Thermodynamic Identities Proof - Gibbs and Helmholtz

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SUMMARY

The discussion focuses on proving the Gibbs-Helmholtz equations using the definitions of Helmholtz free energy (F) and Gibbs free energy (G). Participants confirm that the entropy (S) can be expressed as S = -(\frac{\partial F}{\partial T})_V and S = -(\frac{\partial F}{\partial T})_P. They derive expressions for internal energy (U) and enthalpy (H) as U = -T^2 \left(\frac{\partial (F/T)}{\partial T} \right)_V and H = -T^2 \left(\frac{\partial (G/T)}{\partial T} \right)_P. The discussion emphasizes the importance of using the thermodynamic identity dU = Tds - pdV in these derivations.

PREREQUISITES
  • Understanding of Helmholtz free energy (F) and Gibbs free energy (G)
  • Familiarity with thermodynamic identities, specifically dU = Tds - pdV
  • Knowledge of partial derivatives and their application in thermodynamics
  • Ability to apply the chain rule and quotient rule in calculus
NEXT STEPS
  • Study the derivation of the thermodynamic identity dU = Tds - pdV in detail
  • Learn about the implications of the Gibbs-Helmholtz equations in chemical thermodynamics
  • Explore advanced applications of partial derivatives in thermodynamic systems
  • Investigate the relationship between Helmholtz and Gibbs free energies in various thermodynamic processes
USEFUL FOR

Students and professionals in thermodynamics, particularly those studying chemical thermodynamics, as well as educators looking to enhance their understanding of Gibbs and Helmholtz free energies.

  • #31
TFM said:
okay,

G=U+PV-TS

G/T = U/T + PV/T - S

H = -T^2 \frac{\partial (U/T +PV/T - S)}{\partial T}

Okay let's go from here;

G/T=U/T+PV/T-S

d(G/T)/dT=\frac{\frac{dU}{dT}T-U}{T^{2}}+\frac{T(V\frac{dP}{dT}+P\frac{dV}{dT})}{T^{2}}-\frac{dS}{dT}

Now let dU equal it's identity again let P be constant, simplify and rearrange. Should get you to the answer. All my d's should be partial d's.
 
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  • #32
So:

d(G/T}/dT=\frac{\frac{dU}{dT}T-U}{T^{2}}+\frac{T(V\frac{dP}{dT}+P\frac{dV}{dT}}{T ^{2}}-\frac{dS}{dT}

dU = Tds + pdv

d(G/T}/dT=\frac{\frac{Tds + pdv}{dT}T-U}{T^{2}}+\frac{T(V\frac{dP}{dT}+P\frac{dV}{dT}}{T ^{2}}-\frac{dS}{dT}

so:


d(G/T}/dT=\frac{T\frac{ds}{dT} + p\frac{dv}{dT}T-U}{T^{2}}+\frac{TV\frac{dP}{dT} + TP\frac{dV}{dT}}{T^{2}}-\frac{dS}{dT}

Pressure is constant, so can get rid of dP

d(G/T}/dT=\frac{T\frac{ds}{dT} + p\frac{dv}{dT}T-U}{T^{2}}+ TP\frac{dV}{dT}}{T^{2}}-\frac{dS}{dT}

Does this look okay?

d(G/T}/dT= T\frac{ds}{dT} + p\frac{dv}{dT}T-\frac{U}{T^{2}}+T(V\frac{dP}{dT}+P\frac{dV}{dT}}-\frac{dS}{dT}
 
  • #33
Vuldoraq said:
Okay let's go from here;

G/T=U/T+PV/T-S

d(G/T)/dT=\frac{\frac{dU}{dT}T-U}{T^{2}}+\frac{T(V\frac{dP}{dT}+P\frac{dV}{dT})}{T^{2}}-\frac{dS}{dT}

Now let dU equal it's identity again let P be constant, simplify and rearrange. Should get you to the answer. All my d's should be partial d's.

Sorry I made an error,

This,

\frac{T(V\frac{dP}{dT}+P\frac{dV}{dT})}{T^{2}}

Should be this,

\frac{T(V\frac{dP}{dT}+P\frac{dV}{dT})-PV}{T^{2}}
 

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