On a ship selling in pacific ocean where temperature is `23^@C` a ballon is filled with 2L air what will be the volume of the ballon when the ship reaches indian ocean where temperature is `26.1^@C`?

Options

A. 2.018L
B. 2.8L
C. 3.5L
D. 1.5L

Answer

To solve the problem of finding the volume of a balloon when the temperature changes, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature when pressure is held constant.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Initial volume of the balloon, \( V_1 = 2 \, \text{L} \)
- Initial temperature, \( T_1 = 23^\circ C \)
- Final temperature, \( T_2 = 26.1^\circ C \)

2. **Convert Temperatures to Kelvin:**
- To use Charles's Law, we need to convert the temperatures from Celsius to Kelvin.
- \( T_1 = 23 + 273 = 296 \, \text{K} \)
- \( T_2 = 26.1 + 273 = 299.1 \, \text{K} \)

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

4. **Substitute the Values:**
- Now, substituting the known values into the equation:
\[ V_2 = 2 \, \text{L} \times \frac{299.1 \, \text{K}}{296 \, \text{K}} \]

5. **Calculate \( V_2 \):**
- Performing the calculation:
\[ V_2 = 2 \times 1.010135135 = 2.02027027 \, \text{L} \]
- Rounding off, we find:
\[ V_2 \approx 2.018 \, \text{L} \]

6. **Conclusion:**
- The volume of the balloon when the ship reaches the Indian Ocean is approximately \( 2.018 \, \text{L} \).