# What is the rate of effusion of N_2(g) in relation to that of Cl_2(g) under the same conditions?

Mar 14, 2017

Well, this could be determined from root-mean-square speeds.

${\upsilon}_{R M S} = \sqrt{\frac{3 R T}{M}}$,

where $M$ is the molar mass in $\text{kg/mol}$, $R = \text{8.314472 J/mol"cdot"K}$, and $T$ is temperature in $\text{K}$.

Taking the ratio of ${\upsilon}_{R M S}$ for each gas, we effectively obtain Graham's law of effusion:

${\upsilon}_{R M S , {N}_{2}} / \left({\upsilon}_{R M S , C {l}_{2}}\right) = \frac{\sqrt{\frac{\cancel{3 R T}}{M} _ \left({N}_{2}\right)}}{\sqrt{\frac{\cancel{3 R T}}{M} _ \left(C {l}_{2}\right)}} \propto {r}_{{N}_{2}} / \left({r}_{C {l}_{2}}\right)$

where ${r}_{i}$ is the rate of effusion of gas $i$.

Thus:

${r}_{{N}_{2}} / \left({r}_{C {l}_{2}}\right) = \sqrt{{M}_{C {l}_{2}} / \left({M}_{{N}_{2}}\right)}$

or, more generally,

$\boldsymbol{{r}_{A} / \left({r}_{B}\right) = \sqrt{{M}_{B} / \left({M}_{A}\right)}}$

In this equation, it does not matter what units of molar mass you use, as long as they are the same on the righthand side numerator as in the denominator.

Thus, the rate of effusion of ${\text{N}}_{2}$ relative to ${\text{Cl}}_{2}$ is:

$\textcolor{b l u e}{{r}_{{N}_{2}} / \left({r}_{C {l}_{2}}\right)} = \sqrt{\frac{35.453 \cdot 2}{14.007 \cdot 2}}$

$= \textcolor{b l u e}{1.59}$

So, gaseous nitrogen molecule effuses about 1.59 times as quickly as gaseous chlorine molecule. This should make sense because nitrogen is lighter.