1 mole of `H_(2)` gas is contained in box of volume `V= 1.00 m^(3) at T = 300 K`. The gas is heated to a temperature of T = 3000 K and the gas gets converted to a gas of hydrogen atoms. The final pressure would be (considering all gases to be ideal)

Options

A. same as the pressure initially
B. 2 times the pressure initially
C. 10 times the pressure initially
D. 20 times the pressure initially

Explanation

To solve the problem, we will use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles
- \( R \) = Ideal gas constant
- \( T \) = Temperature in Kelvin

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Initial number of moles of \( H_2 \) gas, \( n_1 = 1 \, \text{mol} \)
- Initial temperature, \( T_1 = 300 \, \text{K} \)
- Volume, \( V = 1.00 \, \text{m}^3 \)

2. **Determine Final Conditions**:
- The gas is heated to a final temperature, \( T_2 = 3000 \, \text{K} \).
- When \( H_2 \) is heated, it dissociates into hydrogen atoms, so the final number of moles, \( n_2 = 2 \times n_1 = 2 \, \text{mol} \).

3. **Use the Ideal Gas Law**:
- Since the volume is constant, we can relate the initial and final states using the Ideal Gas Law:
\[ P_1 V = n_1 R T_1 \quad \text{(initial state)} \]
\[ P_2 V = n_2 R T_2 \quad \text{(final state)} \]

4. **Divide the Two Equations**:
- By dividing the final equation by the initial equation, we get:
\[ \frac{P_2}{P_1} = \frac{n_2 T_2}{n_1 T_1} \]

5. **Substitute Known Values**:
- Substitute \( n_1 = 1 \, \text{mol} \), \( n_2 = 2 \, \text{mol} \), \( T_1 = 300 \, \text{K} \), and \( T_2 = 3000 \, \text{K} \):
\[ \frac{P_2}{P_1} = \frac{2 \times 3000}{1 \times 300} \]

6. **Calculate the Pressure Ratio**:
- Simplifying the right side:
\[ \frac{P_2}{P_1} = \frac{6000}{300} = 20 \]

7. **Final Pressure**:
- Therefore, the final pressure \( P_2 \) is:
\[ P_2 = 20 P_1 \]

### Conclusion:
The final pressure of the hydrogen atoms in the gas is 20 times the initial pressure.