Why is dG = dH − TdS?

I get that $G=H-TS$ because then: $$\begin{align}\mathrm dG&=\mathrm dH-T\,\mathrm dS-S\mathrm dT\\&=T\,\mathrm dS+V\,\mathrm dp-T\,\mathrm dS-S\,\mathrm dT\end{align}$$
Therefore, by cancelling: $\mathrm dG=V\,\mathrm dp-S\,\mathrm dT$ which is the equations for $\mathrm dG$. However, I can’t get this result from using $\mathrm dG=\mathrm dH-T\,\mathrm dS$.


$$\mathrm dH=S\,\mathrm dT+V\,\mathrm dp$$
$$\mathrm dG=\left(S\,\mathrm dT+V\,\mathrm dp\right)-T\,\mathrm dS$$


This does not give me the right equation. Where am I going wrong?





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> Why is dG=dH-TdS?
> 
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It isn't. That would only be true at constant temperature. In general:


$$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S - S\,\mathrm{d}T$$


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If you write $\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S$, you drop off one term of the differentiation of $G$ which is $S\mathrm{d}T$.
By definition $G = H - TS$. So, $$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S-S\,\mathrm{d}T$$
Then, you should write $\mathrm{d}H = T\,\mathrm{d}S + V\,\mathrm{d}p$ (not as you have written $\mathrm{d}H = S\,\mathrm{d}T + V\,\mathrm{d}p$ ).


Finally, $$\mathrm{d}G = T\,\mathrm{d}S + V\,\mathrm{d}p - T\,\mathrm{d}S - S\,\mathrm{d}T$$
 $$\mathrm{d}G = V\,\mathrm{d}p - S\,\mathrm{d}T$$


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