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

Question

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?

Answer 1

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 if 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.