$$\mathrm{d}U<T\,\mathrm{d}S-p\,\mathrm{d}V= \left(\frac{\partial U}{\partial S}\right)_{\!V}\,\mathrm{d}S +\left(\frac{\partial U}{\partial V}\right)_{\!S}\,\mathrm{d}V$$

So for a change to occur at a constant entropy and volume, the internal energy must decrease. I get that this statement isn't useful because you can't control entropy in a lab. Furthermore, given that: $$\mathrm{d}G<V\,\mathrm{d}p-S\,\mathrm{d}T= \left(\frac{\partial G}{\partial p}\right)_{\!T}\,\mathrm{d}p +\left(\frac{\partial G}{\partial T}\right)_{\!p}\,\mathrm{d}T$$ this shows that a spontaneous change at a constant pressure and temperature is accompanied by a decrease in Gibbs free energy – far more useful as these variables can be controlled.

However, since, for a spontaneous process, $\mathrm{d}U$ must decrease at a constant entropy and volume which is expressed by $\mathrm dU<T\,\mathrm dS-p\,\mathrm dV$. However, expressions for $\mathrm dS$ and $\mathrm dV$ can be expressed as a function of $p$ and $T$ making this inequality a function of pressure an temperature. \begin{align} \mathrm{d}U &<T\,\mathrm{d}S-p\,\mathrm{d}V\\ S&=S(p,T)\\ V&=V(p,T)\\ \end{align}

Sub in the total derivatives and factorize: $$\mathrm{d}U<\left[T\left(\frac{\partial S}{\partial T}\right)_{\!p} - p\left(\frac{\partial V}{\partial T}\right)_{\!p}\right]\,\mathrm{d}T + \left[T\left(\frac{\partial S}{\partial p}\right)_{\!T} - p\left(\frac{\partial V}{\partial p}\right)_{\!T}\right]\,\mathrm{d}p$$ Thus, $U=U(p,T)$

This implies that for a spontaneous change at constant pressure and temperature the internal energy must decrease however this is not the case clearly. Does making the substitutions invalidate the inequality? Where is the fault in my reasoning?

So, what's the need for Gibbs free energy. Surely a spontaneous change at constant pressure and temperature is accompanied by a decrease in internal energy? There must be a fault in my reasoning, but where and why?


2 Answers 2


Thanks for the edit - I see now what you meant.

The short answer for why we "need" Gibbs energy is that the internal energy for a spontaneous process at constant temperature $T$ and pressure $p$ does not necessarily decrease. Internal energy is the thermodynamic potential (wiki link) for constant $S$ and $V$ - this makes sense if you think about it:

If the internal energy decreases, then heat must be released by the system, and/or work must be being done on the surroundings. However, real spontaneous processes happen where heat is absorbed and work is not done on the surroundings - the reason is that if the net entropy increase is greater than the enthalpy decrease, the process will still be spontaneous.

The equation for $\mathrm{d}U$ includes entropy terms, but not in a way that reflects spontaneity - the equation for Gibbs free energy does.

The other reason we "need" Gibbs energy is that it also tells us how much "thermodynamically useful" work we can get out of a process.

Mathematical Explanation:

The two inequalities that you started with only represent spontaneous process under very specific conditions:

$$\mathrm{d}U < 0$$

is true for spontaneous conditions only when $S$ and $V$ are constant. The reason for this is that we define spontaneity as the direction in which total entropy of the universe increases - it is at the root of our concept of "spontaneous" vs. "nonspontaneous."

For example, to arrive at this expression for $\mathrm{d}U$, we start with the Clausius inequality:

$$\mathrm{d}S - \frac{\mathrm{d}q}{T} \geq 0$$

This is just a way of saying that the net change of the entropy of the universe must be greater than or equal to zero for any process, and that the net entropy change of the universe is given by the change for the system ($\mathrm{d}S$) plus the change in surroundings, $-\frac{\mathrm{d}q}{T}$.

To turn this into something useful, we make a restriction: constant volume. This lets us assume no $pV$ work, and we can therefore write:

$$\mathrm{d}S - \frac{\mathrm{d}U}{T} \geq 0$$


$$T\,\mathrm{d}S \geq \mathrm{d}U$$

since $\mathrm{d}U = \mathrm{d}q$ at constant $V$.

If we then assume no entropy change for the system, we can also write:

$$\mathrm{d}U \leq 0$$

In other words, the restriction of constant $S$ and $V$ are built-in into the inequality. You can substitute in other thermodynamic variables to get $\mathrm{d}U$ in terms of $T$ and $p$, but it doesn't tell you anything about spontaneity under conditions other than constant $S$ and $V$.

Gibbs energy is useful because the inequality $\mathrm{d}G < 0$ is true for spontaneous processes under constant $T$ and $p$ conditions, which as you mentioned, are easy to achieve in a lab.

The other thermodynamic potentials are derived in similar ways - by holding different variables constant and seeing what happens when change in entropy is greater than zero.

  • $\begingroup$ What do you mean by "The equation for dU includes entropy terms, but not in a way that reflects spontaneity - the equation for Gibbs free energy does."? I have shown that the internal energy can be expressed in terms of pressure and temperature alone. Am I missing the point somewhere? $\endgroup$
    – RobChem
    Jan 7, 2015 at 15:16
  • 4
    $\begingroup$ I think what is happening is that you are assuming that a spontaneous change will be accompanied by a decrease in internal energy - this is only true at constant S and V, not constant T and P. You have an equation for U(T,P), but it won't tell you anything about spontaneity - just how U changes with T and P. I'll update my answer with some math to see if it explains it more clearly. $\endgroup$
    – thomij
    Jan 7, 2015 at 15:22
  • $\begingroup$ Yes please, I am just unsure as to why my equation for U=U(T,p) does not help to indicate spontaneity. $\endgroup$
    – RobChem
    Jan 7, 2015 at 16:02
  • $\begingroup$ I added a more mathematical explanation - I hope this clears it up for you. Also - I noticed a small error in your dU equation - dV in the second term should be dp. $\endgroup$
    – thomij
    Jan 8, 2015 at 0:54
  • $\begingroup$ @thomij how is it possible to get G from this? Or is it more complex? $\endgroup$
    – RobChem
    Jan 8, 2015 at 16:54

This is oversimplifying. Changes in Energy, enthalpy, and entropy, are based on the First Law and are the changes for a fixed amount of reaction, usually one equation unit, from a given state to completion. The Gibbs and Helmholtz free energies combine the opposing effects of energy change and entropy change to give the change between a given state and EQUILIBRIUM.


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