Physical chemistry


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Physical chemistry

  1. 1. Physicalchemistry
  2. 2. Gas Laws
  3. 3. Elements that exist as gases at 250C and 1 atmosphere
  4. 4. Physical Characteristics of GasesGases assume the volume and shape of their containers. •Gases are the most compressible state of matter. •Gases will mix evenly and completely when confined to •the same container.Gases have much lower densities than liquids and solids. •
  5. 5. Force Pressure = Area Units of Pressure1 pascal (Pa) = 1 N/m21 atm = 760 mmHg = 760 torr1 atm = 101,325 Pa Barometer
  6. 6. 10 miles 0.2 atm 4 miles 0.5 atmSea level 1 atm
  7. 7. BOYLE‘S LAW:At a constant temperature,the volume occupied by a fixed quantity of gas is inversely proportional to the applied pressure.
  8. 8. This can be expressedmathematically as:• V α 1/P• V = constant . 1/p.• PV = constant.
  9. 9. As P (h) increases V decreases
  10. 10. Boyle’s Law P 1/V Constant temperatureP x V = constant Constant amount of gasP1 x V1 = P2 x V2
  11. 11. A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 726 mmHg x 946 mLP2 = = = 4460 mmHg V2 154 mL
  12. 12. Charles law: At a constant pressure, thevolume of a fixed mass of gas is proportional to its temperature. V α T (P constant) or, V/ T = constantIt is most important to use Kelvin scale here.
  13. 13. As T increases V increases
  14. 14. Variation of gas volume with temperatureat constant pressure. Charles’ & Gay- Lussac’s Law V T Temperature must be in Kelvin V = constant x T V1/T1 = V2/T2 T (K) = t (0C) + 273.15
  15. 15. We can use -273 oC point on the graph to define the zero of new scale of temperature (Kelvin scale (K)).• We can convert between oC and K• T (K) = t (oC) + 273 K• We can combine Boyle`s and Charle`s laws by using the equation:• PV/T = constant.• The name is given constant is the gas constant, (R). Its value is 8.314 J K-1 mole-1 (standard value)
  16. 16. A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1/T1 = V2/T2 V1 = 3.20 L V2 = 1.54 L T1 = 398.15 K T2 = ? V2 x T1 1.54 L x 398.15 KT2 = = = 192 K V1 3.20 L
  17. 17. Avogadros law:Under conditions of constant temperature and pressure equal volumes of gas contain equal number of molecules. (Since equal numbers of molecules means equal numbers of moles, the number of moles of any gas is related directly to its volume. Vαn
  18. 18. Avogadro’s LawV number of moles (n) Constant temperatureV = constant x n Constant pressureV1/n1 = V2/n2
  19. 19. Ammonia burns in oxygen to form nitric oxide (NO)and water vapor. How many volumes of NO areobtained from one volume of ammonia at the sametemperature and pressure? 4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P1 volume NH3 1 volume NO
  20. 20. Ideal Gas Equation 1Boyle’s law: V (at constant n and T) PCharles’ law: V T (at constant n and P)Avogadro’s law: V n (at constant P and T) nTV P nT nTV = constant x = R R is the gas constant P P PV = nRT
  21. 21. The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Experiments show that at STP, 1 mole of an ideal gas occupies 22.414 L.PV = nRT PV (1 atm)(22.414L) R= = nT (1 mol)(273.15 K) R = 0.082057 L • atm / (mol • K)
  22. 22. What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = 273.15 K P = 1 atmPV = nRT 1 mol HCl nRT n = 49.8 g x = 1.37 mol V= 36.45 g HCl P 1.37 mol x 0.0821 L•atm x 273.15 K V= mol•K 1 atm V = 30.6 L
  23. 23. Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)?PV = nRT n, V and R are constantnR P = = constantV T P1 = 1.20 atm P2 = ? T1 = 291 K T2 = 358 KP1 P2 =T1 T2 T2 358 K P2 = P1 x = 1.20 atm x = 1.48 atm T1 291 K
  24. 24. Density (d) Calculations m PM m is the mass of the gas in g d= = V RT M is the molar mass of the gasMolar Mass (M ) of a Gaseous Substance dRT M= d is the density of the gas in g/L P
  25. 25. Combined gas law:• In ideal gas equation we assume we have a gas under two different sets of conditions Pi, Vi, Ti and Pf, Vf, Tf• Pi Vi/ Ti = Pf Vf/ Tf
  26. 26. Ex(3): what would be the volume of a gasat (S.T.P) if it was found to occupy a volume of 255 ml at 25oC and 650 torr? • (i) (f) • V 255 ml ? • P 650 torr 760 torr • T 298 k 273 k • Vf =255 mlx(650torr/760torr)x(273 k/298k) • The volume at ST.P ~ 200 ml.
