Helmholtz Energy Proof (thermodynamics)

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SUMMARY

The discussion focuses on the derivation of the internal energy equation from the Helmholtz free energy definition, F=E-TS. The key equation established is E=-T²(∂/∂T)(F/T)V, which is derived through manipulation of the Helmholtz free energy and its relationship to entropy, S=(∂F/∂T)V. The participants clarify the use of the product rule for partial derivatives and confirm the correctness of the final equation.

PREREQUISITES
  • Understanding of thermodynamic concepts such as Helmholtz free energy and internal energy.
  • Familiarity with partial derivatives and their application in thermodynamics.
  • Knowledge of the relationship between temperature, entropy, and free energy.
  • Basic proficiency in calculus, particularly the product rule for differentiation.
NEXT STEPS
  • Study the derivation of the Gibbs free energy and its relation to Helmholtz free energy.
  • Learn about the implications of the Maxwell relations in thermodynamics.
  • Explore advanced topics in statistical mechanics related to thermodynamic potentials.
  • Review applications of the Helmholtz free energy in phase transitions and chemical reactions.
USEFUL FOR

Students and professionals in physics and engineering, particularly those specializing in thermodynamics and statistical mechanics, will benefit from this discussion.

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Homework Statement


Define the Helmholtz free energy as F=E-TS.
Show that the internal energy E=-T2\frac{∂}{∂T}(\frac{F}{T})V

Homework Equations


S=(\frac{∂F}{∂T})V

Perhaps \beta=\frac{1}{\tau}
and \tau=kBT

The Attempt at a Solution


E = F+TS
E = F+T(\frac{∂F}{∂T})V
.
.
. (some steps to final equation)
.
.
E=-T2\frac{∂}{∂T}(\frac{F}{T})V


Any help/hints would be greatly appreciated. My partial derivatives are a bit rusty. Thanks
 
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Note \frac{∂}{∂T}(\frac{F}{T})V = \frac{∂}{∂T}(\frac{1}{T} \cdot F)V and use the product rule to write out the partial derivative.

Also, check to see if there's a sign error in your equation S=(\frac{∂F}{∂T})V
 
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Okay! I got it..

So

E=-T2[-\frac{1}{T^2}F + \frac{∂F}{∂T}\frac{1}{T}]

E= F + (\frac{∂F}{∂T})(-T)
E= F+(-S)(-T)
E= F+TS

F=E-TS

Thank you!
 

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