Inverting a quartic equation of state

Question

I have the following equation (which is an adaptation of the Beattie-Bridgeman Equation of State):

$$ P = \frac{RT}{V} + \frac{B}{V^2} + \frac{C}{V^3} + \frac{D}{V^4} $$

This is a function of the form $P = f(V)$ as R, T, B, C and D are all constant with respect to P and V.

From this, my ultimate goal is to derive an equation for $dV/dP$. My first step is therefore to write the equation explicit in V, however I get stuck at:

$$ PV^4 - RTV^3 - BV^2 - CV - D = 0 $$

What steps am I missing in order to be able to write this in the form $V = f(P)$?

Answer 1

You have: $$ P = {RT}{V^{-1}} + {B}{V^{-2}} + {C}{V^{-3}} + {D}{V^{-4}} $$ Now diffrentiate wrt $P$: $$ 1 = {R{d{(TV^{-1}})\over dP}} + {B}{dV^{-2}\over dP} + {C}{dV^{-3}\over dP} + {D}{dV^{-4}\over dP} $$ On the RHS the last three terms are fine and you can take $dV\over dP$ common but the first term will also introduce a $dT\over dP$
So the approach to get $dV\over dP$ will be:

• Take the original equation $$ P = {RT}{V^{-1}} + {B}{V^{-2}} + {C}{V^{-3}} + {D}{V^{-4}} $$ and diffrentiate it wrt $P,V,T$ to get three equations.
• These three equations will have 3 variables $dT\over dP$or $dP\over dT$ , $dP\over dV$ or $dV\over dP$ , $dT\over dV$or $dV\over dT$
• Solve the three equations to get $dV\over dP$

Answer 2

Look up "implicit differentiation".

There are formulas for solving quartic polynomials, but they are so complicated that I doubt they would be helpful to you.

Answer 3

Applying the technique of implicit differentiation, as Robert Israel recommended, you will get something like: $$ V^4 + 4PV^3\frac{dV}{dP} + 3RTV^2\frac{dV}{dP} + 2BV\frac{dV}{dP} - C\frac{dV}{dP} = 0 $$ which you can solve to get $\frac{dV}{dP}$.

This assumes that $T$ is constant with respect to $P$, which others have suggested might be unrealistic.