Entropy, free energy and chemical potential of mixtures

It sounds like you're trying to solve for the internal energy and Helmholtz free energy of the gas mixture, but you're not sure how to start. You might want to look at the equations for those properties and see if you can find a way to solve for them explicitly.f
  • #1
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Hi,

Homework Statement


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.


Homework Equations





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 http://books.google.com/books?id=12... thermodynamics&pg=PA42#v=onepage&q=&f=false" 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
 
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  • #3
ok folks, let's 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
 

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