Solution Thermodynamics - Chemical Engineering Thermodynamics II, Slides of Thermodynamics

Download Solution Thermodynamics - Chemical Engineering Thermodynamics II and more Slides Thermodynamics in PDF only on Docsity! CH2351 Chemical Engineering Thermodynamics II Unit — |, Il Solution Thermodynamics Dr. M. Subramanian Associate Professor Department of Chemical Engineering Sri Sivasubramaniya Nadar College of Engineering Kalavakkam - 603 110, Kanchipuram (Dist) Tamil Nadu, India msubbu.in[AT]gmail.com ss Contents e UNIT I: PROPERTIES OF SOLUTIONS - Partial molar properties, ideal and non-ideal solutions, standard states definition and choice, Gibbs-Duhem equation, excess properties of mixtures. e UNIT II: PHASE EQUILIBRIA - Criteria for equilibrium between phases in multi component non- reacting systems in terms of chemical potential and fugacity NE Te-0h ae VS TOLe Te Ua) Multicomponent Systems - Basic Relations e Single component system: — Intensive properties: depends on Pressure, Temperature — Extensive properties: depends on Pressure, Temperature, and amount e Multicomponent system: — Intensive properties: depends on Pressure, Temperature, and composition — Extensive properties: depends on Pressure, Temperature, amount of each component SS? NE Te-0h ae VS TOLe Te Ua) Composition Mole fraction n. ht r= SS ny Sor=1 For binary solution m+22=1 dx, = —dirg In dealing with dilute solutions it is convenient to speak of the component present in the largest amount as the solvent, while the diluted component is called the solute. SS? Jan-2012 M Subramanian Other Measures of Composition e Mass fraction — preferable where the definition of molecular weight is ambiguous (eg. Polymer molecules) e Molarity — moles per litre of solution e Molality — moles per kilogram of solvent. The molality is usually preferred, since it does not depend on temperature or pressure, whereas any concentration unit is so dependent. e Volume fraction e Mole ratio or volume ratio (for binary systems) NE Te-0h ae VS TOLe Te Ua) , Jow ptPAHOFH'O) A “OUEUPS JO SUENIOA JEFOW jee” cos fel B re Os Mole fraction of ethanal, “(CH,OH) 0.8 o4 O2 0 L o © + - - PU .WOKO%H) A SIEM JO SmuMPOA JePOW JeIVeR : D5)? 58 mL/mole a fasere 0502 Oo Pure Solute oO OO OGOO. COA 2 = USZsos —s ceBooso FORGO OOgGS° O96SS° ISOS0 000000 °0 00 09 000000 Solution at x," 0.2 Solution at x," 0.1 tely Dilute Solution 000006 18 mL/mole 00000 000000 000006 Pure Liquid Solvent 54 mL/mole = Pure Solute Behaving As Though Infinitely Dilute VE = Abang — AV nega = AV neg Data taken from Dortmund Data Bank Excess Volume Mixture of Ethanol and Water T=25 °C 0 O17 O02 03 04 05 06 OF 08 O98 1 Pure Mole Fraction of Ethanol [mol/mol] Pure Water Ethanol 1 liter of ethanol and 1 liter of water are mixed at constant temperature and pressure. What is the expected volume of the resultant mixture ? : D5)? Partial Molar Properties e The partial molar property of a given component in solution is defined as the differential change in that property with respect to a differential change in the amount of a given component under conditions of constant pressure and temperature, and constant number of moles of all components other than the one under consideration. i = po Ong | T.Pyng Fa where M is any thermodynamic property. e The concept of partial molar quantity can be applied to any extensive state function. NE Te-0h ae VS TOLe Te Ua) S 5 Partial Molar Volume e Benzene-Toluene: Benzene and toluene form