Question

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?

Answer 1

`Z_("eff",N_2) > Z_("eff",NO) > Z_("eff",Ar) > Z_("eff",N_2O_5)`

Clearly, the rate is the reciprocal of the time it takes, since the rate is in `"s"^(-1)` and the time is in `"s"`. Therefore:

`t_("eff",N_2) < t_("eff",NO) < t_("eff",Ar) < t_("eff",N_2O_5)`

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The rate of effusion, which I will denote as `Z_"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_("eff",i)` is the rate of effusion of compound `i`.

Since `Z_("eff",1) prop 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,NO) = 14.007 + 15.999 = "30.006 g/mol"`
`M_(m,Ar) = "39.948 g/mol"`
`M_(m,N_2) = 2xx14.007 = "28.014 g/mol"`
`M_(m,N_2O_5) = 2xx14.007 + 5xx15.999 = "108.009 g/mol"`

So, the ranking of molar masses goes as:

`M_(m,N_2) < M_(m,NO) < M_(m,Ar) < M_(m,N_2O_5)`

which means that:

`color(blue)(Z_("eff",N_2) > Z_("eff",NO) > Z_("eff",Ar) > Z_("eff",N_2O_5))`