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 P: 5,462 Confused about thermodynamic quantities and their minimums. Not quite sure of your symbols here. E is not normally used for enthalpy. Assuming you are using F for what is normally called the Helmoltz free energy (or work function) and H for enthalpy, equation 38 should read. G = H - TS = F + PV. since there is some doubt as to the meaning of E I shall avoid it and use U for internal energy. So for a change at constant temperature along a reversible path dU = TdS + dwrev (Gibbs equation) At constant temp TdS = d(TS) so we can rewrite the above as d(U-TS) = dwrev (at constant T) Let F = A-TS then dF = dwrev (at const T) Now the expression wrev refers to all possible kinds of work. If only pressure/volume work is envisaged dwrev = PdV At const volume dV = 0 so dF = 0 so F is a minimum, at equilibrium. So the condition for equilibrium at constant temp and volume is that the Helmholtz free energy is a minimum and dF = 0. Now to develop a similar discussion for constant pressure Note that H = U + PV dH = dU + PdV + VdP Substitute from the above dH = TdS + dwrev + PdV + VdP or at constant T and P d(H-TS) = PdV + dwrev This new function is called the Gibbs free energy and is G = H - TS = F+PV So dG = PdV + dwrev Again restricting this to PV work dwrev, dwrev = -PdV So at const T and P dG = 0, Thus G is a minimum, at equilibrium. Now applying all that to my melting ice example. dG= 0 = d(H-TS) = dH - d(TS) = dH - TdS dS = dH / T The entropy of melting (fusion) can be calculated by dividing the enthalpy (=latent heat) by the melting point. You might like to look at this thread and in particular post#11 (Which no longer contains a typo) http://www.physicsforums.com/showthr...lmholtz&page=2 does this help?