Question

The pressure in a vessel that contained pure oxygen dropped from $2000torr$ to $1500torr$ in $40\min $ as the oxygen leaked through a small hole into a vacuum. When the same vessel was filled with another gas, the pressure dropped from $2000torr$ to $1500torr$ in \[80min\] . What is the molecular weight of the second gas?

Answer 1

Hint: To solve this question, first we will find the remaining pressures of both the cases separately. And then we will relate the relationship between the molecular weight, diffusion and the time taken to diffuse of the gas. [gas leaked means the occurring of the diffusion]

Complete step by step answer:
According to the question, the given pressure of the ${O_2}$ gas vessel from $2000torr$ to $1500torr$ in $40\min $ .
So, the pressure remains in the ${O_2}$ gas vessel in $40\min $ $ = 2000torr - 1500torr = 500torr$ .
Again, in the second gas , the given pressure remains $ = 2000torr - 1500torr = 500torr$ , but in \[80min\] .
Now, as we can see that in question, oxygen leaked through a small hole into a vacuum that means diffusion is occurring here.
Now, we have to find the molecular weight of the second gas.
Therefore, according to the concept of diffusion, Diffusion is directly proportional to the Molecular weight of the gas that means if the diffusion increases then the molecular weight also increases:
$Rate\,of\,diffusion\,\propto \,Molecular\,Weight$
or, ${R_d}\alpha M$
And, also according to the concept of diffusion, Diffusion is inversely proportional to the time taken by the gas for leakage:
${R_d}\alpha \dfrac{1}{t}$
Now, the rate of change in the diffusion between both the gases:
$\therefore \dfrac{{{R_{d1}}}}{{{R_{d2}}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} $
here, ${R_{d1}}$ is the rate of diffusion of the first case; and ${R_{d2}}$ is the rate of diffusion of the second case.
${M_2}$ is the molecular weight of second gas and ${M_1}$ is the molecular weight of first gas.
And, by combining both the concepts or equations of Diffusion, there is a relation between Rate of diffusion, Molecular Weight and the time:
$\therefore \dfrac{{{R_{d1}}}}{{{R_{d2}}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} = \dfrac{{{T_2}}}{{{T_1}}}$ or,
$ \because \dfrac{{{T_2}}}{{{T_1}}} = \sqrt {\dfrac{{{M_2}}}{{{M_1}}}} \\
   \Rightarrow \dfrac{{80}}{{40}} = \sqrt {\dfrac{{{M_2}}}{{32}}} \\
   \Rightarrow \dfrac{{80 \times 80}}{{40 \times 40}} \times 32 = {M_2} \\
   \Rightarrow {M_2} = 128gm \\ $
(32 is the molecular weight of ${O_2}$ )

Hence, the molecular weight of the second gas, ${M_2}$ is $128gm$ .

Note: Molecular weight will affect the rate of diffusion. All other things being constant, lighter weight molecules will move faster or diffuse faster than heavier molecules. This property could be used to separate heavier molecules from lighter weight molecules.