Thermodynamic Identities Proof - Gibbs and Helmholtz

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Homework Help Overview

The discussion revolves around proving relationships involving Helmholtz free energy (F) and Gibbs free energy (G) within the context of thermodynamics. Participants are tasked with demonstrating specific equations related to entropy (S) and internal energy (U), as well as exploring the Gibbs-Helmholtz equations, which are significant in chemical thermodynamics.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss using definitions of free energies and thermodynamic identities to derive relationships. There is exploration of differentiation techniques, including the chain rule and quotient rule, to manipulate expressions involving F and G. Questions arise regarding the handling of terms and the implications of differentiating with respect to temperature.

Discussion Status

The conversation is ongoing, with various participants offering different approaches and suggestions. Some participants express uncertainty about their calculations and seek clarification on specific steps. There is no explicit consensus, but multiple lines of reasoning are being explored, indicating a productive exchange of ideas.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the information they can access or share. There are discussions about the proper application of thermodynamic identities and the need for careful differentiation in the context of the equations being derived.

  • #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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