Entropy Question

[Jerzey2Atl]

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.

[shrumeo]

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 :)

[Rodajc]

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.