Thermodynamics: deriving expression for S = S(T, V, N) - constant problems

[laser1]

Homework Statement

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Relevant Equations

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I have an issue with (b). What I did was simply integrate $dS$. It's a perfect gas, so, $$\left(\frac{\partial E}{\partial T}\right)_V=NC_V$$ and $$\left(\frac{\partial E}{\partial V}\right)_T=0$$ Next I used the relation that $PV=NkT$ to get $\frac{P}{T}=\frac{Nk}{T}$, and after integrating I got $$S=NC_V \ln T + Nk \ln V + const$$ I understand that I could substitute $Nk\ln V = Nk\ln\frac{V}{N}+Nk\ln N$, but then the constant depends on $N$ and in the problem statement, it says that $C$ is a constant for the particle type of particle, so it shouldn't depend on the number of particles in my opinion ($N$). Thank you

 

[haruspex]

but then the constant depends on $N$ and in the problem statement, it says that $C$ is a constant for the particle type of particle, so it shouldn't depend on the number of particles

Look carefully. Your $const$ is not the same as the given $C$.

 

[laser1]

Look carefully. Your $const$ is not the same as the given $C$.

I know but as said in my last paragraph, then $C$ will depend on $N$. To show more explicitly what I mean: I let $$C=k\ln N + \frac{const}{N}$$

 

[haruspex]

I got $$S=NC_V \ln T + Nk \ln V + const$$ I understand that I could substitute $Nk\ln V = Nk\ln\frac{V}{N}+Nk\ln N$, but then the constant depends on $N$

It is not a problem if your integration constant depends on $N$.
So substitute $const=Cf(N)$, equate your answer to the given answer and see what happens.

 

[laser1]

It is not a problem if your integration constant depends on $N$.
So substitute $const=Cf(N)$, equate your answer to the given answer and see what happens.

Yes as stated, I got the answer as required in the problem with my C defined as above. My issue was that the C depended on N. You say it is not a problem, but... the question said C is a constant "for a particular type of particle". With the C I got, C depends on N, the number of particles. So this doesn't seem to be "for a particular type of particle". Also, S is a function of N, so the constant shouldn't include N, in my opinion. Thanks

 

[haruspex]

Yes as stated, I got the answer as required in the problem with my C defined as above. My issue was that the C depended on N. You say it is not a problem, but... the question said C is a constant "for a particular type of particle". With the C I got, C depends on N, the number of particles. So this doesn't seem to be "for a particular type of particle". Also, S is a function of N, so the constant shouldn't include N, in my opinion. Thanks

No, you are still missing it.
It is ok that your $const$ depends on $N$. It is perfectly reasonable that the integration constant involves it. It is the $C$ in the given form that needs to be independent of $N$.
I claim that if you follow the procedure I specified in post #4 you will find you can arrive at the given answer by plugging in the right $f(N)$. Of course, you will need to justify that choice.
If you do not find that works for you, please post your attempt.

 

[laser1]

No, you are still missing it.
It is ok that your $const$ depends on $N$. It is perfectly reasonable that the integration constant involves it. It is the $C$ in the given form that needs to be independent of $N$.
I claim that if you follow the procedure I specified in post #4 you will find you can arrive at the given answer by plugging in the right $f(N)$. Of course, you will need to justify that choice.
If you do not find that works for you, please post your attempt.

Ah okay got it. So my last paragraph in the OP was correct. My bad, usually N is not a variable so I just automatically assumed that constant was constant when in fact it was a partial so you have to add function of N afterwards! I think my last paragraph method is easier than yours, but both work out well (yours is more general which is better probably). Thanks!

 

[haruspex]

Ah okay got it. So my last paragraph in the OP was correct. My bad, usually N is not a variable so I just automatically assumed that constant was constant when in fact it was a partial so you have to add function of N afterwards! I think my last paragraph method is easier than yours, but both work out well (yours is more general which is better probably). Thanks!

View attachment 352816

Good, but as I noted, you should provide an argument for $f$ being of that form.

 

[laser1]

Good, but as I noted, you should provide an argument for $f$ being of that form.

Is that because when integrating partials there was no dN despite it being a variable? I wouldn't even say C*f(N) I would just say f(N) tbh but it works out in the question. perhaps I am missing something

 

[haruspex]

Is that because when integrating partials there was no dN despite it being a variable? I wouldn't even say C*f(N) I would just say f(N) tbh but it works out in the question. perhaps I am missing something

You can reason that, keeping all else the same, $S$ must be proportional to $N$, therefore your $const$ in post #1 is too.

 

[laser1]

You can reason that, keeping all else the same, $S$ must be proportional to $N$, therefore your $const$ in post #1 is too.

Hi

I thought I understood it, then I returned to this question now, and I am still struggling to understand why f(N) in my expression is in this form. Why is it in this form? Or is my f(N) incorrect? Thank you

 

[haruspex]

Hi
View attachment 353953
I thought I understood it, then I returned to this question now, and I am still struggling to understand why f(N) in my expression is in this form. Why is it in this form? Or is my f(N) incorrect? Thank you

Is it reasonable to assume that $S(T, V, M+ N)=S(T, V, M)+S(T, V, N)$?

Edit: I meant
$S(T, V/(M+N), M+ N)=S(T, V/M, M)+S(T, V/N, N)$

 

[laser1]

Is it reasonable to assume that $S(T, V, M+ N)=S(T, V, M)+S(T, V, N)$?

Yes I believe that is reasonable.

 

[haruspex]

Yes I believe that is reasonable.

then do you see how that leads to the result?

 

[laser1]

then do you see how that leads to the result?

no :(

 

[haruspex]

no :(

It implies $f(M+N)=f(M)+f(N)$, so f is linear.

 

[laser1]

It implies $f(M+N)=f(M)+f(N)$, so f is linear.

what about the $-Nk\ln N$ term?

 

[haruspex]

what about the $-Nk\ln N$ term?

Ah, sorry. I was not paying attention to V versus V/N.
Please see my edit to post #12.

 

[laser1]

Ah, sorry. I was not paying attention to V versus V/N.
Please see my edit to post #12.

Am I correct in saying that what we choose to be $f(N)$ is not deducible by itself, but only as a result of seeing how entropy should scale with $N$, and choosing $f(N)$ so the equation works?

 

[haruspex]

Am I correct in saying that what we choose to be $f(N)$ is not deducible by itself, but only as a result of seeing how entropy should scale with $N$, and choosing $f(N)$ so the equation works?

Yes.