`6 xx 10^(22)` gas molecules each of mass `10^(-24) kg` are taken in a vessel of 10 litre. What is the pressure exerted by gas molecules? The root mean square speed of gas molecules is 100 m/s.

Options

A. 20 Pa
B. `2 xx 10^(4)` Pa
C. `2 xx 10^(5)` Pa
D. `2 xx 10^(7)` Pa

Explanation

To find the pressure exerted by the gas molecules in the vessel, we can use the kinetic theory of gases. The formula we will use is derived from the ideal gas law and is given as:

\[ P = \frac{1}{3} \cdot \frac{n \cdot m \cdot v_{rms}^2}{V} \]

Where:
- \( P \) = pressure (in Pascals)
- \( n \) = number of gas molecules
- \( m \) = mass of one gas molecule (in kg)
- \( v_{rms} \) = root mean square speed (in m/s)
- \( V \) = volume of the gas (in cubic meters)

### Step-by-step Solution:

1. **Identify the given values:**
- Number of gas molecules, \( n = 6 \times 10^{22} \)
- Mass of one gas molecule, \( m = 10^{-24} \, \text{kg} \)
- Volume of the vessel, \( V = 10 \, \text{liters} = 10 \times 10^{-3} \, \text{m}^3 = 0.01 \, \text{m}^3 \)
- Root mean square speed, \( v_{rms} = 100 \, \text{m/s} \)

2. **Substitute the values into the formula:**
\[ P = \frac{1}{3} \cdot \frac{(6 \times 10^{22}) \cdot (10^{-24}) \cdot (100)^2}{0.01} \]

3. **Calculate \( v_{rms}^2 \):**
\[ v_{rms}^2 = (100)^2 = 10000 \, \text{m}^2/\text{s}^2 \]

4. **Now substitute \( v_{rms}^2 \) into the equation:**
\[ P = \frac{1}{3} \cdot \frac{(6 \times 10^{22}) \cdot (10^{-24}) \cdot (10000)}{0.01} \]

5. **Simplify the equation:**
\[ P = \frac{1}{3} \cdot \frac{(6 \times 10^{22}) \cdot (10^{-24}) \cdot (10^4)}{10^{-2}} \]
\[ = \frac{1}{3} \cdot \frac{6 \times 10^{22-24+4}}{10^{-2}} \]
\[ = \frac{1}{3} \cdot \frac{6 \times 10^{2}}{10^{-2}} \]
\[ = \frac{1}{3} \cdot (6 \times 10^{4}) \]

6. **Calculate the final pressure:**
\[ P = \frac{6 \times 10^{4}}{3} = 2 \times 10^{4} \, \text{Pa} \]

### Final Answer:
The pressure exerted by the gas molecules is \( 2 \times 10^{4} \, \text{Pa} \).