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In Klotz, Introduction to Chemical Thermodynamics, Ex. 8.2 requires me to derive $$ dG = V \left( \frac{\partial p}{\partial V} \right)_V dV + \left[ V \left( \frac{\partial p}{\partial T} \right)_V - S \right]dT, $$ where $G$ is the Gibbs free energy, $S$ is the entropy and all the other variables have their ordinary common-sense meaning.

I'd appreciate feedback on my solution, which I find just very simple, maybe too simple to be right.

First of all, when recognizing that $G$ is a function of $V$ and $T$, we can write the total differential as

$$ dG = \left( \frac{\partial G}{\partial V} \right)_T dV + \left( \frac{\partial G}{\partial T} \right)_V dT $$

In Eq. 8.19, we learn that

$$ dG = V dp - S dT $$

and so, when multiplying both sides by $1/\partial V$ at constant temperature, we find

$$ \left( \frac{\partial G}{\partial V} \right)_T = V \left( \frac{\partial p}{\partial V} \right)_T - S \left( \frac{\partial T}{\partial V} \right)_T $$

where $(\partial T /\partial V)_T$ is zero, because we're at constant $T$. A similar procedure yields the bracket expression, this time instead we just multiply both sides with $1/\partial T$ and using constant $V$, we arrive at

$$ \left( \frac{\partial G}{\partial T} \right)_V = V \left( \frac{\partial p}{\partial T} \right)_V - S \left( \frac{\partial T}{\partial T} \right)_V = V \left( \frac{\partial p}{\partial T} \right)_V - S $$

and from comparison with the total differential expression, one immediately recognizes the above very first expression.

Is this a valid solution/Ansatz?

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Comment 1: Thanks for the edit. Comment 2: Can we add equation numbers? – TMOTTM Sep 22 '13 at 18:24

Seems OK to me. Although I would just write $dp$ in $dG=Vdp−SdT$ in terms of $dV$ and $dT$.

EDIT (09/22/2013): In more detail: substitute $dp=(\frac{\partial p}{\partial V})_T dV+(\frac{\partial p}{\partial T})_V dT$ into $dG=Vdp-SdT$.

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I'm not sure I'm following. Can you elaborate what you would do? – TMOTTM Sep 22 '13 at 18:30
@TMOTTM: We start with what you have to proof an rewrite it once, just by setting the brackets differently: $dG = V \left( \frac{\partial p}{\partial V} \right)_V dV + \left[ V \left( \frac{\partial p}{\partial T} \right)_V - S \right]dT $$ $$= V\ \left[ \left( \frac{\partial p}{\partial V} \right)_V dV + \left( \frac{\partial p}{\partial T} \right)_V \ dT \right] - S \ dT$. Username akhmeteli points out that the expression in the bracket is $dp$, and then you're done. – NikolajK Oct 22 '13 at 20:48

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