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

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• EE18
In summary: I am trying to use the concepts developed in Callen up to this point and not to use information from outside.
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?

Which parts are you having trouble with?

hutchphd said:
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.

EE18 said:
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.

hutchphd and Lord Jestocost
Chestermiller said:
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.

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EE18 said:
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.

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

hutchphd said:
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).

TSny said:
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 :)

hutchphd said:
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.

EE18 said:
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

EE18 said:
$$dS_j = N_j\frac{c_{vj}}{T}dT + \left(\frac{\partial S}{\partial V}\right)_{T,N}dV = N_j\frac{c_{vj}}{T}dT + \left(\frac{\partial P}{\partial T}\right)_{V,N}dV$$

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.

TSny said:
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.

TSny said:
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.

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Sorry, fiducial state?

EE18 said:
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."

hutchphd said:
Sorry, fiducial state?
Reference state is what's meant by that I believe.

TSny said:
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!

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

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

PhDeezNutz and EE18
EE18 said:
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.

TSny said:
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.

berkeman

1. What is Callen's entropy expression for a general ideal gas?

Callen's entropy expression for a general ideal gas is derived from the fundamental thermodynamic relation and typically takes the form $$S = Nk_B \left( \ln \left[ \frac{V}{N} \left( \frac{U}{N} \right)^{3/2} \right] + \frac{5}{2} \right)$$, where $$S$$ is entropy, $$N$$ is the number of particles, $$k_B$$ is Boltzmann's constant, $$V$$ is the volume, and $$U$$ is the internal energy.

2. What kind of error is being discussed in Callen's entropy expression?

The error in question usually pertains to potential inconsistencies or inaccuracies in the derivation or application of the entropy expression, such as incorrect assumptions, mathematical mistakes, or misinterpretations of the physical parameters involved.

3. How can one verify the accuracy of Callen's entropy expression?

To verify the accuracy of Callen's entropy expression, one can cross-check the derivation steps with standard thermodynamic principles, compare the results with empirical data, and consult other reliable sources or textbooks that discuss the entropy of ideal gases.

4. Are there any known corrections or modifications to Callen's entropy expression?

While Callen's expression is widely accepted, there have been discussions and papers suggesting minor corrections or alternative formulations to address specific cases or to improve the theoretical rigor. These corrections often involve more precise definitions of thermodynamic quantities or adjustments for specific conditions.

5. Why is it important to scrutinize Callen's entropy expression for errors?

Scrutinizing Callen's entropy expression for errors is crucial because entropy is a fundamental concept in thermodynamics that affects our understanding of energy distribution, efficiency of processes, and the behavior of gases. Ensuring the accuracy of its expression helps maintain the integrity of theoretical predictions and practical applications in science and engineering.

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