3
$\begingroup$

I've been working through the derivation of quantities like Gibb's free energy and internal energy, and I realised that I couldn't easily justify one of the final steps in the derivation.

For instance, in deriving the formula for Gibb's Free Energy, we first found the differential equation:

$$dG = dH - TdS$$

which has the property that, for spontaneous processes, $dG \leq 0$. We then went from there to defining the state function:

$$G = H - TS$$

and claimed that this had the analagous property that $\Delta G\leq0$ for all spontaneous processes. Apparently we can reason this way because the second equation can be obtained from the first by integration. But I'm not entirely sure of this. For instance, temperature is not necessarily independent of entropy, so I'm not convinced that $TS$ must be the integral of $TdS$. Physically I'm not convinced because the derivative refers to small changes at constant temperature, while the state function applies at all temperatures.

Wikipedia's Gibbs free energy page said that this part of the derivation is justified by 'Euler's Homogenous Function Theorem'. Now, I've done some work with ODE's before, but I've never seen this theorem, and I've been having trouble seeing how it applies to the derivation at hand.

If anyone can shed any light on the matter or recommend some further reading I'd appreciate it. Thanks

Improve this question
$\endgroup$
3
  • $\begingroup$ A quick clarification: it's an oversimplification to say that $\Delta G$ < 0 for spontaneous processes. The equilibrium constant for the process is exp(-$\Delta G/RT$), so if $\Delta G$ is negative the process goes almost to completion. $\endgroup$
    – John Rennie
    Mar 20, 2012 at 8:38
  • $\begingroup$ I don't see in which part of Wikipedia article it is stated that for integrating $G$ as you want you need Euler's theorem. You just integrates between initial and final states directly, as Gibbs free energy is defined for isothermal and isobaric processes. Entropy depends on the temperature, but as long as it is constant you can integrate without any problem. $\endgroup$
    – Dani
    Mar 20, 2012 at 8:55
  • $\begingroup$ @DaniH Yes, I may have made a mistake here. Euler's theorem is invoked in the integration of the internal energy formula on this page, and I, perhaps incorrectly, extrapolated this logic to the similar - looking gibbs-free energy derivation. $\endgroup$
    – tom
    Mar 20, 2012 at 10:57

3 Answers 3

Reset to default
2
$\begingroup$

In deriving the Gibbs free energy, the system is assumed to be in contact with a heat reservoir that maintains its temperature and pressure at constant values. This is why the integral of $TdS$ is simply $TS$ in the derivation, and the correct conclusion from the integration is that $\Delta G\le 0$ for all spontaneous processes that take place at constant temperature and pressure. (Or just constant temperature in the case of the Helmholtz free energy.) This important caveat is often not emphasised enough.

For systems not connected to a heat bath (i.e. isolated systems), finding the minimum of a free energy function is no longer the correct procedure. One must instead maximise the entropy. In fact, minimising the Gibbs free energy of a system at constant temperature and pressure is just equivalent to maximising the total entropy of the system and its environment.

Improve this answer
$\endgroup$
1
$\begingroup$

An elementary, mathematically precise derivation of the whole thermodynamic formalism on 17 pages is given in Chapter 7: Phenomenological thermodynamics of my book Classical and Quantum Mechanics via Lie algebras.

You can read this chapter completely independent of the rest of the book. In particular, you don't need any knowledge of quantum mechanics or Lie algebras to read that chapter.

Thus you don't be deterred by the title of the book!

Improve this answer
$\endgroup$
0
$\begingroup$

Euler's theorem for homogeneous functions states that $f(x)$ is an homogeneous function of degree $k>0$

$f(\{\lambda x_i \})= \lambda^k f(\{x_i\})$

for $i=1,\dots,N$ iff

$\mathbf{x} \cdot \nabla f(\{x_i\}) = k f(\{x_i\})$ [1]

Euler's homogeneous function theorem allows you the integration of differential quantities when your differentials correspond to infinitesimal extensive quantities.

First notice that your definition of $dG$ is not the most general, as the term $dN$ has already been dropped. This means that in your derivation you are working just with closed systems which do not interchange particles across their boundaries. Since the term associated to the natural intensive variable has vanished you can integrate using [1] for $k=1$ and get

$G=H-TS$

Hence, you get the thermodynamical function from the differentials although this argument is not general as we have ruled out the $\mu dN$ term.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.