A container is divided into two chambers by a partition. The volume of first chamber is 4.5 litre and second chamber is 5.5 litre. The first chamber contain 3.0 moles of gas at pressure 2.0 atm and second chamber contain 4.0 moles of gas at pressure 3.0 atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is `"x"xx10^(-1)` atm. Value of x is`"___"`.

Answer

To solve the problem, we need to find the common equilibrium pressure after the partition is removed. We will use the ideal gas law, which states that \( PV = nRT \). However, since the temperature and the gas constant \( R \) remain constant, we can simplify our calculations.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Chamber 1:
- Volume \( V_1 = 4.5 \) L
- Moles \( n_1 = 3.0 \) moles
- Pressure \( P_1 = 2.0 \) atm
- Chamber 2:
- Volume \( V_2 = 5.5 \) L
- Moles \( n_2 = 4.0 \) moles
- Pressure \( P_2 = 3.0 \) atm

2. **Calculate the Initial Conditions:**
- For Chamber 1, using the ideal gas equation:
\[ P_1 V_1 = n_1 RT \implies 2.0 \times 4.5 = 3.0 \times R \times T \implies 9 = 3RT \implies RT = 3 \text{ atm L} \]
- For Chamber 2:
\[ P_2 V_2 = n_2 RT \implies 3.0 \times 5.5 = 4.0 \times R \times T \implies 16.5 = 4RT \implies RT = 4.125 \text{ atm L} \]

3. **Find the Total Moles and Total Volume:**
- Total moles after removing the partition:
\[ n_{total} = n_1 + n_2 = 3.0 + 4.0 = 7.0 \text{ moles} \]
- Total volume after removing the partition:
\[ V_{total} = V_1 + V_2 = 4.5 + 5.5 = 10.0 \text{ L} \]

4. **Calculate the Equilibrium Pressure:**
- Using the ideal gas law for the entire system at equilibrium:
\[ P_{equilibrium} V_{total} = n_{total} RT \]
- We can substitute \( RT \) from either chamber (let's use Chamber 1):
\[ P_{equilibrium} \times 10.0 = 7.0 \times 3 \]
\[ P_{equilibrium} \times 10.0 = 21 \implies P_{equilibrium} = \frac{21}{10} = 2.1 \text{ atm} \]

5. **Express the Final Pressure in the Required Form:**
- The problem states that the common equilibrium pressure is \( x \times 10^{-1} \) atm. Thus:
\[ 2.1 \text{ atm} = 21 \times 10^{-1} \text{ atm} \]
- Therefore, \( x = 21 \).

### Final Answer:
The value of \( x \) is **21**.