简单混合物 Simple mixtures 偏摩尔体积 Partial molar Volume V≠V1+V 1+1≠2 Add 1.0 mol ho Add 1.0 mol Ho Volume increases Volume increases by 18 cm'mol Large H○ EtoH by 14 cm'mol volumes Molar volume of ho: Partial molar volume of 18 cm mol 1 HO in EtoH: 14 cm mol 1 Partial molar volume of substance a in a mixture is the change in volume per mole of a added to the large volume of the mixture
简单混合物 Simple Mixtures 偏摩尔体积 Partial Molar Volume Partial molar volume of substance A in a mixture is the change in volume per mole of A added to the large volume of the mixture. V V1 + V2 1 + 1 2
偏摩尔体积 Partial molar Volume The partial molar volume of components of a mixture vary as the mixture goes from pure a to pure B- that is because the molecular environments of each molecule change (i.e, packing, solvation, etc. 58 56 18 54 85019>56556E页 物质J的偏摩尔体积V PT.n' a14 2 04 06 Mole fraction of ethand, X(C HS OH
偏摩尔体积 Partial Molar Volume 物质 J 的偏摩尔体积 VJ P,T ,n' J J n V V =
偏摩尔体积 Partial molar volume va ∨(b) J/pTn )偏摩尔体积与组成有关. (2)b点处的偏摩尔体积为负(此时, 日 b 加物质A会使总体积变小) Composition, n For a binary mixture, the composition can be changed by addition of dna of a and dng of B with the total volume changing by on P.7,B A′A BB
偏摩尔体积 Partial Molar Volume (1) 偏摩尔体积与组成有关. (2) b点处的偏摩尔体积为负 (此时, 加物质A会使总体积变小). P,T ,n' J J n V V =
化学势 Chemical potential G 化学势μ p, I,n 偏摩尔 Gibbs能 (a) u(b) Total Gibbs energy Is G nR u B B Composition n
化学势 Chemical Potential 化学势 偏摩尔Gibbs能
化学势 Chemical potential Gibbs energy depends on composition, pressure and temperature(so, G may change if any of these variables change - which they may )For a system with components A, B dG= vdp ST+A如A+HB2mB+… which is the fundamental equation of classical thermodynamics At constant temperature and pressure, BB Recall e max dG △G e. max in+ e. max NA B What does this mean? Non-expansion work can happen just from changing system composition at constant pressure and temperature (e.g., battery, chemical rxn in two sites called electrodes, and the work the battery produces comes from reactants going to products
化学势 Chemical Potential Recall:
More on chemical potential Chemical potential tells us even more than just about variation in G U+ pV- TS so for an infinitesimal change in u, we can write pdv-vap+ sdT+ Tas+ dG p-中p+Sd+7ds+(dp-Sa+uAhn+ eDna+…) P ∥+TS B-B and at constant volume and entropy dU an)s and if that' s not enough. what about H and A??? These too also depend upon the composition of a mixture! Chemical potential is IMPORTANT aH dA a) (b) n Dn
化学势 Chemical potential 纯物质G=n×Gm ang G O T P 化学势=摩尔 Gibbs能 理想气体的化学势 u o In P, standard state at p H=p°+R7h|2 P Pressure
化学势 Chemical Potential 理想气体的化学势 纯物质 G = n Gm m T P m G n nG = = , μ 化学势 = 摩尔Gibbs能
Thermodynamics of Mixing The simplest example of mixing: What is the Gibbs free energy if we take two pure ideal gases and mix them together? Consider gas a and gas B, both in separate containers at pressure p at temperature T The chemical potentials are at their "pure values"at this point. Gibbs energy is ∵ μA e矿4 RT In nUb+ RT In A B P We can simplify things by letting p denote the 3 pressure relative to p, writing + RT In pi+bLub+ RT In p
Gibbs Energy of Mixing After mixing, the partial pressures of the gases are pa and pB, where the total pressure is p= pA pB. The total Gibbs energy is then nAHA +RT In pa+nBluB+RT In pal The difference in Gibbs energies, Gr -Gi, is the Gibbs energy of mixing n,rt In +nRT II IX We use mole fractions, replacing ny with Xn IX nRT( In B Since the mole fractions are never greater than 1 the In terms are negative and Ami g<o This allows is to conclude that mixing processes are spontaneous, and gases mix spontaneously in 08 all proportions Molo fraction of A. x
简单混合物 Simple mⅸ tures一液体 纯液体A的化学势μ*(1) u,+ RT In 气体A的化学势μ=μ°+ RTInPa* 两相平衡,化学势相等 液体A的化学势[世*三p°+RTmP(1) 液体混合物中A的化学势μA() A(g)+B(g μA=HA°+ RT InP(2) uA(g P vapour pressure Is PA Equal at 由(1)、(2)得 equilibrium A=以A*+RF(P A()+B( D
简单混合物 Simple Mixtures — 液体 纯液体A的化学势 A*(l) 气体A的化学势 = ° + RT lnPA* 两相平衡, 化学势相等: 液体A的化学势 * = ° + RT lnPA* 液体混合物中A的化学势 A(l) = + * μA μA * ln A A p p RT A = A ° + RT lnPA (1) (2) 由(1),(2) 得
