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The Gas Laws

The Gas Laws. An Overview of Gases. Like solids and liquids, gases have a temperature, volume, and an amount that can be measured in moles. Unlike solids and liquids, gases have a pressure that will affect the nature of the gas.

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The Gas Laws

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  1. The Gas Laws

  2. An Overview of Gases Like solids and liquids, gases have a temperature, volume, and an amount that can be measured in moles. Unlike solids and liquids, gases have a pressure that will affect the nature of the gas. Gases have many units: atmospheres, mm Hg (torr), kPa, Pa, lbs/in2, bar, mbar, and more.

  3. Pressure • Pressure = Force per unit area. • Gas molecules will always fill any container they are in. • Molecules move around and hit the insides of the container. These collisions cause the pressure. • Measured with a barometer or a manometer.

  4. Barometer Vacuum • The pressure of the atmosphere at sea level will raise a column of mercury 760 mm Hg. • 1 atm = 760 mm Hg 760 mm Hg 1 atm of Air Pressure

  5. Manometer • Column of mercury to measure pressure. • ‘h’ is how much lower the pressure is than outside. h Gas

  6. Manometer • h is how much higher the gas pressure is than the atmosphere. • The Hg is pushed by the gas with the greater pressure. h Gas

  7. The difference in the height of mercury on both sides is the difference in pressure between the two gases. The gas with the higher pressure will be able to move the mercury more than the gas with the lower pressure. Remember our “Law of Phyics” The stronger force always wins !! Pressure Readings

  8. Units of pressure • These values are Standard Pressure and need to be known. • 1 atmosphere = 760 mm Hg = 760 torr • 1 atm = 101,325 Pascals = 101.325 kPa • They also need to be able to be converted from one unit to another.

  9. Converting Pressure Units • Convert 3.5 atm to ___ kPa • Like any conversion, set up a proportion. • The known values are on the left. 1 atm / 101.325 kPa = 3.5 atm / x kPa x = 354.64 kPa

  10. The Gas Laws Our first look at Gas Laws will be a group of laws that focus on simple relationships between two variables. When these two variables are studied, it implies that the other variables are constant. So Boyle’s Law of P1V1 = P2V2 means that while Pressure and Volume are being examined, moles and temperature remain constant.

  11. The Gas Laws • Boyle’s Law • Pressure and volume are inversely related at constant temperature. OR • As one goes up, the other goes down with the same factor. • PV= k was Boyle’s original work and it turned into P1V1 = P2 V2

  12. Boyle’s Law The following two graphs show the data when Pressure v. Volume are plotted. In the first graph, it shows Pressure graphed v. Volume. The second graph shows the straight line plot of Volume v. 1 / Pressure

  13. Pressure v. Volume V P (at constant T)

  14. Pressure v. 1 / Volume Slope = k V 1/P (at constant T)

  15. Looking at those graphs The second graph uses the reciprocal and makes a straight line. Straight lines allow us to calculate a slope easily and form relationships and enables us to calculate much more easily.

  16. Examples • 20.5 L of nitrogen at 742 torr are compressed to 9.8 atm. What is the new volume ? • Pressure is changing and a new volume is needed = Boyle’s Law • Make sure the units match up as torr does not equal atm. • (20.5 L) (742 torr) = (7448 torr) V2 • V2 = 2.042 L

  17. More examples • 30.6 mL of carbon dioxide at 740 torr is expanded to 750 mL What is the final pressure in kPa ? (30.6 mL) (740 torr) = (750 mL) P2 P2 = 30.19 torr (needs to be in kPa) 30.19 torr = 4.025 kPa

  18. Charles’ Law Before we explore Charles’ Law, lets work a simple problem, given the Charles’ Law formula of V1 = V2 T1 T2 If V1 = 10 L and T1 = -10 what is the new volume when the temperature is raised to 30 ?

  19. Charles’ Law We see we have a problem as we can’t have a negative volume. For quite some time, scientists knew that there was a positive relationship between the temperature of a gas and the volume it would occupy. The problem was that they couldn’t determine how to mathematically and accurately express this relationship.

