Does Callen's Entropy Expression for a General Ideal Gas Contain an Error?

[EE18]

In Ch. 13.1 of the second edition, Callen defines a general ideal gas as follows:

(1) The mechanical equation of state is of the form $PV = NRT$.

(2)For a single-component ideal gas the temperature is a function only
of the molar energy (and inversely). $u = u(T,v) = u(T)$ in particular.

(3) The Helmholtz potential $F(T, V,N_1, N_2, \dots , N_r)$ of a multicomponent ideal gas is additive over the components ("Gibbs's Theorem"):

$$F(T, V, N_1, ... , N_r)= F_1(T, V, N_1)+ F_2 (T, V, N_2)
+ \dots +F_r(T,V,N_r).$$

Of course, all of these can be proved as a theorem of statistical mechanics given a no-interaction assumption.

At any rate, my claim is about Callen's claim that a single component $j$ of general ideal gas has the following expression for its entropy:

$$S_j = N_js_{j0} + N_j\int_{T_0}^T \frac{c_{vj}}{T'} \, dT' + N_j R \ln \left(\frac{v}{v_0}\frac{N_0}{N_j}\right).$$

I supply a derivation below but want to confirm:

Let $S_{j0} = N_js_{j0}$ be the entropy in some reference state, and consider taking this substance to some other $T,V$ (at fixed $N_j$). Then we have
$$dS_j = N_j\frac{c_{vj}}{T}dT + \left(\frac{\partial S}{\partial V}\right)_{T,N}dV = N_j\frac{c_{vj}}{T}dT + \left(\frac{\partial P}{\partial T}\right)_{V,N}dV $$
$$ = N_j\frac{c_{vj}}{T}dT + \left(\frac{\partial (N_jRT/V)}{\partial T}\right)_{V,N}dV = N_j\frac{c_{vj}}{T}dT + (N_jR/V)dV$$
so that
$$\int dS_j = S_j - S_{j0} = S_J - N_js_{j0} = \int_{T_0}^T N_j\frac{c_{vj}}{T'}dT' + \int_{V_0}^V(N_jR/V')dV' \implies S_j = N_js_{j0} + N_j\int_{T_0}^T \frac{c_{vj}}{T'} \, dT' + N_j R \ln \left(\frac{V}{V_0}\right).$$

But this is surely different than what Callen gets. Does Callen perhaps write $v \equiv V/N_j$ and $v_0 \equiv V_0/N_0$ (where $N_0$ is some randomly chosen number of moles)? Then his result is just
$$S_j = N_js_{j0} + N_j\int_{T_0}^T \frac{c_{vj}}{T'} \, dT' + N_j R \ln \left(\frac{v}{v_0}\frac{N_j}{N_0}\right).$$

But things are similar enough that I wonder if I've made an error?

 

[hutchphd]

Which parts are you having trouble with?

 

[EE18]

Which parts are you having trouble with?

Hi! The two parts that I'm struggling with are (1) that I am getting a different argument in the logarithm than Callen (see the $N_j/N_0$ in particular) and (2) understanding how this $N_0$ is defined. I assumed the reference state had the same $N$ as the actual state of interest, but perhaps that's not correct? But, if not, then I'd expect another term involving $\mu_j$ somehow, but that's not present.

 

[Chestermiller]

In Ch. 13.1 of the second edition, Callen defines a general ideal gas as follows:
Of course, all of these can be proved as a theorem of statistical mechanics given a no-interaction assumption.

At any rate, my claim is about Callen's claim that a single component $j$ of general ideal gas has the following expression for its entropy:I supply a derivation below but want to confirm:

Let $S_{j0} = N_js_{j0}$ be the entropy in some reference state, and consider taking this substance to some other $T,V$ (at fixed $N_j$). Then we have
$$dS_j = N_j\frac{c_{vj}}{T}dT + \left(\frac{\partial S}{\partial V}\right)_{T,N}dV = N_j\frac{c_{vj}}{T}dT + \left(\frac{\partial P}{\partial T}\right)_{V,N}dV $$
$$ = N_j\frac{c_{vj}}{T}dT + \left(\frac{\partial (N_jRT/V)}{\partial T}\right)_{V,N}dV = N_j\frac{c_{vj}}{T}dT + (N_jR/V)dV$$
so that
$$\int dS_j = S_j - S_{j0} = S_J - N_js_{j0} = \int_{T_0}^T N_j\frac{c_{vj}}{T'}dT' + \int_{V_0}^V(N_jR/V')dV' \implies S_j = N_js_{j0} + N_j\int_{T_0}^T \frac{c_{vj}}{T'} \, dT' + N_j R \ln \left(\frac{V}{V_0}\right).$$

In a mixture, the partial molar entropy of a component is equal to that of the pure component at the same temperature and at a pressure equal to its partial pressure in the mixture.

 

[EE18]

In a mixture, the partial molar entropy of a component is equal to that of the pure component at the same temperature and at a pressure equal to its partial pressure in the mixture.

