A non-conducting container is divided into two parts of volume 4.5 Litre and 5.5 Litre, pressure 2 atmosphere and 3 atmosphere, number of moles 3 and 4. If partition valve is opened then find out common pressure (in atmosphere). (In both parts ideal gases are identical)

Answer

To find the common pressure when the partition valve is opened, we can use the concept of conservation of energy and the ideal gas law. Here’s a step-by-step solution:

### Step 1: Identify the given data
- Volume of part 1, \( V_1 = 4.5 \, \text{L} \)
- Volume of part 2, \( V_2 = 5.5 \, \text{L} \)
- Pressure in part 1, \( P_1 = 2 \, \text{atm} \)
- Pressure in part 2, \( P_2 = 3 \, \text{atm} \)
- Number of moles in part 1, \( n_1 = 3 \)
- Number of moles in part 2, \( n_2 = 4 \)

### Step 2: Calculate the total volume
The total volume \( V \) when the partition is opened is:
\[ V = V_1 + V_2 = 4.5 \, \text{L} + 5.5 \, \text{L} = 10 \, \text{L} \]

### Step 3: Calculate the total number of moles
The total number of moles \( n \) is:
\[ n = n_1 + n_2 = 3 + 4 = 7 \]

### Step 4: Use the ideal gas law to find the common pressure
Using the ideal gas law, we can express the pressure in terms of the total number of moles, total volume, and the temperature (which remains constant). The formula is:
\[ PV = nRT \]
When the partition is opened, the common pressure \( P \) can be calculated as:
\[ P = \frac{nRT}{V} \]
However, since we are looking for the common pressure after the gases mix, we can use the weighted average of the pressures based on the volumes and moles.

### Step 5: Calculate the weighted average pressure
The common pressure \( P \) can be calculated using the formula:
\[ P = \frac{P_1 V_1 + P_2 V_2}{V_1 + V_2} \]
Substituting the values:
\[ P = \frac{(2 \, \text{atm} \times 4.5 \, \text{L}) + (3 \, \text{atm} \times 5.5 \, \text{L})}{4.5 \, \text{L} + 5.5 \, \text{L}} \]
Calculating the numerator:
\[ = \frac{(9 \, \text{atm} \cdot \text{L}) + (16.5 \, \text{atm} \cdot \text{L})}{10 \, \text{L}} = \frac{25.5 \, \text{atm} \cdot \text{L}}{10 \, \text{L}} = 2.55 \, \text{atm} \]

### Final Answer
The common pressure after opening the partition valve is:
\[ \boxed{2.55 \, \text{atm}} \]