# 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?

• I got 546.0 K, but I did the calculations in Mathematica. What's the answer supposed to be? – theorist Aug 17 at 23:39
• I do not have the answer. 546.0 K seems like a reasonable answer, did the problem boil down to pV^3 -aV + 2ab = 0. – Stab Reberie Aug 18 at 4:37

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. • @StabReberie Yes, that's easy enough; I just change (in the Solve statement at the end) dPdV2 > 0 to dPdV2< 0, and then rerun the notebook. When I do this, I get T = 1.79 K [essentially the same as yours; you may have a round-off error; note that Mathematica uses R =8.31446 J/(mol K) = 0.0820574 L atm/(mol K)], and V = Vm = 0.0744 m^3. – theorist Aug 18 at 17:36