Question

From a certain apparatus, the diffusion rate of hydrogen has an average value of $28.7c{m}^3{s}^{-1}.$ The diffusion of another gas under the same conditions is measured to have an average rate of $28.7c{m}^3{s}^{-1}.$ Identify the gas.

[Hint : Use Graham’s law of diffusion: $R_1/R_2={\left(\frac{M_2}{M_1}\right)}^{1/2}$, where $R_1,R_2$ are diffusion rates of gases $1$ and $2,$ and $M_1$ and $M_2$ their respective molecular masses. The law is a simple consequence of kinetic theory.]

Answer 1

Given, rate of diffusion of hydrogen $R_1=28.7c{m}^3/s$.

Rate of diffusion of another gas $R_2=7.2c{m}^3/s$.

Molecular weight of hydrogen $M_1=2.02u$.

Let molecular weight of unknown gas be $M_2$.

Using, Graham’s law

$\frac{R_1}{R_2}={\left(\frac{M_2}{M_1}\right)}^{1/2}$

${\left(\frac{R_1}{R_2}\right)}^2=\frac{M_2}{M_1}$

$M_2={\left(\frac{R_1}{R_2}\right)}^2\times M_1$

$M_2={\left(\frac{28.7}{7.2}\right)}^2\times 2.02=32.09u$

As molecular weight of another gas is $32u$, so the unknown gas is $\text{oxygen}.$