# Homework Help: Silly doubt: ideal gas eq. of state entropic representation

1. Jan 30, 2012

### Telemachus

Hi there. I have a silly doubt about the entropy of mixing for ideal gases

The entropy of mixing is this (Eq. [1]):
$$S_{mix}=\sum_j N_j s_{j0}+\left (\sum_j N_j c_j \right) R \ln{\frac{T}{T_0}}+\sum_j N_j R \ln {\frac{V}{N_jv_0}}$$
Now I don't know what identity the book uses to rewrite this on this form (Eq. [2]):
$$S_{mix}=\sum_j N_j s_{j0}+\left (\sum_j N_j c_j \right) R \ln{\frac{T}{T_0}}+ N R \ln {\frac{V}{Nv_0}} - \sum_j N_j R \ln {\frac{N_j}{N}}$$

The book (Callen, page 69) also mentions the Gibbs's theorem: "The entropy of a mixture of ideal gases is the sum of the entropies that each gas would have if it alone were to occupy the volume V ate temperature T."

"The last term in equation [2] is known as the "entropy of mixing". It represents the difference in entropies between that of a mixture of gases and that of a collection of separate gases each at the same temperature and the same density as the original mixture Nj/Vj=N/V, (and hence at the same pressure as the original mixture)".

I mention the last, because I thought that perhaps it was using the equality Nj/Vj=N/V as the identity.

I thought that perhaps he was using that Nj/Vj=N/V→Nj/N=Vj/V

Then replacing, subtracting the first term in the sum, and using properties of logarithms perhaps I could get the equality, but I don't think that's the case. Anyway, this is what I mean:
$$\sum_j N_j R \ln {\frac{V}{N_jv_0}}=N R \ln {\frac{V}{Nv_0}} - \sum_j N_j R \ln {\frac{N_jv_0}{V}}$$
Then using Nj/N=Vj/V
$$\sum_j N_j R \ln {\frac{V}{N_jv_0}}=N R \ln {\frac{V}{Nv_0}}-\sum_j N_j R \ln {\frac{N_j}{N}}$$

I don't think this is the case because of many reasons. In the first place, there is no v0 in the term corresponding to the entropy of mixing. It doesn't specify in the sum for the entropy of mixing that it must exclude the term that I assumed as subtracted from the sum neither. And the identity I'm using is mentioned by callen only to illustrate and distinguishes between the two different situations that correspond to Gibbs's theorem. And the case in which the identity is mentioned is not the case in concern, as in the mixing actually I don't think all gases would have the same density, because they have different mole numbers, and I think it's not necessary to have them in the same proportion in a mixing. And a term is missing even assuming the identity is valid, there should be a product inside the logarithm of mixing, and the fraction should be NjNj/N (and actually now that I wrote all this I see that I confused vj with v0. So only if v=v0 it would be valid, but the equality isn't taken in account in the positive logarithm involving the volume.

Any help will be appreciated.

Bye there.

Last edited: Jan 30, 2012
2. Jan 31, 2012

### Telemachus

Alright, I completely misunderstand what N was. It is the sum of the total mole numbers:
$$\sum_j N_j=N_1+N_2+...N_n=N$$
I defined j=1,2,...,n
The previous analysis is totally wrong, so I'll probably delete it later.
This is what I'm trying to get:
$$\sum_j N_j R \ln {\frac{V}{N_jv_0}}=N R \ln {\frac{V}{Nv_0}} - \sum_j N_j R \ln {\frac{N_j}{N}}$$

Now I've tried this, I think it's wrong again, because it would imply that the first term in the right side of the last equality would be zero, as I show below.
$$\sum_j N_j R \ln {\frac{V}{N_jv_0}}=N_1 R \ln {\frac{V}{N_1v_0}}+N_2 R \ln {\frac{V}{N_2v_0}}+...+N_n R \ln {\frac{V}{N_nv_0}}$$
Then I assumed that the total volume is actually:
$$V=N_1v_0+N_2v_0+...+N_nv_0=(N_1+N_2+...+N_n)v_0=Nv_0$$
and
$$N_jv_0=V_j$$
Then
$$\frac{V}{N_jv_0}=\frac{V}{V_j}=\frac{N}{N_j}=\left ( \frac{N_j}{N} \right )^{-1}$$

But as:
$$V=Nv_0$$
This would imply
$$NR\ln\left (\frac{V}{Nv_0} \right )=0$$
and this would give:
$$\sum_j N_j R \ln {\frac{V}{N_jv_0}}=- \sum_j N_j R \ln {\frac{N_j}{N}}$$

So I'm assuming something wrong here, I don't think that term should vanish. I'm sure of what N represents now, but perhaps this assumption is wrong:
$$V=Nv_0$$

Any help?

