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Entropy, free energy and chemical potential of mixtures

by Derivator
Tags: chemical, energy, entropy, free, mixtures, potential
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Mar28-10, 11:33 AM
P: 144

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 i-th component and [tex]\mu_{i,0}(p,T)[/tex] 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:


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.


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...


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)

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Mar30-10, 01:47 PM
P: 144
Mar31-10, 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

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?


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