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

In summary: I am trying to use the concepts developed in Callen up to this point and not to use information from outside.
  • #1
EE18
112
13
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?
 
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  • #2
Which parts are you having trouble with?
 
  • #3
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.
 
  • #4
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.
 
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  • #5
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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  • #6
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:

1690743791430.png
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.
 
  • #7
I still don't see the difference. One is per mole, one is not. How is the result incorrect?
 
  • #8
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).
 
  • #9
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 :)
 
  • #10
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.
 
  • #11
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.
 
  • #12
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.
 
  • #13
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.
 
Last edited:
  • #14
Sorry, fiducial state?
 
  • #15
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."
 
  • #16
hutchphd said:
Sorry, fiducial state?
Reference state is what's meant by that I believe.
 
  • #17
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!
 
  • #18
I think the answer is yes......(truth: but thermo is my shakiest subject)

PS Just saw @TSny post.....listen to him
 
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  • #19
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.
 
  • #20
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.
 
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