lecture 16

lecture 16 - In the previous lecture we have...

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Unformatted text preview: In the previous lecture we have concluded that two phases (e.g. liquid and solid) can coexist when their chemical potentials are equal. Some properties of the chemical potential. S = S/n, V = V/n. A different symbol is used for G/n, which is the chemical potential: µ = G/n We have µ = H − TS and dµ = VdP − SdT We can use this expression to calculate the change of the chemical potential as a function of temperature and/or pressure. Change of the chemical potential in a process where T is held constant. dµ = VdP or µ ( P, T ) = µ ( P0 , T ) + V dP P0 ! P Example 1: incompressible solid or liquid. V is constant (i.e., independent of pressure) so that µ(P,T)  ­ µ(P0,T) = V(T,p)dp = V ( p − p0 ) p0 ∫ p Example 2: Ideal gas. µ(P,T)  ­ µ(P0,T) = V(T,p)dp = (RT/p)dp =RT ln(P/P0 p0 p0 ∫ p ∫ p µ(P,T) = µ(P0,T) +RT ln(P/P0) The Clapeyron equation. Back to the coexistence curve. Suppose the equation µl(P,T) = µg(P,T) is satisfied at some (P,T). Let us calculate how P will change if we increase the temperature by dT: dµl(P,T) = dµg(P,T) or so that Vl dP − Sl dT = Vg dP − Sg dT dP / dT = (S g − Sl ) /(Vg − Vl ) Using S g − Sl = ΔH vap /T , with !H vap being the vaporization heat we get dP/dT = Since the volume of gas is greater than that of liquid, then dP/dT > 0 Then the coexistence curve between liquid and gas looks like this: H g - Hl T (Vg - Vl ) = ΔH vap T (Vg - Vl ) The points on the curve correspond to P and T at which liquid and gas are in equilibrium with one another. The points to the left correspond to liquid, to the right, to gas. To see this consider any point on the coexistence curve, for which we have µ g = µl If we raise the temperature by dT (i.e., move to the right from the original point) then the chemical potential of the gas (liquid) will become µ g (l ) − S g (l ) dT . Since S g > Sl , the chemical potential is lower for the gas phase for positive dT, which means that gas is the stable phase. The above curve shows that there are two ways to convert gas into liquid: by cooling it or by compressing it. The Clausius ­ Clapeyron equation. This equation is an approximation that applies to solid ­to gas and liquid ­to ­gas transitions (but NOT liquid to solid). Start with the Clapeyron equation and make 3 approximations: 1. Neglect Vl compared to Vg 2. Assume that vapor is an ideal gas so that Vg = RT/P 3. Neglect temperature and pressure dependence of ΔHvap Then the Clapeyron equation becomes: dP/dT = ΔHvap/(T (RT/P)) dP/P = dT ΔH vap /RT2 If we know that P=P0 at T=T0 then by integrating the above equation we get: ln(P/P0) =  ­ ( ΔH vap /R) (T ­1 – T0 ­1) Or P = P0 exp[ ΔH vap RT0 exp[ − ΔH vap RT A similar equation can also be written down for solid ­gas equilibrium but not for liquid ­solid equilibrium. Example. Estimate the boiling temperature at the summit of Mt. Whitney. Solution: we know that water boils at T0 = 373K. We also know how the atmospheric pressure changes with h: P = P0 exp( ­Mgh/RTatm) so that ln(P/P0) =  ­Mgh/RTatm =  ­ ( ΔH vap /R) (T ­1 – T0 ­1) We will assume that the temperature of atmosphere does not change with h and is 300K. Then we get: (T ­1 – T0 ­1) = Mgh/( ΔH vap Tatm) and T = 356.045K ~ 86 0C The case of more than two phases We know from experience that most materials can assume three forms: solid, liquid, and, gas. This is a highly oversimplified picture. For example, many solids can have different phases (al solid) corresponding to different crystalline lattices. Anyway, for now let us assume that only three phases are possible. The stable form of a chemical at any (P,T) is the one that has the lowest chemical potential. Consider how the chemical potentials of different phases change with temperature at constant P: Fig. 1 The slope of the curve µ(T) is the entropy with the minus sign. In the situation shown in Fig. 1, as the temperature is raised the solid first melts then becomes vapor. In the situation shown in Fig. 2 the solid goes directly into the vapor phase. Fig. 2 By changing pressure, one can move from one scenario to the other. Question: if I have the situation shown in Fig. 1, do I need to reduce or raise pressure to go to that in Fig. 2? Answer: the pressure has to be lowered. The complete phase diagram in the (P,T) plane looks like this: Each of these curves is described by the Clapeyron Eq. Since the molar volume of liquid can be lower than that of the solid (e.g., for water), dP/dT for the liquid ­solid coexistence can be either positive or negative. The above picture corresponds to the case of dP/dT<0. T is called the triple point and C is the critical point. For water, the triple point is at P~ 611 Pa and T ~ 0.010C, and the critical point is at 647K and ~220 bar = 220 x 105 Pa A gas can be converted to liquid either by lowering T or via compression (i.e. by raising P). It is possible to go from gas to liquid without experiencing a phase transition If we fix P and raise T our process is s →l→g (melting then vaporization) if P > PT or s →g (sublimation) if P < PT Question: why no critical point on the liquid ­solid coexistence curve? ...
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