an ideal solution. The volume of 1 mole pure benzene is 88.9 ml; the volume of 1 mole pure toluene is 106.4 ml. 88.9 ml benzene mixed with 106.4 ml toluene results in 88.9 ml + 106.4 ml, or 195.3 ml of solution. (ideal solution) e Ethanol-Water: — The volume of 1 mole pure ethanol is 58.0 ml and the volume of 1 mole pure water is 18.0 ml. However, 1 mole water mixed with 1 mole ethanol does not result in 58.0 ml + 18.0 ml, or 76.0 ml, but rather 74.3 ml. — When the mole fraction is 0.5, the partial molal volume of ethanol is 57.4 ml and the partial molal volume of water is 16.9 ml. (non-ideal solution) SS? NE Te-0h ae VS TOLe Te Ua) Fundamental Equations of Solution Thermodynamics For any extensive thermodynamic property nif with a molar value of A’, the partial molar property JW; is defined as M; = - po (1) Oni T.P.njHi ‘Thermodynamic properties of f:omogeneous phase are functions of pressure, temperature, and the nuzober of moles of the individual species which comprise the phase. ‘Therefore, for a thermody- namic property 7, we can write nM = M(P,T,n1,no,ng,...) (2) The total differential of nJAZ is, d{nM) = [ena aps] a D om Ler “Ss? dn; Jan-2012 M Subramanian Taking derivative of Eqn.(9), we get dM = > x,dM;+ 5° Midx; (10) From Eqns.(8) and (10), we get OM OM _ — dP + (— dT = dM, (ir)... (or). ym ‘ 1e., aM aM’ - (Se) are (Sr) ot Lona =0 (11) This equation is known as Gibbs-Duhem equation. At constant T and P, the above equation becomes, SS vdM, = 0 (12) SS? Jan-2012 M Subramanian Gibbs-Duhem Equation e This equation is very useful in deriving certain relationships between the partial molar quantity for a solute and that for the solvent. NE Te-0h ae VS TOLe Te Ua) ¢« Partial Molar Properties in Binary Solutions Me x,M, 11.11) Binary system (x, + x, =1) — — 2 Cee amr \M=xM,+x,M@, Whence, dM = x,dM, + M, dx, + x,dM, +M,dx, (B) The Gibbs/Duhem Equation at const (T, P) for binary solutions is: x,dM, +x,dM, =0 (C) Substituting (C) into (B) and noting tat x, + x, =1 and dx, = — dx,, yield: dM _ vf M, (D) dx, Two equivalent forms of Eq. (A), noting that x, + x, =1, are: M=M,-x,(M@,-M,) and M=M,+x,(M,-M,) Substituting Eq. (D) into the above two eqs. we have the following equations used for calculation of M, from the M: M, =M+x, 7 (11.15) and M, =M-x, dM % dx, Jan-2012 M Subramanian S 5 (11.16) Me Xi Partial molar properties in binary solution e For binary system M =x,M,+x,M, — ‘WZ — dM = x,dM , +M dx, +x,dM,+M,dx, Const. P and T, using Gibbs/Duhem equation v dM = M,dx,+M,dx, xX,+x,=1 —_— — IM | |— IM dM -M.-M,|___.,-=u+x,4 d dx, dx, dx, Jan-2012 M Subramanian Partial Molar Quantities — Physical Interpretation e The partial molar volume of component j in a system is equal to the infinitesimal increase or decrease in the volume, divided by the infinitesimal number of moles of the substance which is added, while maintaining T, P and quantities of all other components constant. e Another way to visualize this is to think of the change in volume on adding a small amount of coreponent / to an infinite total volume of the system. e Note: partial molar quantities can be positive or negative! NE Te-0h ae VS TOLe Te Ua) 700 650 + oo 600 4 ts & “ —— H-mix S550) 0 H-ideal £ —<e— Hibar 2 —a— H2bar x 500 5 450 400 + 350 1 1 1 0 0.2 0.4 0.6 0.8 1 x1 (-) Vi Subramanian It is required to prepare 3 m? of a 60 mole% ethanol(1)-water(2) mixture. Determine the volumes of ethanol and water to be mixed in order to prepare the required