A system consists of two types of gas molecules A and B having same number density 2 x 10^25 /m^3. T

Q: Why do we use $\sqrt{2}$ in the mean free path formula?

The $\sqrt{2}$ factor arises because we must consider the *relative* velocity of the gas molecules. Since all molecules are moving randomly, the relative speed between any two colliding molecules is, on average, $\sqrt{2}$ times their individual average speed.

Q: Why didn't we convert the diameters from Angstroms to meters?

Because we are calculating a ratio! When you divide $\lambda_A$ by $\lambda_B$, the units of diameter in the numerator and denominator perfectly cancel each other out. Converting to meters would just add unnecessary calculation steps and increase the chance of a silly mistake.

Q: What exactly is 'number density'?

Number density, denoted by $n$, is the total number of molecules $N$ divided by the total volume $V$ of the container ($n = N/V$). It tells us how crowded the gas is. In this problem, both gases are equally crowded.

Q: Does the temperature of the room affect the answer here?

Not in this specific setup! The formula $\lambda = \frac{1}{\sqrt{2} \pi d^2 n}$ only depends on the diameter $d$ and the number density $n$. Since $n$ is explicitly given as a constant, the temperature doesn't change our calculated ratio.

Q: Why is the final answer 25 and not 0.25?

This is a classic JEE trap! The question asks for the answer in the specific format of $...... \times 10^{-2}$. Since our ratio is $0.25$, we must rewrite it as $25 \times 10^{-2}$. Therefore, the integer that fills the blank is exactly 25.

Question

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

Sigma Percentile · Accepted Answer

### The Dance of Molecules

Imagine two types of gas molecules, A and B, zooming around chaotically in a container. Molecule A is quite large, with a diameter of $10 	ext{ \AA}$, while molecule B is smaller, measuring just $5 	ext{ \AA}$ across. We are told that they both have the exact same number density, $n = 2 	imes 10^{25} 	ext{ m}^{-3}$.

The question asks for the ratio of the average distance covered by these molecules between two successive collisions. In the world of kinetic theory, this average distance is famously known as the **mean free path**.

### The Master Equation

To solve this, we need to recall the formula for the mean free path of a gas molecule. The mean free path $\lambda$ is given by:

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

Here, $d$ is the diameter of the molecule, and $n$ is the number density (the number of molecules per unit volume). The $\sqrt{2}$ factor accounts for the relative velocity of the colliding molecules, assuming they are all in random motion.

### Setting Up the Ratio

We need to find the ratio of the mean free path of gas A to that of gas B. Let's set up the division:

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

Since the number density $n$ is the same for both gases ($n_A = n_B$), and the constants $\sqrt{2}$ and $\pi$ are present in both the numerator and the denominator, they perfectly cancel out. This beautiful cancellation leaves us with an inverse square relationship:

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

### Final Calculation

Now, we simply substitute the given diameters into our simplified ratio. The diameter of B is $5 	ext{ \AA}$, and the diameter of A is $10 	ext{ \AA}$:

$$\frac{\lambda_A}{\lambda_B} = \left(\frac{5}{10}ight)^2$$

This fraction simplifies to $\frac{1}{2}$. Squaring it gives us:

$$\frac{\lambda_A}{\lambda_B} = \frac{1}{4} = 0.25$$

Finally, we must be careful to format our answer exactly as the question demands. The question asks for the value that fills the blank in $...... 	imes 10^{-2}$.

Since $0.25$ is mathematically identical to $25 	imes 10^{-2}$, the integer we need is **25**.

Enter Numerical Value:

Visualized Solution

$n_{A} = n_{B} = 2 \times 1 0^{25} m^{- 3}$

$λ = \frac{1}{2 π d ^{2} n}$

$\frac{λ _{A}}{λ _{B}} = \frac{\frac{1}{2 π d _{A}^{2} n _{A}}}{\frac{1}{2 π d _{B}^{2} n _{B}}}$

$\frac{λ _{A}}{λ _{B}} = \frac{d _{B}^{2}}{d _{A}^{2}}$

$\frac{λ _{A}}{λ _{B}} = (\frac{5}{10})^{2}$

$\frac{λ _{A}}{λ _{B}} = 0.25$

$Answer = 25$

$λ = \frac{k _{B} T}{2 π d ^{2} p}$

The Sigma Insight: Kinetic Theory of Gases

The Dance of Molecules

The Master Equation

Setting Up the Ratio

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } nA​=nB​=2×1025 m−3

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=2​πd2n1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λB​λA​​=2​πdB2​nB​1​2​πdA2​nA​1​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λB​λA​​=dA2​dB2​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λB​λA​​=(105​)2

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λB​λA​​=0.25

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Answer=25

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ=2​πd2pkB​T​