Why is dG = dH − TdS?

Question

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?

Answer 1

Why is dG=dH-TdS?

It isn't. That would only be true at constant temperature. In general:

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

Answer 2

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