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Chapter 8 Thermodynamic Properties of Mixtures

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Chapter 8 Thermodynamic Properties of Mixtures Chapter 8 Thermodynamic Properties of Mixtures 2012/3/29 1 Abstract The thermodynamic description of mixtures, extended from pure fluids. The equations of change, i.e., energy and entropy balance, for mixtures are developed. The criteria for phase and chemical equilibrium in mixtures 2012/3/29 2 8.1 THE THERMODYNAMIC DESCRIPTION OF MIXTURES Thermodynamic property for pure fluids, θθ=()TPN , , where N is the number of moles. θθ=()TP , where the number of mole equals to 1. Thermodynamic property for mixtures, θθ=()TPN , ,12 , N ,L , Nc where Ni is the number of moles of the ith component. θθ=()TPx , ,12 , x ,L , xci where x is the mole fraction of the ith component. For example UUTPNN=() , ,12 , ,LL , Ncc or UUTPxx=() , , 12 , , , x VVTPNN=() , ,12 , ,L , Nc or VVTPxx=() , ,12 , ,L , xc 2012/3/29 3 Summation of the properties of pure fluids (before mixing at TP and ) C UTPxx(), ,12 , ,L , xci− 1= ∑ xUTPi () , (8.1-1) i=1 where UU is the molar internal energy, i is the internal energy of the pure i-th component at TP and . C ˆˆ UTPww()(), ,12 , ,L , wcii− 1= ∑ wUTP , (8.1-2) i=1 where wi is the mass fraction of component i. 2012/3/29 4 At the same T and P 50 cc 25 cc + 25 cc H2O H2O or 52 cc 25 cc + 25 cc 48 cc + 2 cc A B -2 cc Attractive Repulsive 2012/3/29 5 Property change upon mixing (at constantTP and ) C Δ=mixθθ()TPx,,ii −∑ x θi () TP , i=1 Volume change upon mixing C Δ=mixVTP(),,,, VTPx()ii −∑ xVTPi () i=1 Enthalpy change upon mixing C Δ=mix HTP(),,,, HTPx()ii −∑ xHTPi () i=1 2012/3/29 6 Experimental data : properties changes upon mixing (H and V) Figure 8.1-1 Enthalpy-concentration diagram for aqueous sulfuric acid at 0.1 MPa. The surfuric acid percentage Is by weight. Reference states: The Enthalpies of pure liquids at 0oC And their vapor pressures are zero. 2012/3/29 8 Figure 8.1-2 (a) Volume change on mixing at 298.15 K: o methyl formate + methanol, zmethy formate + ethanol. (b) Enthalpy change on mixing at 298.15 K for mixtures of benzene (C6H6) and aromatic fluorocarbons (C6F5Y), with Y = H, F, Cl, Br, and I. 2012/3/29 9 Equations relating molar and partial molar properties NTPNNNθφ= (),,12 , ,L , c At constant TP and , NNNNθφ= ()12,,,L c Total differential of Nθ is: ⎛⎞∂∂()NNθθ ⎛⎞() ⎛⎞ ∂() N θ d() Nθ =+⎜⎟ dN12 ⎜⎟ dN ++L ⎜⎟ dNc ∂∂NN12 ∂ Nc ⎝⎠TPN,,jj≠≠12 ⎝⎠ TPN ,, ⎝⎠ TPN ,, j≠ c cc⎛⎞∂()Nθ ==∑∑⎜⎟dNi θ i () T,, P x dNi ii==11∂Ni ⎝⎠TPN,, ji≠ Definition of the partial molar property ⎛⎞∂()Nθ θ i ()TPx, ,= ⎜⎟ (8.1-12) ∂Ni ⎝⎠TPN,, ji≠ c dN()θθ= ∑ i dNi i=1 2012/3/29 10 At a constant T and P θ 50 cc + 1 mol of A θ + Δθ A + B Δθ is a partial molar property (at Constant T, P, and NB) 2012/3/29 11 Total property and partial molar property c dN()θθ= ∑ i dNi i=1 dN()θθθ=+ Nd dN (a) dNiiii==+ d() Nx Ndx x dN (b) cc (a) (b) Ndθθ+= dN ←⎯⎯ d() N θ∑∑ θii dNiii⎯⎯→ θ() Ndx + x dN ii==11 cc Nddxxθθ+=dN N∑∑ θiiii +dN θ ii==11 ⎡⎤c ⎡⎤c ⎢dθθ−+−∑ idxi ⎥ N ⎢⎥ θ∑θ i xdNi = 0 ⎣ i=1 ⎦ ⎣⎦i=1 The NdN and are the number of moles in system and changing in number of moles, respectively. Thus, NdN and are arbitrary and not equal to zero. 2012/3/29 12 cc ddxθθ−==∑ iiii0; ddx θθ∑ resulted from 1st term of LHS (c) ii==1 1 c c θθ− ∑ i xi = 0; θθ= ∑ i ()TPx,, xi resulted from 2nd term of LHS (8.1-13) i=1 i=1 Euler Theorem defined by (8.1-13) means θ can be calculated from θ i ()TPx,, . c Since: θθ= ∑ i xi i=1 cc ddxxdθθ=+∑∑iiii θ ii==11 ccc (c) ∑θθiiidxiii ←+⎯⎯ddxdθθ = ∑∑x i===11i 1 i c ∑ xidθ i = 0 (8.2-9b) i=1 ⇒ Gibbs Duhem equation at constant TP and . 2012/3/29 13 2012/3/29 14 Property change upon mixing at constant T and P 50 cc 25 cc + 25 cc H2O H2O or 52 cc 25 cc + 25 cc 48 cc + 2 cc A B -2 cc Attractive Repulsive 2012/3/29 15 N1 θ1 Constant T and P Before After mixing mixing N2 θ2 θi θi θ = xθ θ = ∑ xiiθ ∑ ii 2012/3/29 16 Properties change upon mixing (i) The total volume and enthalpy of the unmixed pure components are C VNVTP= ∑ i i (), i C HNHTP= ∑ i i (), i whereas the volume and enthalpy()TPx , ,i of the mixture at the same temperature and pressure are, from Eq. 8.1-1 3, C VT(),,PN12 , N ,L == NVTPx() ,,∑ NVi i () TPx ,, i C HTPNN(),,12 , ,L == NHTPx() ,,∑ NHi i () TPx ,, i 2012/3/29 17 Properties change upon mixing (ii) The isothermal volume change on mixing, C Δ=−mixVNNVNNNV()()TP,,,,, 1 2LL TPT ,,, 1 2 ∑ i i () ,P i C =−NVTPx⎡⎤i , , VTP , (8.1-14) ∑ i ⎣⎦()i () i C Δ=−mixHTPNN()(),,,, 1 2LL HTPNN ,,,, 1 2 ∑ NHi i () T , P i C = NH⎡⎤i TPx, ,− H TP , (8.1-15) ∑ i ⎣⎦()i () i 2012/3/29 18 Thermodynamic relations for partial molar properties AUTS=− ⎛⎞∂∂∂NA ⎛⎞ NU ⎛ NTS ⎞ ⎜⎟=− ⎜⎟ ⎜ ⎟ ∂∂NNii ∂ N i ⎝⎠TPN,,ji≠≠ ⎝⎠ TPN ,, ji ⎝ ⎠ TPN ,, ji ≠ AUTSi =−ii Also, GHTSii=−i For each relationship among the thermodynamic variables in a pure fluid, there exists an identical relationship for the partial molar thermodynamic properties in a mixture. 2012/3/29 19 ILLUSTRATION 8.1-l Calculating the Energy Release of an Exothernuc Mixing Process Three moles of water and one mole of sulfuric acid are mixed isothermally at 0 oC. How much heat must be absorbed or released to keep the mixture at 0 oC? SOLUTION Water has a molecular weight of 18.015, and that of sulfuric acid is 98.078. Therefore, the mixture will contain 3 ×× 18.0154 + 1 98.078 = 152.12 g, and will have a (mass ) composition of 98.078 g ×=100% 64.5 wt% of sulfuric acid 152.12 g From Fig. 8.1-1 the enthalpy of the mixture is about − 315 kJ/kg. Therefore, when 3 mol water and 1 mol sulfuric acid are mix ed isothermally, kJ Δ=HwHˆˆˆHˆ − −wH =−315 mixmix 1 1 2 2 kg ˆˆoo since HT1 ()==0 C = 0 and H2 ()T 0 C , so that a total of −×315 kJ/kg 0.152 kg=− 47.9 kJ of energy must be removed to keep the mixture at a constant temperature of 0o C. 2012/3/29 20 Figure 8.1-1 Enthalpy-concentration diagram for aqueous sulfuric acid at 0.1 MPa. The sulfuric acid percentage