What is the necessary and sufficient condition for a mixture to be ideal?

I am trying to understand some concepts from solution thermodynamics related to ideal mixtures and fugacity. **My main question is what is the necessary and sufficient condition for a mixture to be an ideal mixture?** Below is my attempt to answer this and some related follow up questions.


The fugacity of component $i$ in a mixture is typically defined by two requirements:


1. $f\_{i}(T,p,\{x\})=p\_{i}\exp\left(\frac{1}{RT}\left[\mu\_{i}(T,p,\{x\})-\mu\_{i}^{IGM}(T,p,\{x\})\right]\right)$
2. $\lim\_\limits{p \to 0}\left(\frac{f\_{i}(T,p,\{x\})}{p\_{i}}\right)=1$


Where IGM is an ideal gas mixture. Using the first part of the definition, we can rewrite the second condition as:
$$\lim\_\limits{p \to 0}\left( \mu\_{i}(T,p,\{x\})-\mu\_{i}^{IGM}(T,p,\{x\}) \right)=0$$


From this we can derive an expression for fugacity that we can calculate explicitly:
$$f\_{i}(T,p,\{x\})=p\_{i}\exp\left(\frac{1}{RT} \lim\_\limits{p’ \to 0} \int\_{p’}^{p} \left[ \bar{V}\_{i}(T,p,\{x\}) - \frac{RT}{p} \right] dp \right)$$


And from this we can show:
$$\frac{f\_{i}(T,p,\{x\})}{x\_{i} f\_{i}(T,p)}=\exp\left(\frac{1}{RT} \lim\_\limits{p’ \to 0} \int\_{p’}^{p} \left[ \bar{V}\_{i}(T,p,\{x\}) - \bar{V}\_{i}(T,p) \right] dp \right)$$


An ideal mixture is defined as a mixture where the chemical potential for any component $i$ is given by:
$$\mu\_{i}^{IM}(T,p,\{x\})=\mu\_{i}(T,p)+RT\ln x\_{i}$$


All other properties of an ideal mixture follow from this definition, since the chemical potential is equal to the partial molar Gibbs free energy. One key property is that $\bar{V}\_{i}(T,p,\{x\}) = \bar{V}\_{i}(T,p)$ at all conditions in an ideal mixture. That is, the volume change of mixing is zero. In turn this means the integrand above is always zero and the expression reduces to the Lewis-Randall rule:
$$f\_{i}^{IM}(T,p,\{x\})=x\_{i}f\_{i}(T,p)$$


From the first part of the definition of fugacity we can also express the chemical potential as:
$$\mu\_{i}(T,p,\{x\})=\mu\_{i}(T,p) +RT \ln \left( \frac{f\_{i}(T,p,\{x\})}{f\_{i}(T,p)} \right)$$


Then if we substitute the LR rule we obtain the definition of an ideal mixture:
$$\mu\_{i}^{IM}(T,p,\{x\})=\mu\_{i}(T,p) +RT \ln x\_{i}$$


In summary, the two requirements below seem to ensure that a mixture is ideal:


1. $\bar{V}\_{i}(T,p,\{x\}) = \bar{V}\_{i}(T,p)$
2. $\lim\_\limits{p \to 0}\left( \mu\_{i}(T,p,\{x\})-\mu\_{i}^{IGM}(T,p,\{x\}) \right)=0$


I have two follow up questions:


* Is this actually true? Does the volume condition always ensure the mixture is ideal? Or are there ideal mixtures where the volume of mixing is nonzero (i.e. $\bar{V}\_{i}(T,p,\{x\}) \ne \bar{V}\_{i}(T,p)$)?
* Is the logic of this argument backwards? I don’t understand the
mathematical reason why we are free to define fugacity with its
second condition. If we didn’t assume that then the volume condition
would not be enough to derive the the chemical potential definition
of an ideal mixture. Is the fugacity limit requirement actually some
kind of consistency condition due to the definition of an ideal
mixture?


