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  1. 1. Principles of Chemistry I GASES QBA Miguel A. Castro Ramírez
  2. 2. CHARACTERISTICS OF GASESThe Distinction of Gases from Liquids and Solids • Expand to fill their containers • Are highly compressible • Have relatively low densities under normal condition- (g/L) : When a gas is cooled, its density increases because its volume is decreases. • Gas volume changes greatly with pressure • Gas volume changes greatly with temperature • Gases have relatively low viscosity– flow much more freely • Gases are miscible – Form homogenous mixtures with each other regardless of the identities or relative proportions of the component gases.
  3. 3. CHARACTERISTICS OF GASESThe Distinction of Gases from Liquids and Solids The chemical behavior of a gas depends on its composition, while all gases have very similar physical behavior. This is because the individual molecules are relatively far apart. Thus, each molecule behaves largely as though the others were not present. For instance, although the particular gases differ, the same physical behaviors are at work in the operation of a car and in the baking of bread.
  4. 4. PRESSURE• Pressure conveys the idea of a force, a push that tends tomove something in a given direction. Pressure (P) is in fact, theforce (F) that acts on a given area (A). F P= A• Atmospheric pressure is the weight of air per unit of area• The pressure on the outside of the body is equalized by thepressure on the inside.
  5. 5. PRESSURE Effect of atmospheric pressure on object at Earth surface• The pressure on the outside of the body is equalized by thepressure on the inside.
  6. 6. PRESSURE• Standard atmospheric pressure which correspond to thetypical pressure at sea level, is the pressure sufficient to supporta column of mercury 760mm high. 1 atm = 760 mm Hg = 760 torr = 1.01325 x 105 Pa = 101.325 kPa• Must able to convert between one and another• Manometer: Devices used to measure the pressure of a gas inexperiment.
  7. 7. PRESSUREclosed-end • A) equal pressure • B) Insert gas, it pushes the mercury , so the mercury level rises. • The different in height = the gas pressure open-end
  8. 8. PRESSURE Converting UnitsEXAMPLE:A geochemist heats a limestone (CaCO3) sample and collectsthe CO2 released in an evacuated flask attached to a closed-end manometer. After the system comes to roomtemperature, Dh = 291.4 mm Hg. Calculate the CO2 pressurein torrs, atmospheres, and kilopascals.
  9. 9. THE GAS LAWThe Pressure-Volume Relationship: Boyle’s Law• Boyle’s law: The volume of a fixed quantity of gas maintainedat constant temperature is inversely proportional to the pressure. • The total pressure applied to the trapped air was the pressure of the atmosphere (measured by barometer) plus that of height mercury column. • By adding mercury, the air volume decreased (T and amount of air is constant)
  10. 10. THE GAS LAW• The result as shown in the graph.• When the pressure gets larger, thevolume of the gas become lower.• So, it is inversely proportional. 1• V P α• This relationship also, can beexpressed as PV = constant (k) or V = k (1/P) • This means a plot of V versus 1/P will be a straight line.
  11. 11. THE GAS LAW The Temperature-Volume Relationship: Charles’s Law• Charles found that the volume of afixed quantity of gas at constantpressure increases linearly withtemperature.• V α T So, V = constant(k) x T or V T =k
  12. 12. THE GAS LAW• Extrapolated of the graph: passes though – 273 °C (0 K).• Gas predicted to have 0 volume at this temperature.• However, absolute zero never reached because no matter canhave zero volume.• Gas will liquefy before reached this temperature.• Charles Law: The volume of a fixed amount of gas maintained atconstant pressure is directly proportional to its absolute (Kelvin)temperature.
  13. 13. THE GAS LAWOther Relationship Based on Boyle’s and Charles Law Amontons’s V and n are fixed P α T Law P = constant P = constant x T TCombined gas T law V V = constant x T/P P α PV = constant T
  14. 14. THE GAS LAW The Quantity – Volume Relationship: Avogadro’s Law• Different mol of gas will occupy different volume.• Double up the quantity (mol) of gas will double up the volumeoccupied by the particular gas. (refer B)
  15. 15. THE GAS LAW
  16. 16. THE IDEAL-GAS EQUATION• So far we’ve seen that V ∝ 1/P (Boyle’s law) V ∝ T (Charles’s law) V ∝ n (Avogadro’s law)• Combining these, we get V ∝ nT nRT P PV = nRT or V = P R = gas constant © 2009, Prentice-Hall, Inc.
  17. 17. THE IDEAL-GAS EQUATION PV = nRT or V = nRT P fixed n and T fixed n and P fixed P and TBoyle’s Law Charles’s Law Avogadro’s LawV = constant V = constant X T V = constant X n PTemperature: Absolute Temperature (K)Quantity of Gas: molesPressure: atmVolume: Liters
  18. 18. THE IDEAL-GAS EQUATION• The value and unit for R are depend on the units of P,V,n and T.• In working problems with the ideal-gas equation, the units ofP,V,n & T must agree with the units in gas constant.• We often use: R = 0.08206 L-atm/mol-K or 0.0821 L-atm/mol-K• Use the value R = 8.314 J/mol-K consistent with the unit Pa forpressure is also very common.
