Minimizing Gibbs free energy with Helmholtz free energy expressions

Question

If I have an expression for the Helmholtz free energy (from statistical associating fluid theory for a polymer system), can I still find the Gibbs free energy minimum under constant $T,P$ conditions?

Seems that since

$\mu_i = \left(\frac{\partial A}{\partial n_i}\right)_{T,V,n_j} = \left(\frac{\partial G}{\partial n_i}\right)_{T,P,n_j}$

where $A$ is the Helmholtz free energy and $G$ is the Gibbs free energy and $\mu_i$ is the chemical potential of component $i$.

If I can just find the expressions for $\mu_i=\left(\frac{\partial A}{\partial n_i}\right)_{T,V,n_j}$ I should be able to use this in the expression for $G = \sum_i \mu_i n_i$ and find its minimum, without having to worry about an equation of state to relate change in volume ($dV$) with respect to constant pressure ($P$)?

Edit I should clarify that I wonder about the Gibbs free energy because I'm interested in the constant $T,P$ (isothermal, isobaric case), and in the case especially for liquids. If you needed to introduce an equation of state to parameterize the change in V as a function of $T,P$ this would not be too difficult for an ideal gas, but for liquids it would be a challenge. So in this case, I wonder how to handle the $PV$ term (for constant $P$) in the Helmholtz free energy which does not appear in the Gibbs.

Answer 1

In short, I agree with Curt F.'s answer above. In general, you need a suitable equation of state (EoS) to handle this calculation.

From the differential form of the fundamental equation of thermodynamics (i.e. the differential form of the internal energy $U(S,V,N)$)

$$\mathrm{d}U = T\mathrm{d}S - p\mathrm{d}V + \sum_{i=1}^{m} \mu_i \mathrm{d}n_i$$

other relevant thermodynamic potentials may be found using the Legendre transform. Helmholtz free energy is defined as a transformation from $(S\rightarrow T)$ and Gibbs free energy by a transformation from $(S\rightarrow T)$ and $(V\rightarrow p)$

$$ A(T,V,N) \triangleq U - \left(\frac{\partial U}{\partial S}\right)_{V,N} = U - TS \\ G(T,p,N) \triangleq U - \left(\frac{\partial U}{\partial V}\right)_{S,N} - \left(\frac{\partial U}{\partial S}\right)_{V,N} = U + pV - TS = A + pV $$

If an expression for the Helmholtz free energy (that is, $A(T,V,N)$) is known, along with a suitable equation of state (EoS) (e.g. $p = p(T,V,N)$), the Gibbs free energy may be calculated. Without the EoS, it is impossible to evaluate the $pV$-term.

If both the volume and pressure may be assumed to be constant, they will not affect the phase equilibrium calculations. An alternative may be to assume that the liquid phase volume is constant or that it follows some (simple) empirical relation. Thus, the last term may be evaluated and the Gibbs free energy may be minimized.

Answer 2

I don't think you can use the approach you are outlining without an equation of state.

$G = \sum_i \mu_i n_i$ is valid only at constant $T$ and $P$.

$A = \sum_i \mu_i n_i$ is valid only at constant $T$ and $V$.

If $P$ and $V$ are both constant, then you can equate the two and use $G=A= \sum_i \mu_i n_i$. Actually, that assumption is probably not horrible for most liquids at "normal" pressures: I think it amounts to assuming the liquid is incompressible.