# How to get equation of state from fugacity

Given that the fugacity of a substance is $$f=Pe^{bT}$$, how do I obtain the equation of state of the substance?

This is my approach, but I have not made good headway: $$RT d (\ln f) = VdP \\ d(\ln f) = \frac{V}{RT} dP \\ bdT = \left( \frac{V}{RT} - \frac{1}{P}\right)dP$$

But what next? I see that $$\left( \frac{V}{RT} - \frac{1}{P}\right)dP = \frac{(Z-1)}{P}dP$$ But I don't see how I can reduce this further.

My end goal is to find derivatives of pressure wrt temperature keeping volume constant. How do I go about this problem?

• That first equation is supposed to be at constant temperature. – Chet Miller Oct 4 '20 at 20:04
• @ChetMiller If i change it to $VdP - SdT$, and I equate both sides, I get $V = RT/P, -S = RTb$. What can i do next? – megamence Oct 4 '20 at 20:16
• You are supposed to do the integration at constant T. – Chet Miller Oct 4 '20 at 20:21

That first equation is supposed to be at constant T. You can rewrite it as $$RTd\ln{(f/P)}=RT(z-1)\frac{dP}{P}$$or $$\ln{\frac{f}{P}}=bT=\int_0^P{(z-1)\frac{dP'}{P'}}$$So if z behaves like z = 1+aPT, this becomes: $$bT=aPT$$ or b = aP
• Thank you for your answer @ChetMiller. I was trying this problem with $b$ being a constant... Is it possible to for $b$ to be a constant? – megamence Oct 4 '20 at 20:59