CHE 330 Thermodynamics February 22 Instructors: Prof. Nicholas A. Kotov, kotov@umich.edu Bean Getsoian, getsoian@umich.edu Raghu Kainkaryam, raghu@umich.edu Chapter 6/8: Thermodynamics of Multicomponent Mixtures What happens when the pure fluids are mixed? U = (T, P. N1,…NL) ΔVmix = V(T, P, x) – x1V1(T, P) – x2V2(T, P) U = (T, P, x1,…xL-1) ΔHmix = H(T, P, x) – x1H1(T, P) – x2H2(T, P) Equation of State for mixtures Thermodynamics of Sulfuric Acid and Water Solutions Substantial Emission of Heat + - Dipole-dipole interactions, hydrogen bonding, etc Partial Molar Properties Nθ=(N1, N2, ….NL) at T and P constant N is the total number of moles in the system 2Nθ=2(N1, N2, ….NL) 2Nθ=(2N1, 2N2, ….2NL) 2(N1, N2, ….NL) =(2N1, 2N2, …2NL) θ(T, P,N1, N2, ….NL) = θ(T, P, 2N1, 2N2, …2NL) Dependence of θ on the number of moles Nθ=N11(N2/N1, ….NL/N1) L θ = ∑xi ∂(Nθ)/∂Ni |T, P, N, j≠i i=1 Molar properties are pretty simple functions of the molar derivatives θi = θi(T, P, x) = ∂(Nθ)/∂Ni |T, P, N, j≠i L θ = ∑xiθi(T, P, x) i=1 Partial molar properties Representation of the effects of Mixing θi(T, P, x) θi(T, P) These could be very different properties Molar volume Hydrogen alone Hydrogen and platinum L V = ∑NiVi(T, P) i L H = ∑NiHi(T, P) i ΔVmix(T, P, N1, N2,…) = V(T, P, N1, N2, …) - L ∑NiVi(T, P) i ΔHmix(T, P, N1, N2,…) = H(T, P, N1, N2, …) - L ∑NiHi(T, P) i L θ = ∑xiθi(T, P, x) i=1 Molar thermodynamic property for other parameters U = ∑ xiUi Same for V, H, S, G, A A = U – TS G = H - TS θi = θi(T, P, x) = ∂(Nθ)/∂Ni |T, P, N, j≠i L θ = ∑xiθi(T, P, x) i=1 L H = ∑xiHi(T, P) i L CP = ∑xiCp I (T, P) i Derivative Quantities Same can be said about other derivatives obtained in the equations (∂Gi/∂T)P, N = -Si How does the differentials of G, A, H, U…etc depend on the number of moles in the mixture? Chemical Potential Partial molar Gibbs free energy is called chemical potential μ = (∂G/∂Ni)T, P, N j≠I = Gi Chemical potential is in all expressions for partial molar quantities μ = (∂G/∂Ni)T, P, N j≠I = (∂H/∂Ni)S, P, N j≠I = (∂U/∂Ni)S, V, N j≠I = (∂A/∂Ni)T, V, N j≠I Example of change in chemical potential is hydrogen in platinum