Take the 2-minute tour ×
Chemistry Stack Exchange is a question and answer site for scientists, academics, teachers and students. It's 100% free, no registration required.

The Wikipedia Article on Fugacity contains statement

The contribution of nonideality to the chemical potential of a real gas is equal to RT ln $\phi$

Here $\phi$ is fugacity coefficient.I think it is wrong it should be fugacity coefficient multiplied by partial pressure of gas divided by standard pressure.So for the example of nitrogen gas at 100 atm it's chemical potential should be μ = μ$_{id}$ + RT ln 97.03 and not μ = μ$_{id}$ + RT ln 0.9703 as given in Wikipedia Article.

I am confused, who is wrong me or wikipedia ?

share|improve this question
add comment

1 Answer 1

up vote 7 down vote accepted

I think you have misinterpreted the passage on Wikipedia a little. Of course, you are right that the chemical potential is given by the formula

\begin{align} \mu_{i} (p_{i}, T) &= \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \log a_{i} \\ &= \mu_{i}^{0} + RT \ln \left(\phi_{i} \frac{p_{i}}{p^0} \right) \end{align}

where $R$ is the universal gas constant, $T$ is the temperature, and $\phi_{i}$, $p_{i}$, $a_{i}$, and $\mu_{i}$ are the fugacity coefficient, the partial pressure, the activity and the chemical potential of the $i$th component in the system, respectively. But the passage you are quoting says that the contribution of nonideality to the chemical potential of a real gas is equal to $RT \ln \phi$ and in the equation

\begin{align} \mu = \mu_{\text{id}} + RT \ln \left(\phi \right) \end{align}

$\mu_{\text{id}}$ is not the same as $\mu^{0}$. Rather it is meant in the following way: For an ideal gas the fugacity coefficient is $1$ by definition and so the chemical potential of an ideal gas, $\mu_{\text{id}}$, is given by

\begin{align} \mu_{\text{id}} = \mu^{\text{id}}_{i} (p_{i}, T) = \mu_{i}^{0} + RT \ln \left(\frac{p_{i}}{p^0} \right) \end{align}

If you use the mathematical property of logarithms that $\log(a \cdot b) = \log(a) + \log(b)$ you can rewrite the equation for the chemical potential of real gases to

\begin{align} \mu_{i} (p_{i}, T) &= \underbrace{\mu_{i}^{0} + RT \ln \left(\frac{p_{i}}{p^0} \right)}_{= \, \mu_{\text{id}}} + RT \ln \left(\phi_{i} \right) \\ &= \mu_{\text{id}} + RT \ln \left(\phi_{i} \right) \end{align}

and so you see, that the contribution of nonideality to the chemical potential of a real gas is indeed equal to $RT \ln \phi$ and you get to the equation from the Wikipedia article.

share|improve this answer
Thanks i didn't look at $\mu_{id}$.I am ashamed of such silly mistake. –  Vishvajeet Patil Apr 22 at 9:33
I can't vote up answer due to lack of reputation –  Vishvajeet Patil Apr 22 at 9:36
@VishvajeetPatil If you keep participating in this site, I'm sure you'll get the chance :) –  Philipp Apr 22 at 9:39
@VishvajeetPatil Mistakes like that happen very easily, nothing to be ashamed of. It shows you that it is important to always define the symbols in your equations. If the Wikipedia article had spelled out what $\mu_{\text{id}}$ was explicitly in the first place your confusion couldn't have arisen. –  Philipp Apr 22 at 9:41
I have another question about chemical potential can you solve that Question –  Vishvajeet Patil Apr 22 at 10:30
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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