Calculate the pressure exerted (in atm) by 1 mole of $\ce{CH4}$ gas at a temperature of $\displaystyle\pu{\frac{24}{0.0821} K}$ if volume occupied by $\ce{CH4}$ molecules is negligible. Given $a = \pu{2 L^2 atm mol^{-2}}$.


$$ \begin{align} \left(p + \frac{a}{V_\mathrm{m}^2}\right)V_\mathrm{m} &= RT\\ pV_\mathrm{m} + \frac{a}{V_\mathrm{m}} &= RT\\ pV_\mathrm{m}^2 - RTV_\mathrm{m} + a &= 0\\ pV_\mathrm{m}^2 -24V_\mathrm{m} + 2 &= 0\\ \end{align} $$

For real gas this equation must have only one root:

$$D = 0\quad\implies\quad b^2 - 4ac = 0$$ $$24^2 - 4×p×2 = 0$$ $$p = \pu{72 atm}$$

My thoughts

I had a doubt whether this solution was really true or not. When I first encountered it, I was convinced that it was true. But, now, I doubt it. I don't have any appropriate reason why and that's what I seek to ask. Is this approach correct? Why?

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    $\begingroup$ I don't understand why the quadratic should have one root. $\endgroup$ – Zhe Feb 25 '19 at 18:47
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    $\begingroup$ Mathematically the equation has two roots, one of which is the root that corresponds to actual methane (or at least, this problem's version of it) and the other is just junk. You have to identify the right root out of the two that come mathematically, which in this case should be easy. $\endgroup$ – Oscar Lanzi Feb 25 '19 at 21:18
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    $\begingroup$ @Polarbear Be aware that VdW equation is just a simplified model of real gas behavior and does not guarantee the physical/chemical sense of its output outside of its applicability range. E.g. Some quantity may be modeled by a quadratic formula with the negative quadratic coefficient. That does not mean the model is wrong, when the value reaches negative values, if they must be positive. It just means that region is out of the model applicability. $\endgroup$ – Poutnik Mar 26 '20 at 7:32

Assuming the Van der Waals equation with $b=0$, I agree with the solution to the equation:

$$pV_\mathrm{m}^2 - 24V_\mathrm{m} + 2 = 0\label{eqn:1}\tag{1}$$

I'll point out that I'm following the notation of the given solution, but $V_\mathrm{m}$ seems odd to me. I'd think that $V_\mathrm{m}$ would be reserved to mean the molar volume at STP.

But equation \eqref{eqn:1} has two unknowns, $p$ and $V_\mathrm{m}$. Letting $a' = p$, $b' = -24$, and $c' = 2$ then the appropriate quadratic equation is of course

\begin{alignat} 2V_\mathrm{m} &= \dfrac{-b' \pm\sqrt{b'^2 - 4a'c'}}{2a'}\tag{2}\\ V_\mathrm{m} &= \dfrac{-(-24) \pm\sqrt{(-24)^2 - 4×p×2}}{2p}\tag{3}\\ V_\mathrm{m} &= \dfrac{24 \pm\sqrt{(-24)^2 - 8p}}{2p}\tag{4}\\ V_\mathrm{m} &= \dfrac{12 \pm\sqrt{144 - 2p}}{p}\label{eqn:5}\tag{5} \end{alignat}

So we have two unknowns and one equation. Thus the problem is unsolvable without additional information. Notice that would still be the case if the simple ideal gas equation, $PV=nRT$, had been used. The ideal gas equation could be solved only to $pV_\mathrm{m}=24$.

For equation \eqref{eqn:5} there are three cases for the sqrt term $\sqrt{144 - 2p}$:

  1. The term is negative.

    If $144 - 2p$ is negative, then both roots would be imaginary.

  2. The term is equal to zero.

    We have $\sqrt{144 - 2p} = 0\implies 144 - 2p = 0,$ so $p = 72$. Therefore, $V_\mathrm{m} = 12/72 = 0.167.$

  3. The term is greater than zero.

    Then $72 > p$ also, and mathematically $V_\mathrm{m}$ has two roots.

    Now, again there are three possibilities.

    (a) $\sqrt{144 - 2p} < 12$

    This is impossible since $V_\mathrm{m}$ would have two valid values.

    (b) $\sqrt{144 - 2p} = 12$

    This is impossible since $V_\mathrm{m} = 0$ and the gas volume can't be $0$.

    (c) $\sqrt{144 - 2p} > 12$

    This would yield one negative value and one positive value for $V_\mathrm{m}$, which is OK for $V_\mathrm{m}$. However it also means that $p < 0,$ which is nonsensical.

Thus the only reasonable solution is if the square root term is equal to zero.

P.S. The comment by user Poutnik made me look at this again, and I now understand the book's solution…

  • $\begingroup$ Vm=V/n=f(T,p), why should it be at STP ?? $\endgroup$ – Poutnik Mar 26 '20 at 5:55
  • $\begingroup$ @Poutnik - Huh? STP isn't mentioned anywhere. I was just disagreeing as to how the quadratic was solved. There is no reason that I can see that 144-2P = 0. It could also be that 144-2P > 0. $\endgroup$ – MaxW Mar 26 '20 at 6:27
  • $\begingroup$ It is mentioned in a quote in your answer, but is unclear what you have quoted. Perhaps a piece of question, deleted later. $\endgroup$ – Poutnik Mar 26 '20 at 6:30
  • $\begingroup$ @Poutnik - ah yes, sorry. I was just commenting that the notation that was used in the OP's book solution was odd. I don't understand why $\mathrm{V_m}$ was used instead of just $\mathrm{V}$. $\endgroup$ – MaxW Mar 26 '20 at 6:35
  • $\begingroup$ VdW equation is commonly used in both general and molar forms. They used the latter, so they had to use V_m. $\endgroup$ – Poutnik Mar 26 '20 at 6:38

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