An open mouthed bottle contains a gas at `60^(@)C` the temperature to which the bottle should be heated so that 1/4 of the mass of the gas may leave is,

Options

A. `171^(@)C`
B. `250^(@)C`
C. `300^(@)C`
D. `342^(@)C`

Explanation

To solve the problem, we need to determine the temperature to which the bottle should be heated so that one-fourth of the mass of the gas may leave. We will use the ideal gas law and the relationship between temperature and the number of moles of gas.

### Step-by-Step Solution:

1. **Understand the problem**: We have a gas in an open-mouthed bottle at an initial temperature of \( T_1 = 60^\circ C \). We need to find the new temperature \( T_2 \) at which one-fourth of the gas mass escapes.

2. **Convert the initial temperature to Kelvin**:
\[ T_1 = 60^\circ C + 273 = 333 \, K \]

3. **Identify the relationship between the number of moles and temperature**: Since the volume and pressure are constant, we can use the relation:
\[ \frac{T_1}{T_2} = \frac{n_2}{n_1} \]
where \( n_1 \) is the initial number of moles and \( n_2 \) is the number of moles after one-fourth has escaped.

4. **Determine the number of moles after one-fourth has escaped**:
If \( n_1 \) is the initial number of moles, then:
\[ n_2 = n_1 - \frac{1}{4}n_1 = \frac{3}{4}n_1 \]

5. **Substitute \( n_2 \) into the equation**:
\[ \frac{T_1}{T_2} = \frac{3/4n_1}{n_1} = \frac{3}{4} \]

6. **Rearranging to find \( T_2 \)**:
\[ T_2 = T_1 \cdot \frac{4}{3} \]

7. **Calculate \( T_2 \)**:
\[ T_2 = 333 \, K \cdot \frac{4}{3} = 444 \, K \]

8. **Convert \( T_2 \) back to Celsius**:
\[ T_2 = 444 \, K - 273 = 171^\circ C \]

9. **Final Answer**: The temperature to which the bottle should be heated is \( 171^\circ C \).