A system consists of two types of gas molecules $$A$$ and $$B$$ having the same number density $$2 \times 10^{25}$$ m$$^{-3}$$. The diameter of $$A$$ and $$B$$ are 10A and 5A respectively. They suffer collisions at room temperature. The ratio of average distance covered by the molecule $$A$$ to that of $$B$$ between two successive collisions is _________ $$\times 10^{-2}$$

We are asked to compare the average distance travelled between two successive collisions, that is, the mean free path of the two kinds of molecules present in the mixture.

For a dilute gas the mean free path is given by the hard-sphere formula

$$\lambda \;=\;\dfrac{1}{\sqrt{2}\,\pi d^{2} n}\,,$$

where $$d$$ is the molecular diameter and $$n$$ is the number density of that species. We are told that both gases have the same number density

$$n_A \;=\; n_B \;=\; 2\times10^{25}\ \text{m}^{-3}\,.$$

Hence any dependence on $$n$$ will cancel out when we form the ratio of the two mean free paths.

Let

$$d_A = 10\ \text{\AA}=10\times10^{-10}\ \text{m}, \qquad d_B = 5\ \text{\AA}=5\times10^{-10}\ \text{m}.$$

Using the formula separately for the two kinds of molecules we have

$$\lambda_A = \dfrac{1}{\sqrt{2}\,\pi d_A^{2} n_A}, \qquad \lambda_B = \dfrac{1}{\sqrt{2}\,\pi d_B^{2} n_B}.$$

Taking the ratio of $$\lambda_A$$ to $$\lambda_B$$ we get

$$ \dfrac{\lambda_A}{\lambda_B} \;=\; \dfrac{\dfrac{1}{\sqrt{2}\,\pi d_A^{2} n_A}} {\dfrac{1}{\sqrt{2}\,\pi d_B^{2} n_B}} \;=\; \dfrac{d_B^{2}}{d_A^{2}}, $$

because the common factors $$\sqrt{2}\,\pi$$ and the equal number densities cancel out.

Now substituting the diameters,

$$ \dfrac{\lambda_A}{\lambda_B} \;=\; \dfrac{(5\times10^{-10})^{2}}{(10\times10^{-10})^{2}} \;=\; \dfrac{25\times10^{-20}}{100\times10^{-20}} \;=\; \dfrac{25}{100} \;=\; 0.25. $$

We can rewrite $$0.25$$ as

$$0.25 \;=\; 25 \times 10^{-2}.$$

Thus the required ratio, expressed in the form “_____ × 10−2”, has the blank filled by 25.

So, the answer is $$25$$.