Molar heat capacity (Thermodynamics)

[Telemachus]

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.

 

[Mapes]

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?

 

[Telemachus]

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 :)

 

[Telemachus]

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.

 

[Telemachus]

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

 

[Mapes]

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).

 

[Telemachus]

Thanks.