# Gibbs Free Energy and dependence on temperature and pressure

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?

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

• 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. Sep 19 at 5:49
• 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
– bm27
Sep 19 at 5:56
• 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. Sep 19 at 5:59
• 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.
– bm27
Sep 19 at 6:02
• 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. Sep 19 at 6:03

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.