Can entropy of universe remain constant in a process where P and T is constant?

We know $$\Delta S(\text{univ})=\Delta S(\text{sys})+\Delta S(\text{surr})$$ at constant $P$, $T$.


$$\Delta S(\text{sys})=\Delta H(\text{sys})/T$$
$$\Delta S(\text{surr})=\Delta H(\text{surr})/T$$


As $\Delta H(\text{univ})=0$, $\Delta S(\text{sys})+\Delta S(\text{surr})=0$
Or $\Delta S(\text{univ})=0$


But we have $\Delta S(\text{univ})>0$ is a must for spontaneous process. That would mean no process is spontaneous at constant $P$, $T$.


What is the fallacy here?




An irreversible process at constant temperature and pressure means that the temperature of the surroundings (assumed ideal) is held at the same temperature T as the initial temperature of the system throughout the process. So, for the surroundings, $$\Delta S\_{surr}=\frac{-\Delta H\_{syst}}{T}$$. For the system, according to the Clausius inequality, $$\Delta S\_{syst}=\frac{\Delta H\_{syst}}{T}+\sigma$$where $\sigma$ is the amount of generated entropy due to irreversibility (always > 0). If we add these two equations together, we obtain: $$\Delta S\_{universe}=\Delta S\_{syst}+\Delta S\_{surr}=\sigma>0$$If the process is carried out reversibly, then $\sigma=0$.


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