A box of 1 L capacity is divided into two equal compartments by a thin partition which are filled with 2g `H_(2)` and 16 g`CH_(4)` respectively. The pressure in each compartment is reorded as P atm. The total pressure when partition is removed will be:

Options

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

Explanation

To solve the problem step by step, we will use the ideal gas law and the concept of partial pressures.

### Step 1: Determine the moles of each gas in the compartments.
1. **For Hydrogen (H₂)**:
- Given mass = 2 g
- Molar mass of H₂ = 2 g/mol
- Moles of H₂ (n₁) = mass / molar mass = 2 g / 2 g/mol = 1 mol

2. **For Methane (CH₄)**:
- Given mass = 16 g
- Molar mass of CH₄ = 16 g/mol
- Moles of CH₄ (n₂) = mass / molar mass = 16 g / 16 g/mol = 1 mol

### Step 2: Calculate the pressure in each compartment before the partition is removed.
- The volume of each compartment = 1 L / 2 = 0.5 L

Using the ideal gas law, \( PV = nRT \):
- For H₂:
\[ P_1 \cdot 0.5 = n_1 \cdot R \cdot T \implies P_1 = \frac{n_1 \cdot R \cdot T}{0.5} = \frac{1 \cdot R \cdot T}{0.5} = 2 \cdot \frac{R \cdot T}{1} \]
Thus, \( P_1 = 2P \) (where P is a constant derived from \( R \cdot T \)).

- For CH₄:
\[ P_2 \cdot 0.5 = n_2 \cdot R \cdot T \implies P_2 = \frac{n_2 \cdot R \cdot T}{0.5} = \frac{1 \cdot R \cdot T}{0.5} = 2 \cdot \frac{R \cdot T}{1} \]
Thus, \( P_2 = 2P \).

### Step 3: Calculate the total pressure after the partition is removed.
- When the partition is removed, the total volume becomes 1 L, and the total number of moles of gas is:
\[ n_{total} = n_1 + n_2 = 1 + 1 = 2 \text{ moles} \]

Using the ideal gas law again for the total pressure \( P_{total} \):
\[ P_{total} \cdot 1 = n_{total} \cdot R \cdot T \implies P_{total} = n_{total} \cdot R \cdot T = 2 \cdot R \cdot T \]

### Step 4: Relate the total pressure back to the individual pressures.
Since we previously established that \( P_1 = 2P \) and \( P_2 = 2P \), the total pressure can also be expressed as:
\[ P_{total} = P_1 + P_2 = 2P + 2P = 4P \]

### Conclusion:
Thus, the total pressure when the partition is removed is:
\[ P_{total} = 4P \]

### Final Answer:
The total pressure when the partition is removed will be \( 4P \) atm.

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