# Expression for fugacity coefficient derived from a pressure-explicite EOS

In the book, Thermodynamics for Process Simulation, the authors propose to derive an expression for the fugacity coefficient from a pressure-explicit equation of state, as an example among many others.

Assuming, $P = P(T, v)$ or $z =\frac{Pv}{RT} = z(T, v)$, the calculations lead to,

$$\left(\frac{\partial \ln \phi}{\partial v}\right)_T = \left(\frac{\partial z}{\partial v}\right)_T-\left(\frac{\partial \ln z}{\partial v}\right)_T + \left(\frac{1}{v}-\frac{P}{RT}\right) \tag1$$

where $\phi$ is the fugacity coefficient.

What I don't understand is, when they integrate, they get,

$$\ln \phi = z - 1 - \ln z + \frac{1}{RT}\int_{\infty}^v \left(\frac{RT}{v}-P\right)\mathrm{d}v \tag2$$

Can someone explain me how they find $z-1-\ln z$ by integrating the first two terms of $(1)$ between $\infty$ and $v$?

When integrating the first two terms, it leads to $$\left[z - \ln z \right]_{\infty}^v$$ which I understand, but then I don't get the final result.

I suppose it's linked to the values of $z$ when the volume is very large and when the volume is "$v$" but it does't make sense for me.

I would have assume to get $$\left[z - \ln z \right]_{\infty}^v = v - \ln v - \left[z(v^{\infty}) - \ln z(v^{\infty}) \right]$$ and nothing else.

But I am missing something probably obvious.

• When $v\rightarrow \infty$ you get the ideal gas eos, so $z\rightarrow 1$. This gives you the result, i.e. $1$ at the lower limit (subtracted) and $z-\ln z$ at the upper one. – user64968 Sep 4 '18 at 12:20

Mindful of the guideline "don't give answers in comments", I've converted my comment to an answer, trivial though it is.

When $v\rightarrow \infty$ you get the ideal gas eos, so $z\rightarrow 1$. This gives you the result, i.e. $1$ at the lower limit (subtracted) and $z-\ln z$ at the upper one.

When you write down the integrated form, $z-\ln z$, you need to remember that it means that you are evaluating the function $z(v)$ at the two limits $v=\infty$ and $v=v$, and using those values in that expression; you should not be trying to set $z=v$ at the limits, which seems to be (partly) what you have written at the end of your question.

Edit following OP comment.

More details. At constant $T$, $z=z(v)$ is a function of molar volume $v$, determined by the equation of state, $P(v)$ at the given value of $T$. This equation is unknown, but we can be sure that, in the ideal gas limit $v\rightarrow\infty$, $z(v)\rightarrow 1$.

Integrating both sides of the equation from the ideal gas limit to the desired volume $v$, and using $v'$ as the integration variable, gives eqn (2) of the question in its full form: $$\left| \ln \phi(v')\right|_{v'=\infty}^{v'=v} = \left| z(v') - \ln z(v') \right|_{v'=\infty}^{v'=v} + \frac{1}{RT}\int_{\infty}^v \left(\frac{RT}{v'}-P(v')\right)\mathrm{d}v'$$ I think it's clearer to distinguish between the integration variable $v'$ and the "upper" limit of integration $v$, but many people would be happy just to use $v$ instead of $v'$.

Anyway, on the left, we know that, in the ideal gas limit, the fugacity coefficient $\phi(\infty)=1$, so $\ln\phi(\infty)=0$ and we are just left with $\ln\phi(v)$. On the right, similarly, we substitute in the upper and lower limits for $v'$. We know $z(\infty)=1$, so the function being evaluated is $z(\infty)-\ln z(\infty)=1$ at the lower limit, and $z(v)-\ln z(v)$ at the upper limit. So the final answer is $$\ln \phi(v) = z(v) - \ln z(v) - 1 + \frac{1}{RT}\int_{\infty}^v \left(\frac{RT}{v'}-P(v')\right)\mathrm{d}v'$$ An important point is that the integration variable is the molar volume $v'$ (or $v$ if you prefer), not $z$. In evaluating the result, the integration limits $v$ and $\infty$ are substituted for $v'$, the argument of the function being evaluated. It is incorrect to set $z=v$, or $z=\infty$. (This should be even more clear in this case, since $z$ is a dimensionless quantity, whereas $v$ is not).

• Why shouldn't I try to set $z = v$ ? I am just reasoning as I always did using integration. And I understand now what you pointed out that I didn't get but then if $z \rightarrow 1$ why not to write $[z - \ln z]_{\infty}^{v} = [v - \ln v] - [1 - \ln 1] = v - \ln v - 1 + 0$ ? Because then it doesn't make sense to subtract two infinities. I just don't understand why it is then correct to assume the relation with the $z$ instead of $v$ correct. – ParaH2 Sep 4 '18 at 13:21
• You are not integrating the variable $z$ between two limits $\infty$ and $v$; you are integrating a function of $v$ between these limits. And actually, you already did the difficult part, namely the integration, so you just need to evaluate the resulting function at those limits. In this case $$\left | z(x)-\ln z(x) \right |_{x=\infty}^{x=v}$$ where I'm deliberately using a dummy variable $x$ for clarity. If this is still not clear, I can edit my answer to go into more detail. – user64968 Sep 4 '18 at 13:35
• I've added more detail to my answer anyway. Let me know if things are still not clear. – user64968 Sep 4 '18 at 15:20
• Thank you for your answer. I'm still a little confused but I will work on it and I will probably get it. – ParaH2 Sep 4 '18 at 18:48