Molar heat capacity (Thermodynamics)

In summary, the system satisfies the Nernst theorem, c_v is proportional to T^3 at low temperature, and c_v\rightarrow 3k_b at high temperatures.
  • #1
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:

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

The Attempt at a Solution


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

[tex]\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][/tex]

[tex]\therefore T \rightarrow 0 \Longleftrightarrow s \rightarrow 0[/tex]

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.

[tex]c_v=T\left(\displaystyle\frac{\partial s}{\partial T}\right)_v[/tex]

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 [tex]du=Tds[/tex], but I'm not pretty much sure about this.

Help please :)

Bye there.
 
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  • #2
You're doing great; your solution for (a) looks fine. For part (b), it is helpful to remember that [itex](\partial s/\partial T)_v=(\partial T/\partial s)^{-1}_v[/itex]; 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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  • #3
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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  • #4
Ok, this is what I did.

[tex]\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)[/tex]

so,

[tex]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)[/tex]

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

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

I think there is something wrong with this. Perhaps I've made some mistakes with the derivatives or something.
 
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  • #5
Couldn't get the relation with [tex]T^3[/tex]. Anyway, how do I get the equation for higher temperatures? it doesn't seem to work just making [tex]s \rightarrow \infty [/tex]
 
  • #6
Telemachus said:
Now, when T approaches to zero s approaches to zero, then remains the constants. But I don't see the [tex]T^3[/tex]

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

A similar approach works for part (c).
 
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  • #7
Thanks.
 

What is molar heat capacity?

Molar heat capacity, also known as specific heat capacity, is the amount of heat energy required to raise the temperature of one mole of a substance by one degree Celsius.

What is the difference between molar heat capacity and specific heat capacity?

Molar heat capacity is the amount of heat energy needed to raise the temperature of one mole of a substance, while specific heat capacity is the amount of heat energy needed to raise the temperature of one gram of a substance. Molar heat capacity is a more useful measurement for scientists working with large quantities of substances.

How is molar heat capacity measured?

Molar heat capacity is typically measured in units of joules per mole per degree Celsius (J/mol·K) using calorimetry experiments. This involves measuring the change in temperature of a substance when a known amount of heat is added or removed.

Why is molar heat capacity important in thermodynamics?

Molar heat capacity is important in thermodynamics because it helps us understand how much energy is required to change the temperature of a substance. It is also used in calculations to determine the amount of heat transferred during a chemical reaction or a physical change.

How does molar heat capacity vary with temperature?

Molar heat capacity can vary with temperature due to changes in the molecular structure of a substance. In general, molar heat capacity increases with temperature for solids and decreases with temperature for gases. This is because at higher temperatures, molecules have more freedom to vibrate and store heat energy, while at lower temperatures, molecules have less freedom and therefore require less energy to change their temperature.

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