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From what I have found online the equation $dH = TdS +Vdp$ holds as all the variables are state variables and the derivation of this just requires $dU = TdS - pdV$ which is valid in all processes (assuming that the number of particles is constant). If we assume that the pressure is constant then $dH = TdS$ (1). However, I can see that there is an apparent contradiction with equation 1. I have also been taught that Gibbs free energy is $G = H -TS$ so assuming constant temperature and pressure $dG = dH -TdS $ Now, we know that $G\leq0$ for a spontaneous change and $G= 0$ for a reversible change only, then $dG\leq 0$ so $0 \leq dH -TdS$, as equality holds only for reversible processes, for a non-reversible process $0<dH -TdS$ which leads to $dH < TdS$, which contradicts (1).

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If you are keeping your system at constant pressure and temperature (as usually assumed when someone claims that $\Delta G<0$), and the change of Gibbs free energy is $$ dG=-SdT+VdP+\sum_i\mu_i dN_i, $$ then the Gibbs free energy can change only by changing the number of particles of different species, which you neglected.

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  • $\begingroup$ Ok I get that part, so continuing on the part of the question that asks how can this hold in all cases$dH=TdS+Vdp$, if the pressure is constant then $dH = TdS$, am I right to assume that this differential equation cannot be solved trivially ($\Delta H = T\Delta S$) as the temperature is not constant? $\endgroup$ Apr 24 at 18:29
  • $\begingroup$ The Helmholtz free energy is just defined as $H\equiv U+PV$, and that's why your equation $dH=TdS+VdP$ always holds. You are right about not being able to solve the equation because temperature is a function of entropy at constant pressure. As a side note, sometimes it is also useful to know that $U=ST+\mu N -PV$ $\endgroup$
    – Pavlo. B.
    Apr 24 at 18:51

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