in equilibrium. You may use without proof the relationship dG = Vdp – SdT , in which the symbols have their usual meanings. [3] b) Develop the result in a) for the case in which one of the phases is gaseous, to establish the Clausius-Clapeyron equation. Identify any assumptions you make. [3], The two quantities are interrelated through the fundamental relationship , dG = VdP – SdT , such that at constant temperature, the pressure derivative of the free energy of solvation is equal to V º (eq 1). 1,2 From a computational standpoint, one can then envision computation of V º via either, 8/10/2020 · ( dG = Vdp – SdT + sum_i mu_i dn_i ) (mu_i = left( dfrac{partial G}{partial n_i} right) _{p,T,n_jneq i}) The last definition, in which the chemical potential is defined as the partial molar Gibbs function is the most commonly used, and perhaps the most useful (Equation ref{eq1}).
1/1/2019 · Injecting this last equation into the previous one, after simplifying, we obtain: dG = Vdp ? SdT + ? k = 1 N ? k d n k Comparing these two equations for d G , we immediately see that. ? G ? p T , n k = V ? G ? T p , n k = S ? G ? n k T , p , n j ? n k = ? k, Special Relations The Legendre transformation › = ›+~ PV (24) leads to the di?erential form d› =~ ¡SdT +V dP ¡Nd„ : (25) But we have established earlier that › · ¡PV, and thus ›~ · 0, and ¡SdT +V dP ¡Nd„ = 0: (26) This formula is called Gibbs-Duhem relation . If we introduce the entropy density s = S=V (27), 11/13/2015 · Thus, we can say dT = 0 and cross out -SdT. color (green) ( ( (delG)/ (delP))_T = V) Next, at a constant temperature, let us find an expression to use volume and pressure (two MEASURABLE variables) to calculate the Gibbs’ Free Energy: dG = VdP.
