Entropy Question: Debye's Law & Heat Capacity of Diamond

  • Thread starter Jerzey2Atl
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In summary: I'm not sure if this is right, but it seems to me that the entropy would go down as the temperature rises, because the internal energy would go down.
  • #1
Jerzey2Atl
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Hi, this is my first time using this forum. I have a question dealing with entropy and thermodynamics.

Here goes:

According to Debye’s law, the molar heat capacity at constant volume of a
diamond varies with temperature according to the relation

Cv = (12*pi^4*R / 5) x (T / TD)^3

where TD = 2230 K. For a diamond of mass 1. 20 g, what is its entropy change when it is heatedat constant volume from 10. 0 K to 350 K? The molar mass of carbon is 12. 0 g/mol.

I know Ihave to first fidn the molar mass, but I am not sure of which form of the entropy equation to use. Any help is appreciated.
 
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  • #2
it's been a long time since i took pchem, but I googled and might have found an equation that deals with what you are talking about

http://www.grc.nasa.gov/WWW/K-12/airplane/entropy.html
has this down the page

s2 - s1 = cv * ln ( T2 / T1) + R * ln ( v2 / v1)

i don't know why they say "constant volume" then have two volumes in the equation
just like the similar one for "constant pressure" has two pressures

s2 - s1 = cp * ln ( T2 / T1) - R * ln ( p2 / p1)

hmm, well it all goes to zero anyway so why bother?
http://www.physics.carleton.ca/~hardy/75342/Lect_16/Lect_16.html

here they have [delta]S = Cv * ln(T1/T2)

i doubt i helped :)
 
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  • #3
This may be very wrong, but this is how I'd do it:

T = Temperature
S = Entropy
U = Internal Energy
dq = Heat transferrd to the system
dw = Work done on the system
Cv = Constant volume heat capacity

dU = dq + dw , but since no work is being done on the system dU = dq.
At constant volume, Cv = (dU/dT) so dU = Cv*dT

At the same time, dS = (dU/T) (at constant volume) so dS = (Cv*dT)/T. If you integrate this between T=10 and T=350 then I reckon that's the answer (not forgetting to divide the molar heat capacity you're given by ten, as this is the number of mols of carbon in the question.)

However, bear in mind that I'm an idiot.
 

Related to Entropy Question: Debye's Law & Heat Capacity of Diamond

1. What is Debye's Law?

Debye's Law, also known as the Debye model, is a theoretical model that describes the heat capacity of solid materials at low temperatures. It was developed by Peter Debye in 1912 and is based on the assumption that all atoms in a solid vibrate at a certain frequency, which is dependent on the temperature and properties of the material.

2. How does Debye's Law relate to heat capacity?

Debye's Law states that the heat capacity of a solid material is directly proportional to the temperature and the number of atoms in the material. This means that as the temperature increases, the heat capacity also increases. It also explains the behavior of heat capacity at low temperatures, where it approaches zero as the temperature approaches absolute zero.

3. What is the heat capacity of diamond?

The heat capacity of diamond, according to Debye's Law, is approximately 6.01 J/mol*K at room temperature. This value may vary slightly depending on the purity and crystal structure of the diamond.

4. How does Debye's Law affect the properties of diamond?

Debye's Law plays a significant role in determining the properties of diamond, such as its thermal conductivity and melting point. It also explains why diamond is such a good thermal insulator at low temperatures, as its heat capacity approaches zero. Additionally, Debye's Law helps to understand the behavior of diamond at high temperatures, where its heat capacity increases and its thermal conductivity decreases.

5. Does Debye's Law apply to all materials?

No, Debye's Law is specifically applicable to solid materials at low temperatures. It does not apply to liquids or gases, as their heat capacity is affected by factors other than the vibrations of atoms. Additionally, Debye's Law may not accurately describe the behavior of materials with complex crystal structures or at extremely low temperatures.

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