Air is filled at 60 ∘ C in a vessel of open mouth. The vessel is heated to a temperature `T` so that `(1)/(4)` part of air escapes. The value of T(in `overset(0) \C`) is

Answer

To solve the problem, we need to determine the temperature \( T \) at which \( \frac{1}{4} \) of the air escapes from the vessel. We can use the ideal gas law and the concept of gas expansion with temperature to find the solution.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The initial temperature of the air in the vessel is \( T_1 = 60^\circ C \).
- We need to convert this temperature to Kelvin for calculations:
\[ T_1 = 60 + 273.15 = 333.15 \, K \]

2. **Volume and Mass Considerations**:
- Let the initial volume of air in the vessel be \( V \).
- Since \( \frac{1}{4} \) of the air escapes, the remaining volume of air in the vessel after heating is \( \frac{3}{4}V \).

3. **Using the Ideal Gas Law**:
- According to the ideal gas law, \( PV = nRT \), where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( P \) is the pressure.
- For the initial state:
\[ P \cdot V = n_1 \cdot R \cdot T_1 \]
- For the final state after heating, where \( \frac{1}{4} \) of the air has escaped:
\[ P \cdot \left(\frac{3}{4}V\right) = n_2 \cdot R \cdot T \]
- Here, \( n_2 = n_1 - \frac{1}{4}n_1 = \frac{3}{4}n_1 \).

4. **Setting Up the Equations**:
- Substitute \( n_2 \) in the second equation:
\[ P \cdot \left(\frac{3}{4}V\right) = \left(\frac{3}{4}n_1\right) \cdot R \cdot T \]
- Now we can equate the two equations:
\[ P \cdot V = n_1 \cdot R \cdot T_1 \quad \text{and} \quad P \cdot \left(\frac{3}{4}V\right) = \left(\frac{3}{4}n_1\right) \cdot R \cdot T \]

5. **Eliminating Common Terms**:
- From the first equation, we can express \( n_1 \):
\[ n_1 = \frac{PV}{RT_1} \]
- Substitute \( n_1 \) into the second equation:
\[ P \cdot \left(\frac{3}{4}V\right) = \left(\frac{3}{4} \cdot \frac{PV}{RT_1}\right) \cdot R \cdot T \]

6. **Simplifying the Equation**:
- Cancel \( P \) and \( V \) from both sides:
\[ \frac{3}{4} = \frac{3}{4} \cdot \frac{T}{T_1} \]
- This simplifies to:
\[ T = T_1 \]

7. **Finding the Final Temperature**:
- Since \( T_1 = 333.15 \, K \):
\[ T = 333.15 \, K \]
- Convert back to Celsius:
\[ T = 333.15 - 273.15 = 60^\circ C \]

### Final Answer:
The value of \( T \) in degrees Celsius is \( 60^\circ C \).