States of matter

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  • MECHANICS OF BREATHING Gas travels from high pressure to low pressure. This is also responsible for all weather patterns.
  • States of matter

    1. 1. ReviewWhat’s matter?What are the three commonstates of matter? 1) Solids 2) Liquids 3) GasesWhat can the states also be called? PhasesSo, a phase describes a physical state of matter.
    2. 2. Comparison of States gas liquid solid Do not have Take the shape Definite shapeSHAPE definite shape of the container in which it is poured lots of free little free space little free spaceDENSITY space between between between particles particles particlesFLOW particles can particles can rigid - particles move past one move/slide past cannot another one another move/slide past one another
    3. 3. THREE GAS LAWS BOYLE’SLAW : RELATE PRESSURE WITH VOLUME CHARLES’ LAW RELATE VOLUME WITH TEMPERATURE GAY LUSSAC’S LAW RELATE PRESSURE WITH TEMPERATURE
    4. 4. Robert Boyle (1627-1691)• Boyle had the good fortune to have Robert Hooke as an assistant and together they made an air pump.In 1662, Boyle published what is now known as Boyles law: At constant temperature the volume of a gas isinversely proportional to the pressure
    5. 5. Boyle’s Law : The volume of definite quantity of gas is inversely proportional to it’s pressure at constant temperature. Mathematically expressed as V ∝ 1 (Constant temp.) P ∴ V = K/P where K = constant ∴ PV = KThus it can also be stated as“At constant temperature the product of volume andpressure of definite quantity of gas is constant. If P1 and V1 is initial pressure and volume of gas at constant temp and P2, V2 at final state then above equation can be written as : P1V1 = P2V2
    6. 6. Boyle’s LawTimberlake, Chemistry 7th Edition, page 253
    7. 7. Boyle’s Law1 atm • As the pressure on a2 atm gas increases - the volume decreases 4 Liters 2 • Pressure and volume are inversely related
    8. 8. Boyle’s Law illustrated
    9. 9. Jacques CharlesIn the centuryfollowing Boyle, aFrench physicist,Jacques Charles(1746-1823), wasthe first personto fill a balloonwith hydrogengas and whomade the firstsolo balloon
    10. 10. Volume vs. Temperature: Charles’ Law • Notice the linear relationship. This relationship between temperature and volume describes a “direct relationship”. This means when temperature increases, so does the volume.
    11. 11. Charles Law :• ‘The volume of definite quantity of gas is directly proportional to its absolute temperature at constant pressure.’• Mathematically expressed as V ∝ T (constant pressure) V = KT (K = constant) V/T = K• If V1 and T1 are the volume and temperature of gas in initial state and V2, T2 at final state then above equation can be written as : V1 = V2 or V1 = T1
    12. 12. GAY LUSACC LAW“At constant volume the pressure of the given quantity 0f the gas is directly proportional to it’s absolute temperature”. Mathematically expressed as P ∝ T (constant pressure) P = KT (K = constant) P/T = K If P1 and T1 are the volume and temperature of gas in initial state and P2, T2 at final state then above equation can be written as : P1 = P2 or P1 = T1 T1 T2 P2 T2
    13. 13. Mechanics of BreathingTimberlake, Chemistry 7th Edition, page 254
    14. 14. Simple gas equation P1 VI T I P2 V 2 T 2 P2 V X T 2HereV T step –volume isVx T instep – 2 stepsP V T P1 change in 1 P2 done two 1 1 1 2 2 2 In first step, according to Boyle’s Law P1V1 = P2Vx (constant temp) ∴Vx = P1V1 P2In second step, according to Charle’s Law Vx = V2 ∴ Vx = V2T1 T1 T2 T2 On combining both above steps, P1V1 = V2T1 ∴P1V1 = P2V2
    15. 15. COMBINED GAS EQUATION P1V1 = P2 V2 . T1 T2In simple form combined gas equation can be written as PV = K (constant) T ∴ PV = KT Value of constant K depends on quantity of gas Here putting K = nR KαV;Vα where n = quantity of gas in mole n R = gas constant (does not depend on quantity of Kαngas) ∴ PV = nRT Above equation is called simple gas equation
    16. 16. Derive the value of R According to Ideal gas equation : PV = nRT ∴ R = PV = pressure x volume nT no. of moles x temp. force x volume = area ______________ no. of moles x temp. force x (length)3 = (length)2 ___________ no. of moles x temp. = force x length__________
    17. 17. • But, force x length = work energy∴R = work energy________ no. of moles x temp.• Thus unit of R is work energy/Kelvin mole.• It is proved by experiment that volume of one mole of any gas at 0°C and 1 atm pressure is 22.4 litre.
