# Thermodynamics ($dE<0, dH<0, dG<0$ )

I didn't understand the derivation of why Gibbs free energy becomes negative when change is spontaneous.

Our chemistry teacher's derivation was like this.

$$dE = TdS - PdV \leq TdS - P_{ext} \cdot dV$$ (equality holds when change is reversible)

Then for constant $$S$$ and $$V$$, $$dE \leq0$$.

But as $$dE$$ always equals to $$TdS - PdV$$(between two any equilibrium states we can always choose a reversible process), isn't $$dE$$ always zero when $$dS, dV = 0$$?

$$dE = TdS - PdV = 0\ \text{for constant }S, V$$ $$dH = TdS + VdP = 0\ \text{for constant }S, P$$ $$dG = VdP - SdT = 0\ \text{for constant }P, T$$

He said that this equality holds when the state is in equilibrium, and when change is irreversible then $$dE<0$$, $$dH<0$$, and $$dG<0$$ because of Clausius inequality. But I can't understand this logic also.

Is this derivation right?

And can you explain how to derive $$dG < 0$$ when change is spontaneous?

• It seems to me that you are right and your teacher is wrong. – Chet Miller Nov 17 '18 at 12:29

But I can't understand this logic also.

Of course you can't. Nobody can.

$$\let\D=\Delta$$ First of all, it's meaningless to write infinitesimal changes for an irreversible transformation, where you may not even trust thermodynamical quantities are defined for the system (which is out of equilibrium).

Let's see the right argument for $$G$$ (others are analogous). $$G$$ is important for chemists since many chemical reactions take place at constant external $$T$$ and $$P$$. It's only needed to consider initial and final states, when we may assume our system is in equilibrium with external $$T$$ and $$P$$. I will use subscript $$_1$$ for initial state, $$_2$$ for the final one.

By definition $$G = U - T\,S + P\,V.$$ Initial state: $$G_1 = U_1 - T\,S_1 + P\,V_1.\tag1$$ Final state: $$G_2 = U_2 - T\,S_2 + P\,V_2\tag2$$ ($$T$$ and $$P$$ are the same, equal to external ones).

Subtracting (1) from (2) $$\D G = \D U - T\,\D S + P\,\D V$$ (I've used $$\D$$ to mean variation: $$\D G = G_2 - G_1$$ and so on).

Clausius' inequality says $$T\,\D S \ge Q$$ (equal it reversible). Then $$\D G \le \D U - Q + P\,\D V = 0.$$ Note that $$\D U = Q - P\,\D V$$ (first principle) as $$-P\,\D V$$ is the work done on the system. In computing work you must always use the external pressure. For instance, in an expansion into vacuum work vanishes because external pressure is zero, even if expanding gas could be thought of having some pressure.

• ΔG=ΔU−TΔS+PΔV (for this equation, P = internal pressure according to the definition of Gibbs free energy right?). But you said that ΔU−Q+PΔV = 0, but isn't it false if the process isn't reversible(because if process is irreversible, work = -Pext*ΔV which doesn't equal to -PΔV ). – 이재용 Nov 17 '18 at 16:01
• Is P constant for the whole change? – 이재용 Nov 17 '18 at 16:01
• "$\Delta G=\Delta U - T \Delta S + P \Delta V$ (for this equation, $P$ = internal pressure according to the definition of Gibbs free energy right?)." Right. But note that I did apply this only to initial and final states, not at all along transformation. It would be impossible, because in an irrev trasf all those quantities could even be undefined. Please read more carefully what I wrote. The above also applies to your second comment. – Elio Fabri Nov 17 '18 at 16:12
• TΔS≥Q. This holds when T = constant because original equation is TdS>= dQ right?? And T is internal temperature. – 이재용 Nov 18 '18 at 11:58
• You write "This holds when $T$ = constant because original equation is $T\,dS \ge dQ$ right?? And $T$ is internal temperature." Sorry, you're all wrong. Clausius' inequality can't be written as you do. First, do not use $dQ$, since $Q$ is not a function of thermodynamic state. Some try to save one's souls writing $\delta Q$, but this is a poor remedy. Be it $d$ or $\delta$, it always conveys the idea of a variation, an increment, and this just what cannot be said about heat. Since bodies do not "possess" some heat, it cannot be changed. – Elio Fabri Nov 18 '18 at 16:29