Molar heat capacity (Thermodynamics)

AI Thread Summary
The discussion revolves around solving a thermodynamics exercise related to molar heat capacity using a fundamental equation for crystalline solids. The user successfully tackled part (a) by deriving the temperature relation but struggles with parts (b) and (c), specifically showing that heat capacity is proportional to T^3 at low temperatures and approaches 3k_b at high temperatures. Guidance is provided on using the inverse of the derivative of temperature with respect to entropy to progress in the solution. The user is encouraged to simplify the equation for low temperatures to find the desired relationships, indicating a need for further clarification on the calculations.
Telemachus
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Homework Statement


Hi there. I'm having some trouble on solving this exercise, which you can find on Callen 2nd edition.

A simple fundamental equation that exhibits some of the qualitative properties of typical crystaline solids is:

u=Ae^{b(v-v_0)^2}s^{4/3}e^{s/3R}
Where A,b, and v0 are positive constants.
a)Show that the system satisfies the Nernst theorem.
b)Show that c_v is proportional to T^3 at low temperature.
c)Show that c_v\rightarrow 3k_b at high temperatures.

The Attempt at a Solution


Well, I think I've solved a. And this is what I did:

\displaystyle\frac{\partial u}{\partial s}=T=Ae^{b(v-v_0)^2} \left[\displaystyle\frac{4}{3}s^{1/3}e^{s/3R}+\displaystyle\frac{1}{3R}s^{4/3}e^{s/3R}\right]

\therefore T \rightarrow 0 \Longleftrightarrow s \rightarrow 0

I'm not sure if this is right. If there's another simple way of doing this I'd like to know.

Then I've tried with b) but I didn't get too far.

c_v=T\left(\displaystyle\frac{\partial s}{\partial T}\right)_v

I don't know what to do from here, I've tried to get the entropic representation for the fundamental equation, but I couldn't, and I think it doesn't help. I think that I should use that for a constant volume du=Tds, but I'm not pretty much sure about this.

Help please :)

Bye there.
 
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You're doing great; your solution for (a) looks fine. For part (b), it is helpful to remember that (\partial s/\partial T)_v=(\partial T/\partial s)^{-1}_v; after all, you have T as a function of s. I also found it useful to simplify T(s) for the low temperature case, when s approaches 0. Which term(s) will dominate?
 
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Thanks Mapes, I didn't see your response before because my email has been hacked. And I was waiting for the advice in my new email :P

Now I'll take my time to analyze your response, it's not comlpetly clear to me yet, and as I left the problem behind because I wasn't making any progress I have to get on it again.

Lets see. You're saying that I should use the inverse, which would be the same than the derivative of T respect to s, the equation of state I get before, right?

Thank you very much sir :)
 
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Ok, this is what I did.

\left(\displaystyle\frac{\partial T}{\partial S}\right)^{-1}_v=\left(\displaystyle\frac{\partial S}{\partial T}\right)=\displaystyle\frac{9}{Ae^{b(v-v_0)^2}}\left(\displaystyle\frac{s^{2/3}}{4e^{s/3R}}+\displaystyle\frac{R}{8s^{1/3}e^{s/3R}}+\displaystyle\frac{R^2}{s^{4/3}e^{s/3R}}\right)

so,

c_v=T\left(\displaystyle\frac{\partial S}{\partial T}\right)_v=9\left(\displaystyle\frac{s}{R}+\displaystyle\frac{R}{12}+\displaystyle\frac{4R^2}{3}+\displaystyle\frac{s^2}{12R}+\displaystyle\frac{s}{24}+\displaystyle\frac{RS}{3} \right)

Now, when T approaches to zero s approaches to zero, then remains the constants. But I don't see the T^3

If s\rightarrow 0, then
c_v \rightarrow 9\left(\displaystyle\frac{R}{12}+\displaystyle\frac{4R^2}{3} \right)

I think there is something wrong with this. Perhaps I've made some mistakes with the derivatives or something.
 
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Couldn't get the relation with T^3. Anyway, how do I get the equation for higher temperatures? it doesn't seem to work just making s \rightarrow \infty
 
Telemachus said:
Now, when T approaches to zero s approaches to zero, then remains the constants. But I don't see the T^3

As T becomes small, s becomes small. As s becomes small, T\approx Ae^{b(v-v_0)^2} 4s^{1/3}/3 because higher powers of s become negligible and the exponential becomes approximately one. Now try finding c_V again.

A similar approach works for part (c).
 
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Thanks.
 
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