once again I am dealing with the Dieterici equation. As I am conducting a peer review with my fellow students, it would really be nice to know the exact result of the question. It is asked to find the linear Taylor expansion of following expression, assuming a<<RT:
I just assumed it would be correct to Taylor-expand the exponential function at the origin point, that means using the MacLaurin expansion of e^z: My result is the equation at the bottom.
But I am unsure if this is correct. My thoughts were: Linear Taylor expansion would mean terminating the sequence after the first n which is not zero. Furthermore, I thought that expanding the e-function is the way to go. But I am unsure where a<<RT would play into this to be honest (maybe it is the reason why we could expand in the first place). Any tips and/or advice or real solutions would really be appreciated, thanks!
Best Answer
When you're told that $a\ll RT$, what this really means is that $\frac{a}{RT}$ is small (i.e. close to $0$). This is precisely why you can do a Taylor expansion near zero, since the expression is well approximated by its (truncated) Taylor expansion near zero.
Your expansion looks correct to me. The only issue is that when you "deleted" the higher order terms, you no longer have an exact equality but an approximation, so you should indicate it as such. Hence, $$ p(V_m-b) \approx RT - \frac{a}{V_m}.$$ (I'm assuming the quantities $V_m,b,R$ are constant with respect to $a/RT$.)