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 the number of moles, temperature, and pressure. The steps are as follows:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Initial mass of gas, \( m_1 = 13 \, \text{g} \)
- Initial temperature, \( T_1 = 27^\circ C = 27 + 273 = 300 \, \text{K} \)
- Final temperature, \( T_2 = 52^\circ C = 52 + 273 = 325 \, \text{K} \)

2. **Use the Ideal Gas Law:**
The ideal gas law can be expressed as:
\[ PV = nRT \]
Since the pressure \( P \) and the gas constant \( R \) remain constant, we can write:
\[ n_1 T_1 = n_2 T_2 \]
where \( n_1 \) and \( n_2 \) are the number of moles of gas before and after releasing some gas.

3. **Express the Number of Moles:**
The number of moles \( n \) is given by:
\[ n = \frac{m}{M} \]
where \( m \) is the mass of the gas and \( M \) is the molar mass of the gas. Therefore, we can write:
\[ n_1 = \frac{m_1}{M} = \frac{13}{M} \]
and after releasing \( x \) grams of gas, the new mass will be \( m_2 = 13 - x \), giving:
\[ n_2 = \frac{m_2}{M} = \frac{13 - x}{M} \]

4. **Set Up the Equation:**
Substitute \( n_1 \) and \( n_2 \) into the equation:
\[ \frac{13}{M} \cdot T_1 = \frac{13 - x}{M} \cdot T_2 \]
Since \( M \) is common on both sides, we can cancel it out:
\[ 13 \cdot T_1 = (13 - x) \cdot T_2 \]

5. **Substitute the Temperatures:**
Substitute \( T_1 = 300 \, \text{K} \) and \( T_2 = 325 \, \text{K} \):
\[ 13 \cdot 300 = (13 - x) \cdot 325 \]

6. **Solve for \( x \):**
Expanding the equation:
\[ 3900 = 4250 - 325x \]
Rearranging gives:
\[ 325x = 4250 - 3900 \]
\[ 325x = 350 \]
\[ x = \frac{350}{325} \approx 1.08 \, \text{g} \]

7. **Final Answer:**
The mass of the gas that has to be released to maintain the temperature at \( 52^\circ C \) while keeping the pressure constant is approximately \( 1 \, \text{g} \).