Air is filled at `60^(@)C` in a vessel of open mouth. The vessel 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. `444^(@)C`
C. `333^(@)C`
D. `171^(@)C`

Answer

To solve the problem, we will use the Ideal Gas Law and the concept of moles of gas. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Initial Conditions
- The initial temperature of the air in the vessel is given as \( T_i = 60^\circ C \).
- We need to convert this temperature to Kelvin for our calculations:
\[ T_i = 60 + 273.15 = 333.15 \, K \]

### Step 2: Define the Initial and Final Moles of Air
- Let the initial number of moles of air in the vessel be \( n_0 \).
- Since one-fourth of the air escapes, the remaining number of moles after heating is:
\[ n_f = \frac{3}{4} n_0 \]

### Step 3: Apply the Ideal Gas Law
- The Ideal Gas Law states that \( PV = nRT \). Since the vessel is open and the volume and pressure remain constant, we can express the relationship between the initial and final states:
\[ n_i T_i = n_f T_f \]
- Substituting the values we have:
\[ n_0 \cdot T_i = \left(\frac{3}{4} n_0\right) \cdot T_f \]

### Step 4: Simplify the Equation
- We can cancel \( n_0 \) from both sides (assuming \( n_0 \neq 0 \)):
\[ T_i = \frac{3}{4} T_f \]

### Step 5: Solve for the Final Temperature \( T_f \)
- Rearranging the equation gives:
\[ T_f = \frac{4}{3} T_i \]
- Now substitute \( T_i = 333.15 \, K \):
\[ T_f = \frac{4}{3} \cdot 333.15 = 444.2 \, K \]

### Step 6: Convert the Final Temperature Back to Celsius
- To convert \( T_f \) back to Celsius:
\[ T_f = 444.2 - 273.15 = 171.05^\circ C \]

### Final Result
Thus, the final temperature \( T \) at which one-fourth of the air escapes is approximately:
\[ T \approx 171.05^\circ C \]