    18. 18. It is proved by experiment that volume of one mole ofany gas at 0°C and 1 atm pressure is 22.4 litre According to simple gas equation R = PV/nT where P = 1 atm n = 1 mole V = 22.4 litre T = 0°C = 273 Kelvin ∴ R = 1 atm x 22.4 litre 1 mole x 273 Kelvin R = 0.082 litre atm/Kelvin mole
    19. 19. Value of gas constant R indifferent unit Value Unit0.082 litre – atm / Kelvin mole } work1.987 calorie/Kelvin mole in heat1.987 x 10-3 Kcal / Kelvin mole energy8.314 x 107 erg / Kelvin mole (CGS)8.314 joule / Kelvin mole (MKS)
    20. 20. Standard temperature and pressure :• The temperature of 0°C or 273 Kelvin and pressure of 1 atmosphere or 760 mm is called standard temp. and pressure.
    21. 21. Dalton’s Law of Partial Pressure : The pressure of gaseous mixture is sum of partial pressure of each component gas’. Suppose A and B are the gases filled in two different vessel of same size and kept at same temperature. Let PA = partial pressure of a gm of gas A PB = partial pressure of b gm of gas B If both this gas are filled in the third container of same volume and kept at same temperature then total pressure of gases, according to Dalton law of partial pressure would be. PTotal = PA + PB. Here gas A and gas B donot react with each other.
    22. 22. Dalton’s Law of Partial Pressure :
    23. 23. Application :• For the gases collected over water, the total pressure of the gas is equal to partial pressure of dry gas as well as partial pressure of water vapour.• Eg : For oxygen gas collected over water, acc. to Dalton’s law• PTotal = Pgas + PH2O Here Pgas = PO2∀ ∴PO2 = PTotal – PH2O where PTotal = pressure of gases PH2O = vapour pressure of water at 25°C• When volume percentage composition is given, then partial pressure of gas P,• = Percentage of volume x total pressure 100
    24. 24. Graham’s law of gaseous difusion  Graham in 1928 presented a relation between diffusion rate of gas and its density in the name of Graham’s law of diffusion of gases.  ‘The rate of diffusion of various gases at same conditions of temperature and pressure is inversely proportional to the square root of their densities.’  Suppose the density of any gas is (d) and its rate of diffusion is (r) then, r α 1/ d  The diffusion rate of two gases are compared after carrying out the experiment at same temperature and pressure.
    25. 25. • Suppose r1 and r2 are the diffusion rate of gas-1 and gas-2 respectively.• The densities of these two gases at the same temperature and pressure are d1 and d2 respectively.• Acc. to Graham’s law of diffusion of gases. r1/r2 = dαM =• The ratio of densities of any two gases is equal to the ratio of the molecular weight of those two gases.
    26. 26. • The diffusion rate of a gas means the volume of the diffused gas in one second.• diffusion rate (r) = Volume of gas diffused (V) Time required for diffusion (t) r= v/t• For two gases at the same temperature and pressure, r1 = V1/t1 and r2 = V2/t2• During the experiment, for convenience, the times required for diffusion of same volumes of two gases diffusing in the same time are measured.
