A system consists of two types of gas molecules A and B having same number density 2 \times 1025/m3. The diameter of A and B are 10 \mathop A\limits^o and 5 \mathop A\limits^o respectively. They suffer collision at room temperature. The ratio of average distance covered by the molecule A to that of B between two successive collision is \underline\hspace2cm \times 10 - 2

Heat and Thermodynamics PYQ — JEE Main 2021 · 25 Jul · Shift 2 · Q64 with Solution

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Correct answer: 25

The “average distance covered by a molecule between two successive collisions” is the mean free path $\lambda$ .

For a gas,

\lambda = \frac{1}{\sqrt{2}\,\pi d^2 n}

where:

Both gases have the same number density:

n_A = n_B = 2\times 10^{25}\,\text{m}^{-3}

Diameters are:

d_A = 10\,\text{\AA}, \qquad d_B = 5\,\text{\AA}

Since the number density is same, mean free path is inversely proportional to $d^2$ :

\lambda \propto \frac{1}{d^2}

\frac{\lambda_A}{\lambda_B} = \frac{d_B^2}{d_A^2}

Substitute values:

\frac{\lambda_A}{\lambda_B} = \frac{5^2}{10^2} = \frac{25}{100} = \frac{1}{4}

So,

\frac{\lambda_A}{\lambda_B} = 0.25 = 25\times 10^{-2}

The question asks for the blank in

\underline{\hspace{2cm}}\times 10^{-2}

Hence the required number is:

25

Stored correct answer = 25

This matches our derived answer.