Two gases Ar (40) and Xe (131) at same temperature have same number density. Their diameters are `0.07 nm` and `0.10 nm` respectively. Find the ratio of their mean free time

Options

A. 1.03
B. 2.04
C. 3.04
D. 2.4

Answer

To find the ratio of the mean free time of Argon (Ar) and Xenon (Xe) gases, we can use the formula for mean free time (τ):

\[ \tau = \frac{1}{\sqrt{2} n \pi d^2 v_{rm}} \]

Where:
- \( n \) is the number density of the gas,
- \( d \) is the diameter of the gas molecules,
- \( v_{rm} \) is the root mean square speed of the gas molecules.

Since both gases have the same number density and are at the same temperature, we can simplify the comparison of their mean free times by focusing on the diameters of the gases.

### Step 1: Write the Mean Free Time for Both Gases

Let:
- \( \tau_{Ar} \) be the mean free time for Argon,
- \( \tau_{Xe} \) be the mean free time for Xenon.

Using the formula, we have:

\[ \tau_{Ar} = \frac{1}{\sqrt{2} n \pi d_{Ar}^2 v_{rm,Ar}} \]

\[ \tau_{Xe} = \frac{1}{\sqrt{2} n \pi d_{Xe}^2 v_{rm,Xe}} \]

### Step 2: Take the Ratio of Mean Free Times

Now, we can take the ratio of the mean free times:

\[ \frac{\tau_{Ar}}{\tau_{Xe}} = \frac{d_{Xe}^2 v_{rm,Xe}}{d_{Ar}^2 v_{rm,Ar}} \]

### Step 3: Simplify the Expression

Since both gases are at the same temperature, the root mean square speed \( v_{rm} \) can be expressed as:

\[ v_{rm} = \sqrt{\frac{3kT}{m}} \]

Where \( m \) is the mass of the gas molecules. However, since we are looking for the ratio, we can ignore \( v_{rm} \) for now, as it will cancel out when we take the ratio.

Thus, we have:

\[ \frac{\tau_{Ar}}{\tau_{Xe}} = \frac{d_{Xe}^2}{d_{Ar}^2} \]

### Step 4: Substitute the Given Values

Given:
- Diameter of Argon, \( d_{Ar} = 0.07 \, \text{nm} \)
- Diameter of Xenon, \( d_{Xe} = 0.10 \, \text{nm} \)

Now substituting these values into the ratio:

\[ \frac{\tau_{Ar}}{\tau_{Xe}} = \frac{(0.10)^2}{(0.07)^2} \]

Calculating the squares:

\[ = \frac{0.01}{0.0049} \]

### Step 5: Calculate the Final Ratio

Now, calculate the ratio:

\[ = \frac{0.01}{0.0049} \approx 2.04 \]

### Conclusion

Thus, the ratio of the mean free time of Argon to that of Xenon is:

\[ \frac{\tau_{Ar}}{\tau_{Xe}} \approx 2.04 \]