Expert: Scotto - 5/10/2009

Question
Dear Sir,

My book has an exercise that says:

Find dV/dp for this equation. Suppose that T and n are constants.

pV= nRT(1+((nB)/V))

The book says that the answer is (-V^3)/(nRT(V+2nB)) but I do not know how can you obtain that result. Could you please explain me what procedure would you use?

Thank you very much.

Answer
The formula given is PV = nRT(1 + nB/V).
Now I remember PV = nRT,
but there is a little extra multiple in this equation.

Since we are looking for dV/dp,
this say we are looking for the derivative of
pV(p) = nRT(1 + nB/V(t)).

The function is V, the varialbe is p,
and n, R, T, and B are all treated as constants.

Now that this is known, put the function back as it was
and remember V is the function and p is the variable.
That is, pV = nRT(1 + nB/V).

To differentiate this, the left side is the product rule
and the right side is the derivative of 1/V(p).
This would make the 1 in the '(1+' disappear.

The derivative is pV' + V = nRT(-nB/V²)V'.
That is, pV' + V = -(n²RTB/V²)V'.

The first thing to do is add the expression with V' to both sides and subtract V from both sides so it is by itself on the right.
In this way, the V' expressions are on one side and those without V' are on the other.  This gives us
pV' + (n²RTB/V²)V' = -V which is the same as (p + (n²RTB/V²))V' = -V.

We can now divide both sides by (p + (n²RTB/V²)), giving
V' = -V/(p + (n²RTB/V²)).  Multiply the top and bottom on the right side by V².  This gives V' = -V^3/(pV² + n²RTB).  

From the basic equation we know that PV = nRT(1 + nVB/V), so
V' = -V^3/((nRT(1 + nVB/V))V + n²RTB).

This can be multiplied out, giving V' = -V^3/(nRTV + n²RTB + n²RTB).

The like terms can be combined, giving V' = -V^3/(nRTV + 2n²RTB).
Note the nRT on the bottom can be factored out, so we now have
V' = -V^3/((nRT(V + 2nB).