  19. 19. THE IDEAL-GAS EQUATION Applying the Volume-Pressure RelationshipEXAMPLEBoyle’s apprentice finds that the air trapped in a J tubeoccupies 24.8 cm3 at 1.12 atm. By adding mercury to thetube, he increases the pressure on the trapped air to2.64 atm. Assuming constant temperature, what is thenew volume of air (in L)?
  20. 20. THE IDEAL-GAS EQUATION Applying the Pressure-Temperature RelationshipEXAMPLEA steel tank used for fuel delivery is fitted with a safety valvethat opens when the internal pressure exceeds 1.00x103 torr.It is filled with methane at 230C and 0.991 atm and placed inboiling water at exactly 1000C. Will the safety valve open?
  21. 21. THE IDEAL-GAS EQUATION Solving for an Unknown Gas Variable at Fixed ConditionsEXAMPLEA steel tank has a volume of 438 L and is filled with 0.885 kg ofO2. Calculate the pressure of O2 at 21oC.
  22. 22. FURTHER APPLICATION OF THE IDEAL-GAS EQUATION Gas Densities and Molar Mass• The ideal-gas equation allows us to calculate gas densityfrom molar mass, pressure and temperature of the gas.• Density = mass/volume (m/V)• Rearrange the gas equation to obtain similar units: moles perunit volume, (n/V): n P = RT n P x M( molar V V = RT mass) Mn MP mass MP V = RT V = RT Density = MP (d) RT
  23. 23. FURTHER APPLICATION OF THE IDEAL-GAS EQUATION Density (d) = MP RT• The density of the gas depends on its pressure, molar mass andtemperature.• The higher the molar mass and pressure the more dense thegas.• The higher the temperature, the less dense the gas.• The above equation can be rearrange : dRT Μ= P• Thus, we can use the experimentally measured density of a gasto determine the molar mass of the gas molecules.
  24. 24. FURTHER APPLICATION OF THE IDEAL-GAS EQUATION Finding DensityEXAMPLEFind the density (in g/L) of CO2 and the number of molecules(a) at STP (0oC and 1 atm) and (b) at room conditions (20.0 °Cand 1.00 atm).
  25. 25. FURTHER APPLICATION OF THE IDEAL-GAS EQUATIONVolumes of Gases in Chemical Reaction• The ideal gas equation relates the number of moles ofa gas to P, V, and T.• Thus, the volume of the gases consumed or producedduring the reaction can be calculated.
  26. 26. GAS MISTURES AND PARTIAL PRESSURE Dalton’s Law of Partial Pressure The total pressure of a mixture of gases equals the sum of the pressures that each would exert if it were present alone.Partial Pressure: The pressure exerted by a particularcomponent of a mixture of a gas.
  28. 28. GAS MISTURES AND PARTIAL PRESSURE Partial Pressures and Mole Fractions• Because each gas in a mixture behave independently, we canrelate the amount of a given gas in a mixture to its partialpressure. (P1/Pt) = (n1 RT/V) / (nt RT/V) = n1/nt• n1/nt is called mole fraction of gas 1. – denoted as X1• The mole fraction , X is a dimensionless number that expressthe ratio of the number of moles• Rearrange the equation: P = (n1/n2) Pt = X1Pt• Thus, the partial pressure of a gas in a mixture is its molesfraction times the total pressure
  29. 29. GAS MISTURES AND PARTIAL PRESSURE Collecting Gases over Water• When one collects a gas over water, there is water vapormixed in with the gas.• To find only the pressure of the desired gas, one must subtractthe vapor pressure of water from the total pressure. Ptotal = Pgas + PH20• Refer appendix b
  30. 30. GAS MISTURES AND PARTIAL PRESSUREEXAMPLE:Acetylene (C2H2) is produced in the laboratory when calciumcarbide (CaC2) reacts with water: CaC2(s) + 2H2O(l) C2H2(g) + Ca(OH)2(aq)A collected sample of acetylene has a total gas pressure of738 torr and a volume of 523 mL. At the temperature of thegas (23oC), the vapor pressure of water is 21 torr. Howmany grams of acetylene are collected?
  31. 31. KINETIC-MOLECULAR THEORY • This is a model that aids in our understanding of what happens to gas particles as environmental conditions change
  32. 32. KINETIC-MOLECULAR THEORY• The kinetic-molecular theory is summarized by the followingstatements:1. Gases consist of large numbers of molecules that are incontinuous, random motion2. The combined volume of all the molecules of the gas isnegligible relative to the total volume in which the gas iscontained.3. Attractive and repulsive forces between gas molecules arenegligible.
