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I don't get why fugacity coefficients, $\phi = f/p$, of pure components are usually calculated via integrating an eos over a pressure or volume range. For example, when using a pressure explicit eos (such as the Virial-Eos for example), one can write: $$ RT \ln \phi = \int_0^p(v-\frac{RT}{p})dp $$ I was wondering, if we already have the Virial-Coefficients, why not calculate the actual pressure straight away? Isn't the fugacity some kind of "real pressure", with $\phi$ serving as a conversion factor, $f = \phi p$. And isn't the pressure calculated from eos (PR, VdW, Virial etc.) also some kind of "real pressure" aswell. But why are they not equal?

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The pressure calculated from an equation of state is just the pressure (for a real gas, if applicable). One useful definition is $P\equiv\left(\frac{\partial G}{\partial V}\right)_{T,N}$.

The fugacity, in contrast, is an effective pressure that takes chemical interactions into account.

This is analogous to the comparison of concentration vs. the chemical potential, where the latter acts an effective concentration that again incorporates chemical interactions. (We sometimes say that matter flows to eliminate concentration gradients, but matter really flows to eliminate chemical potential gradients.)

Indeed, the chemical potential relative to that at a reference state is $\mu=RT\ln a$, where the dimensionless activity $a$ for a gas is the fugacity divided by a reference pressure. For ideal mixtures, the activity is the concentration (specifically, the molar fraction). For ideal gases, the fugacity is the pressure.

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    $\begingroup$ But using a real gas eos does also consider molecular interactions via the interaction parameter. E.g. the VdW-eos uses the parameter "a" to account for attractive interactions. So why should the fugacity, obtained from the VdW-eos, yield a better or different pressure? Your statement about equal chemical potentials as a requirement for chemical equilibrium is true. However, i don't get why it is analogous to pressure/fugacity example, since mechanical equilibrium requires equal pressure (not equal fugacities) $\endgroup$
    – AimLow
    Oct 13, 2022 at 10:45

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