Thermodynamics - Rods, bubbles and magnets

AI Thread Summary
The discussion focuses on deriving the relationship between the heat capacity at constant length (C_L) of an elastic rod and the second derivative of the Helmholtz free energy function (f) with respect to temperature. The key equation to demonstrate is that the partial derivative of C_L with respect to length at constant temperature equals negative temperature times the second derivative of f with respect to temperature. Participants clarify the use of thermodynamic identities, specifically the relationships between internal energy (U), entropy (S), and the Helmholtz free energy (F). The conversation also touches on the correct application of partial differentiation and the symmetry of partial derivatives in thermodynamic equations. The thread concludes with a resolution on how to proceed with the derivation.
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Homework Statement


For an elastic rod, show that

\[ \frac{\partial C_{L}}{\partial L}_{T} = -T*\frac{\partial^2 f}{\partial T^2}

where C_{L} is the heat capacity at constant length L.

Homework Equations


dU = Tds + fdL
F = U - TS

The Attempt at a Solution


I've used
dF = dU - Tds - SdT = fdL - SdT

Therefore, S = -\frac{\partial F}{\partial T}_{L} and
f = \frac{\partial F}{\partial L}_T
Hence, \frac{\partial S}{\partial L}_T = -\frac{\partial f}{\partial T}_L

I've also found that C_{L} = T*\frac{\partial S}{\partial T}_L

Just not sure what conditions there are when I partially differentiate C_{L}

EDIT: Sorry, I've seen posts here where the TeX code gets translated directly. Is there anyway I can write in TeX language and have it appear as it were compiled in TeX?
 
Last edited:
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Homework Statement


For an elastic rod, show that

\[ \frac{\partial C_{L}}{\partial L}_{T} = -T*\frac{\partial^2 f}{\partial T^2}

where C_{L} is the heat capacity at constant length L.

Homework Equations


dU = Tds + fdL
F = U - TS

The Attempt at a Solution


I've used
dF = dU - Tds - SdT = fdL - SdT

Therefore, S = -\frac{\partial F}{\partial T}_{L} and
f = \frac{\partial F}{\partial L}_T
Hence, \frac{\partial S}{\partial L}_T = -\frac{\partial f}{\partial T}_L

I've also found that C_{L} = T*\frac{\partial S}{\partial T}_L

Just not sure what conditions there are when I partially differentiate C_{L}
 
Last edited:
OK, what exactly do you mean "not sure what conditions there are when you partially differentiate C_{L}"?

I did, \frac{\partial C_{L}}{\partial L_{T}} = \frac{\partial}{\partial L_{T}} \left( T \frac{\partial S}{\partial T_{L}} \right) = T \frac{\partial}{\partial L_{T}} \left( \frac{\partial S}{\partial T_{L}} \right) , and from there on had an apparently clear path to the solution.

I can't write down all the steps of the solution. If I do the admins will kick me out.
 
Hello omoplata,

Thank you very much for your reply and help with TeX. I'll re-write it out to make everything clear:

1. Homework Statement
For an elastic rod, show that

\left( \frac{\partial C_{L}}{\partial L}\right)_{T} = -T \left( \frac{\partial^2 f}{\partial T^2}\right)_{L}

where C_{L} is the heat capacity at constant length L.

2. Homework Equations
dU = Tds + fdL
F = U - TS


3. The Attempt at a Solution
I've used
dF = dU - Tds - SdT = fdL - SdT

Therefore, S = -\left( \frac{\partial F}{\partial T}\right) _{L} and
f = \left( \frac{\partial F}{\partial L} \right)_T
Hence, \left( \frac{\partial S}{\partial L}\right) _T = -\left( \frac{\partial f}{\partial T}\right)_L

I've also found that C_{L} = T\left(\frac{\partial S}{\partial T}\right)_L

Just not sure what conditions there are when I partially differentiate C_{L}
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New comments:

Oh symmetry of partial derivatives, you got me again :(.

Thank you very much, omoplata! I see what to do now.
 
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