    27. 27.     Hence the above equation can be written as follows :  r1/r2 = V1/t1 = M2 OR V1•t2 = = d1 M2 V2/t2 M1 V 2•t1 d2 M1 V1 = V2    Now if t1 = t2  M2 M1 then t2 t1 =     But if V1 = V2 then M2 M1
    28. 28. Importance of Graham’s law of gaseous diffusion Uranium metal has two isotopes : U235 and U238. U235 is very important in production of atomic energy. The proportion of U235 in uranium metal is only 0.7%. As uranium hexafluoride (UF₆) is a volatile compound uranium hexafluoride is prepared from uranium metal. The difference of molecular weight between 235UF₆ and 238 UF₆ is much less. Hence, the ratio of rate of diffusion of these gases will be 1.0047. Now, if uranium hexafluoride gas is filled in a porous vessel allowed to have the diffusion, the amount of 235 UF₆ of less density will diffused somewhat more. Because of the small difference in diffusion rate, a series of a number of experiments is constructed.
    29. 29. This type of work is carried out in a laboratoryextended to kilometer at oak-Ridge in tenessystate of america.The experiment of diffusion of this gas throughporous membranes distributed (extended) toabout a kilometer.After a long time pure 235UF₆ is obtained which isdecomposed to get pure 235U. In short, the isotopes of uranium 235U and 238Ucan be separated.The importance of this law is in finding themolecular weights of gases and the densities.The components gases can also be separatedfrom the mixture of gases.
    30. 30. Avogadro’s Hypothesis Avogadro gave a principle in 1811 A.D. According to it, “Equal volume of the gases contain equal number of molecules at standard temperature and pressure . Simple gas equation is one of the methods to presents the Avogadro’s principle. One important dimension resulting from Avogadro’s principle is molar volume. Molar volume means the volume occupied by molecular weight expressed in gram of gas. The volume of 1 mole at 273 kelvin and 1 atmosphere pressure can be found out by general gas equation.
    31. 31. • PV = nRT where P = 1 atmosphere V = ? litre V = nRT n = 1 mole gas P R = 0.082 lit.a tm./k.mol. T = 273 K. V = 1 mole x 0.082 lit.atm./k.mol. 273 ‫ ג‬kelvin 1 atmosphere V = 22.4 litre Thus molar volume is also called gram molar
    32. 32. Thus molar volume is also called gram molar volume.The presentation of Avogadro’s principle on the basisof molar volume can be done as follows : “In 22.4 litre of any gas at 273 kelvin 1 atmospherecontains 1 mole molecules.”This statement can be given alternatively as, “Theweight of 22.4 litre of any gas at 273 kelvintemperature and 1 atmosphere pressure, is itsmolecular weight.”According to Avogadro’s principle, “ the number ofmolecules in 1 molar volume of any gas is 6.022 10 23.”The weight of one mole of any substance is itsmolecular weight.”
    33. 33. 1 mole 6.022 x 102322.4 litre Gram molecular weight
    34. 34. Kinetic molecular theory All the gases are composed of innumerable microscopic particles (atoms or molecules). The volume of molecules is negligible in comparison with the volume (volume of the vessel) occupied by the gas. All the molecules in each have same volume and weight. The molecules of the gas are in continuous motion. The molecules of the gas created (develops) pressure on the wall by striking with the walls of the vessel.The molecules in the gas have no attraction or repulsion for each other. When the continuously moving molecules strike with one another, theyexchange the kinetic energyAs this process is continuously going on, the molecules in the gas do notmove with uniform velocity. At any time and any temperature, the velocity ofcertain molecules will be very less, some will have moderate and will have veryhigh. Really the average velocity of each molecule
    35. 35. What are the different types of solids? There are four types of crystalline solids -- Ionic solids-- These substances have a definite melting point and contain ionic bonds. An example would be sodium chloride (NaCl). View the 3-D structure of a salt crystal. Covalent solids -- These substance appear as a single giant molecule made up of an almost endless number of covalent bonds. An example would be graphite. View the 3-D structure of graphite). Molecular solids are represented as repeating units made up of molecules. An example would be ice. View the 3-D structure of ice.Metallic solids are repeating units made up of metal atoms. Thevalence electrons in metals are able to jump from atom to atom.
