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 time 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 relationship between the initial and final states of the gas. Here are the steps to find the final pressure \( P_2 \):

### Step 1: Identify the Given Values
- Number of moles of \( H_2 \) gas, \( n_1 = 1 \) mole
- Initial volume, \( V = 1.00 \, m^3 \)
- Initial temperature, \( T_1 = 300 \, K \)
- Final temperature, \( T_2 = 3000 \, K \)

### Step 2: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant (approximately \( 8.314 \, J/(mol \cdot K) \))
- \( T \) = temperature in Kelvin

### Step 3: Write the Initial and Final States
For the initial state:
\[ P_1 V = n_1 R T_1 \]

For the final state:
\[ P_2 V = n_2 R T_2 \]

### Step 4: Relate the Initial and Final States
Since the volume \( V \) and the gas constant \( R \) remain constant, we can relate the pressures and temperatures:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]

### Step 5: Determine the Number of Moles After Heating
When \( H_2 \) is heated to \( 3000 \, K \), it dissociates into hydrogen atoms. Each mole of \( H_2 \) produces 2 moles of \( H \):
\[ n_2 = 2 n_1 = 2 \times 1 = 2 \, \text{moles} \]

### Step 6: Substitute Values into the Equation
Now we can express \( P_2 \) in terms of \( P_1 \):
\[ P_2 = P_1 \cdot \frac{T_2}{T_1} \cdot \frac{n_2}{n_1} \]

### Step 7: Substitute Known Values
Substituting the values we have:
\[ P_2 = P_1 \cdot \frac{3000}{300} \cdot \frac{2}{1} \]
\[ P_2 = P_1 \cdot 10 \cdot 2 \]
\[ P_2 = 20 P_1 \]

### Step 8: Conclusion
The final pressure \( P_2 \) is 20 times the initial pressure \( P_1 \).