  27. 27. Dalton’s Law of Partial Pressures V and T are constantP1 P2 Ptotal = P1 + P2
  28. 28. Dalton`s law of partial pressures:• Total pressure in a container is equal to the sum of the partial pressures of the component gases. P total = Pa + Pb + Pc + ……..• Pa, Pb, Pc is the partial pressure gases a, b, c respectively
  29. 29. when two or more gases that do not react chemically areplaced in the same container,the pressure exerted by each gas in the mixture is thesame as it would be if it werethe only gas in the container.
  30. 30. Ex (2) : A ten liter flask at 25oC contains a gaseous solution of carbon monoxide and carbon dioxide at a total pressure of 2.0 atm. If 2 mole of carbon monoxide is present, find its partial pressure and also that of the carbon dioxide.•By Dalton`s law Pt = Pco + Pco2 = 2.0 atm• Pco = n RT/V =•2 mol x 0.0821 lit atm mol-1 K-1 x 298K/10 lit• = 0.49 atm• Pco2= Pt- Pco = 2.0 - 0.49 = 1.51atm
  31. 31. Ex (1): If 200 ml of nitrogen gas at 25 oCand a pressure of 250 torr are mixed with350 ml of oxygen at 25 oC and a pressure of 300 torr, so that the resulting volume is 300ml, what would be the final pressure of the mixture at 25 oC?. There is no temperature change, we simply a Boyle`s law calculation for each gas.
  32. 32. For N2 for O2 (i ) (f) (i) (f)P 250 torr ? 300 torr ?V 200 ml 300 ml 350 ml 300ml for N2 : P = 250 torr x 200 ml 300 ml = 167 torr forO2: P =300 torr x 350 ml 300 ml = 350 torrthe total pressure of the mixture is the sum of partial pressures Pt = P O2+ P N2 = 167 + 350 = 517 torr.
  33. 33. Physical Propertiesof Solutions
  34. 34. A solution is a homogenous mixture of 2 ormore substancesThe solute is(are) the substance(s) present in thesmaller amount(s)The solvent is the substance present in the largeramount 12.1
  35. 35. A saturated solution contains the maximum amount of asolute that will dissolve in a given solvent at a specifictemperature.An unsaturated solution contains less solute than thesolvent has the capacity to dissolve at a specifictemperature.A supersaturated solution contains more solute than ispresent in a saturated solution at a specific temperature. Sodium acetate crystals rapidly form when a seed crystal is added to a supersaturated solution of sodium acetate. 12.1
  36. 36. Colligative Properties are those properties of aliquid that may be altered by the presence of asolute.Examples•vapor pressure•melting point•boiling point•osmotic pressure.
  37. 37. VAPOUR PRESSUREThe term "vapour" is applied to the gas of anycompound that would normally be found as aliquid at room temperature and pressureFor example, water, gasoline, rubbing alcohol, andfinger nail polish remover (ethyl acetate) are allnormally liquids, but they all evaporate to give agas.
  38. 38. The molecules that are found on the upper side have only about half as many neighbors as do molecules inside the liquid so to begin with the forces holding them in the liquidare slightly lower than molecules in the bulk liquid. The rate at which a given volume of liquid will evaporate is determined by the surface area.
  39. 39. Equilibrium Vapour PressureIf we put the beaker in a closed chamber a slightly differentphenomenon occurs. Now molecules from the liquid evaporate asbefore, but some of these evaporated molecules may also return tothe liquid.
  40. 40. When a nonvolatile soluteis added to a liquid to forma solution, the vapourpressure above thesolution decreases
  41. 41. On the surface of the pure solvent (shown on the left)there are more solvent molecules at the surface thanin the right-hand solution flask. Therefore, it is morelikely that solvent molecules escape into the gasphase on the left than on the right. Therefore, thesolution should have a lower vapour pressure thanthe pure solvent.
  42. 42. The French chemist Francois Raoult discoveredthe law that mathematically describes the vaporpressure lowering phenomenon.Raoults law states that the vapor pressure ofa solution, P, equals the mole fraction of thesolvent, solvent, multiplied by the vaporpressure of the pure solvent, Po. P= solvent * Po
  43. 43. Raoult`s law: (Vapour pressureof solutions).the vapour pressure of anysolution (P total) is the sum ofthe partial pressure of thecomponents(PA, PB).
