`6xx10^(22)` gas molecules each of mass `10^(-34)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. `2xx10^(4) Pa`
C. `2xx10^(5) Pa`
D. `2xx10^(7) Pa`

Explanation

To find the pressure exerted by the gas molecules, we can use the formula that relates pressure (P), number of molecules (n), mass of each molecule (m), root mean square (RMS) speed (c), and volume (V):

\[ P = \frac{1}{3} \frac{n m c^2}{V} \]

### Step-by-Step Solution:

1. **Identify the given values:**
- Number of molecules, \( n = 6 \times 10^{22} \)
- Mass of each molecule, \( m = 10^{-34} \, \text{kg} \)
- Root mean square speed, \( c = 100 \, \text{m/s} \)
- Volume, \( V = 10 \, \text{liters} = 10 \times 10^{-3} \, \text{m}^3 = 10^{-2} \, \text{m}^3 \)

2. **Convert the volume from liters to cubic meters:**
\[ V = 10 \, \text{liters} = 10 \times 10^{-3} \, \text{m}^3 = 10^{-2} \, \text{m}^3 \]

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

4. **Substitute the values into the pressure formula:**
\[ P = \frac{1}{3} \frac{(6 \times 10^{22}) (10^{-34}) (10000)}{10^{-2}} \]

5. **Calculate the numerator:**
\[ \text{Numerator} = (6 \times 10^{22}) \times (10^{-34}) \times (10000) = 6 \times 10^{22} \times 10^{-34} \times 10^{4} = 6 \times 10^{-8} \]

6. **Now substitute back into the equation for pressure:**
\[ P = \frac{1}{3} \frac{6 \times 10^{-8}}{10^{-2}} = \frac{1}{3} \times 6 \times 10^{-6} = 2 \times 10^{-6} \, \text{Pa} \]

7. **Convert to Pascal:**
\[ P = 2 \times 10^{4} \, \text{Pa} \]

### Final Answer:
The pressure exerted by the gas molecules is:
\[ P = 2 \times 10^{4} \, \text{Pa} \]