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So, I have read that change in Gibbs free energy is defined under constant pressure and temperature, i.e $$\Delta G= \Delta H -T \Delta S $$

but for an ideal gas change in entropy of system can be written as $$\Delta S= nC_pln(T_2/T_1) + nRln(p_1/p_2)$$ this formula implies that change in entropy at constant temperature and pressure is 0. Also for an ideal gas change in enthalpy is only a function of temperature so for an ideal gas at constant temperature and pressure change in enthalpy would be 0. But as I said above $\Delta G$ is defined at constant temperature and pressure, so wouldn't change in gibbs free enregy for an ideal gas always be 0?

PS I have read some related answers but they have not helped me solve this query. Some Read answers - https://physics.stackexchange.com/questions/716364/how-can-the-gibbs-free-energy-equation-be-at-constant-temperature-and-pressure

Why at constant pressure and temperature Gibbs energy change of a process can be negative?

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    $\begingroup$ change in Gibbs free energy is defined under constant pressure G is defined regardless of conditions as G = H - TS, but only at T=const (p=const is not required) it can be written as ΔG=ΔH-Δ(TS)=ΔH-TΔS. $\endgroup$
    – Poutnik
    Sep 19 at 5:49
  • $\begingroup$ But finally for us to write $\Delta G= \Delta H -T \Delta S$ doesn't your statement imply that the conditions should be constant pressure and temperature $\endgroup$
    – bm27
    Sep 19 at 5:56
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    $\begingroup$ No, it does not. G as well as H are especially useful and usually used at isobaric conditions (like A and U at isochoric ones), but the above formula does not have isobaric requirement, being directly derived from G definitions and isothermic requirement. $\endgroup$
    – Poutnik
    Sep 19 at 5:59
  • $\begingroup$ Ok, so the equation is valid for isothermal conditions, I guess many places it is written wrong then, couple books I have, have written it as both constant temperature and pressure. $\endgroup$
    – bm27
    Sep 19 at 6:02
  • $\begingroup$ Well, it is valid for that too. If something is said to be valid for 2 simultaneous conditions, it does not generally imply it cannot be valid for just one of the conditions. $\endgroup$
    – Poutnik
    Sep 19 at 6:03

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There is happening no change in a (not reacting) ideal gas system at isothermic and isobaric conditions.

Therefore, unless no chemical reaction is happening, $ΔG = 0$

Similarly, for reversible processes at isothermic and isobaric conditions: \begin{align} ΔS&=Q/T\\ ΔH&=Q_p\\ ΔG&=ΔH-TΔS=Q_p-Q_p=0 \end{align}

If there is ongoing spontaneous process, like chemical reaction, $ΔG \lt 0$. We then say the process is exergonic. Note that a process need not to be exothermic ($ΔH \lt 0$) to be exergonic, if balanced by the system entropy increase. See e.g. dissolving of $\ce{KNO3}$ or $\ce{KClO3}$, which is endothermic but spontaneous.

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