    36. 36. CRYSTAL LATTICE• The definate arrangement of constituent particle (atoms,ions or molecules) shown by dots in three dimension in crystal is known as crystal lattice.UNIT CELL :• A tiny or smallest part of the crystal lattice lattice which bears all the characteristics of the crystal and when repeated in three dimension forms complete crystal structure
    37. 37. UNIT CELL OF NaCl
    38. 38. Unit cell of NaCl It is be observed that each Na+ is surrounded octahedrally bysix chloride icons and similarly each chloride icon by six Na+ion.Here the ionic size of Na+ ions is similar and cannot bearranged in a manner that each ions touches its neighbourion.In this configuration Na+ and Cl- ions are arranged in such away that they remain as near as possible with each other.In this construction Na+ - Na+ ion is maximum. Because of asuch a arrangement, the distance between Cl- ions increasedautomatically.The co-ordination number of Na+ in NaCl crystal is six and theratio of Na+ /Cl- radii is 0.53.
    39. 39. UNIT CELL OF CsCl
    40. 40. Unit cell of CsClIf we examine the configuration of unit cell of CsCl,it will be found that each Cs+ is surrounded by eightCl- and similarly each Cl- ions is surrounded by eightCs+ ions.If the co-ordination number of metal ion is more in anionic crystal, the stability is also more.Hence the stability of CsCl is more than that ofNaCl .The co-ordination number of Cs+ in CsCl crystal iseight and the ratio of Cs+ in CsCl crystal is eight andthe ratio of Cs+ / Cl- radii is 0.92.It has body centered cubic arrangement.
    41. 41. UNIT CELL OF CsCI
    42. 42. Unit cell of LiIIn LiI, negative charge possessing I- is much larger insize as compared to the positive charge possessingLi+ ion. Hence the negatively charged ions can bearranged very near to each other.The positively charged ion can be arranged verynear to each other.The positively charged ions can be easily placed inthe vacant space formed between these ions. Thistype situation arises in the crystal of LiI.The cross section of layer containing ions is shownin figure. In this four I- ions are arranged almosttouching each other. I- ions are also above and belowof this central void of this configuration.
    43. 43. UNIT CELL OF LiI• Due to eight I- ions arranged in a manner touching each other, the shape that evolves is octahedral.• As I- ion is big, the size of octahedral configuration is comparatively big.• Li+ ion being smaller in size can be arranged easily in the central void (space). In this configuration, similarly charged I- ions are arranged near each other in such a manner that the attraction between them is less and repulsion is more.• Thus this configuration possesses relatively less stability. Because of this, the melting point of LiI is less than that of NaCl.
    44. 44. Information about different co-ordination numbers and ratio of radiiRadii Ratio Co-ordination Arrangement of Examples(r+/r-) number of positive-negative positive ions ionsUpto 0.155 2 Linear0.155 to 0225 3 Planer triangle0.225 to 0.414 4 Tetrahedral ZnS0.414 to 0.73 4 Square planer0.414 to 0.73 6 FCC NaCl0.73 to 1.0 8 Octahedral-BCC CsClAbove 1.0 12 HCP
    45. 45. CHARACTERISTICS OF THE LIQUID  Fix volume  Fluidity  Non-compressibility  Diffusion  Evaporation  Vapour pressure  Surface tension  Viscosity
    46. 46. Clearity of the term Diffusion : The property of the liquid to spread in another liquid. Evaporation : The property of liquids to get converted of its own into gaseous state at normal temperature
    47. 47.  Vapour pressure : The vapour exert pressure on the surface of the liquid at equillibrium. Surface tension : The force exerted by the molecules on the hypothetical line of unit length parellel to the surface of the liquid and perpendicular to the molecules on the other side of the molecules.

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