which is identical to the equation obtained before we

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Unformatted text preview: µB dnB At the same time we know that G = nA µA + nB µB therefore: dG = µA dnA + µB dnB + nA dµA + nB dµB Therefore, we must have: µA dnA + µB dnB + nA dµA + nB dµB = µA dnA + µB dnB which leads to: nA dµA + nB dµB = 0 which we can generalize for any number of substances in the mixture: Σ nJ dµ J = 0 or Σ xJ dµ J = 0 (Gibbs- Duhem Equation) So we if know experimentally the variation of the chemical potential of substance A with composition in the mixture, then we can estimate the variation in the chemical potential of substance B with composition. Marand’s Notes: Chapter 5 - The Properties of Simple Mixtures 162 Mixing of Ideal Gases Consider the following set up consisting of two gas containers connected via a pipe, which can be closed by a valve. At the beginning of the experiment, the first container holds nA moles of molecules of an ideal gas A, at pressure P, temperature T in a volume VA. The second container holds nB moles of molecules of a gas B, at pressure P, temperature T in a volume VB. The valve between the two containers is initially closed. Now, we open the valve and the gases mix, so that, when equilibrium is reached, molecules of type A are under a partial pressure PA and molecules of type B are under a vapor pressure PB. The total pressure P has not changed and the temperature T also remains the same. Let us calculate the free energy change for this process. This free energy change can be calculated by considering the fact that gas A expands isothermally from VA to VA + VB and that gas B expands isothermally from VB to VA + VB. We note that dG = VdP = nRT dP/P = - nRT dV/V (since nRT = PV = cst, then dV/V + dP/P = 0) Marand’s Notes: Chapter 5 - The Properties of Simple Mixtures 163 We apply this relationship to both A and B and get for th...
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