Deriving Helmholtz Thermodynamic Potential & Corresponding Maxwell Relation

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SUMMARY

The discussion focuses on deriving the Helmholtz thermodynamic potential and its corresponding Maxwell relation. The Helmholtz function is defined as F = U - TS, leading to the differential form dF = -PdV - SdT. By comparing coefficients, the relationships P = -(\partial F/\partial V)_{T} and S = -(\partial F/\partial T)_{V} are established. The final Maxwell relation derived is (\partial P/\partial T)_{V} = (\partial S/\partial V)_{T}, confirming the accuracy of the derivation.

PREREQUISITES
  • Understanding of thermodynamic potentials, specifically Helmholtz free energy.
  • Familiarity with differential calculus in thermodynamics.
  • Knowledge of Maxwell relations and their significance in thermodynamics.
  • Basic concepts of state functions and exact differentials.
NEXT STEPS
  • Study the implications of Helmholtz free energy in various thermodynamic processes.
  • Explore additional Maxwell relations and their applications in physical chemistry.
  • Learn about the role of state functions in thermodynamic systems.
  • Investigate the relationship between Helmholtz potential and other thermodynamic potentials, such as Gibbs free energy.
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Students and professionals in physics and physical chemistry, particularly those focusing on thermodynamics and statistical mechanics.

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Homework Statement



To state the differential form of the Helmholtz thermodynamic potential and
derive the corresponding Maxwell's relation.

Homework Equations



Stated within the solution attempt.

The Attempt at a Solution



  • Helmholtz function: [tex]F = U - TS[/tex]


  • Calculating the differential form:

    For infinitesimal change: [tex]dF = dU - tdS - SdT[/tex]

    Then using: [tex]TdS = dU + PdV[/tex] ,

    Therefore:

    [tex]dF = -PdV - SdT[/tex]


  • Which then follows that can write: [tex]F = F(V,T)[/tex]

    Hence:

    [tex]dF = \left(\frac{\partial F}{\partial V}\right)_{T}dV + \left(\frac{\partial F}{\partial T}\right)_{V}dT[/tex]


  • Comparing coefficients in the previous two stated expressions for [tex]dF[/tex]:

    [tex]P = -\left(\frac{\partial F}{\partial V}\right)_{T}[/tex]

    and

    [tex]S = -\left(\frac{\partial F}{\partial T}\right)_{V}[/tex]


  • Then as F is a function of state, then dF is an exact differential and
    the condition for an exact differential gives:


    [tex]\left(\frac{\partial P}{\partial T}\right)_{v} = \left(\frac{\partial S}{\partial V}\right)_{T}[/tex]

    Which is the Maxwell relation, derived as required.
 
Physics news on Phys.org
Looks good to me.
 
Brill.. :biggrin: .. just wanted to check it through with someone else!
 

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