Can we show that the helmholtz free energy at equilibrium is minimized from its second...

kislotd

kislotd

Answered

2022-07-14

Can we show that the helmholtz free energy at equilibrium is minimized from its second derivative?

Answer & Explanation

Damarion Pierce

Damarion Pierce

Expert

2022-07-15Added 11 answers

The first derivative at an extremum point x 0 is 0 only when evaluated at x 0 . Same goes with the second derivative. So you first have to evaluate the second derivative, then you plug in the specific x = x 0
For example, take y = x 2
d y d x = 2 x and d 2 y d 2 x = 2
At x = 0, the minimum, d y d x | 0 = 0, and d 2 y d 2 x | 0 = 2
If we were to apply your logic however, we would have d 2 y d 2 x = d d x ( d y d x ) = 0 = 0.
So, you have to expand your last line.
d 2 F = d P d V P d 2 V d S d T S d 2 T .
Now you apply the condition that you are at equilibrium, so d T | e q = d V | e q = d P | e q = d S | e q = 0, so that:
d 2 F | e q = P d 2 V | e q S d 2 T | e q .
Then I guess that if pressure is positive, then the volume is a maximum so d 2 V < 0 so that the first term is positive.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?