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- Is [itex]\Delta G[/itex] useful for situations other than constant temperature and pressure?

If this belongs in classical physics, please move it there. But it seems like the kind of question chemistry people would know so I'm putting it here.

I was reading a textbook on chemical thermodynamics, and it says to raise the partial molar Gibbs free energy of [itex]n[/itex] moles a substance from [itex]\overline{G}_1[/itex] to [itex]\overline{G}_2[/itex] you have to do work [itex]W = n(\overline{G}_2 -\overline{G}_1)[/itex]. Sounds reasonable enough. But then they give as an example moving [itex]n[/itex] moles of an ideal gas from a tank at pressure [itex]p_1[/itex] to a higher pressure [itex]p_2[/itex] at constant temperature. This indeed requires compression work [itex]W = nRT(\ln(p_2) - ln(p_1))[/itex], which happens to be equal to [itex]n(\overline{G}_2 -\overline{G}_1)[/itex].

But this seems like kind of a fluke to me. For an isothermal process, it's the change in Helmholtz free energy, [itex]\Delta A[/itex] that measures the work required to bring about the change of state. In this case, [itex]\Delta A = \Delta G[/itex] because we happen to have an ideal gas kept at constant temperature, so the product [itex]pV[/itex] doesn't change. But in general, I thought change in Gibbs free energy was useful only when both temperature and pressure are constant, and you want to know how much non-pV work is required. So it doesn't seem like you'd use it for the case of compressing a gas. Am I wrong?

I was reading a textbook on chemical thermodynamics, and it says to raise the partial molar Gibbs free energy of [itex]n[/itex] moles a substance from [itex]\overline{G}_1[/itex] to [itex]\overline{G}_2[/itex] you have to do work [itex]W = n(\overline{G}_2 -\overline{G}_1)[/itex]. Sounds reasonable enough. But then they give as an example moving [itex]n[/itex] moles of an ideal gas from a tank at pressure [itex]p_1[/itex] to a higher pressure [itex]p_2[/itex] at constant temperature. This indeed requires compression work [itex]W = nRT(\ln(p_2) - ln(p_1))[/itex], which happens to be equal to [itex]n(\overline{G}_2 -\overline{G}_1)[/itex].

But this seems like kind of a fluke to me. For an isothermal process, it's the change in Helmholtz free energy, [itex]\Delta A[/itex] that measures the work required to bring about the change of state. In this case, [itex]\Delta A = \Delta G[/itex] because we happen to have an ideal gas kept at constant temperature, so the product [itex]pV[/itex] doesn't change. But in general, I thought change in Gibbs free energy was useful only when both temperature and pressure are constant, and you want to know how much non-pV work is required. So it doesn't seem like you'd use it for the case of compressing a gas. Am I wrong?