A flask is filled with `13 g` of an ideal gas at `27^(@)C` and its temperature is raised to `52^(@)C`. The mass of the gas that has to be released to maintain the temperature of the gas in the flask at `52^(@)C`, the pressure remaining the same is

Options

A. `2.5 g`
B. `2.0 g`
C. `1.5 g`
D. `1.0 g`

Explanation

To solve the problem, we will use the ideal gas law and the relationship between mass, temperature, and the number of moles of an ideal gas.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- Initial mass of gas, \( W_1 = 13 \, \text{g} \)
- Initial temperature, \( T_1 = 27^\circ C = 27 + 273 = 300 \, \text{K} \)

2. **Identify the Final Conditions:**
- Final temperature, \( T_2 = 52^\circ C = 52 + 273 = 325 \, \text{K} \)

3. **Use the Ideal Gas Law Relationship:**
- According to the ideal gas law, for a constant pressure and volume, the product of mass and temperature remains constant:
\[ W_1 \cdot T_1 = W_2 \cdot T_2 \]
- Here, \( W_2 \) is the final mass of the gas we need to find.

4. **Substitute the Known Values:**
- Substitute \( W_1 \), \( T_1 \), and \( T_2 \) into the equation:
\[ 13 \, \text{g} \cdot 300 \, \text{K} = W_2 \cdot 325 \, \text{K} \]

5. **Solve for \( W_2 \):**
- Rearranging the equation to solve for \( W_2 \):
\[ W_2 = \frac{13 \, \text{g} \cdot 300 \, \text{K}}{325 \, \text{K}} \]
- Calculate \( W_2 \):
\[ W_2 = \frac{3900 \, \text{g} \cdot \text{K}}{325 \, \text{K}} \approx 12 \, \text{g} \]

6. **Calculate the Mass of Gas to be Released:**
- The mass of the gas that has to be released to maintain the temperature at \( 52^\circ C \):
\[ \text{Mass released} = W_1 - W_2 = 13 \, \text{g} - 12 \, \text{g} = 1 \, \text{g} \]

### Final Answer:
The mass of the gas that has to be released is \( 1 \, \text{g} \).

---