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 step by step, we will use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure is held constant. The formula can be expressed as:

\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]

### Step 1: Convert Temperatures to Kelvin
First, we need to convert the given temperatures from Celsius to Kelvin:
- Initial temperature \( T_1 = 27^\circ C = 27 + 273 = 300 \, K \)
- Final temperature \( T_2 = 227^\circ C = 227 + 273 = 500 \, K \)

### Step 2: Define Variables
Let:
- \( V \) = volume of the bulb in liters
- The volume of gas expelled = 2 liters

When the gas is heated to \( 227^\circ C \), the total volume of gas in the bulb becomes \( V + 2 \) liters.

### Step 3: Set Up the Equation Using Charles's Law
According to Charles's Law:
\[ \frac{V + 2}{V} = \frac{T_2}{T_1} \]
Substituting the temperatures we converted:
\[ \frac{V + 2}{V} = \frac{500}{300} \]

### Step 4: Simplify the Equation
Cross-multiplying gives:
\[ 300(V + 2) = 500V \]
Expanding the left side:
\[ 300V + 600 = 500V \]

### Step 5: Rearranging the Equation
Rearranging the equation to isolate \( V \):
\[ 500V - 300V = 600 \]
\[ 200V = 600 \]

### Step 6: Solve for \( V \)
Dividing both sides by 200:
\[ V = \frac{600}{200} = 3 \, \text{liters} \]

### Final Answer
The volume of the bulb is **3 liters**.

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