An open vessel at `37^(@)`C is heated until `3//5` of the air in it has been expelled. Assuming that the volume of the vessel remains constant, the temperature to which the vessel is heated is :

Options

A. `502^(@)`C
B. 502 K
C. `243.67^(@)`
D. `92.5^(@)`C

Explanation

To solve the problem, we will use the ideal gas law concept, specifically the relation between the number of moles (n), temperature (T), and the fact that the volume of the vessel remains constant.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have an open vessel at an initial temperature of \( T_1 = 37^\circ C \).
- We need to find the final temperature \( T_2 \) after \( \frac{3}{5} \) of the air has been expelled from the vessel.

2. **Convert Initial Temperature to Kelvin**:
- The initial temperature in Kelvin is calculated as:
\[ T_1 = 37 + 273 = 310 \, K \]

3. **Determine the Moles of Air**:
- Let the initial number of moles of air in the vessel be \( n_1 = n \).
- After heating, \( \frac{3}{5} \) of the air is expelled, which means \( \frac{2}{5} \) of the air remains.
- Therefore, the number of moles remaining is:
\[ n_2 = \frac{2}{5} n \]

4. **Apply the Ideal Gas Law Relation**:
- For an open vessel, we can use the relation:
\[ n_1 T_1 = n_2 T_2 \]
- Substituting the values we have:
\[ n \cdot 310 = \left(\frac{2}{5} n\right) T_2 \]

5. **Cancel Out the Moles**:
- Since \( n \) is common on both sides, we can cancel it out:
\[ 310 = \frac{2}{5} T_2 \]

6. **Solve for \( T_2 \)**:
- Rearranging the equation gives:
\[ T_2 = 310 \cdot \frac{5}{2} \]
- Calculating \( T_2 \):
\[ T_2 = 310 \cdot 2.5 = 775 \, K \]

7. **Convert \( T_2 \) Back to Celsius**:
- To convert Kelvin back to Celsius:
\[ T_2 = 775 - 273 = 502^\circ C \]

### Final Answer:
The temperature to which the vessel is heated is \( 775 \, K \) or \( 502^\circ C \).

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