Is Constant Temperature Required in the Proof of Helmholtz Free Energy?

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Discussion Overview

The discussion revolves around the requirement of constant temperature in the proof of the inequality \( dw \leq -dA \) for isothermal work in thermodynamics, particularly in the context of Helmholtz free energy. Participants explore whether this condition is necessary and the implications of temperature variations during the process.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • Some participants assert that isothermal work implies constant temperature, while others question the necessity of this assumption for deriving the inequality \( dw \leq -dA \).
  • One participant suggests that the inequality can be derived using the first law of thermodynamics and the relation \( dQ \leq TdS \), indicating that it should hold regardless of whether temperature is constant.
  • Another participant points out that if temperature is not constant, the term \(-S dT\) in the differential equality for Helmholtz free energy could vary in sign, complicating the derivation.
  • It is noted that the relationship \( dA = dU - dTS \) simplifies to \( dA = dU - TdS \) only under the assumption of constant temperature.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of constant temperature in the proof, indicating that multiple competing perspectives remain unresolved.

Contextual Notes

The discussion highlights potential limitations in the assumptions made regarding temperature constancy and the implications for the derivation of the inequality, but does not resolve these issues.

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Isothermal work = T is constant!
 
yes, but you could get the equation ##dw \leq-dA## without assuming T=constant anywhere?

I mean, from what I can see, all they do is use the 1st law of thermodynamics and the general fact that ##dQ \leq TdS##. So shouldn't the result they get, ie ##dw \leq-dA##, apply regardless if T is constant or not during the whole process?
 
I also don't see why constant temperature should be necessary.
 
Nikitin said:
yes, but you could get the equation ##dw \leq-dA## without assuming T=constant anywhere?

I mean, from what I can see, all they do is use the 1st law of thermodynamics and the general fact that ##dQ \leq TdS##. So shouldn't the result they get, ie ##dw \leq-dA##, apply regardless if T is constant or not during the whole process?

Well, the differential equality for Helmholtz free energy is:

dA = -S dT -p dV = -S dT - dW

It seems to me that if T is not constant, then the term -S dT could be either positive or negative, depending on whether the temperature is increasing or decreasing.
 
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stevendaryl said:
Well, the differential equality for Helmholtz free energy is:

dA = -S dT -p dV = -S dT - dW

It seems to me that if T is not constant, then the term -S dT could be either positive or negative, depending on whether the temperature is increasing or decreasing.

You are right. The point is that in general, dA =dU -dTS. This only reduces to dA=dU-TdS if T is constant.
 
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Ahh, but of course. thank you.
 

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