solution. The partial molar volumes of ethanol and water in 60 mole% ethanol-water mixture are: Vi =57.5x 107° m3/mol ~—- V2 = 16 x 10° m3 /mol The molar volumes of pure components «re: Ethanol = 57.9 x 10-§ m*/mol; = Water = i8 x 10-® m*/mol. (Anna University, May-2006, 10 marks) Solution: Molar volume of mixture = > 2;V; = 21V1 + toVo = 0.6 x 57.5 x 107° + 0.4 x 16 x 10-® = 40.9 x 10-6 m3 /mol 3 40.9 x 10-6 Moles of ethanol in mixture = 0.6 x 73350 = 44,010 mol Volume of pure ethanol required = 44010 x 57.9 x 107° = 2.548 m® Moles of water in mixture = 0.4 x 73350 = 29, 340 mol Volume of pure water required = 29340 x 18 x 10-° = 0.528 m® Number of moles in 3 m? of mixture = = 73,350 mol : D5)? 1 1 1 _ (8.69) y= I+. 1+(1—-2) 2-7 qe (2-m)x0-(-1)_ 1 dey => (2 _ 4) — (2 _ v4) (8.70) Using Eqn.(8.69) and (8.70) in Eqn.(8.6%), = 1 Tr 1 1l—2 = + ag + a (Q—2)? 2-2, (2—2,)" ae 11 VPP = Vil. 0 = gh eee av’ If the partial volume of species | in a binary solution at constant T’ and P is given by Y=Vt+ars find the corresponding equation for V2. What equation for V is consistent with these equations for the partial volumes? The following table gives the partial molar volumes at 298.15 K of ethyl acetate (1) and carbon tetra chloride (2) in solutions of the two. (a) What is the volume of the solution when 3 moles of ethyl acetate are mixed with 7 moles of carbon tetra choloride? (b) Calculate the change in volume when 0.6 moles of ethyle acetate are mixed with 0.4 moles of carbon tetrachloride. zy Vi/(em?.mol!) V/(em*.mol-') 1.0 97.81 96.74 0.9 97.81 96.68 0.8 97.82 96.63 0.7 97.83 GE.59 0.6 97.87 96.52 0.5 97.87 96.52 0.4 97.91 96.49 0.3 97.96 96.47 0.2 98.03 96.45 0.1 98.13 96.44 0.0 98.25 96.43 : D5)? 0.045 0.04 4 ~_ V wees V-ideal = 0.035 | g = 0.03 4 "E 2 0.025 | SD eo) > 0.02 4 0.015 | 0.01 ) 0 0.2 0.4 0.6 0.8 1 x1 Jan-2012 M Subramanian AVimix 0.00E+00 -2.00E-04 | -4.00E-04 5 -6.00E-04 | -8.00E-04 | -1.00E-03 + -1.20E-03 x1 Jan-2012 M Subra Weight Peroent H.50, in Resulling Solution 9) 35 60 66 70 75 80 8 30 95 1M | #40 Ss Heel Evntved, Btu per lb cé H,S0,,al 7° eB SSS 8 es 5 1 20 2 ao as wo as tl Weight Parcect 1,50, in Aesulling Solution Jan-2012 M Subramanian Redlich-Kister Model e Also known as Guggenheim-Scatchard Equation e Fits well the data of AM,,,, vS. xX; NE Te-0h ae VS TOLe Te Ua) 0.40 0.60 -00 AHmix oo = Data RK Model fit x1 ao -55.9287 al 27.0094 Redlich-Kister model fits well the data of AM, VS. X; SS? Vi Subramanian Weight % Ethanol Density (g/mL) at 22 T 0 0.99799 10 0.98061 20 0.96808 30 0.95155 40 0.93521 50 0.91778 60 _ 0.89532 70 - 0.86838 80 0.84248 90 0.81570 100 0.78808 Calculate the partial molar volume of Ethanol and Water as a function of composition. : D5)? UM its et Pa Fundamental Equation for Closed System e The basic relation connecting the Gibbs energy to the temperature and pressure in any closed system: d(nG') = (nV )dP — (nS)dT — applied to a single-phase fluid in a closed system wherein no chemical reactions occur. | =nVv and | =—ns Pin NE Te-0h ae VS TOLe Te Ua) Fundamental Equation for Open System e Consider a single-phase, open system: nG = G(P,T.ni,na,na,...) d(nG) = oo ap+| 5 oP d(nG) T \. dT +E [GO dn, i Nn; i e Definition of chemical potential: _| 9G) ne| dn, I. (The partial derivative of G with respect to the mole number n, at constant T and P and mole numbers n, # n; ) e The fundamental property relation for single-phase fluid systems of constant or variable composition: NE Te-0h