An open vessel containing air is heated from `27^(@)C` to `127^(@)C`. The fraction of air originally present which goes out of it is

Options

A. `3//4`
B. `1//4`
C. `2//3`
D. `1//8`

Explanation

To solve the problem of how much air goes out of an open vessel when heated from 27°C to 127°C, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature (in Kelvin) when pressure is constant.

### Step-by-Step Solution:

1. **Convert Temperatures to Kelvin**:
- The initial temperature \(T_1\) is \(27°C\).
- To convert to Kelvin:
\[ T_1 = 27 + 273 = 300 \, K \]
- The final temperature \(T_2\) is \(127°C\).
- To convert to Kelvin:
\[ T_2 = 127 + 273 = 400 \, K \]

2. **Apply Charles's Law**:
- According to Charles's Law:
\[ \frac{V_1}{T_1} = \frac{V_2}{T_2} \]
- Rearranging gives:
\[ \frac{V_1}{V_2} = \frac{T_1}{T_2} \]

3. **Substitute the Values**:
- We know \(T_1 = 300 \, K\) and \(T_2 = 400 \, K\):
\[ \frac{V_1}{V_2} = \frac{300}{400} = \frac{3}{4} \]

4. **Determine the Volume of Air Remaining**:
- Let’s assume the initial volume of air \(V_1\) is 1 unit.
- The volume of air remaining \(V_2\) after heating is:
\[ V_2 = \frac{3}{4} \times V_1 = \frac{3}{4} \times 1 = \frac{3}{4} \, \text{units} \]

5. **Calculate the Volume of Air that Escapes**:
- The volume of air that has escaped is:
\[ V_{\text{escaped}} = V_1 - V_2 = 1 - \frac{3}{4} = \frac{1}{4} \, \text{units} \]

6. **Find the Fraction of Air that Escaped**:
- The fraction of air that has escaped is:
\[ \text{Fraction escaped} = \frac{V_{\text{escaped}}}{V_1} = \frac{\frac{1}{4}}{1} = \frac{1}{4} \]

### Final Answer:
The fraction of air originally present which goes out of the vessel is \(\frac{1}{4}\).