0
$\begingroup$

I am a little confused about fugacity and pressure of a gas. In ideal gases, pressure is same as fugacity. But for real gases this is not the case. Here is the link of the question that is the source of my confusion :

https://chemistry.stackexchange.com/questions/135002/what-is-fugacity

The answer by @AChem hints that the pressure gauge measures the pressure of the gas and not its fugacity.

My question is:

Is the value corrected with the help of fugacity coefficient ($f=$$\Phi$$P$,where $P$ is the pressure as measured by manometer and $\Phi$ is fugacity coefficient ) is the actual pressure felt by the tires?

Please leave reasons for your downvote, so that I can improve in writing questions.

$\endgroup$
7
  • 1
    $\begingroup$ Fugacity is not the actual pressure. It is used to account for the real behaviour of gasses and also to overcome the assumption that chemical potential for a real gas can be based on using that based on an ideal gas $\mu=\mu^\text{o}+RT\ln(P(atm)/1(atm))$ rather than starting over again with, say, the van der Waals eqn. $\endgroup$
    – porphyrin
    Commented Jul 23, 2023 at 8:15
  • $\begingroup$ So if fugacity is not the actual pressure, then what does $\Phi$$P$$=$$f$ tell us about? Isn't fugacity just corrected pressure ? $\endgroup$
    – Natasha J
    Commented Jul 23, 2023 at 9:10
  • $\begingroup$ yes, but usually we use dimensionless activity $a$ instead of fugacity ratio $a=f/f^\text{o}=f/1$ such that for a real gas $a=\gamma P$ with $\gamma$ as activity coefficient and so $\mu=\mu^\text{o}+RT\ln(a)$ or $\mu=\mu^\text{o}+RT\ln(P(atm)/1(atm))+RT\ln(\gamma)$ $\endgroup$
    – porphyrin
    Commented Jul 23, 2023 at 9:46
  • 1
    $\begingroup$ It could be said the fugacity is the pressure a real gas would have had if ( it had behaved ideally ) AND ( had had the same chemical potential as it has as a real gas). $\endgroup$
    – Poutnik
    Commented Jul 23, 2023 at 9:52
  • $\begingroup$ @Poutnik, then fugacity is corrected pressure? Is it right to think like that? $\endgroup$
    – Natasha J
    Commented Jul 23, 2023 at 12:42

1 Answer 1

1
$\begingroup$

For an ideal gas, the mechanical pressure (the one you can measure and that is relevant for the behavior of tires) and the fugacity is the same.

For a real gas, which quantity you use depends on the context. For mechanical questions (whether your reaction vessel can withstand a pressure difference, for example), you use the pressure. If you are interested in chemical reactions, the fugacity is the relevant quantity (e.g. in equilibrium calculations, or any other equilibrium thermodynamics like reduction potential or definition of standard state).

Wikipedia has a succinct definition that captures those two contexts:

In chemical thermodynamics, the fugacity of a real gas is an effective partial pressure which replaces the mechanical partial pressure in an accurate computation of the chemical equilibrium constant. It is equal to the pressure of an ideal gas which has the same temperature and molar Gibbs free energy as the real gas.

Other definitions spell out the "molar Gibbs energy" as the chemical potential $\mu$, which is at the core of many formulas (equilibrium constant, Nernst law, entropy of mixing) that describe changes when you change partial pressure or concentration.

The same concept (to deal with non-ideal solutions, in this case) exists for concentration, where concentration is replaced by thermodynamic activity $a$ to account for non-ideal behavior. However, activity is a dimensionless quantity while fugacity has the same dimensions as pressure.

[OP] Is the value corrected with the help of fugacity coefficient (f=ΦP,where P is the pressure as measured by manometer and Φ is fugacity coefficient ) is the actual pressure felt by the tires?

No, fugacity is the appropriate quantity for questions of chemical reactions or transport, i.e. "the actual pressure felt by a chemical reaction". For mechanical questions, the measured pressure (or partial pressure) is relevant. For example, in a gas mixture, individual fugacities do not add up to the total pressure of the mixture.

$\endgroup$
1
  • $\begingroup$ I see... so fugacity is relevant for chemical reactions whereas pressure is relevant for mechanical questions. But if fugacity considers all the non-idealness of real gases, then wouldn't thermodynamical models using pressure(instead of fugacity) lead to erroneous results? Thank you for answering!! $\endgroup$
    – Natasha J
    Commented Jul 23, 2023 at 11:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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