I agree, but I am trying to use the concepts developed in Callen up to this point and not to use information from outside.

To some extent, I am trying to prove the statement you give, although I note that this question involves a so-called "general" ideal gas (Callen Ch 13) rather than a "simple" ideal gas (Callen Ch 3.4) wherein I've seen the derivation which leads to the statement you give. The statement holds for a general ideal gas too, but that's not the crux of my question.

 

[TSny]

At any rate, my claim is about Callen's claim that a single component $j$ of general ideal gas has the following expression for its entropy:

I believe there is an error in Callen's expression for $S_j$. The molar volumes in the last term should be actual volumes:

$$S_j = N_j s_{j0} + N_j \int_{T_0}^T \frac{c_{vj}}{T'}dT' + N_j R \ln\left(\frac{V}{V_0}\frac{N_0}{N} \right)$$

This is how Callen writes it in the first edition of his textbook and it agrees with the result from statistical mechanics.

I think this result can be obtained by going back to equation (3.36) in the second edition: $$ds = \left(\frac 1 T \right) du + \left( \frac P T \right) dv,$$ where $s, u$ and $v$ are molar quantities. This is for a single-component ideal gas. Let $du = c_v dT$ where $c_v$ is the molar heat capacity at constant volume. Also, let $P = \frac{NRT}{V} = \frac{RT}{v}$. So, $$ds = \frac{c_vdT}{T} + R \frac{dv}{v}$$ Integrate to get $$s = s_0 + \int_{T_0}^T \frac{c_vdT}{T} +R \ln \frac{v}{v_0}$$ The total entropy is then $$S = Ns_0 + N\int_{T_0}^T \frac{c_vdT}{T} + NR \ln \frac{v}{v_0} = Ns_0 + N\int_{T_0}^T \frac{c_vdT}{T} + NR \ln\left(\frac{V}{V_0}\frac{N_0}{N} \right)$$

I hope I'm not overlooking something. It seems a bit odd to me that Callen makes the same mistake in equations (13.4) and (13.6) of the 2nd edition.

 

[hutchphd]

I still don't see the difference. One is per mole, one is not. How is the result incorrect?

 

[EE18]

I still don't see the difference. One is per mole, one is not. How is the result incorrect?

Callen's result is not extensive (see the logarithm argument).

 

[EE18]

I believe there is an error in Callen's expression for $S_j$. The molar volumes in the last term should be actual volumes:

$$S_j = N_j s_{j0} + N_j \int_{T_0}^T \frac{c_{vj}}{T'}dT' + N_j R \ln\left(\frac{V}{V_0}\frac{N_0}{N} \right)$$

This is how Callen writes it in the first edition of his textbook and it agrees with the result from statistical mechanics.

I think this result can be obtained by going back to equation (3.36) in the second edition: $$ds = \left(\frac 1 T \right) du + \left( \frac P T \right) dv,$$ where $s, u$ and $v$ are molar quantities. This is for a single-component ideal gas. Let $du = c_v dT$ where $c_v$ is the molar heat capacity at constant volume. Also, let $P = \frac{NRT}{V} = \frac{RT}{v}$. So, $$ds = \frac{c_vdT}{T} + R \frac{dv}{v}$$ Integrate to get $$s = s_0 + \int_{T_0}^T \frac{c_vdT}{T} +R \ln \frac{v}{v_0}$$ The total entropy is then $$S = Ns_0 + N\int_{T_0}^T \frac{c_vdT}{T} + NR \ln \frac{v}{v_0} = Ns_0 + N\int_{T_0}^T \frac{c_vdT}{T} + NR \ln\left(\frac{V}{V_0}\frac{N_0}{N} \right)$$

I hope I'm not overlooking something. It seems a bit odd to me that Callen makes the same mistake in equations (13.4) and (13.6) of the 2nd edition.

I agree with your derivation, thank you! I am extremely confused in where I have erred in my derivation. If you see any slips I would greatly appreciate it, but I'll roll with this at any rate :)

 

[TSny]

I still don't see the difference. One is per mole, one is not. How is the result incorrect?

It's my understanding that the fiducial (reference) state is some arbitrarily chosen $S_0$, $V_0$, and $N_0$ and that $v = V/N$ while $v_0 = V_0/N_0$. The state of interest has values $S$, $V$, and $N$.Then, $v/v_0 \neq V/V_0$.

But maybe in the second edition, Callen is defining $v_0$ as $V_0/N$ instead of $V_0/N_0$. Then $v/v_0$ would equal $V/V_0$. That seems to me to be equivalent to assuming that the fiducial state has the same number of moles $N$ as the state of interest. But then I don't see why Callen bothers to use the notation $N_0$ at all.

Anyway, maybe I'm misinterpreting some of the notation.

