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

[Nikitin]

http://en.wikipedia.org/wiki/Helmho...Isothermal_Process_and_Equilibrium_Conditions

Hi. In the proof that dw <= -dA for isothermal work, where is it assumed T must be constant during the whole process?

 

[UltrafastPED]

Isothermal work = T is constant!

 

[Nikitin]

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?

 

[DrDu]

I also don't see why constant temperature should be necessary.

 

[stevendaryl]

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.

 

[DrDu]

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.

 

[Nikitin]

Ahh, but of course. thank you.