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

Answer

To solve the problem, we need to calculate the total pressure in a box divided into two compartments after the partition is removed. The box has a total volume of 1 L, with each compartment having a volume of 0.5 L. One compartment contains 2 g of \( H_2 \) and the other contains 16 g of \( CH_4 \).

### Step-by-Step Solution:

**Step 1: Calculate the number of moles of \( H_2 \) and \( CH_4 \)**

1. **For \( H_2 \)**:
- Molar mass of \( H_2 \) = 2 g/mol
- Mass of \( H_2 \) = 2 g
- Number of moles of \( H_2 \) (\( n_{H_2} \)):
\[ n_{H_2} = \frac{\text{mass}}{\text{molar mass}} = \frac{2 \, \text{g}}{2 \, \text{g/mol}} = 1 \, \text{mol} \]

2. **For \( CH_4 \)**:
- Molar mass of \( CH_4 \) = 16 g/mol
- Mass of \( CH_4 \) = 16 g
- Number of moles of \( CH_4 \) (\( n_{CH_4} \)):
\[ n_{CH_4} = \frac{\text{mass}}{\text{molar mass}} = \frac{16 \, \text{g}}{16 \, \text{g/mol}} = 1 \, \text{mol} \]

**Step 2: Calculate the pressure in each compartment before the partition is removed**

Using the ideal gas law \( PV = nRT \):

- For each compartment:
- Volume of each compartment \( V = 0.5 \, \text{L} = 0.5 \, \text{dm}^3 \)
- Number of moles in each compartment \( n = 1 \, \text{mol} \)
- Using \( R = 0.0821 \, \text{L atm/(K mol)} \) and assuming temperature \( T \) is constant.

For \( H_2 \):
\[ P_{H_2} \cdot 0.5 = 1 \cdot R \cdot T \implies P_{H_2} = \frac{2RT}{1} = 2RT \]

For \( CH_4 \):
\[ P_{CH_4} \cdot 0.5 = 1 \cdot R \cdot T \implies P_{CH_4} = \frac{2RT}{1} = 2RT \]

**Step 3: Calculate the total pressure after removing the partition**

When the partition is removed, the total number of moles is:
\[ n_{total} = n_{H_2} + n_{CH_4} = 1 + 1 = 2 \, \text{mol} \]

The total volume is now 1 L. Using the ideal gas law again:
\[ P' \cdot 1 = 2 \cdot R \cdot T \implies P' = 2RT \]

**Step 4: Conclusion**

From the calculations, we see that:
\[ P' = 2RT \]
This means that the total pressure when the partition is removed is equal to the pressure recorded in each compartment before the partition was removed, which is \( P \).

Thus, the total pressure when the partition is removed is:
\[ \text{Total Pressure} = P \]