Presentation on theme: "Physical Transformations of Pure Substances"— Presentation transcript:
1Physical Transformations of Pure Substances Chapter 4Physical Transformations of Pure Substances
2Phase Diagrams and Phase Boundaries The phase diagram of a substance showsthe regions of pressure and temperatureat which the various phases are stable.A typical phase diagram is shown to the right.Phase BoundariesThese are the lines (combination of p and T) on which two phasesare in equilibrium
3The Critical Point The Triple Point The critical temperature (Tc) is the highesttemperature at which a substance canbe liquified.The vapor pressure at Tc is called thecritical pressure (pc).The Triple PointThe triple point temperature and pressure,T3 and p3, is the one point in a phase diagramat which all three phases (solid, liquid and vapor)are in equilibrium.
4Liquid + Gas (boiling point) The Boiling PointThe boiling point is the temperature at whichliquid and vapor are in equilibrium.The boiling point depends upon theapplied (external) pressure.PextLiquidorGasPext > Vap. Press.LiquidPext = Vap. Press.Liquid + Gas (boiling point)Pext < Vap. Press.GasThe curve, representing the combinations of p and T at which solidand liquid are in equilibrium is often called the vaporization curve.
5The Melting Point The Sublimation Curve The melting (or freezing) point is thetemperature at which solid and liquidare in equilibrium.The melting point depends upon theapplied (external) pressure, but to amuch smaller extent than the boiling point.The Sublimation CurveThis represents the combination of temperatures and pressures atwhich solid and vapor are in equilibrium
6A Typical Phase Diagram Pressure (bar)Melting point increases with PressureBoiling point increases with PressureSolidLiquidVaporp1Melting + Vaporizationp3T3p2SublimationTemperature (oC)
7Pressure dependence of the Boiling Point Temperature (oC)Pressure (bar)Liquid VaporLiquidVapord(liq) >> d(vap)2120V(liq) << V(vap)Increased pressure shiftsequilibrium in directionof lower volume.1100
8Pressure dependence of the Melting Point in “normal” substances Temperature (oC)Pressure (bar)SolidLiquidSolid Liquid100Tmpd(sol) > d(liq)V(sol) < V(liq)Increased pressure shiftsequilibrium in directionof lower volume.1Tmpo
9Water melts and then boils (i.e. it has a liquid phase) Question:Water melts and then boils (i.e. it has a liquid phase)Benzene melts and then boilsIron melts and then boilsIodomethane melts and then boilsetc., etc., etc.Why does CO2 sublime directly from solid to vapor ???Answer:Because the triple point pressure of CO2 is 5.1 bar (~5 atm.),which is higher than the pressure of the atmosphere on Earth.
10The Phase Diagram of CO2 Temperature (oC) Pressure (bar) Solid Liquid Vaporp2Melting + VaporizationSublimation5.1bar-56 oCMelting point increases with PressureBoiling point increases with Pressure
11Phase Diagram of H2O Temperature (oC) Pressure (bar) Solid Liquid VaporMelting point decreases with PressureP1Melting + Vaporization4.6torr0.01P2Sublimation
12Pressure dependence of the Melting Point in Water Temperature (oC)Pressure (bar)SolidLiquidSolid (ice) Liquid140-1d(ice) < d(liq)V(ice) > V(liq)Increased pressure shiftsequilibrium in directionof lower volume.1
13Some More Complex Phase Diagrams (FYI)WaterThis is a phase diagram of water over anextended range of pressure and temperature.It shows the regions of stability of the variousdifferent crystal structures of ice.
14Helium Even Helium has a fairly sophisticated phase diagram. However, all of the "action" occursat very low temperature; Tc 5 K
15Phase Stability and Phase Transitions Chemical PotentialAs discussed in Chapter 3, the chemical potential of a substance isdefined by:However, in a pure (one component) system, the chemical potentialis simply the Molar Gibbs Energy: = GmPhase EquilibriumConsider two phases, and (e.g. solid and liquid). The two phaseswill be in equilibrium when: = When the pressure and temperature are varied, they will stay inequilibrium when: d = d
16The Temperature Dependence of Phase Stability From Chapter 3, the temperature and pressure dependenceof the Gibbs Energy is given by:Therefore, the change in chemical potential (Molar Gibbs Energy) withrising temperature is:NotesThe entropies of all substances in all phases is positive, Sm > 0.Therefore always decreases with rising temperature.Relative entropies of different phases are: Sm(gas) >> Sm(liq) > Sm(sol)Therefore, the decrease in with increasing temperature is greatestfor the gas phase, and smallest for the solid.
