A bulb is having ideal gas at `27^(@)C`. On heating the bulb to `227^(@)C`, 2 litre of gas measured at `227^(@)C` is expelled out. The volume of bulb in litre is ____________.

Explanation

To solve the problem, we will use the ideal gas law and the relationship between volume and temperature for an ideal gas.

### Step-by-Step Solution:

1. **Convert temperatures from Celsius to Kelvin**:
- The initial temperature \( T_1 = 27^\circ C = 27 + 273 = 300 \, K \)
- The final temperature \( T_2 = 227^\circ C = 227 + 273 = 500 \, K \)

2. **Define the initial volume of the gas in the bulb**:
- Let the volume of the bulb be \( V \) liters.

3. **Determine the volume of gas expelled**:
- When heated, 2 liters of gas is expelled at \( 227^\circ C \) (or 500 K). Therefore, the total volume of gas at this temperature is \( V + 2 \) liters.

4. **Use the relationship between volume and temperature**:
- According to Charles's Law, the volume of a gas is directly proportional to its temperature when pressure and the amount of gas are constant:
\[ \frac{V_1}{V_2} = \frac{T_1}{T_2} \]
- Here, \( V_1 = V + 2 \) (total volume at \( 500 \, K \)) and \( V_2 = V \) (initial volume at \( 300 \, K \)):
\[ \frac{V + 2}{V} = \frac{500}{300} \]

5. **Cross-multiply to solve for \( V \)**:
\[ 300(V + 2) = 500V \]
\[ 300V + 600 = 500V \]
\[ 600 = 500V - 300V \]
\[ 600 = 200V \]
\[ V = \frac{600}{200} = 3 \, \text{liters} \]

6. **Final Answer**:
- The volume of the bulb is \( 3 \) liters.