Question

Two gases A and B are kept in two cylinders and allowed to diffuse through a small hole. If the time taken by gas A is four times the time taken by gas B for the diffusion of an equal volume of gas under similar conditions of temperature and pressure, what will be the ratio of their molecular masses?

Answer 1

• From Graham's Law of diffusion, the rate of diffusion of a gas is inversely proportional to the square root of its molecular weight.
• r = $\frac{1}{\sqrt{\mathrm{M}}}$
• let the time of diffusion of B is t.

Rate of diffusion of B= $\frac{\mathrm{Volume}\mathrm{of}\mathrm{gas}\mathrm{B}\mathrm{diffused}\left({\mathrm{V}}_1\right)}{\mathrm{t}}$

Time for diffusion of A = 4t

Rate of diffusion of A = $\frac{{\mathrm{V}}_2}{4\mathrm{t}}$

Since V₁=V₂ , $\frac{{\mathrm{r}}_{\mathrm{A}}}{{\mathrm{r}}_{\mathrm{B}}}$= $\frac{{\mathrm{V}}_1}{4\mathrm{t}}\times \frac{t}{V_2}$=$\sqrt{\frac{{\mathrm{M}}_{\mathrm{B}}}{{\mathrm{M}}_{\mathrm{A}}}}$ Square the terms

then $\frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{M}}_{\mathrm{B}}}$=$\frac{16}{1}$ i.e., the ratio is 16:1.

Hence the ratio of their molecular masses is 16:1