A cylinder of capacity `20 L` is filled with `H_(2)` gas. The total average kinetic energy of translatory motion of its molecules is `1,5 xx 10^(5) J`. The pressure of hydrogen in the cylinder is

Options

A. `2 xx 10^(6) N//m^(2)`
B. `3 xx 10^(6) N//m^(2)`
C. `4 xx 10^(6) N//m^(2)`
D. `5 xx 10^(6) N//m^(2)`

Explanation

To find the pressure of hydrogen gas in the cylinder, we can use the relationship between the average kinetic energy of gas molecules and pressure. The formula we will use is:

\[ KE = \frac{3}{2} PV \]

Where:
- \( KE \) is the total average kinetic energy,
- \( P \) is the pressure,
- \( V \) is the volume of the gas.

### Step-by-Step Solution:

1. **Identify the given values**:
- Total average kinetic energy, \( KE = 1.5 \times 10^5 \, J \)
- Volume of the cylinder, \( V = 20 \, L \)

2. **Convert the volume from liters to cubic meters**:
- Since \( 1 \, L = 10^{-3} \, m^3 \),
- Therefore, \( V = 20 \, L = 20 \times 10^{-3} \, m^3 = 0.02 \, m^3 \).

3. **Rearrange the kinetic energy formula to solve for pressure \( P \)**:
\[ P = \frac{2 \times KE}{3V} \]

4. **Substitute the values into the formula**:
\[ P = \frac{2 \times (1.5 \times 10^5)}{3 \times (0.02)} \]

5. **Calculate the numerator**:
\[ 2 \times (1.5 \times 10^5) = 3.0 \times 10^5 \]

6. **Calculate the denominator**:
\[ 3 \times (0.02) = 0.06 \]

7. **Now divide the numerator by the denominator**:
\[ P = \frac{3.0 \times 10^5}{0.06} \]

8. **Perform the division**:
\[ P = 5.0 \times 10^6 \, \text{N/m}^2 \]

9. **Final result**:
The pressure of hydrogen in the cylinder is:
\[ P = 5.0 \times 10^6 \, \text{Pa} \, \text{(or N/m}^2\text{)} \]