# Solving a quadratic of for pressure of real gas with unknown volume

I had this question on my test:

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

The solution they presented for it was:

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?

• I don't understand why the quadratic should have one root. – Zhe Feb 25 at 18:47
• 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. – Oscar Lanzi Feb 25 at 21:18

The "solution" to the quadratic equation makes no sense to me.

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

$$\mathrm{PV}_m^2 - 24\mathrm{V}_m + 2 = 0\tag{1}$$

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

But Equation 1 has two unknowns, $$\mathrm{P}$$ and $$\mathrm{V}_m$$. Letting $$a' = P$$, $$b' = -24$$, and $$c' = 2$$ then the appropriate quadratic equation is of course

$$\mathrm{V}_m = \dfrac{-b' \pm\sqrt{b'^2 - 4a'c'}}{2a'}\tag{2}$$

so

$$\mathrm{V}_m = \dfrac{-(-24) \pm\sqrt{(-24)^2 - 4(\mathrm{P})(2)}}{2(\mathrm{P})}\tag{3}$$

$$\mathrm{V}_m = \dfrac{24 \pm\sqrt{(-24)^2 - 8\mathrm{P}}}{2\mathrm{P}}\tag{4}$$

$$\mathrm{V}_m = \dfrac{12 \pm\sqrt{144 - 2\mathrm{P}}}{\mathrm{P}}\tag{5}$$

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 $$\mathrm{PV}_m=24$$.

For equation 5 there are three cases for the sqrt term $$\sqrt{144 - 2\mathrm{P}}$$:

(1) If the term is negative:

If $$144 - 2\mathrm{P}$$ is negative then both roots would be imaginary.

(2) If the sqrt term is equal to zero:

$$\sqrt{144 - 2\mathrm{P}} = 0$$

$$144 - 2\mathrm{P} = 0$$

$$\mathrm{P} = 72$$

and therefore

$$\mathrm{V}_m = \dfrac{12}{72} = 0.167$$

(3) If the sqrt term is greater than 0:,

then 72 > P also, and $$\mathrm{V}_m$$ has two roots.