  20. Charles’ Law A scientist named Lord Kelvin was working on the ideas of temperature and determined that the lowest possible temperature that an object can reach is - 273.15 ˚C. Charles took this idea and did the rest.

  21. Charles’ Law • Volume of a gas varies directly with the absolute temperature at constant pressure. • V = kT (if T is in Kelvin) • V1 = V2 T1 T2 • Graphically it looks like this.

  22. Charles’ Law Graph He H2O CH4 V (L) H2 -273.15 ºC T (ºC)

  23. Charles’ Law Given the following formula V1 = V2 T1 T2 If V1 = 10 L and T1 = -10 what is the new volume when the temperature is raised to 30 ?

  24. Examples • What would the final volume be if 247 mL of gas at 22ºC is heated to 98 ºC ? • 247 / 295 = V2 / 371 V2 = 310.6 mL • Remember that all Gas Law temperatures must be in Kelvin

  25. Avogadro's Law • At constant temperature and pressure, the volume of gas is directly related to the number of moles. When air is added to a ball to pump it up, it is putting more moles of air in it !! • V = k n (n is the number of moles) • V1 = V2 n1 = n2

  26. Gay- Lussac Law • At constant volume, pressure and absolute temperature are directly related. • P = k T • P1 = P2 T1 T2

  27. Combined Gas Law • If the moles of gas remains constant, use this formula and cancel out the other things that don’t change. • P1 V1 = P2 V2 . T1 T2

  28. Examples • A cylinder has a volume of 175 mL and a pressure of 3.8 atm at 22 ºC. What would the pressure be if the cylinder’s volume was increased to 250 mL as it was heated to 100 ºC ? • (175)(3.8) / 295 = V2 (250) / 373 V2 = 3.36 atm.

  29. Molar Volume Standard Pressure = 1 atmosphere Standard Temperature = 273 K Together they from STP conditions 1 mole of any gas at STP occupies 22.4 Liters of Volume

  30. What have we seen so far ? We have looked at several gas law formulas so far. Have you noticed something peculiar about them so far ? Is there anything about them that might make it difficult to work with from a practical point of view.

  31. What have we seen so far ? What has to be happening in our situations to be able to solve ? So far, we are only working on problems where a change in conditions are taking place.

  32. Ideal Gas Law • PV = nRT • R is the ideal gas law constant. • R = ? It depends on the pressure unit. .08205 L atm / mol K OR 62.36 L torr / mol K OR 8.314 J / mol K • The other laws tell you about a gas when it changes.

  33. Examples 1. A 47.3 L container containing 1.62 mol of He is heated until the pressure reaches 1.85 atm. What is the temp. ? 2. Kr gas in a 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8 ºC. What is the mass of Kr ?

  34. Solved Answers 1. PV = nRT (1.85)(47.3) = 1.62 R T T = 658.32 K or 385.32 °C 2. PV = nRT (8.61) (18.5) = n (.08205)(297.8) n = 6.52 moles = 546.28 grams

  35. Gas Density and Molar Mass • Sometimes we have to combine our gas laws with formulas that we learned earlier, such as : • Density = m / V OR moles = g / M • PV = nRT substitute for n. • M = g RT / PV Another algebra manipulation is • P x M = gRT/ V = P x M = DRT

  36. Gas Density and Molar Volume It is strongly recommended that you don’t memorize the two derived formulas from the last slide. You will be much better off by learning the Ideal Gas Law and the formulas for Density and n = g / M which you should already know very well. These are also given to you on tests.

  37. Density Example • What is the density of ammonia (NH3) at 23 ºC and 735 torr ? • For this we need to remember that Density is an intensive property. • This means that it doesn’t matter what the amount of ammonia we use. • So we might as well use a convenient amount.