 

[TSny]

I agree with your derivation, thank you! I am extremely confused in where I have erred in my derivation. If you see any slips I would greatly appreciate it, but I'll roll with this at any rate :)

If $N_0$ for the fiducial state differs from $N$ for the state of interest, then I think we need an additional term in your equation

$$dS_j = N_j\frac{c_{vj}}{T}dT + \left(\frac{\partial S}{\partial V}\right)_{T,N}dV = N_j\frac{c_{vj}}{T}dT + \left(\frac{\partial P}{\partial T}\right)_{V,N}dV $$

That is, $$dS_j = N_j\frac{c_{vj}}{T}dT + \left(\frac{\partial S}{\partial V}\right)_{T,N}dV + \left(\frac{\partial S}{\partial N}\right)_{T,V}dN$$ I'm not sure if this would fix things.

 

[EE18]

I believe there is an error in Callen's expression for $S_j$. The molar volumes in the last term should be actual volumes:

$$S_j = N_j s_{j0} + N_j \int_{T_0}^T \frac{c_{vj}}{T'}dT' + N_j R \ln\left(\frac{V}{V_0}\frac{N_0}{N} \right)$$

This is how Callen writes it in the first edition of his textbook and it agrees with the result from statistical mechanics.

I think this result can be obtained by going back to equation (3.36) in the second edition: $$ds = \left(\frac 1 T \right) du + \left( \frac P T \right) dv,$$ where $s, u$ and $v$ are molar quantities. This is for a single-component ideal gas. Let $du = c_v dT$ where $c_v$ is the molar heat capacity at constant volume. Also, let $P = \frac{NRT}{V} = \frac{RT}{v}$. So, $$ds = \frac{c_vdT}{T} + R \frac{dv}{v}$$ Integrate to get $$s = s_0 + \int_{T_0}^T \frac{c_vdT}{T} +R \ln \frac{v}{v_0}$$ The total entropy is then $$S = Ns_0 + N\int_{T_0}^T \frac{c_vdT}{T} + NR \ln \frac{v}{v_0} = Ns_0 + N\int_{T_0}^T \frac{c_vdT}{T} + NR \ln\left(\frac{V}{V_0}\frac{N_0}{N} \right)$$

I hope I'm not overlooking something. It seems a bit odd to me that Callen makes the same mistake in equations (13.4) and (13.6) of the 2nd edition.

A question on this derivation for you. If we are indeed starting at some fiducial state with arbitrary $N_0$, should not the very first term in your expression for $S$ be $N_0s_0$ rather than $Ns_0$?

Edit: scratch that, that's wrong.

 

[EE18]

If $N_0$ for the fiducial state differs from $N$ for the state of interest, then I think we need an additional term in your equation

I see. Is it a correct assessment then to say that my conclusion is correct if we take the fiducial state with $N_0 = N$? Extensively would hold in this case since $V_0 = v_0N_0$ would scale with $N = N_0$ just as $V$ would.

 

[hutchphd]

Sorry, fiducial state?

 

[TSny]

I see. Is it a correct assessment then to say that my conclusion is correct if we take the fiducial state with $N_0 = N$?

I think so. However, when dealing with a mixture of gases, the different components will generally have different mole numbers $N_j$. Then, $N_0$ for the fiducial state cannot be chosen to equal $N_j$ for all of the components. In the first edition, Callen makes the following statement:

"It will be noted that in specifying a fiducial state for each [component] gas we have adopted the convention that $T_0$, $V_0$, and $N_0$ are the same for each gas. However, $U_0$ and $S_0$ vary from gas to gas and consequently carry the subscript j in the foregoing equations."

 

[EE18]

Sorry, fiducial state?

Reference state is what's meant by that I believe.

 

[EE18]

I think so. However, when dealing with a mixture of gases, the different components will generally have different mole numbers $N_j$. Then, $N_0$ for the fiducial state cannot be chosen to equal $N_j$ for all of the components. In the first edition, Callen makes the following statement:

"It will be noted that in specifying a fiducial state for each [component] gas we have adopted the convention that $T_0$, $V_0$, and $N_0$ are the same for each gas. However, $U_0$ and $S_0$ vary from gas to gas and consequently carry the subscript j in the foregoing equations."

I've just purchased the first edition. It seems much more in depth from a glance online. It seems Callen sacrificed a great deal to include thermostatistics in the second edition!

 

[hutchphd]

I think the answer is yes......(truth: but thermo is my shakiest subject)

PS Just saw @TSny post.....listen to him

 

[TSny]

I've just purchased the first edition. It seems much more in depth from a glance online. It seems Callen sacrificed a great deal to include thermostatistics in the second edition!

I think both editions are worth having access to. The edition that I own is the 1st edition. It was the textbook for the thermo course that I took back in 1969! For the 2nd edition, I have online access. I like both editions. The 2nd edition has more worked-out examples and some of the homework problems seem more interesting than in the 1st edition.

 

[EE18]

I think both editions are worth having access to. The edition that I own is the 1st edition. It was the textbook for the thermo course that I took back in 1969! For the 2nd edition, I have online access. I like both editions. The 2nd edition has more worked-out examples and some of the homework problems seem more interesting than in the 1st edition.

Amazing, it seems like it has stood the test of time :)

Thank you as always for your help! It has been invaluable to me as I work through Callen.