  33. 33. KINETIC-MOLECULAR THEORY4. Energy can be transferred between molecules during collisions, but the average kinetic energy of the molecules does not change with time, as long as the temperature of the gas remains constant.5. The average kinetic energy of the molecules is proportional to the absolute temperature.• This theory explains both pressure and temperature at molecular level.• Pressure: caused by collisions of the molecules with the wall of the container.• Absolute Temperature: measure of kinetic energy of its molecule. If 2 different gases are at the same T, theit molecules have the same average kinetic energy (stat 5)
  34. 34. KINETIC-MOLECULAR THEORY Distribution of Molecular Speed• The curve shows the fraction of molecules moving at each speed.• Higher T, a larger fraction of molecules moves at greater speeds; the distribution curve has shifted to the right toward higher speed and hence average kinetic energy.
  35. 35. KINETIC-MOLECULAR THEORY Distribution of Molecular Speed• Root-mean-square (rms) speed- u: The speed of molecules possessing average kinetic energy.• rms speed is important bcoz: The average kinetic energy of the gas molecules in a sample, ε, related directly to u2 : ε = ½ mu2 m = mass of individual molecules• Mass does not change with T, thus the increase in the averagekinetic energy as the T increases implies that the rms speed (alsothe average speed ) of molecules likewise increases with theincrease of T.
  36. 36. KINETIC-MOLECULAR THEORY Application to the Gas Laws• Effect of a volume increase at constant temperature: -Constant T: A.K.E unchanged. rms speed,u, unchanged. - Increased Volume: molecules move a longer distance between collision. So, fewer collisions per unit time with container walls & the pressure decreases. – BOYLE’S LAW
  37. 37. KINETIC-MOLECULAR THEORY• Effect of a temperature increase at constant volume:-Increase T: increased A.K.E thus increase in u.- If no change in volume, more collision between moleculesandwith the walls.- Thus, pressure increase.- CHARLES’S LAW
  38. 38. MOLECULAR EFFUSION AND DIFFUSIONGraham’s Law of Effusion • Effusion (a process by which a gas escapes from its container) rate of a gas is inversely proportional to the square root of its molar mass. • If we have 2 gases at the same T & P in containers with identical pinholes. • If the rates of effusion of the two substances are r1 and r2 and their respective molar masses are M1 & M2, Graham’s Law states: √ r1 M2 = r2 M1
  39. 39. MOLECULAR EFFUSION AND DIFFUSION Graham’s Law of Effusion √ r1 M2 = r2 M1• Above equation comparesthe rates of effusion of twodifferent gases underidentical conditions; itindicates that the lighter gaseffuses more rapidly.• To escape, molecules have to hit the hall. The faster they move,the greater they will hit the wall.• Thus, rate of effusion is directly proportional to the rms speed.
  40. 40. MOLECULAR EFFUSION AND DIFFUSION √ 3RT/ M1 √ r1 u1 M2 = = r2 u2 3RT / M2 = M1u= √ 3RT /M
  41. 41. MOLECULAR EFFUSION AND DIFFUSION Diffusion and Mean Free Path• Diffusion is the spread of one substance throughout aspace or throughout a second substance.• Faster for lower mass molecules.
  42. 42. REAL GASES: DEVIATIONS FROM IDEAL BEHAVIOUR • In the real world, the behavior of gases only conforms to the ideal-gas equation at relatively high temperature and low pressure. • Real gases do not behave ideally at high pressure.
  43. 43. REAL GASES: DEVIATIONS FROM IDEAL BEHAVIOUR • The deviation also depends on T. • We can see that, as the P increased, the behavior of the gas more nearly approaches the ideal gas • Thus, the deviations from ideal behavior increase as T decrease and becoming significant near the T at which the gas is converted into liquid.
  44. 44. REAL GASES: DEVIATIONS FROM IDEAL BEHAVIOUR Why Real Gases Deviate From Ideal-Gas• Assumptions: Molecules of ideal gas are assumed to occupy no space and have no attractions for one another.• However: Real molecules do have finite volume, and attract one another.• Fig: (low P) The free unoccupied space in which molecules can move is less than the container volume.• Thus, the free volume of the available to the molecules is essentially the entire volume of the container.
  45. 45. REAL GASES: DEVIATIONS FROM IDEAL BEHAVIOUR• Increase P: The free space in which the molecules can move become smaller fraction of the container volume.• Thus, the gas volumes tend to be slightly greater than those predicted by the ideal-gas equation.• In addition, the attractive forces between molecules also play at short distance.• The impact is increased.• However, the attraction between molecules also increased due to the short distance.• As a result, the pressure is less than the ideal gas.
  46. 46. REAL GASES: DEVIATIONS FROM IDEAL BEHAVIOUR The van der Waals Equation• The ideal-gas equation can be adjusted to take thesedeviations from ideal behavior into account.• The corrected ideal-gas equation is known as the van derWaals equation.• The van der Waals Equation n2a ) (V − nb) = nRT (P + 2 V