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