Good evening everyone.
After studying some equations of state, I've read about the mathematical steps formulated to model some. In the particular case of Van der Waals, where $$P=\frac{RT}{v-b}-\frac{a}{v^2}$$
For those not familiar with this expression, this equation describes general characteristics of a real gas, P standing for pressure, R for the universal gas constant, T for temperature, v for molar volume (Yep, I know the little line is missing, sorry!), and a and b are characteristic constants for the gas .
There are some special values for Temperature, Pressure and Molar volume. Finding them, apparently, helps me in my quest for a and b. I am told the following steps to do so in VdW equation:
1) Derive the function twice.
2) Solve RT for each derivative and make them equal to zero.Make an equation of of the derivative and the second derivative, then solve for v
3) Plug the value of V in the first derivative, and solve for T, now called critical temperature. 3
4) Plug the values of V and T in the original function, thus obtaining the critical pressure. 4
5) Solve "a" for critical T and critical P, then making the equation and obtaining "b"[5]
6) Plug "b" in critical T, now obtaining "a"[6]
Now, back to my problem, I'm looking for "a" and "b" again, but now in the Redlich Kwong equation: $$P=\frac{RT}{v-b}-\frac{a}{\sqrt(T)*(v^2+vb)}$$ I've already derived the function twice and solved for RT: $$P'=\frac{-RT}{(v-b)^2}+\frac{a(2v+b)}{\sqrt(T)*((v^2+vb)^-2)}=0$$ $$RT=\frac{a(2v+b)*((v-b)^2}{\sqrt(T)*((v^2+vb)^-2)}$$
$$P''=\frac{2RT}{(v-b)^-3}+\frac{2a}{\sqrt(T)*((v^2+vb)^-2)}-\frac{2a*(2v+b)^2}{\sqrt(T)*((v^2+vb)^-3)}=0$$ $$RT=\frac{a}{\sqrt(T)}*((v-b)^3)*(\frac{(2v+b)^2}{((v^2)+vb)^3}-\frac{1}{(v^2)+vb)^2})$$
After putting the two =0, I've obtained this far: $$3v+2b=(v-b)*((2v+b)^2)((v^2)+vb)$$
As it isn't as simple as in the VdW case, I certainly don't know where to go now in order to obtain a value of v based on b. I would be so grateful if someone checked this, thank you!
$\endgroup$ 12 Answers
$\begingroup$I put a second answer to show you how the work can be done in a simpler manner.
Start with the equation of state $$P=\frac{R T}{V-b}-\frac{a}{\sqrt{T}\, V (b+V)}\tag 1$$ and develop it in terms of $V$; this leads to a cubic polynomial in $V$ $$V^3-\frac{R T}{P} V^2+ \left(\frac{a}{P \sqrt{T}}-b^2-\frac{b R T}{P}\right)V-\frac{a b}{P \sqrt{T}}=0$$ Now, at the critical point $(T=T_c, P=P_c)$, this must be identical to the development of $$(V-V_c)^3=V^3-3V_c V^2+3V_c^2V-V_c^3$$ Now, by identification $$3V_c=\frac{R T_c}{P_c}\tag 2$$ $$3V_c^2=\frac{a}{P_c \sqrt{T_c}}-b^2-\frac{b R T_c}{P_c}\tag 3$$ $$V_c^3=\frac{a b}{P_c \sqrt{T_c}}\tag 4$$ So, we can only use equations $(3)$ and $(4)$ to identify $a$ and $b$.
From $(4)$ we can express $a$ as a function of $b$ and plug it in $(3)$; this gives now a cubic equation in $b$; solve it and go back to the definition we made of $a$ as a function of $b$.
Edit
More details. From $(2)$ we have $V_c=\frac{R T_c}{3 P_c}$ this makes the cubic in $b$ to be $$b^3+\frac{b^2 R T_c}{P_c}+\frac{b R^2 T_c^2}{3 P_c^2}-\frac{R^3 T_c^3}{27 P_c^3}=0$$ Define now $b=\frac{B RT_c}{P_c}$; this makes the equation $$27 B^3+27B^2+9B-1=0$$ Using Cardano method, there is only one real root which is $$B=\frac{1}{3} \left(\sqrt[3]{2}-1\right)\implies b=\frac{1}{3} \left(\sqrt[3]{2}-1\right)\frac{ RT_c}{P_c}$$ Now, $(4)$ becomes $$\Big(\frac{R T_c}{3 P_c}\Big)^3=\frac{a b}{P_c \sqrt{T_c}}=\frac{a }{P_c \sqrt{T_c}}\frac{1}{3} \left(\sqrt[3]{2}-1\right)\frac{ RT_c}{P_c} \implies a=\frac{R^2 T_c^{5/2}}{9 \left(\sqrt[3]{2}-1\right) P_c}$$
$\endgroup$ 3 $\begingroup$Even if there exist much simpler procedures to get the characteristic constants of cubic equations of state, let us do it using almost what you have been suggested.
Using $$P=\frac{R T}{V-b}-\frac{a}{\sqrt{T} V (b+V)}\tag 1$$ $$P'=\frac 1{\sqrt T}\Big(\frac{a (b+2 V)}{V^2 (b+V)^2}-\frac{R T^{3/2}}{(b-V)^2}\Big)\tag2$$ $$P''=\frac 1{\sqrt T}\Big(\frac{a \left(\frac{1}{(b+V)^3}-\frac{1}{V^3}\right)}{b}+\frac{R T^{3/2}}{(V-b)^3}\Big)\tag3$$ SInce, at the critical point, $P'=P''=0$, equations $(2)$ and $(3)$ can rewrite as $$\frac{R T^{3/2}}{(V-b)^2}=\frac{a (b+2 V)}{V^2 (b+V)^2}\tag 4$$ $$\frac{R T^{3/2}}{(V-b)^3}=-\frac{a \left(\frac{1}{(b+V)^3}-\frac{1}{V^3}\right)}{b}\tag 5$$ Now, making the ratio ($(4)$ divided by $(5)$) and simplifying, we then get $$V-b=\frac{V (b+V) (b+2 V)}{b^2+3 b V+3 V^2}\tag 6$$ Cross multiplying and simplifying, we then get $$-b^3-3 b^2 V-3 b V^2+V^3=0$$ which is an homogeneous polynomial of degree $3$. Setting $V=x b$, we then need to solve $$x^3-3 x^2-3 x-1=0$$ Using Cardano method, the only real solution is given by $$x=1+2^{2/3}+2^{1/3}$$
I am sure that you can continue from here.
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