A `20` litre container at `400 K` contains `CO_(2) (g)` at pressure `0.4 atm` and an excess of `SrO` (neglect the volume of solid `SrO`). The volume of the container, when pressure of `CO_(2)` attains its maximum value, will be:
(Given that: `SrCO_(3)(s) hArr SrO(s) +CO_(2)(g)K_(p) =1.6 atm`)

Options

A. `10 litre`
B. `4 litre`
C. `2 litre`
D. `5 litre`

Explanation

To solve the problem step by step, we will use the ideal gas law and the concept of equilibrium in the given reaction.

### Step 1: Identify the given values
- Initial volume of the container, \( V_1 = 20 \) liters
- Initial pressure of \( CO_2 \), \( P_1 = 0.4 \) atm
- Equilibrium pressure of \( CO_2 \) at maximum value, \( P_2 = 1.6 \) atm (given by \( K_p \))

### Step 2: Use the relationship between pressure and volume
Since the number of moles of gas and temperature are constant, we can use the relationship:
\[ P_1 V_1 = P_2 V_2 \]
Where:
- \( V_2 \) is the final volume we need to find.

### Step 3: Substitute the known values into the equation
Substituting the known values into the equation:
\[ 0.4 \, \text{atm} \times 20 \, \text{L} = 1.6 \, \text{atm} \times V_2 \]

### Step 4: Solve for \( V_2 \)
Now, we can rearrange the equation to solve for \( V_2 \):
\[ V_2 = \frac{0.4 \times 20}{1.6} \]

### Step 5: Calculate \( V_2 \)
Calculating the right-hand side:
\[ V_2 = \frac{8}{1.6} = 5 \, \text{L} \]

### Conclusion
The volume of the container when the pressure of \( CO_2 \) attains its maximum value is \( 5 \) liters.