One mole of `O_(2)` gas is contained in a box of volume `V=2m^(3)` at pressure `P_(0)` and temperature 300K The gas is now heated to 600 K and the molecules now get dissociated into oxygen atoms. The new pressure of the gas is

Options

A. `P_(0)`
B. `2P_(0)`
C. `4P_(0)`
D. `8P_(0)`

Answer

To solve the problem, we will use the Ideal Gas Law and the relationship between pressure, volume, temperature, and the number of moles of gas.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial moles of gas, \( N_1 = 1 \) mole (for \( O_2 \))
- Initial temperature, \( T_1 = 300 \, K \)
- Initial pressure, \( P_1 = P_0 \)
- Volume, \( V = 2 \, m^3 \)

2. **Identify Final Conditions:**
- Final temperature, \( T_2 = 600 \, K \)
- Since each \( O_2 \) molecule dissociates into 2 \( O \) atoms, the final moles of gas, \( N_2 = 2 \) moles (for \( O \))
- Volume remains the same, \( V = 2 \, m^3 \)

3. **Use the Ideal Gas Law:**
- The relationship between pressure, volume, temperature, and number of moles can be expressed as:
\[ \frac{P_1 \cdot N_1}{T_1} = \frac{P_2 \cdot N_2}{T_2} \]

4. **Substitute Known Values:**
- Substitute \( P_1 = P_0 \), \( N_1 = 1 \), \( T_1 = 300 \), \( N_2 = 2 \), and \( T_2 = 600 \):
\[ \frac{P_0 \cdot 1}{300} = \frac{P_2 \cdot 2}{600} \]

5. **Simplify the Equation:**
- Rearranging gives:
\[ P_0 \cdot 1 \cdot 600 = P_2 \cdot 2 \cdot 300 \]
- This simplifies to:
\[ 600 P_0 = 600 P_2 \]

6. **Solve for \( P_2 \):**
- Dividing both sides by 300:
\[ P_2 = \frac{600}{300} P_0 = 2 P_0 \]

7. **Final Calculation:**
- Since we have \( N_2 = 2 \) moles, we need to account for the doubling of moles due to dissociation:
\[ P_2 = 2 \cdot 2 P_0 = 4 P_0 \]

### Conclusion:
The new pressure of the gas after heating and dissociation is \( P_2 = 4 P_0 \).