An open vessel at `27^@C` is heated untill two fifth of the air (assumed as an ideal gas) in it has escaped from the vessel. Assuming that the volume of the vessel remains constant, the temperature at which the vessel has been heated is :

Options

A. `500^(@)C`
B. `750^(@)C`
C. 750 K
D. 500 K

Explanation

To solve the problem step by step, we can use the Ideal Gas Law and the information provided in the question.

### Step 1: Understand the Given Information
- Initial temperature (T1) = 27°C = 300 K (since K = °C + 273)
- Two-fifths of the air has escaped, which means three-fifths of the air remains.

### Step 2: Set Up the Ideal Gas Law
The Ideal Gas Law is given by:
\[ PV = nRT \]
Since the volume (V) and the universal gas constant (R) are constant, and the pressure (P) is also constant (as it is an open vessel), we can express the relationship between the initial and final states as:
\[ n_1 T_1 = n_2 T_2 \]

### Step 3: Define Initial and Final States
- Let \( n_1 \) be the initial number of moles of air.
- After two-fifths of the air escapes, the remaining moles \( n_2 \) can be expressed as:
\[ n_2 = n_1 - \frac{2}{5}n_1 = \frac{3}{5}n_1 \]

### Step 4: Substitute Values into the Equation
Now substituting the values into the equation:
\[ n_1 T_1 = n_2 T_2 \]
\[ n_1 \cdot 300 \, \text{K} = \left(\frac{3}{5}n_1\right) T_2 \]

### Step 5: Simplify the Equation
We can cancel \( n_1 \) from both sides (assuming \( n_1 \neq 0 \)):
\[ 300 = \frac{3}{5} T_2 \]

### Step 6: Solve for \( T_2 \)
To find \( T_2 \), multiply both sides by \( \frac{5}{3} \):
\[ T_2 = 300 \cdot \frac{5}{3} = 500 \, \text{K} \]

### Final Answer
The temperature at which the vessel has been heated is **500 K**.

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