Tow gases-argon (atomic radius `0.07` nm, atomic weight 40) and xenon (atomic radius `0.1` nm, atomic weight 140) have the same number density and are at the same temperature. The ratio of their respective mean free times is closest to :

Options

A. `1.83`
B. `4.67`
C. `3.67`
D. `2.3`

Answer

To find the ratio of the mean free times of argon and xenon, we will follow these steps:

### Step 1: Understand the Mean Free Time Formula
The mean free time \( T \) can be expressed as:
\[ T = \frac{\lambda}{v} \]
where \( \lambda \) is the mean free path and \( v \) is the average velocity of the gas molecules.

### Step 2: Mean Free Path Expression
The mean free path \( \lambda \) is given by:
\[ \lambda = \frac{1}{\sqrt{2} \pi n d^2} \]
where \( n \) is the number density and \( d \) is the diameter of the gas molecules.

### Step 3: Average Velocity Expression
The average velocity \( v \) of gas molecules can be expressed as:
\[ v = \sqrt{\frac{3RT}{M}} \]
where \( R \) is the universal gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas.

### Step 4: Combine the Expressions
Substituting the expressions for \( \lambda \) and \( v \) into the mean free time formula, we get:
\[ T \propto \frac{\lambda}{v} \propto \frac{1}{\sqrt{2} \pi n d^2} \cdot \sqrt{\frac{M}{3RT}} \]
Since \( n \) and \( R \) are the same for both gases, we can simplify this to:
\[ T \propto \frac{M^{1/2}}{d^2} \]

### Step 5: Set Up the Ratio of Mean Free Times
Let \( T_1 \) be the mean free time for argon and \( T_2 \) for xenon. Then:
\[ \frac{T_1}{T_2} = \frac{\sqrt{M_1}}{d_1^2} \cdot \frac{d_2^2}{\sqrt{M_2}} \]
Substituting the values:
- For argon: \( M_1 = 40 \) g/mol, \( d_1 = 0.07 \) nm
- For xenon: \( M_2 = 140 \) g/mol, \( d_2 = 0.1 \) nm

### Step 6: Calculate the Ratio
Now substituting these values into the ratio:
\[ \frac{T_1}{T_2} = \frac{\sqrt{40}}{(0.07)^2} \cdot \frac{(0.1)^2}{\sqrt{140}} \]
Calculating each part:
- \( \sqrt{40} \approx 6.32 \)
- \( (0.07)^2 = 0.0049 \)
- \( (0.1)^2 = 0.01 \)
- \( \sqrt{140} \approx 11.83 \)

Now substituting these values:
\[ \frac{T_1}{T_2} = \frac{6.32}{0.0049} \cdot \frac{0.01}{11.83} \]

Calculating this gives:
\[ \frac{T_1}{T_2} \approx \frac{6.32 \cdot 0.01}{0.0049 \cdot 11.83} \approx \frac{0.0632}{0.0579} \approx 1.09 \]

### Final Answer
Thus, the ratio of their respective mean free times is closest to:
\[ \frac{T_1}{T_2} \approx 1.09 \]