Nitrogen and an unknown gas is leaked into a common vessel... The gaseous mixture contains 0.4 moles of nitrogen. Molecular mass of the other gas is? For 10 minutes each, from two identical holes, nitrogen and an unknown gas is leaked into a common vessel of 3L capacity at 300K. The resulting pressure is 4.15 bar. If the gaseous mixture contains 0.4 moles of nitrogen, then molecular mass of the unknown gas is?

attempt: According to Graham's law of effusion, $$\frac{r_1}{r_2}=\frac{V_1}{V_2}=\sqrt{\frac{M_2}{M_1}}$$ where $V_i$ is the volume of gas effused. Let $V_1$ and $M_1$ be volume and molecular mass of nitrogen. $$V_1=22.4\times0.4=8.96;M_1=28$$
Substituting, $${V_2}=8.96\sqrt{\frac{28}{M_2}}$$
and $V_1+V_2=3$, so $$8.96(1+\sqrt{\frac{28}{M_2}})=3$$
This is not possible since $\sqrt{x}\ge0$.
What am I doing wrong?
Calculate total moles using the ideal gas law and solving for $n$:

$$n = {4.15\;{\rm bar}\cdot 3\;{\rm L}\over {0.0831\;{\rm L}\,{\rm bar}\,{\rm K}^{-1}\,{\rm mol}^{-1}}\cdot 300\;{\rm K}} = 0.5\;{\rm mol}$$

Given 0.4 mol $\ce{N2}$, moles of unknown gas = 0.5 mol - 0.4 mol = 0.1 mol.

Rate of effusion of $\ce{N2} = r_{\ce{N2}} = 0.4\;{\rm mol}/10\;{\rm min} = 0.04\;{\rm mol}\cdot{\rm min}^{-1}$.

Rate of effusion of unknown ${\rm X} = r_{\rm X} = 0.1\;{\rm mol}/10\;{\rm min} = 0.01\;{\rm mol}\cdot{\rm min}^{-1}$.

Using Graham's law of diffusion:

$${r_{\ce{N2}}\over r_{\rm X}} = \sqrt{M_{\rm X}\over M_{\ce{N2}}}$$

we solve for $M_{\rm X}$:

$$M_{\rm X} = {M_{\ce{N2}}\times (r_{\ce{N2}})^{2}\over (r_{\rm X})^{2}}$$

and plugging in numbers we get the molecular mass of the unknown gas:

$$M_{\rm X} = {(28\,{\rm g}\cdot{\rm mol}^{-1})\times (0.04)^{2}\over (0.01)^{2}} = 448\,{{\rm g}\over{\rm mol}}$$

where the units on the effusion rates have been omitted for clarity and because they cancel.