  44. 44. Pt = PA + PBPA = XA PAO, PB = XB PBOPt = XA PAO + XB PBO(PAO, PBO is the vapourpressure of the componentA&B)
  45. 45. Ex: If heptane and octane formed idealsolution at 40 oC of a solution containing 1.0mole of heptane and 4.0 mole of octane. At40 oC the vapour pressure of heptane is0.121 atm, and the vapour pressure ofoctane is 0.041 atm, what is the vapourpressure of the solution? • X heptane = 1/5 and X octane = 4/5 Hence, • Pt= 1/5(0.121atm) + 4/5(0.041atm) • = 0.057 atm.
  46. 46. The change in the vapor pressure thatoccurs when a solute is added to asolvent is therefore a colligative property.If it depends on the mole fraction of thesolute, then it must depend on the ratioof the number of particles of solute tosolvent in the solution but not the identityof the solute.
  47. 47. Colligative Properties:Colligative properties are properties that depend only on thenumber of solute particles in solution and not on the nature ofthe solute particles.Vapor-Pressure Lowering 0 = vapor pressure of pure solvent P1 = X1 P 0 1 P1 Raoult’s law X1 = mole fraction of the solvent If the solution contains only one solute: X1 = 1 – X2 0 0 P 1 - P1 = P = X2 P 1 X2 = mole fraction of the solute 12.6
  48. 48. Boiling point and freezing point of solutions•The boiling point of a liquid isdefined as the temperature atwhich the vapour pressure of theliquid is equal to the atmosphericpressure.* At the freezing point the vapourpressure of solid and liquid areequal.
  49. 49. Boiling-Point Elevation 0 Tb = Tb – T b 0 T b is the boiling point of the pure solvent T b is the boiling point of the solution 0 Tb > T b Tb > 0 Tb = Kb m m is the molality of the solution Kb is the molal boiling-point elevation constant (0C/m) 12.6
  50. 50. Freezing-Point Depression Tf = T 0 – Tf f T f0is the freezing point of the pure solvent T f is the freezing point of the solution T 0 > Tf f Tf > 0 Tf = Kf m m is the molality of the solution Kf is the molal freezing-point depression constant (0C/m) 12.6
  51. 51. 12.6
  52. 52. What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g. Tf = Kf m Kf water = 1.86 0C/m 1 mol 478 g x moles of solute 62.01 gm = = = 2.41 m mass of solvent (kg) 3.202 kg solvent Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C Tf = T 0 – Tf f Tf = T 0 – Tf = 0.00 0C – 4.48 0C = -4.48 0C f 12.6
  53. 53. Ex: What is the boiling point and freezing point of a solution prepared by dissolving 2.4g of biphenyl(C6H12) in 75g of benzene? The molecular weight of biphenyl is 154 (the molal boiling point elevation and freezing point depression of benzene is:(b = 80.1oC, Kb = 2.53 oC & f = 5.5oC, Kf = -5.12oC)
  54. 54. Solution:• m = 1000 x 2.4/(75 x 154) = 0.208 m• Δtb = m kb• = 0.208 x 2.53 = 0.526 oC• The boiling point of the solution is 80.1 + 0.526 = 80.6 oC• Δtf = m kf• =0.208 x -5.12 = -1.06oC• The freezing point of the solution is 5.5oC – 1.1oC = 4.4oC
  55. 55. Colligative Properties of SolutionsColligative properties are properties that depend only on the number of soluteparticles in solution and not on the nature of the solute particles. 0 Vapor-Pressure Lowering P1 = X1 P 1 Boiling-Point Elevation Tb = Kb m Freezing-Point Depression Tf = Kf m Osmotic Pressure ( ) = MRT 12.6
  56. 56. Osmotic Pressure ( )Osmosis is the selective passage of solvent molecules through a porousmembrane from a dilute solution to a more concentrated one.A semipermeable membrane allows the passage of solvent molecules butblocks the passage of solute molecules.Osmotic pressure ( ) is the pressure required to stop osmosis. more dilute concentrated 12.6
  57. 57. Semi permeable membrane such as cellophane that permits some molecules but not all to pass through it. Osmosis plays an important rule in plants and animal physiologicalprocesses, the passage of substances through the semi permeable walls of living cell. The action of the kidneys and the rise of sap in trees.
  58. 58. Osmotic Pressure ( ) High Low P P = MRTM is the molarity of the solutionR is the gas constantT is the temperature (in K) 12.6
  59. 59. Chemistry In Action:Desalination