Air is filled at `60^(@)C` in a vessel of open mouth. The vessle is heated to a temperature `T` so that `1//4th` of air escapes. Assuming the volume of vessel remaining constant, the value of `T` is

Options

A. `80^(@)C`
B. `440^(@)C`
C. `333^(@)C`
D. `171^(@)C`

Explanation

To solve the problem, we will use the ideal gas law and the concept of constant pressure and volume. Here are the steps to find the temperature \( T \) after heating the vessel:

### Step 1: Understand the Initial Conditions
We have air in a vessel at an initial temperature \( T_1 = 60^\circ C \). We need to convert this temperature to Kelvin for calculations:
\[ T_1 = 60 + 273 = 333 \, K \]

### Step 2: Determine the Initial Amount of Air
Let the initial amount of air in the vessel be \( n_1 \).

### Step 3: Calculate the Final Amount of Air
Since one-fourth of the air escapes when the vessel is heated, the remaining amount of air is:
\[ n_2 = \frac{3}{4} n_1 \]

### Step 4: Set Up the Ideal Gas Equation
According to the ideal gas law, we have:
\[ PV = nRT \]
Since the pressure \( P \) and volume \( V \) are constant, we can relate the initial and final states of the gas:
\[ n_1 T_1 = n_2 T_2 \]

### Step 5: Substitute Known Values
Substituting \( n_2 \) into the equation gives:
\[ n_1 T_1 = \left(\frac{3}{4} n_1\right) T_2 \]
We can cancel \( n_1 \) from both sides (assuming \( n_1 \neq 0 \)):
\[ T_1 = \frac{3}{4} T_2 \]

### Step 6: Solve for \( T_2 \)
Rearranging the equation to solve for \( T_2 \):
\[ T_2 = \frac{4}{3} T_1 \]

### Step 7: Substitute \( T_1 \) into the Equation
Now substitute \( T_1 = 333 \, K \):
\[ T_2 = \frac{4}{3} \times 333 = 444 \, K \]

### Step 8: Convert \( T_2 \) Back to Celsius
To convert \( T_2 \) back to Celsius:
\[ T_2 = 444 - 273 = 171^\circ C \]

### Final Answer
Thus, the final temperature \( T \) after heating the vessel is:
\[ \boxed{171^\circ C} \]

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