# How would you rank the following gases, NO, Ar, N_2, N_2O_5, in order of increasing speed of effusion through a tiny opening and increasing time of effusion?

Oct 16, 2016

${Z}_{\text{eff",N_2) > Z_("eff",NO) > Z_("eff",Ar) > Z_("eff} , {N}_{2} {O}_{5}}$

Clearly, the rate is the reciprocal of the time it takes, since the rate is in ${\text{s}}^{- 1}$ and the time is in $\text{s}$. Therefore:

${t}_{\text{eff",N_2) < t_("eff",NO) < t_("eff",Ar) < t_("eff} , {N}_{2} {O}_{5}}$

The rate of effusion, which I will denote as ${Z}_{\text{eff}}$, is defined in Graham's Law of Effusion as:

Z_("eff",1)/(Z_("eff",2)) = sqrt(M_(m,2)/(M_(m,1))

where ${M}_{m , i}$ is the molar mass of compound $i$ and ${Z}_{\text{eff} , i}$ is the rate of effusion of compound $i$.

Since ${Z}_{\text{eff} , 1} \propto \frac{1}{\sqrt{{M}_{m , 1}}}$ the higher the molar mass, the slower the rate of effusion, which makes sense since the larger the gas, the less often it can get through a small hole.

${M}_{m , N O} = 14.007 + 15.999 = \text{30.006 g/mol}$
${M}_{m , A r} = \text{39.948 g/mol}$
${M}_{m , {N}_{2}} = 2 \times 14.007 = \text{28.014 g/mol}$
${M}_{m , {N}_{2} {O}_{5}} = 2 \times 14.007 + 5 \times 15.999 = \text{108.009 g/mol}$

So, the ranking of molar masses goes as:

${M}_{m , {N}_{2}} < {M}_{m , N O} < {M}_{m , A r} < {M}_{m , {N}_{2} {O}_{5}}$

which means that:

$\textcolor{b l u e}{{Z}_{\text{eff",N_2) > Z_("eff",NO) > Z_("eff",Ar) > Z_("eff} , {N}_{2} {O}_{5}}}$