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 and the relationships between pressure, volume, temperature, and the number of moles of gas.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Initial number of moles, \( N_1 = 1 \, \text{mol} \)
- Initial volume, \( V = 1.00 \, \text{m}^3 \)
- Initial temperature, \( T_1 = 300 \, \text{K} \)

2. **Determine Final Conditions**:
- Final temperature, \( T_2 = 3000 \, \text{K} \)
- When the hydrogen gas is heated, it dissociates into hydrogen atoms. Therefore, the final number of moles, \( N_2 = 2 \times N_1 = 2 \, \text{mol} \).

3. **Use the Ideal Gas Law**:
The ideal gas law states that:
\[ PV = nRT \]
where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the ideal gas constant, and \( T \) is temperature.

4. **Relate Initial and Final States**:
Using the relationship for the initial and final states, we can write:
\[ \frac{P_1}{N_1 T_1} = \frac{P_2}{N_2 T_2} \]
Rearranging gives:
\[ P_2 = P_1 \cdot \frac{N_2}{N_1} \cdot \frac{T_2}{T_1} \]

5. **Substitute Known Values**:
- We know \( N_2 = 2 \, \text{mol} \), \( N_1 = 1 \, \text{mol} \), \( T_2 = 3000 \, \text{K} \), and \( T_1 = 300 \, \text{K} \).
- Substituting these values into the equation:
\[ P_2 = P_1 \cdot \frac{2}{1} \cdot \frac{3000}{300} \]
\[ P_2 = P_1 \cdot 2 \cdot 10 \]
\[ P_2 = 20 P_1 \]

6. **Conclusion**:
The final pressure \( P_2 \) is 20 times the initial pressure \( P_1 \).

### Final Answer:
The final pressure would be \( 20 P_1 \).

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