This is the equation of state

$$ \left(P + \frac{n^2*a}{V^2}\right)(V-nb)=nRT $$

where a, b, n, R, T are constants.

The actual question is to calculate $\frac{dV}{dP}$, so in order to do that I'm first trying to solve the above function for V in terms of P.

What I did so far:

  1. simplify by eliminating brackets
  2. move terms with V to the left and without to the right

What I end up with is a 3rd power polynomial

$$PV^3-(nbP+nRT)V^2 + n^2aV = n^3ab$$

and I don't know how to proceed. Usually I factor out the variable in question and divide by the other factor, but that isn't applicable this time.

Is my whole approach wrong? Please help!

  • 1
    $\begingroup$ This belongs on math.se or possibly physics.se But that approach isn't going to work. Just look at this: wolframalpha.com/input/… $\endgroup$ – Feyre Sep 20 '16 at 17:12
  • $\begingroup$ allright, so that approach is unfeasible. I guess at least I can move on. $\endgroup$ – ttdijkstra Sep 20 '16 at 18:42

\begin{align*} \left(p+\frac{an^2}{V^2} \right)(V-nb) &= nRT \\ \left(p+\frac{an^2}{V^2} \right) \frac{\partial}{\partial V}(V-nb)+ (V-nb) \frac{\partial}{\partial V}\left(p+\frac{an^2}{V^2} \right) &= \frac{\partial}{\partial V} (nRT) \\ \left(p+\frac{an^2}{V^2} \right)+(V-nb) \left[ \left( \frac{\partial p}{\partial V} \right)_{T}- \frac{2an^2}{V^3} \right] &= 0 \\ \left( \frac{\partial p}{\partial V} \right)_{T} &= \frac{2an^2}{V^3}-\frac{nRT}{(V-nb)^2} \\ \left( \frac{\partial V}{\partial p} \right)_{T} &= \frac{1}{\left( \frac{\partial V}{\partial p} \right)_{T}} \\ &= \frac{V^3(V-nb)^2}{2an^2(V-nb)^2-nRTV^3} \end{align*}

| cite | improve this answer | |
  • $\begingroup$ This looks like what I want, but could you elaborate what your reasoning is? I don't follow. $\endgroup$ – ttdijkstra Sep 21 '16 at 6:27

If I have not make any error, here is what I think you are waiting for. First set the constants

SetAttributes[a, Constant]
SetAttributes[n, Constant]
SetAttributes[R, Constant]
SetAttributes[b, Constant]
SetAttributes[T, Constant]

eq := (P[V] + ( a n^2)/V^2) (V - n b) - n R T == 0 

Then Dt[eq, V] is the derivative of the lhs of the equation according to V

Solve[Dt[eq, V], P'[V]]

Dt[eq[V], V]

is the derivative or eq[V] according to V. Have a look at it. Then

sol = Solve[Dt[eq[V], V], P'[V]]

is your answer. You can go further in substituting its value to P that is

FullSimplify[sol[[1, 1, 2]] /. P[V] -> Solve[ (P + ( a n^2)/V^2) (V - n b) - n R T == 0 , P][[1, 1, 2]]]
| cite | improve this answer | |
  • $\begingroup$ If you do not want to write P[V], but only P, you will be oblige to substitute Dt[P,V] to P'[V] $\endgroup$ – cyrille.piatecki Sep 20 '16 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.