Thermodynamic Connection (Helmholtz Energy)

[@melevarck]

Thank you very much for the lecture! I have a question regarding the difference in free energy between two states "a" and "b": I've frequently found in textbooks and papers that one possible way of obtaining this value is by calculating it as dA = -kT*ln(Pa/Pb), where Pi is the probability of finding your system in the state "i", which could in principle be readily computed (assuming a simple system with enough sampling) by the number of conformations found in each state during a molecular dynamics simulation. Considering that the free energy doesn't depend on the probability of a specific microstate and actually depends on the partition function itself (which is constant for a given system, so Qa=Qb=Q), how come this computation is correct? I mean, wouldn't the equation A = -kT*ln(Q) imply that there isn't such a thing as a difference in free energy between two microstates from the same macrostate (for instance two different conformations of a molecule)? Or is it the case that when we compute these differences we actually assume that each conformation (a and b) represents their own macrostate (thus Qa =/= Qb) instead of being a microstate from the same system? Thank you very much for your time, sorry for the long question! I'd really appreciate it if you could suggest to me a book where I can find more on this subject as well, I haven't been able to find this discussion.

[@PhysicalChemistry]

You've mostly answered your own question: you do need to treat the two states a and b as their own macrostates in order to calculate their free energies. Maybe here "meso"state would be more appropriate, since it's not a single microstate and not as large as the full macrostate of the entire system. The key to remember (and the thing that makes it confusing) is that any free energies and entropies are collective properties of some group of states. For example: you can calculate the entropy of a two-sided fair coin (P_H = P_T = 1/2). But you can't calculate the entropy of just "heads". Likewise, you can calculate the free energy of a system that can take on a range of microstates -- and compare it to the free energy of the same system with access to a different set of microstates. But you can't calculate the free energy of just a single, maximally specified microstate.

[@melevarck]

@@PhysicalChemistry I see! Thank you very much for the answer, this really clears things up!

[@sebanthomas9361]

why is helmholtz and gibbs energy called free energy?

[@PhysicalChemistry]

These are the forms of energy that include the entropic term (−TS). So they are called "free" energy because they reflect the amount of energy that is free, or available, to do useful work, after also paying the cost of decrease the temperature-energy product (or getting a bonus for increasing it).