A vessel contains 1 mole of `O_(2)` at `27^(@)C` and 1 atm pressure. A certain amount of the gas was withdrawn and the vessel was heated to `327^(@)C` to maintain the pressure of 1 atm. The amount of gas removed was

Options

A. 0.2 mole
B. 0.5 mole
C. 0.25 mole
D. 0.1 mole

Answer

To solve the problem step by step, we will use the ideal gas law and the relationship between the number of moles of gas and temperature under constant pressure and volume conditions.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Initial number of moles of \( O_2 \) (n1) = 1 mole
- Initial temperature (T1) = \( 27^\circ C \) = \( 27 + 273 = 300 \, K \)
- Initial pressure (P) = 1 atm (constant)

2. **Identify the Final Conditions:**
- Final temperature (T2) = \( 327^\circ C \) = \( 327 + 273 = 600 \, K \)
- Final pressure (P) = 1 atm (constant)

3. **Use the Relationship of Moles and Temperature:**
- Since pressure and volume are constant, we can use the relationship:
\[ \frac{n_1}{n_2} = \frac{T_1}{T_2} \]
- Rearranging gives:
\[ n_2 = n_1 \cdot \frac{T_2}{T_1} \]

4. **Substitute the Known Values:**
- Substitute \( n_1 = 1 \, \text{mole} \), \( T_1 = 300 \, K \), and \( T_2 = 600 \, K \):
\[ n_2 = 1 \cdot \frac{600}{300} = 1 \cdot 2 = 2 \, \text{moles} \]

5. **Calculate the Amount of Gas Removed:**
- The amount of gas removed (n removed) is given by:
\[ n_{\text{removed}} = n_1 - n_2 \]
- Substitute the values:
\[ n_{\text{removed}} = 1 - 0.5 = 0.5 \, \text{moles} \]

6. **Final Answer:**
- The amount of gas removed is **0.5 moles**.