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Hi,

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.

I have no idea at all, how to solve this exercise. Here is my attempt:

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?

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 is given by

[tex]A = U - T\cdot S[/tex]

But how should I give an explicit expression for the mixture.

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)

Best,

derivator

## 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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