# What would be the physical interpretation of the equation dG = Vdp − SdT?

I was trying to understand the derivation of the formula $$G=G^\circ+RT\ln (p/p^\circ)$$ starting with the following relationships:

\begin{align} \mathrm dU&=T\,\mathrm dS-p\,\mathrm dV\tag{1}\\ \mathrm dH&=\mathrm dU+p\,\mathrm dV+V\,\mathrm dp=T\,\mathrm dS+V\,\mathrm dp\tag{2}\\ \mathrm dG&=\mathrm dH-T\,\mathrm dS-S\,\mathrm dT=V\,\mathrm dp-S\,\mathrm dT\tag{3} \end{align}

Under isothermal condition $$\mathrm dG=V\,\mathrm dp$$ $$(\mathrm dT=0)$$. Therefore, the change in $$G$$ from the standard state is given by:

\begin{align} \int\mathrm dG&=\int V\,\mathrm dp=\int\frac{nRT}p\,\mathrm dp\tag{4}\\ G-G^\circ&=\ln p-\ln p^\circ\tag{5}\\ G&=G^\circ+\ln\frac p{p^\circ}\tag{6} \end{align}

But I'm confused about the equation $$\mathrm dG=V\,\mathrm dp-S\,\mathrm dT$$. What does this physically mean?

I'm assuming $$V\,\mathrm dp$$ is some form of work done by the system, but that suggests $$G$$ of the system increases while it loses energy doing the work. I also have the same problem for $$\mathrm dH=T\,\mathrm dS+V\,\mathrm dp.$$ Can someone please explain what they really mean?

• "I'm assuming 𝑉𝑑𝑃 is some form of work done by the system, but that suggests 𝐺 of the system increases while it loses energy doing the work." You have it backwards. G decreases when the system does pV work isothermally: If the system is doing pV work, p is decreasing, which means dp is negative. And since V is always positive, VdP would likewise be negative. – theorist Mar 6 at 23:31
• @theorist The fact that V is always positive doesn't say much about the change in V, and the change in V should be considered, $\delta V$ can be positive or negative depending on the initial and final volumes of the process, here it looks like you are considering V constant which is not always the case for VdP changes. The overall change in the sign of VdP will depend on the V dependence wrt the P and dP together solving the integral W = - \int pdV for any given process. – Ernek Mar 12 at 0:05
• @Ernek Nope. You're confusing the sign of the $p$-dependence of $V(p)$, namely $\frac{dV(p)}{dp}$, with the sign of $V(p)$ itself. If $dp<0$, then $V(p) dp<0$, always. This is independent of the sign of $\frac{dV(p)}{dp}$ (which, incidentally, must always be negative except for certain v. artifical cases). [Also, I am not considering $V$ constant. I simply didn't make the $p$-dependence explicit in my last comment, since it's obvious $V$ is $p$-dependent, and strongly so for the gas-based system described by the OP; I added the explicit $p$-dependence above to make it clearer for you.] – theorist Mar 12 at 7:12

For a continuous flow system operating at steady state in contact with a constant temperature reservoir at the same temperature as that of the entering fluid, and with negligible change in kinetic energy and gravitational potential energy, the reversible shaft work per mole that the system does on the surrounding is equal to the integral of -vdP, where v is the fluid molar volume. So $$\Delta G$$ per mole is equal to minus the shaft work per mole.

• I see what I was misunderstanding. Just out of curiosity - does SdT also refer to some form of work? – taichiii Mar 7 at 13:31
• I don’t think so. – Chet Miller Mar 7 at 15:36
• The SdT term is normally associated with changes in heat ( dq ) . You can think of it as another way of transferring energy - similar to work (i.e also a mean of transfering energy) but with a different microscopic connotation. – Ernek Mar 11 at 23:58

The sign convention you use for work 'done by' and work 'done on' the system is not correct when you attempt to interpret the $$\Delta G$$ equation. The first law is:
$$U = W + q$$ $$dU = dW + dq$$ $$dW$$ in this context is work done 'on the system' and is associated with an external pressure - (not the internal pressure of the gas)
[-> Consider an adiabatic process ($$dq = 0$$). Only by transfering energy from the surroundings to the system ($$dW > 0$$ ) there is an increase in the internal energy of the sytem $$dU > 0$$]
Now, $$PV$$ work is associated with changes in $$P$$ and/or $$V$$ only which means: $$dW = \Delta(P \cdot V)$$

If, for example, the process is at constant Pressure, we get the common expressions:

$$dW_{on-the-system} = -\int P_{external}dV$$ $$dW_{by-the-system} = - W_{on-the-system} = \int P_{ext}dV$$

Only if the system is changing reversibly you can equal the magnitude of $$P_{ext}$$ and $$p_{internal}$$ in a process ($$P_{ext} \sim p_{internal}$$). Now, if the system is expanding (i.e gas pushing outwards), $$dW_{by-the-system} = \int p_{intern}dV > 0$$ ( system doing work) and $$dW_{on-the-system} = - \int p_{intern}dV < 0 \rightarrow dW < 0$$

Henceforth in $$dG = Vdp - SdT$$ , $$\textbf{Vdp}$$ is directly associated with work done 'on the system' ( and not 'by the system' as you state above) with pressure in $$dp$$ directly associated with the internal pressure (keeping track of the sign convention we used from the begining of the derivation).

Thus , if your system is at constant T and there is a positve change in $$Vdp$$ (i.e your equation for an ideal gas: $$\ln \frac{p}{p_0} > 0$$ for $$p_f > p_0$$ ) means that the positive work was done 'on the system', energy was transfered from the surroundings to the system, hence why the system Gibbs Free energy increased. If your $$p_f$$ would have been smaller than $$p_0$$ in $$\ln \frac{p}{p_0}$$ then $$Vdp < 0$$ and now is the system the one that is doing the work $$\rightarrow \Delta G < 0$$ and "loosing" that energy

You can derive the same reasoning using the correct sign convention to interpret energy changes associated with $$\Delta H$$ changes.