# How can The expression $dH = TdS + Vdp$ hold for non-reversible process?

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 $$0 which leads to $$dH < TdS$$, which contradicts (1).

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.
• 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? Apr 24 at 18:29
• 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 $U=ST+\mu N -PV$ Apr 24 at 18:51