# Entropy, free energy and chemical potential of mixtures

by Derivator
Tags: chemical, energy, entropy, free, mixtures, potential
 P: 144 Hi, 1. The problem statement, all variables and given/known data Consider a mixture of different gases with $$N_i$$ molecules each (i=1...k denotes the species). For ideal gases the following relation yields: $$S(T,V,N_1,...N_k)=\sum_{i=1}^k S_i(T,V,N_i)$$ a)Give explicit expressions for the entropy, the internal energy, Helmholtz free energy and Gibbs free energy b) What is the change in entropy, if the k components are initially separated by moveable and diathermal seperators which get removed. c) Calculate for this mixture the chemical potential $$\mu_i$$ for each component and show that the following relation holds: $$\mu_i(p,T,N_1,...,N_k)=\mu_{i,0}(p,T) + kT\ln(c_i).$$ Where $$c_i := N_i/N$$ (with $$N = \sum_i N_i$$) is the concentration of the i-th component and $$\mu_{i,0}(p,T)$$ the chemical potential of the i-th component in unmixed state. 2. Relevant equations 3. The attempt at a solution I have no idea at all, how to solve this exercise. Here is my attempt: a) Entropy: I know from here that the entropy of an ideal gas is given by $$S_i(T,V,N_i) = S_0 + N_i k \ln\left(\left(\frac{T}{T_0}\right)^{3/2}\frac{V}{V_0}\right)$$ So the entropy of the mixture (which we look for) may be given by the sum over this expression. Correct? Internal energy: I know that the internal energy is an extensive property, so $$U = \sum_i U_i$$ with $$U_i = \frac{3}{2}\cdot N_i \cdot k \cdot T$$ But I think, i should derive the internal energy of the mixture from the given equation $$S(T,V,N_1,...N_k)=\sum_{i=1}^k S_i(T,V,N_i)$$. Helmholtz free energy: Helmholtz free energy is given by $$A = U - T\cdot S$$ But how should I give an explicit expression for the mixture. Gibbs free energy: It is given by: [text]G = H - T\cdot S[/tex] Same problem as for the Helmholtz free energy, I don't know how to give an explicit expression for the mixture. b) I think, due to the diathermal seperators, I can assume, that each component has the same temperature. But I don't know how to take into account that the seperators are moveable. To be honest, I have no clue at all, how to solve this part... c) According to the definition in our lecture, the chemical potential is given by: $$\mu_i = \left(\frac{\partial U}{\partial N_m}\right)_{(S,V,N_1,...,N_k)}$$ where U is the internal energy and N_m the number of particles of species m. So i probably should derivate $$U_i = \frac{3}{2}\cdot N_i \cdot k \cdot T$$ with respect to N_i, to get $$\mu_i$$ However, I see to chance how to show with this derivation, that the following relation holds: $$\mu_i(p,T,N_1,...,N_k)=\mu_{i,0}(p,T) + kT\ln(c_i).$$ (Sorry for my english, it's not my native language) Best, derivator
 P: 144 ok folks, lets look at b), please: Entropy will change, because the available volume for one species will change. So I can take my formula for entropy $$S_i(T,V,N_i) = S_0 + N_i k \ln\left(\left(\frac{T}{T_0}\right)^{3/2}\frac{V}{V_0}\right)$$ and calculate $$\Delta S = \sum_i{S_i(T,V,N_i)} -\left(S_0 + N_i k \ln\left(\left(\frac{T}{T_0}\right)^{3/2}\frac{V_i}{V_0}\right)\right)$$ and simplify it. But I didn't took notice of those movable separators. Do I have to take notice of them? --derivator