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

Answer

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

### Step-by-Step Solution:

1. **Understand the Ideal Gas Law**:
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

2. **Convert Temperatures to Kelvin**:
We need to convert the given temperatures from Celsius to Kelvin.
- Initial temperature \( T_1 = 27^\circ C = 27 + 273 = 300 \, K \)
- Final temperature \( T_2 = 52^\circ C = 52 + 273 = 325 \, K \)

3. **Relate Mass and Temperature**:
Since the pressure and volume are constant, we can relate the mass of the gas to its temperature. The mass of the gas is directly proportional to the temperature when pressure and volume are constant.
\[ \frac{m_1}{m_2} = \frac{T_1}{T_2} \]
Where:
- \( m_1 \) = initial mass of the gas = 13 g
- \( m_2 \) = final mass of the gas
- \( T_1 \) = initial temperature = 300 K
- \( T_2 \) = final temperature = 325 K

4. **Substituting Values**:
Substitute the known values into the equation:
\[ \frac{13}{m_2} = \frac{300}{325} \]

5. **Cross-Multiply and Solve for \( m_2 \)**:
Cross-multiplying gives:
\[ 13 \times 325 = 300 \times m_2 \]
\[ 4225 = 300 m_2 \]
Now, solve for \( m_2 \):
\[ m_2 = \frac{4225}{300} \approx 14.0833 \, g \]

6. **Calculate the Mass to be Released**:
The mass of the gas that has to be released to maintain the temperature at \( 52^\circ C \) is:
\[ \text{Mass released} = m_1 - m_2 = 13 \, g - 12.75 \, g \approx 0.25 \, g \]

### Final Answer:
The mass of the gas that has to be released is approximately **0.25 g**.