen gas is collected in a eudiometer by water displacement. The water level inside the tube is 27.2 mm higher than the outside. The temp is 25.0 C. The barometric pressure is 741.0 mm Hg. What is the partial pressure of the dry O2 in the eudiometer? (715.2 mm Hg)

  49. (ex) Calculate the pressure of a dry gas inside a eudiometer collected over water. The temp is at 15.0 C and at an atmospheric pressure of 750. mm Hg if (a) the inside level and outside level are equal (732.2 mm Hg) (b) the inside level is 68 mm higher than the outside level.

  50. (731.5 mm Hg) (3) the inside level is 204 mm lower than the outside level (74.2. mm Hg) (ex) A eudiometer contains 38.4 ml of air, collected by water displacement at a temp of 20.0 C. The water level inside the eudiometer is 140. mm higher than the outside. The barometric

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