# Hemholtz free energy proof

1. Mar 8, 2014

2. Mar 8, 2014

### UltrafastPED

Isothermal work = T is constant!

3. Mar 9, 2014

### 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?

4. Mar 9, 2014

### DrDu

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

5. Mar 9, 2014

### stevendaryl

Staff Emeritus
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.

6. Mar 9, 2014

### DrDu

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

7. Mar 10, 2014

### Nikitin

Ahh, but of course. thank you.