Is by weight. Reference states: The Enthalpies of pure liquids at 0oC And their vapor pressures are zero. -315 2012/3/29 64.5 21 COMMENT Sulfuric acid and water are said to mix exothermically since energy must be released to the environment to mix these two components at constant temperature. The temperature rise that occurs when these two components are mixed adiabalically is considered in Illustration 8.4-1. Note also that to solve this problem we have, in effect, used an energy balance without explicitly writing a detailed balance equation. 2012/3/29 22 Partial molar property θ + 1 mol of A θ + Δθ A + B Δθ is a partial molar property (at Constant T, P, and NB) 2012/3/29 23 Physical meaning of partial molar property For a binary mixture, θθθ()TPN, ,12 , N=+ N 112() TPx , , N 2() TPx , , (8.1-17a) Now suppose we add a small amount of species 1, ΔN1, which is so small compared with the total number of moles of species 1 and 2 that xx12 and are essentially unchanged, In this case we have θθθ()()TPN, ,121+=+ΔΔNN11 , N N1 () TPx , , + N 22 () TPx , , (8.1-17b) Subtracting Eq. 8.1-17a from Eq. 8.1-17b, we find that the change in the property is Δ+−=Δθθ=,,()()TPN112ΔNTP , N θ TPN ,,, 1 N21N θ 1 (), , x Therefore, the amount by which a small addition of a species to a mixture changes the mixture property is equal to the product of the amount added and its partial molar property, that is, how the species behaves in a mixture, and not its pure component property. 2012/3/29 24 8.2 THE PARTIAL MOLAR GIBBS ENERGY AND THE GENERALIZED GIBBS-DUHEM EQUATION Since the Gibbs energy of a multicomponent mixture is a function of temperature, pres- sure, and each species mole number, the total differential of the Gibbs energy function can be written as ⎛ ∂G ⎞⎛⎞∂∂GGC ⎛⎞ dG = ⎜ ⎟⎜⎟dT++ dP∑⎜⎟ dNi ⎝ ∂T ⎠⎝⎠PN,,∂∂PN TN i=1 i ii⎝⎠TPN,, j C =−SdT + VdP +∑ Gi dNi (8.2-1) i=1 The partial molar Gibbs energy, Gi = μi , the chemical potential. ⎛⎞∂∂HH ⎛⎞ C ⎛⎞ ∂ H dH=++⎜⎟ dP ⎜⎟ dS∑⎜⎟ dNi ⎝⎠∂∂PSSN,, ⎝⎠ PN i=1 ∂ Ni ii⎝⎠PSN,, j C ⎛⎞∂H =++VdP TdS∑⎜⎟ dNi (8.2-2) i=1 ∂Ni ⎝⎠PSN,, j ⎛⎞∂∂HH ⎛⎞ ⎜⎟≠= ⎜⎟ H i ∂∂NNii ⎝⎠PSN,,j ⎝⎠ TPN ,, 2012/3/29 25 Chemical Potential HGTS= + C dH=+ dG TdS + SdT =−+SdT VdP +∑GidNi + TdS + SdT i=1 C =++VdPTdSN∑Gdi i (8.2-3) i=1 Comparing Eqs. 8.2-2 and 8.2-3 establishes that ⎛⎞∂∂HG ⎛⎞ ⎜⎟==Gi ⎜⎟ ∂∂NNii ⎝⎠PSN,,ji≠≠ ⎝⎠ TPN ,, ji Using the procedure established here, it is also easily shown that C dH=− TdS PdV +∑ Gi dNi (8.2-4) i=1 C dA=− SdT − PdV +∑ Gi dNi (8.2-5) i=1 ⎛⎞∂∂∂∂HGUA ⎛⎞ ⎛⎞ ⎛⎞ ⎜⎟==Gi ⎜⎟ = ⎜⎟ = ⎜⎟ =μ ∂∂∂∂NNNNi ⎝⎠i PSN,, ⎝⎠i TPN ,, ⎝⎠ii SV,, N ⎝⎠ T ,,V N ji≠ ji≠ ji≠≠ ji 2012/3/29 26 Generalized Gibbs-Duhem equation Any thermodynamic function θ can be written as NNθθ= ∑ i i dN()θθθ=+∑∑ Ndiiii dN (8.2-6) NNTPNθθ= (),, i ⎛⎞∂∂NNθθ ⎛⎞ C ⎛⎞ ∂ N θ dN()θ = ⎜⎟ dT++ ⎜⎟ dP ∑⎜⎟ dNi ⎝⎠∂∂T PN,, ⎝⎠PN TN i ∂ i ii⎝⎠TP,,N ji≠ ⎛⎞∂∂θθ ⎛⎞ C d() Nθθ=++N ⎜⎟ dTN ⎜⎟ dPi dNi (8.2-7) ∂∂TP∑ ⎝⎠PN,,ii ⎝⎠ TN i Subtracting Eq.
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