# Gibbs free energy can be expressed as a function of P,T and n but are enthalpy and internal energy also (partially) functions of n?

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

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?

• 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. Jan 9, 2015 at 19:17
• 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. Jan 9, 2015 at 19:36
• 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. Jan 9, 2015 at 20:05
• 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? Jan 9, 2015 at 20:18
• 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. Jan 9, 2015 at 20:30

• 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. May 2, 2015 at 14:06