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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?

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  • $\begingroup$ I got 546.0 K, but I did the calculations in Mathematica. What's the answer supposed to be? $\endgroup$ – theorist Aug 17 at 23:39
  • $\begingroup$ 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. $\endgroup$ – Stab Reberie Aug 18 at 4:37
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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:

enter image description here

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  • $\begingroup$ Thanks, you seem comfortable with Mathematica, could you solve the same problem but this time for local maxima? Is the answer 1.78 K in this case? $\endgroup$ – Stab Reberie Aug 18 at 10:55
  • $\begingroup$ @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. $\endgroup$ – theorist Aug 18 at 17:36

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