ae VS TOLe Te Ua) d(nG) = (nV )dP — (nS )dT + > y1,dn, SS? Chemical potential and phase equilibria e Consider a closed system consisting of two phases in equilibrium: d(nG )* = (nV )* dP ~(nS)*dT + wane Janay? = (nV )* dP ~(nS)$ dT +>. uP dn? nM = (nM )“ +(nM yf d(nG) = (nV )dP —(nS )dT +/9° w2dn? + uP dn? Since the two-phase system is Mass balance: closed, d(nG) = (nV)aP — (nS)aT Multiple phases at the same T and P are in equilibrium H; =H; Jan-2012- M Subramanian NE Te-0h ae VS TOLe Te Ua) S 5 Partial Molar Energy Properties and Chemical Potentials The partial molar Gibbs free energy is chemical potential; however, the other partial molar energy properties such as that of internal energy, enthalpy, and Helmholtz free energy are not chemical potentials: because chemical potentials are derivatives with respect to the mole numbers with the natura! independent variables held constant. dl = TdS'— Pdv di—T — VdP—Tds dG = —SdT+WVWdP dA = —PdV —sdT Jan-2012 M Subramanian re) os | hia= 3 = ; ‘ Ong PTD .nj i Oni PT nj; nA= A(V,T.ni,n2,n3,...) —— [O(nA) ee Onn ViTyn; nl’ = U(V, S.ni,n2,n3,-..-) _ {O(nt’) ue One I sn, nH = H(P, S.ny,n2,1ng,..-) oe | Me = ¢ Ong PSyny SS? d(nG) = (nV)dP — (n8)dT + S~ Gdn; If F = F(z,y,2) then dF = Mdx + Ndy + Pdz (ar).. - Ge), aM \ ap (a ln Ga), (>) = (3) Oz zy Vay 2,2 Using the above exactness criteria relations, for Eqn.(19) we get, aGi;\ _ fa(nV) - OP Jon | Oni PT ,n; - aG:\ __ [a(nS) _g or P.n — On; PT nj — Exactness Criteria: OQ a (19) (20) (21) From the definition of G, G=H-TS i.e., nG = nH —T(nS) Differentiation with respect to n,; at constant P,T, and n,; yields [ae _ ace i [aces] On: | pr, Oni | pan, Oni | erm, By applying the definition of partial mole; property in the above equation, we get G,;=H,-TS; (22) Rearrangin Eqn.(22), we get TS; + G; = H; (23) Since G; = pj, we can write Eqns.(20) and (21) as Opi 5 e& ).. - " e and, Ofte = ——§, 25 ( oT ) Pn ( Pe: >>) / Variation of ji; with T is S;. However, experimental data are available in terms of V; and H;. Whereas, Oui/T) _ T(Opi/OT) — wi oT T? Using Eqn.(25) in the above equation, we get, Aui/T} -TS:- ui oT 7 Using Eqn.(23) in the above equation, we get, A(ui/T) _ =H OF iE (26) Partial molar volume (V;) and partial molar enthalpy (H;) are useful properties as they represent the variation of chemical potential with pressure and temperature : D5)? Wire = Wa + We VatVe VatVe =— | PdV — | PdV Va Ve VatVe Va+Ve av dV ==— | Tha RT a | LB RT va Ve V ¥ Vat ¥, = —n, RT In-at Ye ~ ng RT In _4—__4 Va Ve fy +H na +H HA yp np na = na RT ln ——— + nig RT In mA na + Mp mB Ha + ip = ng RT InX, +ngRT nXz Jan-2012 M Subramanian As the mixing process is isothermal, and the mixture is an ideal gas, AU=0 Orey = —Wrey = —na RT In X4 — ng RT In Xp AS mixing => = -Ha R lr 45 — Hp R InXz Jan-2012 M Subramanian Entropy Change of Mixing e Consider the process, where n, moles of ideal gas A are confined in a bulb of volume V, at a pressure P and temperature T. This bulb is separated by a valve or stopcock from bulb B of volume V, that contains ng moles of ideal gas B at the same pressure P and temperature T. When the stopcock is opened, the gas molecules mix spontaneously and irreversibly, and an increase in entropy AS, occurs. e The entropy change can be calculated by recognizing that the gas molecules do not interact, since the gases are ideal. AS,,j, is then simply the sum of AS,, the entropy change for the expansion of gas A from V, to (V, + Vg) and AS,, the