17Two phases ( and ) are in equilibrium when = The phase with the lowest chemical potential at a given temperature isthe most stable phase.s = ll = gRemember:Therefore:
18The Effect of Pressure on the Melting Point The chemical potential of all phasesincrease with rising pressure.In most materials, Vm(liq) > Vm(sol)Therefore, (liq) rises more rapidly than(sol) with increasing pressure.This results in a higher melting pointat higher pressure.Low pHigh p
19The Effect of Pressure on the Melting Point High pThe chemical potential of all phasesincrease with rising pressure.In a few materials, most notably water,Vm(sol) > Vm(liq) Therefore, (sol) risesmore rapidly than (liq) with increasingpressure.This results in a lower melting pointat higher pressure.Low p
20Quantitative Treatment of Phase Equilibria: The Location of Phases BoundariesTwo phases, will be in equilibriumwhen their chemical potentials are equal:For the two phases to remain in equilibrium asthe pressure/temperature are changed, theirvariations must be equal:Remembering that:Thus, the criterion for two phases to remain in equilibrium is:
21Thus, the criterion for two phases to remain in equilibrium is: The Clapeyron EquationBecause the two phases are in equilibrium, we can use the relation,trsS = trsH/T, to obtain an alternative, equivalent form of theClapeyron Equation:
22Slopes of the Phase Boundaries The Clapeyron Equation can be used both qualitatively and quantitatively.In the next several slides, we will show how it can be used in aqualitative fashion to explain the relative slopes of variousequilibrium curves in a phase diagram.
25H2O vap vap liq sol sol liq Why is Slope(AC) < 0 in H2O ? Entropy (S)Volume (V)vapvapliqsolsolliqWhy is Slope(AC) < 0 in H2O ?AC: Solid-LiquidH2OV(s-l) = Vliq-Vsoldliq > dsolS(s-l) = Sliq-SsolVliq < VsolVliq-Vsol < 0Because Vliq < Vsol
26Quantitative Application of the Clapeyron Equation The Solid-Liquid Transition: Melting (aka Fusion)Over small ranges of temperature, T, fusH and fusV are approximatelyconstant. Therefore, one can write:orNote: The Clapeyron Equation cannot be used for quantitativecalculations on vaporization or sublimation because itcannot be assumed that V is independent of temperature,pressure for these transitions.
27Side Note: A more rigorous (but unnecessary) equation (FYI Only) It's straightforward to show that if one doesn't make the approximationthat T = Constant, then the result is:However, it can also be shown that if T = T2 -T1 << T1 ,then the more rigorous solution reduces to the one we're using.
28Example: The melting point of benzene at 1 bar is 6.0 oC. What is the melting point of benzene underan applied pressure of 200 bar?fusH = 9.8 kJ/molM = 78 g/mold(sol) = 0.95 g/cm3d(liq) = 0.88 g/cm31 kPa-L = 1 JTmp (200 bar) = 9.7 oC
29Vaporization and Sublimation These two phase transitions involve the gas phase and, therefore,one cannot assume that trsV = Vgas -Vliq (or Vsol) = Constant.However, it is valid to assume that (a) trsV Vgasand (b) Vgas = RT/p (for 1 mole)Using trsV = RT/p, one has:Separate Variables:
30Separate Variables:Now, integrate, assuming that trsH is independent of T:The Clausius-Clapeyron Equation
31The Clausius-Clapeyron Equation For vaporization, trsH = vapHFor sublimation, trsH = subHNote: The text's notation is a little different.ApplicationsIf one is given the enthalpy of vaporization (or sublimation) +the vapor pressure at one temperature, then one can calculate eitherThe vapor pressure at a second temperature orThe temperature at which the vapor pressure reaches a certain value.If one knows the vapor pressure at two temperatures, then theenthalpy of vaporization (or sublimation) can be calculated.
32The Clausius-Clapeyron Equation Example: The "normal boiling point"** of benzene is 80 oC, andthe Enthalpy of Vaporization is 30.8 kJ/mol.Calculate the temperature at which the vapor pressureof benzene is 150 torr.1 bar = 750 torrR = 8.31 J/mol-K**You should know that the "normal" boiling point is the temperatureat which the vapor pressure of a substance is 1 bar.T = 33 oC