Entropy, free energy and chemical potential of mixtures

[Derivator]

Hi,

Homework Statement

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.

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

$$S_i(T,V,N_i) = S_0 + N_i k \ln\left(\left(\frac{T}{T_0}\right)^{3/2}\frac{V}{V_0}\right)$$

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)
derivator

 

[Derivator]

*push*

 

[Derivator]

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

$$S_i(T,V,N_i) = S_0 + N_i k \ln\left(\left(\frac{T}{T_0}\right)^{3/2}\frac{V}{V_0}\right)$$

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