1
$\begingroup$

$$G=G(P,T,n)$$ $$\mathrm dG=V\,\mathrm dP-S\,\mathrm dT+\mu\,\mathrm dn=\left(\frac{\partial G}{\partial P}\right)_{T,n}\,\mathrm dP+\left(\frac{\partial G}{\partial T}\right)_{P,n}\,\mathrm dT+\left(\frac{\partial G}{\partial n}\right)_{T,P}\,\mathrm dn$$

This allows open systems to be considered (where $\mathrm dn$ does not equal zero). However, can enthalpy, internal energy and Helmholtz free energy also be treated in this way to allow for open systems?

$\endgroup$
  • 1
    $\begingroup$ I'm not really sure what you are asking. Do you mean, is it as simple as adding a $(\frac {\partial X}{\partial n})dn$ term, where $X$ is the state function of interest? In that case no .. for example, Helmholz free energy is defined as the useful work that can be obtained from a closed thermodynamic system at constant T. The definition of Gibbs free energy includes the open nature of the system, and the state function has the corresponding "natural" thermodynamic variables, P,T and n. $\endgroup$ – dtmoore1971 Jan 9 '15 at 19:17
  • 1
    $\begingroup$ Because that's specifically how it is defined. Also, it's not that enthalpy (and other function) don't allow for open systems to be considered .. the enthalpy is part of the Gibbs free energy, after all. It's more that open systems are not part of the natural definitions of those other state functions. You can define the exact differential of helmholz free energy for an open systems .. you just add $\sum_i\mu_idn_i$, so the definition is still in terms of chemical potential, which relates to changes in the Gibbs free energy with composition. $\endgroup$ – dtmoore1971 Jan 9 '15 at 19:36
  • 1
    $\begingroup$ What I mean is that the derivation of the Gibbs free energy was specifically done to include the effects of composition changes .. this is not true of the other functions. See here for a short explanation of how the natural thermodynamic variables are obtained. Regarding the other point, I understand .. that's why my original comment said "no, you can't just do that" ... you can still account for open systems, you just can't do it that way. $\endgroup$ – dtmoore1971 Jan 9 '15 at 20:05
  • 1
    $\begingroup$ Thank you. Although it does show that one can do it for internal energy. Also, searching enthalpy in wikipedia I was able to find the expression that involved the $\sum_i\mu_idn_i$ term. With all due respect, are you absolutely sure you are right? $\endgroup$ – RobChem Jan 9 '15 at 20:18
  • 1
    $\begingroup$ I am not sure why you seem to find my comments inconsistent. I have said that you CAN account for the effects of open systems on thermodynamic state functions by adding $\sum_i\mu_idn_i$ to the exact differential. I have also said (although not directly) that $\mu_i=(\frac {\partial G}{\partial n_i})_{T,P}$ is always the definition of the chemical potential. I said you CANNOT add the analogous quantity $\sum_i(\frac {\partial A}{\partial n_i})_{T,V}dn_i$ to the exact differential of Helmholz free energy (for example) to allow for open systems. Yes, I am sure that is all correct. $\endgroup$ – dtmoore1971 Jan 9 '15 at 20:30
2
$\begingroup$

Yes, you can define chemical potentials as partial molar enthalpies, or partial molar internal energies, or partial molar Helmholz free energies. (Sorry for my initial confusion on this point ... see also my answer on your other question on this topic, here).

The definitions as partial molar enthalpies and partial molar internal energies are not particularly useful, since they involve working at constant entropy, which is practically impossible, but they are still mathematically valid.

The definition of chemical potential as partial molar Helmholz free energy at constant temperature and volume may actually have some practical applications, (e,g. perhaps for reactions that evolve gases in closed containers? although I have never come across any.

$\endgroup$
  • $\begingroup$ The definition $\mu_i = \left(\frac{\partial{A}}{\partial{n_i}}\right)_{T,V,n_{j \neq i}}$ is very useful for gases in general - they are frequently described by equations of state (EoS) in (T,V,N)-coordinates (e.g. ideal gas law, SRK etc.), which makes Helmholtz free energy the most suitable thermodynamic potential. A practical example is the calculation of phase equilibria, but there are many others. $\endgroup$ – Kjetil Sonerud May 2 '15 at 14:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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