Why is $\Delta G = -T \Delta S_{\mathrm{total}}$ valid only at constant pressure?

Question

According to my book(Elements of Physical Chemistry by Atkins and de Paula, 5th ed.), $\Delta G = - T \Delta S_{\mathrm{total}}$ is valid only for constant pressure and temperature.

Enthalpy is defined as $H = U + PV$. So, when pressure and volume both are not constant, $\Delta H = \Delta U + \Delta (PV)$. So, enthalpy can be defined even if the pressure is not constant.

Now the criterion for spontaneity is that $\Delta S_{\mathrm{total}} > 0$. But $$\Delta S_{\mathrm{total}} = \Delta S_{\mathrm{sys}} - \frac{\Delta H}{T}$$ where $T$ is the temperature of the surroundings. Now, if any heat enters the surroundings, the temperature of the surroundings does not change provided that the surroundings are large. I think that due to this the temperature of the system has essentially to be the same or infinitesimally greater/lesser than the surrounding temperature as the heat transfer has to take place reversibly.

But now, as $G = H - TS_{\mathrm{sys}}$, $$\Delta G = \Delta H - T\Delta S_{\mathrm{sys}}.$$

Thus, from comparison of equations, $\Delta G = -T \Delta S_{\mathrm{total}}$. So, I did not need to specify that pressure is constant and why does my book do so?

Answer 1

You have actually assumed in your equations that both pressure and temperature are constant. Firstly, we have that

$$\mathrm{d}S_\text{total} = \mathrm{d}S_\text{system} + \mathrm{d}S_\text{surroundings}.$$

By definition, differential of entropy is

$$\mathrm{d}S = \frac{\delta q_\text{rev}}{T}.$$

From the first law of thermodynamics, and the definition of entalphy, for a reversible process

$$\mathrm{d}H = \mathrm{d}U + P\mathrm{d}V + V\mathrm{d}P= \delta q_\text{rev} + \delta w + P\mathrm{d}V + V\mathrm{d}P.$$

When all work done is to expand, then $\delta w = -PdV$, and thus

$$\mathrm{d}H = \delta q_\text{rev} -P\mathrm{d}V+ P\mathrm{d}V + V\mathrm{d}P = \delta q_\text{rev} + V\mathrm{d}P.$$

Only if pressure is constant, do we get the relationship

$$\mathrm{d}H \overset{\mathrm{d}P=0}{=} \delta q_\text{rev} \to \mathrm{d}S = \frac{\mathrm{d} H}{T}.$$

Therefore,

$$\mathrm{d}S_\text{total} = \mathrm{d}S_\text{system} - \frac{\mathrm{d} H}{T}\tag{constant pres., exp. work}.$$

Additionally for constant temperature, or $\mathrm{d}T = 0$,

$$\mathrm{d}G = \mathrm{d}H - S_\text{system}\mathrm{d}T - T\mathrm{d}S_\text{system} \overset{\mathrm{d}T = 0}{=} \mathrm{d}H - T\mathrm{d}S_\text{system}.\tag{constant temp.}$$

Hence the result $$\mathrm{d}G = -T\mathrm{d}S_\text{total}$$ is true when pressure and temperature are constant (in addition to possible other constraints).

---

It is probably not the best practice on my part to write things like

$$\mathrm{d}H = \delta q_\text{rev} $$

which has the impression of sloppy (but probably not wrong) notation. Interpret this as that, given our assumptions, the number $1$ becomes an integrating factor for $\delta q$, equivalent to the inexact differential becoming exact under special conditions.

A better option might be to write this expression with definite integrals, i.e.,

$$\int_{\text{state 1}}^{\text{state 2}} \delta q_\text{rev} = \int_{H_1}^{H_2}\mathrm{d}H.$$