Question

The experiment value of d is found to be smaller than the estimate obtained using Graham's law. This is due to:

• A. larger mean free path for X as compared to that of Y
• B. larger mean free path for Y as compared to that of X
• C. increased collision frequency of Y with the inert gas as compared to that of X with the inert gas
• D. increased collision frequency X with the inert gas as compared of Y with the inert gas

Answer 1

The correct option is D increased collision frequency X with the inert gas as compared of Y with the inert gas

We know,

Rate of diffusion $\alpha \frac{1}{Molecular-weight}$

Let distance covered by X be ${}^{\prime }{d}^{\prime }$ and distance covered by Y is ${}^{\prime }24-d^{\prime }$

Now,

$\frac{r_X}{r_Y}=\frac{d}{24-d}$

$\implies \dfrac{d}{24-d}= \sqrt{\dfra

$⟹\frac{d}{24-d}=\sqrt{\frac{40}{10}}$

$⟹d=16cm$

• Here, the value of d is smaller than the estimate obtained using Graham's law due to the increased collision frequency X with the inert gas as compared of Y with inert gas,