How to get critical constants from Dieterci's real gas equation?

Question

Please, before closing this question as off-topic, consider that this is completely a calculus problem, no chemistry involved

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Dieterci's real gas equation is $$p = {RTe^{-a/(RTV)}\over V - b}$$

I need to find critical constants of this equation.

To do that I need to equation first second derivative of this equation to 0. and find the expression for $V$ and $T$ from them in terms of $a, b$. And then put those values in real equation to obtain a expression for $p$ in terms of $a, b$.

So differentiating the formula with respect to V,

$$p^\prime = RT\left[(e^{-a/(RTV)})^\prime (V -b)^{-1} + (-1)(1)(V -b)^{-2}e^{-a/(RTV)} \right] \tag{1}$$ $$p^\prime = -RTe^{-a/(RTV)}\left[ {a\over RT}(V^{-2})(V -b)^{-1} + (V - b)^{-2} \right] \tag{2}$$ $$p^\prime = -RTe^{-a/(RTV)}(V -b)^{-1}\left[ {a\over RT}(V^{-2}) + (V - b)^{-1} \right] \tag{3}$$ $$\bbox[5px,Border:2px solid black]{p^\prime = -p\left[ {a\over RT}(V^{-2}) + (V - b)^{-1} \right]} \tag{4}$$

And differentiating it again w.r.t to V

$$\bbox[5px,Border:2px solid black]{p^{\prime\prime} = p^\prime\left({a\over RT}(V^{-2} - (V - b)^{-1})\right) + \left[ {a \over RT}(-2)V^{-3} + (V- b)^{-2} \right]p} \tag {5}$$

Now equating $(5)$ and $(4)$ to zero I get,

For $(4)$ $$p^\prime = -p\left[ {a\over RT}(V^{-2}) + (V - b)^{-1} \right] = 0$$ $$\left[ {a\over RT}(V^{-2}) + (V - b)^{-1} \right] = 0$$ $$RTV^2 - aV + ab = 0 \tag{6}$$

For $(5)$ Since we know $p^\prime = 0$

$$0 = \left[ {a \over RT}(-2)V^{-3} + (V- b)^{-2} \right]p$$ $${a\over RTV^3} = \frac1{V^2 + b^2 -2Vb}$$ $${ RTV^3\over a} = {V^2 + b^2 -2Vb} \tag {7}$$

Now to get the critical constants I need to solve for $V$ and $T$ in equations $(6)$ and $(7)$. I tried it but I ended up in a mess.

I need help as to how to solve these equations conviently. Thanks.

Answer 1

This is too long for a comment.

You could have done the first steps faster using logarithmic differentiation $$P = {RT\over V - b}e^{-\frac a {RTV}}\implies \log(P)=\log(RT)-\log(V-b)-\frac a {RTV}$$ So, differentiate to get $$\frac{P'}{P}=\frac{a}{R T V^2}-\frac{1}{V-b}\implies P'=P\left(\frac{a}{R T V^2}-\frac{1}{V-b} \right)\tag 1$$ Differentiate a second time $$P''=P'\left(\frac{a}{R T V^2}-\frac{1}{V-b} \right)+P\left(\frac{a}{R T V^2}-\frac{1}{V-b} \right)'\tag 2$$ and since you will impose later $P'=P''=0$, the equations are $$\frac{a}{R T V^2}-\frac{1}{V-b}=0\tag 3$$ $$\left(\frac{a}{R T V^2}-\frac{1}{V-b} \right)'=\frac{1}{(V-b)^2}-\frac{2 a}{R T V^3}=0\tag 4$$ So $$\frac{1}{V-b}=\frac{a}{R T V^2}\tag 5$$ $$\frac{1}{(V-b)^2}=\frac{2 a}{R T V^3}\tag 6$$ Squaring $(5)$ and computing the ratio to $(6)$ leads to $a=2RTV$; using this result in $(5)$ gives$$\frac {1}{V-b}=\frac{2RTV}{RTV^2}=\frac 2 V\implies V=2b\implies T=\frac{a}{4 b R}$$ Back to the initial expression of $P$, the above give $$P=\frac{a}{4 e^2 b^2}$$ So, as functions of parameters $a,b$, the critical coordinates of Dieterici equation of state are given by $$V_c=2b \qquad T_c=\frac{a}{4 b R} \qquad P_c=\frac{a}{4 e^2 b^2}\qquad Z_c=\frac{P_cV_c}{RT_c}=\frac{2}{e^2}$$

In practice, in this domain of equations of state, since, for components, critical properties are known, we use to compute the $a,b$ parameters from $T_c,P_c$ $$a=\frac{4 R^2 T_c^2}{e^2 P_c}\qquad b=\frac{R T_c}{e^2 P_c}$$

Answer 2

I was reviewing my past documents, and found that first derivative should be

$$p^\prime = e^{-a/(RTV)}\left[-{RT\over(V-b)^{2}}+{a\over(V-b)V^{2}} \right] $$

Can you find second derivative?