At `20^(@)C` temperature, an argon gas at atmospheric pressure is confined in a vessel with a volume of `1 m^(3)` The effective hard spere diameter of argon atom is `3.10xx10^(-10)`m. determine mean free path.

Options

A. 100 nm
B. 90 nm
C. 93.6 nm
D. 95 nm

Answer

To determine the mean free path of argon gas at a temperature of 20°C and atmospheric pressure, we can use the formula for mean free path (λ):

\[ \lambda = \frac{k \cdot T}{\sqrt{2} \cdot \pi \cdot P \cdot D^2} \]

Where:
- \( \lambda \) = mean free path
- \( k \) = Boltzmann constant \( = 1.381 \times 10^{-23} \, \text{J/K} \)
- \( T \) = absolute temperature in Kelvin
- \( P \) = pressure in Pascals
- \( D \) = effective hard sphere diameter in meters

### Step 1: Convert Temperature to Kelvin
The temperature given is 20°C. To convert this to Kelvin:

\[ T = 20 + 273 = 293 \, \text{K} \]

### Step 2: Convert Pressure to Pascals
The atmospheric pressure is given as 1 atm. In Pascals:

\[ P = 1 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa} \]

### Step 3: Use the Diameter of Argon Atom
The effective hard sphere diameter of argon is given as:

\[ D = 3.10 \times 10^{-10} \, \text{m} \]

### Step 4: Substitute Values into the Mean Free Path Formula
Now, we substitute the values into the mean free path formula:

\[ \lambda = \frac{(1.381 \times 10^{-23} \, \text{J/K}) \cdot (293 \, \text{K})}{\sqrt{2} \cdot \pi \cdot (1.013 \times 10^5 \, \text{Pa}) \cdot (3.10 \times 10^{-10} \, \text{m})^2} \]

### Step 5: Calculate the Denominator
First, calculate \( D^2 \):

\[ D^2 = (3.10 \times 10^{-10})^2 = 9.61 \times 10^{-20} \, \text{m}^2 \]

Now calculate the denominator:

\[ \text{Denominator} = \sqrt{2} \cdot \pi \cdot (1.013 \times 10^5) \cdot (9.61 \times 10^{-20}) \]

Calculating this step by step:

1. Calculate \( \sqrt{2} \approx 1.414 \)
2. Calculate \( \pi \approx 3.14 \)
3. Calculate \( \sqrt{2} \cdot \pi \approx 1.414 \cdot 3.14 \approx 4.442 \)
4. Now calculate \( 4.442 \cdot (1.013 \times 10^5) \cdot (9.61 \times 10^{-20}) \)

\[ = 4.442 \cdot 1.013 \cdot 9.61 \times 10^{-15} \approx 4.442 \cdot 9.61 \times 10^{-15} \approx 4.27 \times 10^{-14} \]

### Step 6: Calculate the Mean Free Path
Now substitute back into the mean free path formula:

\[ \lambda = \frac{(1.381 \times 10^{-23}) \cdot (293)}{4.27 \times 10^{-14}} \]

Calculating the numerator:

\[ 1.381 \times 10^{-23} \cdot 293 \approx 4.05 \times 10^{-21} \]

Now, calculate \( \lambda \):

\[ \lambda = \frac{4.05 \times 10^{-21}}{4.27 \times 10^{-14}} \approx 9.48 \times 10^{-8} \, \text{m} = 94.8 \, \text{nm} \]

### Final Answer
The mean free path of argon gas is approximately:

\[ \lambda \approx 93.6 \, \text{nm} \]