Answer:
[tex]\large \boxed{\text{5.9 s}}[/tex]
Explanation:
Graham’s Law applies to the effusion of gases:
The rate of effusion (r) of a gas is inversely proportional to the square root of its molar mass (M).
[tex]r \propto \dfrac{1}{\sqrt{M}}[/tex]
If you have two gases, the ratio of their rates of effusion is
[tex]\dfrac{r_{2}}{r_{1}} = \sqrt{\dfrac{M_{1}}{M_{2}}}[/tex]
The time for diffusion is inversely proportional to the rate.
[tex]\dfrac{t_{2}}{t_{1}} = \sqrt{\dfrac{M_{2}}{M_{1}}}[/tex]
Let CO₂ be Gas 1 and O₂ be Gas 2
Data:
M₁ = 44.01
M₂ = 32.00
Calculation
[tex]\begin{array}{rcl}\dfrac{t_{2}}{t_{1}} & = & \sqrt{\dfrac{M_{2}}{M_{1}}}\\\\\dfrac{t_{2}}{\text{5 s}}& = & \sqrt{\dfrac{44.01}{32.00}}\\\\& = & \sqrt{1.375}\\t_{2}& = & \text{5 s}\times 1.173\\& = & \mathbf{5.9 s} \\\end{array}\\\text{It will take $\large \boxed{\textbf{5.9 s}}$ for the carbon dioxide to effuse.}[/tex]
