# Lecture20A - Phase Equilibria(Chapter 23 We will begin to...

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Unformatted text preview: Phase Equilibria(Chapter 23) We will begin to look at phase diagrams, initially for pure materials to understand the relationships between various phases. Phase diagram for benzene Fig 23.1 From this diagram, we can analyze the degrees of freedom of the system. Within a region of single phase, pressure and temperature can freely vary. Along a phase boundary line, P=P(T), i.e. the two variables are coupled. When two phase boundary lines intersect at a point there are no independent variables. Triple point (3 phases) has f = degrees of freedom = 3 - p ( p =# of phases ) a fixed T and P. Lecture 20 1 Fig. 23.2 Fig. 23.3 Typical behavior for most compounds: slow increase in Tfus with P and rapid increase in Tvap with P. Benzene is typical of most compounds we know in that as we increase its temperature solidliquidvapor under normal atmospheric conditions. This is because the triple point occurs at a temperature where P<1bar, so the liquid phase will exist under typical atmospheric conditions. Lecture 20 2 CO2 (first member of "weird" compounds). Triple Point is 5.11 atm so at 1 atm solid sublimes and goes directly to the vapor phase. Fig. 23.4 CO2 H2O is 2nd member of weird compounds Tfus decreases with increasing pressure. Fig. 23.6 H2O At 200o C, a vertical line shows l and v at equilibrium. Note at the critical point the densities are the same and vap H = 0. Fig. 23.7 Benzene Lecture 20 3 Near the critical point large fluctuations in cause light to scatter (think fog!), which is called critical opalescence. This can be easily done with SF6 (which we do in our P. Chem. Lab). TC=45.55C, PC=3.76MPa=37.6bar, Ts=-63.9C Remember: In solids each atom/molecule (in principle) is distinguishable, quite different from liquids/gases. These are fluids with free motion, so transitions between these two phases can be made gradually. With solids the transition is much more abrupt. G 1 T = -S 1 P The slopes are understandable since S s < Sl < S g Fig. 23.10 Benzene Lecture 20 4 "normal" Fig. 23.10 substance like water G 1 P = V 1 T normally Vg >> Vl > Vs but at % 1 bar water is different Vs > Vl % When we have two phases present in our system at constant T and P: G = G I + G II we can have matter move from one phase to the other. G I 1 dG = I 1 n 14 2 I G II dn I + II dn II P n P T, T, 4 3 1 4 2 43 II = chemical potential (intensive thermo property) Since dn I = -s II dn If I > II If I < II If I = II dG = ( I - II ) dn I Matter moves from high to low potential. 5 matter goes from phase I to II matter goes from phase II to I system is at equilibrium Lecture 20 At equilibrium: I = II d I= d II I I II II 1 or dP + dT = dP + dT 1 T P T P T P T P I I 1 = V I , = -S I P 1 T T P V I dP - S I dT = V II dP - S II dT trans H dP S II - S I trans S = II = = Clapeyron equation I dT V - V transV T transV This relates the slope of phase coexistence curves to trans H and transV . dP Think why is >0 and steep for s p l. dT Lecture 20 6 Consider the coexistence curve between liquid and vapor (or for a compound that sublimes like CO2) trans H trans H trans H dP = = for an ideal gas g g l dT T ( V - V ) TV T ( RT / P ) 1 dP d ln P vap H = = P dT dT RT 2 Clausius-Clapeyron equation vap H R dT T 2 assumes vap H constant T1 T2 d ln P = RT P 1 T1 P2 T2 vap H 2 dT vap H P2 T 1 1 vap H 2 - T1 Easy to determine vapor ln = - = - P R 2 T1 R T1T2 pressures. T 1 For a greater range in temperatures, we need to consider the explicit temp dependence of vap H = A + BT + CT 2 + ... Lecture 20 7 d ln P A B C = + + + ... 2 dT RT RT R A B C ln P = - + ln T + T + constant RT R R If you look in CRC handbook, you will find tables with constant log10 P = + b with specific temperatures of validity. T Now it is time to make the connection with stat. mech. p ln Q 2 ln Q U = k BT , S = k BT + k B ln Q 1T ,V T ,V N N A = U - TS = - k BT ln Q A ( T ,V , N ) since Q ( T ,V , N ) p A A A A dA = dT + dV + dN = - SdT - PdV + dN 1 N T ,V V N N V N V T, T, T, p A A note: dN = dnLecture 20 because N=n A , dN = dn A N N 1 T, N V n V T, 8 p A dG = dA + d ( PV ) = - SdT + VdP + dn 1 T, n V p G G G dG = dT + dP + dn 1 P, T N P N n P T, T, A G so = 1 T, n V n P T, ln Q ln Q This means = - k BT = - RT 1n V N V T, T, N q ( V,T ) for the ideal gas partition function: Q= N! ln Q = N ln q - ( N ln N - N ) q ln Q = ln q - ( ln N + 1 - 1) = ln N 1N q so = - RT ln Lecture 20 N 9 Just a few more manipulations and things will (may?) be clearer. q V V k BT = - RT ln but = N P V N q = - RT ln k BT RT ln P + V P Recall G ( T , P ) = G ( T ) + RT ln 0 , P 0 = 1bar , standard state P 0 q P k BT = - RT ln 0 RT ln 0 + P 4 4 1 4 4 2 V P 3 0 ( T ) =standard state chemical potential P = ( T ) + RT ln 0 P 0 will be a useful expression as you will see! Lecture 20 10 ...
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