Helmholtz Free Energy Legendre Transformation

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SUMMARY

The discussion focuses on the derivation of the Helmholtz free energy A(T,V) using the Legendre transformation from internal energy U(S,V). The key equations include A = U - TS, where T is temperature and S is entropy. The differential expression for A is derived as dA = -SdT - pdV, confirming the relationship between Helmholtz free energy, entropy, and pressure. This derivation is essential for understanding thermodynamic potentials in advanced physics.

PREREQUISITES
  • Understanding of thermodynamic concepts such as internal energy and entropy.
  • Familiarity with Legendre transformations in the context of thermodynamics.
  • Knowledge of differential calculus as applied to physical equations.
  • Basic grasp of thermodynamic potentials, specifically Helmholtz free energy.
NEXT STEPS
  • Study the implications of the Helmholtz free energy in thermodynamic systems.
  • Explore the relationship between Helmholtz free energy and Gibbs free energy.
  • Learn about the applications of Legendre transformations in statistical mechanics.
  • Investigate the role of temperature and pressure in phase transitions using thermodynamic potentials.
USEFUL FOR

Students and professionals in physics, particularly those studying thermodynamics and statistical mechanics, will benefit from this discussion. It is particularly useful for those looking to deepen their understanding of thermodynamic potentials and their applications in physical systems.

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Homework Statement



Show how a Legendre transformation is used to obtain the Helmholtz free energy A(T,V) from the internal energy and derive the general expression for the differential of A.

Homework Equations


Internal Energy is a function of Entropy and Volume.

U Ξ (S, V)

A Ξ (T,V)

A = U - TS

where
U: Internal Energy
S: Entropy
V: Volume
T: Temperature
A: Helmholtz Free Energy

The Attempt at a Solution



A = U - (dU/dS)VS

dU = TdS - pdV

(dU/dS)V = T

A = U - TS (Helmholtz free energy)

dA - dU - TdS-SdT

dA = TdS - pdV - TdS - SdT

dA = -SdT - pdV

A Ξ (T,V)

I just wanted someone to check my derivation as I'm still a bit new to Legendre transformations. Cheers
 
I would post this in the advanced physics forum. It's not introductory material.
 

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