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I have been given a problem to derive a relation for change in entropy from change in Gibbs free energy:

Starting with the isothermal equation

$$G(p_2) = G(p_1) + nRT\ln\frac{p_2}{p_1}\label{eqn:1}\tag{1}$$

derive

$$S(V_2) = S(V_1) + nR\ln\frac{V_2}{V_1}\label{eqn:2}\tag{2}$$

using the relation

$$\mathrm dG = V\,\mathrm dp - S\,\mathrm dT.\label{eqn:3}\tag{3}$$

The derivation wants you to assume that pressure is constant and thus $\mathrm dp = 0,$ and \eqref{eqn:3} can be arranged to

$$\left(\frac{\partial G}{\partial T}\right)_p = -S.\label{eqn:4}\tag{4}$$

Now I understand that then differentiating \eqref{eqn:1} w.r.t $T$ and using the fact $p_1 V_1 = p_2 V_2$ (for constant $T)$ can produce the required equation \eqref{eqn:2}.

However, I don't understand quite how we can justify the derivation of an isobaric equation from an isothermal one as this method appears to. Equation \eqref{eqn:1} describes isothermal processes and is a function of pressure, $p_2.$ How is it valid then to suddenly say now pressure is constant and therefore use \eqref{eqn:4} when in equation \eqref{eqn:1} the pressure is clearly not constant?

Might it have something to do with Gibbs free energy being a state function meaning, once there is a relationship for it, constraints (for different state variables) can be changed to derive new relationships using a different constraint?

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  • $\begingroup$ Who says you are assuming that the pressure is constant? $\endgroup$ – Chet Miller Feb 24 at 0:16
  • $\begingroup$ @ChetMiller The answer to the question says it. Is it wrong? $\endgroup$ – user246795 Feb 24 at 9:16
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For an ideal gas at constant temperature $\Delta H=0$, so $\Delta G=-\Delta (TS)$, and, at constant temperature, $$\Delta G=-T\Delta S$$

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  • $\begingroup$ Is the derivation set out in the question flawed then? I know there are many ways to derive the equation, I'm just unsure about the method behind the original post derivation. $\endgroup$ – user246795 Feb 24 at 9:21
  • $\begingroup$ I don't follow the derivation set out in the question. $\endgroup$ – Chet Miller Feb 24 at 12:27
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Maybe, what they expect you to do is to write: $$-\left(\frac{\partial S}{\partial P}\right)_T=\frac{\partial ^2 G}{\partial P\partial T}$$From Eqn. 1, this gives, for an ideal gas $$-\left(\frac{\partial S}{\partial P}\right)_T=\frac{nR}{P}$$

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