Finding the isotherm with given minima Get the temperature of the isotherm for water for which the local minimum is at $\pu{100 Pa}.$ Use the values of $a$ and $b$ of water.


My approach

In the van der Waals equation, set 

$$\frac{\mathrm dP}{\mathrm dV} = 0$$

and get the value of $T.$ Plug this value in the initial van der Waals equation and and solve for $V$ from the biquadratic.
The calculations get very messy and the values come out to be unrealistic. (For example, temperature of the order of $10^6$)

My doubt is whether the $\pu{100 Pa}$ as provided by the problem setter a realistic value?
Using Mathematica, I obtained $T=546 K$, as follows.  One of the nice things about using Mathematica for physical calculations is that it has the ability to understand/keep track of/cancel out/convert units.

The volume is non-physically small -- about 750,000 x smaller than what we'd expect for a mole of, say, ideal gas at $P=100 Pa, T = 546 K$).  One does expect a non-physically small volume, since this is at a non-physically low point on the isotherm, but I would not have predicted it would be this much smaller.

If you were doing it by hand, you'd probably want to use a different approach: