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I was solving some problems and I had a doubt a problem since the resulting pressure gives me a negative value. The problem was the following:

Could $100~\mathrm{g}$ of $\ce{N2}$ in a $200~\mathrm{mL}$ container exert a pressure of $25~\mathrm{atm}$ at $28~^\circ\mathrm{C}$ if it behave like an ideal gas? If your answer is no, what pressure would exert the gas, knowing that the van der Waals constants for $\ce{N2}$ are $3.11~\mathrm{\frac{L\,atm}{mol^2}}$ and $0.076~\mathrm{\frac{L}{mol}}$.

I have converted the 100 grams of $\ce{N2}$ to moles of $\ce{N2}$, and with that I've replaced the data in the van der Waals formula, using $a = 3.11~\mathrm{\frac{L\,atm)}{mol^2}}$ and $b = 0.076~\mathrm{\frac{L}{mol}}$. The result was a negative pressure, and I am pretty sure that's impossible (or at least that's what my chemistry teacher said).

This is what I did:

$$\begin{align} P &= \frac{nRT}{V-nb} -\frac{an^2}{V^2}\\ &= \frac{(3.57~\mathrm{mol})(0.082\mathrm{\frac{L\,atm}{K\,mol}}) (301.1~\mathrm{K})}{(0.2~\mathrm{L}) -(3.57 ~\mathrm{mol})(0.076\mathrm{\frac{L}{mol}})} -\frac{(3.11~\mathrm{\frac{L~atm}{mol^2}}) (3.57~\mathrm{mol})^2}{(0.2~\mathrm{L})^2} \end{align}$$

The result was: $-1626.19~\mathrm{atm} - 9.9075~\mathrm{atm} = -1636.0975~\mathrm{atm}$

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    $\begingroup$ I think we need more details about what you did. With what you've given I cannot tell where your mistake is. $\endgroup$
    – orthocresol
    Sep 8, 2016 at 2:57
  • $\begingroup$ I just didn't want it to be taken as "homework", but I alredy edited the post. $\endgroup$
    – Mauricio G.
    Sep 8, 2016 at 3:15
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    $\begingroup$ Ok. Firstly, the constant $a$ has units of $\mathrm{L^2~atm~mol^{-2}}$ (not $\mathrm{L~atm~mol^{-2}}$), but I am guessing it is just a typo on your part. The issue seems to arise from the values given for $a$ and $b$. Wikipedia gives the constants as $1.352$ and $0.0387$ respectively (converted to the same units as yours). Using those values I get a pressure of 997.3 atm. $\endgroup$
    – orthocresol
    Sep 8, 2016 at 3:49
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    $\begingroup$ As soon as you got $V<nb$, you should have stopped and asked yourself "What on earth am I doing?" The van der Waals formula is not supposed to work in these conditions, not even approximately. As soon as you are there, you could be dead sure something's seriously amiss (like, a typo in the problem statement, or one of the values is in the wrong units, or whatever). $\endgroup$
    – Ivan Neretin
    Sep 8, 2016 at 5:58

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Let's get the obvious out of the way: The answer to that exercise is yes. Hence it is completely unnecessary to do the second part. With the parameters given in the exercise I calculated a pressure of about 440 atmospheres using the ideal gas law.

But then since we're all here to learn, let's also have a look at the second part. The good news is that I can't see anything wrong with how you used the equation. The problem is already pointed out by orthocresol in the comments, which I will quote here for safekeeping (emphasis by me):

Ok. Firstly, the constant $a$ has units of $\mathrm{L^2~atm~mol^{-2}}$ (not $\mathrm{L~atm~mol^{-2}}$), but I am guessing it is just a typo on your part. The issue seems to arise from the values given for $a$ and $b$. Wikipedia gives the constants as $1.352$ and $0.0387$ respectively (converted to the same units as yours). Using those values I get a pressure of 997.3 atm.

I would assume that nobody caught this mistake earlier since you are not actually supposed to do that part of the question.

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