entropy change for the expansion of gas B from V, to (V, + Vp). That is, VatV AS mix = AS, + AS A 24 +8 AS, =ngRIn ~— 8 mix A B 5°] [®e @5| [o® Ma oS l < @/—>|P°e ~@§ Va + Vp Oo ® gO Oo AS, = npR In Amix Sin = —Rixa In Xa > XB In xpi. Order AS>O Disorder Ve oT NE Te-0h ae VS TOLe Te Ua) Gibbs free energy change of mixing By definition, G = H —TS. Therefore AG = AH —-TAS — SAT For an ideal gas mixture, ; ; ; ig _ ig ig i AGE -— ABS TASS, = SEAT For the changes at constant temperature, AY = 0, and enthalpy change of mixing is zero for ideal gas mixture. Therefore, the above equation reduces to AGE, = -TAS®,, (39) Using Eqn.(37) in (39), we get AGE. = yy y:RT In y; (40) From this we can get G,* = GE + RT Iny; (41) : D5)? Chemical potential of component i in an ideal gas mixture From the fundamental property relation, dG =VdP — SdT At constant T and for an ideal gas i, the above equation reduces to dGi® = Vi#dP Since viz = RT/P, the above equation becomes dG’? = RTdInP (42) Integrating the above equation, we get G = RTInP+-T.(T) (43) Substituting this in Eqn.(41), we get G;® = RT InP +1;(T) + RT Iny; Rearranging the above, we get G;* = RT In(y:P) +T\(T) (44) This equation gives the chemical potential of component 7 in an ideal gas mixture, in terms of measurable quantities (T,P, and y;). We need a similar expression for chemical potential of component 7 in a real gas mixture and any solution. To this need, we will define a property what is called as residual property. —wn Li’ =G,8 =G% +RT Iny, L G® =. (T)+RT InP wi = (T)+RT Iny,? l l l Jan-2012 M Subramanian Chemical potential of component i in a solution in terms of fugacity Partial molar Gibbs free energy of component in a solution (G, or ;) can be written in terms of residual Gibbs free energy (G;") and ideal gas value (2) as hi = Gi = G* ia (50) For a component 7 in a solution, Eqns.(48) and (49) are written as G, = RT In f, +T,(T) (51) and . G;* = RT Ing; (52) where fi is the fugacity of component 7 in solution, and bi is the fugacity coefficient of i in solution. Using the above equations and Eqn.(44), in Eqn.(50}, we get RT In f, +1 j(T) = RT In(y;,P) +1, (T) + BT nd; ie., RT In f; — RT In(y;P) = RT Ing; => , fi i.e., fi=digsP (53) This equation for fugacity coefficient is applicable for a component at any state (gas, liquid, or solid). However, this expression is normally used for gaseous solution. mf | Fugacity and fugacity coefficient: species in solution e For species i in a mixture of real gases or in a solution of liquids: a, =0(T)+ RT inl f, V Fugacity of species / in solution (replacing the particle pressure) e Multiple phases at the same T and P are in equilibrium when the fugacity of each constituent species is the same in all phases: NE Te-0h ae VS TOLe Te Ua) The residual property: |M* =-M—M* The partial residual property: M: LU, =V(T)+RT In f, {ui =0,(L)+ RT In y,P A M,— f = RT nA y,P [220 _z an, PT nj — For ideal gas, Zz = = G,“ =0 G;" = RT In@, , = fi = ‘yp A f, x i =p = y,P . ; ; The fugacity coefficient ies / i in O ion Jan-2012 M Subramanian Fugacity of a pure liquid e The fugacity of pure species i as a compressed liquid: Sa. f; P G,-G/" = RT In fon G,-G*" = i V,dP_ (isothermal process ) Mv’ . 1 fp In f =—|__V.dP fr" RT Pet | Since V; is a weak function of P 1 sa sa In fi = Vi (P — P. ‘) > f, = eo" PR exp V; (P ~~ P. ») fe" RT fin _ os" Pp ' ' ' RT : D5)? 4 ii) be Wl www.shutterstock.com - 13497712 Infinite dilution of girls in boys eo: D5)?