Register to reply 
Entropy, free energy and chemical potential of mixtures 
Share this thread: 
#1
Mar2810, 11:33 AM

P: 144

Hi,
1. The problem statement, all variables and given/known data Consider a mixture of different gases with [tex]N_i[/tex] molecules each (i=1...k denotes the species). For ideal gases the following relation yields: [tex]S(T,V,N_1,...N_k)=\sum_{i=1}^k S_i(T,V,N_i)[/tex] 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 [tex]\mu_i[/tex] for each component and show that the following relation holds: [tex]\mu_i(p,T,N_1,...,N_k)=\mu_{i,0}(p,T) + kT\ln(c_i).[/tex] Where [tex]c_i := N_i/N[/tex] (with [tex]N = \sum_i N_i[/tex]) is the concentration of the ith component and [tex]\mu_{i,0}(p,T)[/tex] the chemical potential of the ith 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 [tex]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)[/tex] 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 [tex]U = \sum_i U_i[/tex] with [tex]U_i = \frac{3}{2}\cdot N_i \cdot k \cdot T[/tex] But I think, i should derive the internal energy of the mixture from the given equation [tex]S(T,V,N_1,...N_k)=\sum_{i=1}^k S_i(T,V,N_i)[/tex]. Helmholtz free energy: Helmholtz free energy is given by [tex]A = U  T\cdot S[/tex] 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: [tex]\mu_i = \left(\frac{\partial U}{\partial N_m}\right)_{(S,V,N_1,...,N_k)}[/tex] where U is the internal energy and N_m the number of particles of species m. So i probably should derivate [tex]U_i = \frac{3}{2}\cdot N_i \cdot k \cdot T[/tex] with respect to N_i, to get [tex]\mu_i[/tex] However, I see to chance how to show with this derivation, that the following relation holds: [tex]\mu_i(p,T,N_1,...,N_k)=\mu_{i,0}(p,T) + kT\ln(c_i).[/tex] (Sorry for my english, it's not my native language) Best, derivator 


#2
Mar3010, 01:47 PM

P: 144

*push*



#3
Mar3110, 07:03 PM

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 [tex] 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) [/tex] and calculate [tex] \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) [/tex] and simplify it. But I didn't took notice of those movable separators. Do I have to take notice of them? derivator 


Register to reply 
Related Discussions  
Chemical Potential Energy and Kinetic Energy  Introductory Physics Homework  5  
Chemical Potential & Fermi Energy  Atomic, Solid State, Comp. Physics  3  
Entropy and Free Energy  Biology, Chemistry & Other Homework  6  
Thermo  Gibbs Free Energy & Entropy  Advanced Physics Homework  2  
What is potential energy?(in chemical)  General Physics  2 