Two gases `A` and `B` having the same temperature `T`, same pressure `P` and same volume `V` are mixed. If the mixture is at the same temperature and occupies a volume `V`, then the pressure of the mixture is

Options

A. `P`
B. `2 P`
C. `4 P`
D. `6 P`

Explanation

To solve the problem, we need to analyze the situation involving the two gases A and B.

### Step-by-Step Solution:

1. **Understanding Initial Conditions**:
- Both gases A and B are at the same temperature \( T \), pressure \( P \), and occupy the same volume \( V \).
- Therefore, we can denote the initial conditions for both gases as:
- Gas A: \( P_A = P, V_A = V, T_A = T \)
- Gas B: \( P_B = P, V_B = V, T_B = T \)

2. **Total Volume Before Mixing**:
- When we mix gases A and B, the total volume before mixing is:
\[ V_{\text{total}} = V_A + V_B = V + V = 2V \]

3. **Conditions After Mixing**:
- After mixing, the problem states that the mixture occupies a volume \( V \) and maintains the same temperature \( T \).

4. **Using Ideal Gas Law**:
- The ideal gas law states that:
\[ PV = nRT \]
- Where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature.

5. **Calculating Moles of Each Gas**:
- For gas A:
\[ n_A = \frac{PV}{RT} \]
- For gas B:
\[ n_B = \frac{PV}{RT} \]
- Total moles after mixing:
\[ n_{\text{total}} = n_A + n_B = \frac{PV}{RT} + \frac{PV}{RT} = \frac{2PV}{RT} \]

6. **Pressure of the Mixture**:
- Now, we apply the ideal gas law to the mixture occupying volume \( V \):
\[ P_{\text{mixture}} \cdot V = n_{\text{total}} \cdot RT \]
- Substituting \( n_{\text{total}} \):
\[ P_{\text{mixture}} \cdot V = \left(\frac{2PV}{RT}\right) \cdot RT \]
- Simplifying this gives:
\[ P_{\text{mixture}} \cdot V = 2PV \]
- Dividing both sides by \( V \):
\[ P_{\text{mixture}} = 2P \]

7. **Final Result**:
- Therefore, the pressure of the mixture is:
\[ P_{\text{mixture}} = 2P \]

### Conclusion:
The pressure of the mixture of gases A and B is \( 2P \).