A balloon is inflated with air in a warm living room `(24^(@)C)` to a volume `2.5 L`. It is taken out on a very cold winter's day `(-30^(@)C)` . Assuming that mass of air and pressure inside the balloon.

Explanation

To solve the problem of the balloon's volume change due to temperature change, we will use Charles's Law, which states that the volume of a gas is directly proportional to its temperature (in Kelvin) when pressure is held constant.

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin**:
- The initial temperature \( T_1 \) is given as \( 24^\circ C \).
\[ T_1 = 24 + 273 = 297 \, K \]
- The final temperature \( T_2 \) is given as \( -30^\circ C \).
\[ T_2 = -30 + 273 = 243 \, K \]

2. **Identify Given Values**:
- The initial volume \( V_1 \) is \( 2.5 \, L \).
- The initial temperature \( T_1 \) is \( 297 \, K \).
- The final temperature \( T_2 \) is \( 243 \, K \).
- We need to find the final volume \( V_2 \).

3. **Apply Charles's Law**:
- According to Charles's Law:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
- Rearranging the equation to solve for \( V_2 \):
\[ V_2 = V_1 \times \frac{T_2}{T_1} \]

4. **Substitute the Values**:
- Now, substitute the known values into the equation:
\[ V_2 = 2.5 \, L \times \frac{243 \, K}{297 \, K} \]

5. **Calculate \( V_2 \)**:
- Perform the calculation:
\[ V_2 = 2.5 \, L \times 0.8185 \approx 2.05 \, L \]

### Final Answer:
The final volume \( V_2 \) of the balloon when taken out into the cold is approximately \( 2.05 \, L \).

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