  38. Solved Answer • What is the density of ammonia (NH3) at 23 ºC and 735 torr ? • Assume 1 mole (17 grams) and PV = nRT to get volume. • PV = nRT (735)V = 1 R (295) Volume = 25.11 Liters Density = 17 / 25.11 = .677 g / L

  39. Gases and Stoichiometry • Reactions happen in moles. • At Standard Temperature and Pressure (STP, 0ºC and 1 atm.) 1 mole of gas occupies 22.42 L. • Recall Molar Volume of a gas. • If not at STP, use the ideal gas law to calculate moles of reactant or volume of product.

  40. Examples • Mercury can be achieved by the following reaction: • 2 HgO  2 Hg + O2 What volume of oxygen gas can be produced from 4.10 g of mercury (II) oxide at STP ? • What volume at 400 ºC and 740 torr ?

  41. Answers First we need to find moles of HgO. n = g/M .0189 moles Use a mole ratio from the balanced eq. 2 HgO / 1 O2 = .0189 / x x = .0095 Proportion: 1 n / 22.4 L = .0095 / x Use molar volume (1 n  22.4 Liters) .212 Liters of Oxygen Gas For the second problem: 4.98 L of gas

  42. Examples • Using the following reaction: • NaHCO3 + HCl NaCl + CO2 (g) + H2O (l) • Calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L of carbon dioxide at 25 ºC and 2.00 atm. • If 27 L of gas are produced at 26 ºC and 745 torr when 2.6 L of HCl are added what is the concentration of HCl ?

  43. Answers PV = nRT 2 (2.87) = n R (298) n = .235 n 1:1 ratio SO .235 moles of NaHCO3 Converting to grams – 19.74 grams PV = nRT (.98) (27) = n R (299) n = 1.079 n 1:1 ratio SO M = n / L  1.079 / 2.6 M = .415 M

  44. Examples • Consider the following reaction: • 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) • What volume of H2O at 1.0 atm and 1000 º C can be produced from 10.0 L of NH3 and excess O2 at the same temperature and pressure ? • What volume of O2 measured at STP will be consumed when 10.0 kg NH3 is reacted ?

  45. Answers PV = nRT 1 (10) = n R (1273) n = .0957 n of NH3 4 NH3 / 6 H2O = .0957 / x x = .144 n H2O Use PV = nRT 15.04 Liters 10 kg = 588.24 moles of NH3 Use molar volume since it is at STP 1 / 22.4 = 588.24 / x x = 735.29 n O2 Convert to Liters n x 22.4 L  16470 L of O2

  46. The Same Reaction • 4 NH3(g) + 5 O2(g) 4 NO(g) + 6 H2O(g) • What mass of H2O will be produced from 65.0 L of O2 and 75.0 L of NH3 both measured at STP ? • What volume of NO would be produced ? • What mass of NO is produced from 500. L of NH3 at 250.0 ºC and 3.00 atm ?

  47. Dalton’s Law • The total pressure in a container is the sum of the pressure each gas would exert if it were alone in the container. • The total pressure is the sum of the partial pressures. • PTotal = P1 + P2 + P3 + P4 + P5 ... • Make sure that the units match up.

  48. Dalton’s Law Problem In a flask filled with a combination of Nitrogen, Oxygen, and Hydrogen, it was found to have a total pressure of 546.3 kPa. If the oxygen is determined to have a partial pressure of 2.1 atm, and the hydrogen is found to have a partial pressure of 138 torr, what is the partial pressure of nitrogen measured in kPa.

  49. Solved Answer • In a flask filled with a combination of Nitrogen, Oxygen, and Hydrogen, it was found to have a total pressure of 546.3 kPa. If the oxygen is determined to have a partial pressure of 2.1 atm, and the hydrogen is found to have a partial pressure of 138 torr, what is the partial pressure of nitrogen measured in kPa. • We need to get all the gases in the same units. 2.1 atm = 212.783 kPa 138 torr = 18.398 kPa Now we have: 546.3 = 212.783 + 18.398 + N2 Solving for N2 = 315.119 kPa

  50. Mole Fraction • Some calculations are done with the concentration term of Mole Fraction. • Similar but not identical to % composition. • DO NOT multiply by 100 % • Ratio of moles of the substance to the total moles.

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