Bye there.

PD: I can't edit the first message.

Last edited: Jan 31, 2012
3. Jan 31, 2012

### rude man

Well, I don't have Callen and I don't know what some of the symbols refer to in your equations, but:

Starting with sk = ∫cpkdT/T + s0k - Rln(p) for the kth ideal gas in its own separate partition before mixing,

where cpk is the kth gas's molar specific heat at constant p,

it follows that Si = RƩnkk - ln(p)}
where σk = (1/R)∫cpk dT/T + s0k/R
and Si is the initial total entropy before mixing;
then, by Gibbs's theorem and the fact that pk = xkp,
Sf = RƩnkk - ln(p) - ln(xk}
where xk is the fraction of gas k in molar terms, i.e. xk = nk/Ʃnk.

Then the difference Sf - Si = -RƩnkln(xk)
which can also be written as
Sf - Si = n1Rln(V/n1v)
+ n2Rln(V/n2v) + .....
where v is molar volume of all the gases.

If you're interested I can fill in some inbetween steps.

4. Jan 31, 2012

### Telemachus

Alright, as I see it you get that the entropy of mix is this term:
$$\sum_j N_j R \ln {\frac{V}{N_jv_0}}=N_1 R \ln {\frac{V}{N_1v_0}}+N_2 R \ln {\frac{V}{N_2v_0}}+...+N_n R \ln {\frac{V}{N_nv_0}}$$

So, perhaps what I did wasn't wrong. But anyway, it calls my attention that the books keeps that term that gives zero.

I don't know where this comes from: sk = ∫cpkdT/T + s0k - Rln(p)

5. Feb 1, 2012

### rude man

Sorry, I should have defined s0:

Change state of 1 mole of gas from (T1, p1) to (T,p) for the kth gas:

dq = cpdT - Vdp ... ideal gas
ds = dq/T = cpdT/T - nRdp/p ..... from pV = nRT
Δs = s - s1 = ∫cpdT/T - R∫dp/p
s = s1 + Δs = s1 + ∫cpdT/T - R∫dp/p
s1 is molar entropy of state 1 (arbitrary constant)

Assume cp constant:
Δs= cpln(T) - cpln (T1) - Rln(p) + Rln(p1)
Let s0 = s1 - cpln(T1) + Rln(p1) = new constant
Then s = cpln(T) - Rln(p) + s0 = ∫cpdT/T - Rln(p) + s0 where ∫cpdT/T is the indefinite integral, i.e. without a constant added.

The first equation needs a couple of explanatory steps but this equation will certainly be found in your textbook under "ideal gases".

Note that this means s depends on what you call s1. It's only when you get to the "3rd law" that s1(T=0) = 0. Until then, we speak only of entropy change from one state to the next.

6. Feb 3, 2012

### Telemachus

Thank you again.

Would you tell me where does this come from?
dq = cpdT - Vdp

7. Feb 3, 2012

### rude man

du = dq - dw but dw = pdv so
dq = du + pdv
Definition of cv = ∂q/∂T at v = constant
So cv = ∂u/∂T at v constant
However, for an ideal gas, du/dT is a fn of T only, so
cv =du/dT or du = cvdT
Thus dq = cvdT + pdV ............. (1)
For ideal gas, pv = RT so
pdv + vdp = RdT or pdv = RdT - vdp
So (1) becomes
dq = cvdT + RdT - vdp
= (cv + R)dT - vdp
=cpdT - vdp since for ideal gas, cp = cv + R
QED

I should have mentioned that I use lower case for molar-specific quantities and upper case for absolute quantities. So u, q, v, w, cv and cp. are for 1 mole of gas wheras U, Q, V, W, Cv and Cp are absolutes, for example Cp = n*cp etc.

8. Feb 4, 2012

Thanks :D