I'm confused about the condition for spontaneity for the Helmholtz energy. My textbook (McQuarrie, "Physical Chemistry") derives the conditions as follows. We start with the combined law of thermodynamics: dU = δq + δw ≤ TdS – PdV since δq/T ≤ dS dU – TdS + PdV ≤ 0 For a process at constant V and T, dU – TdS ≤ 0 since dV = 0 which we can rewrite as d(U – TS) ≤ 0 or dA ≤ 0, where A = U – TS I'm first confused about the inequality in the combined law. For a reversible process, we get dU – TdS + PdV = 0 because δqrev = TdS and δwrev = –PdV. However, I've read that because this leads to an equation where all the variables are state functions, the equality holds for all processes. So when does the inequality occur? Secondly, from the combined law dA can be written as dA = –PdV – SdT. If V and T are constant, then how can dA be nonzero? Wouldn't dV = 0 and dT = 0? Wikipedia has a brief explanation (http://en.wikipedia.org/wiki/Helmholtz_free_energy#Minimum_free_energy_and_maximum_work_principles), but it doesn't make much sense to me. Also, why does only the equality of the combined law apply now (i.e. why not dA ≤ –PdV – SdT)? Thanks in advance for your help!