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?

  • $\begingroup$ What are the temperature and pressure of universe to keep them constant? $\endgroup$
    – Poutnik
    Mar 22, 2022 at 8:02
  • $\begingroup$ @poutnik i don't get what you mean. I have taken P, T for system only, not for universe $\endgroup$
    – Sagnik
    Mar 22, 2022 at 16:40
  • $\begingroup$ It is not about constant pressure or temperature but on reversibility. What ever may be the process the change in entropy for universe is positive. But zero for reversible process. This is popularly called second law of thermodynamics. $\endgroup$
    – Infinite
    Mar 22, 2022 at 16:54
  • $\begingroup$ @Infinite can you elaborate where reversibility comes into play here ? And what is wrong with my proof ? $\endgroup$
    – Sagnik
    Mar 23, 2022 at 3:32
  • 1
    $\begingroup$ The formula that you used ( ΔS = ΔH/T ) is applicable for reversible process but not for irreversible process. And in reality no process is reversible. $\endgroup$
    – Infinite
    Mar 23, 2022 at 14:09

1 Answer 1


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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