`28 g "of" N_2` gas is contained in a flask at a pressure of `10 atm` and at a temperature of `57^@ C`. It is found that due to leakage in the flask, the pressure is reduced to half and the temperature to `27^@ C`. The quantity of `N_(2)` gas that leaked out is.

Options

A. `11//20 g`
B. `20//11 g`
C. `5//63 g`
D. `63//5 g`

Explanation

To solve the problem of finding the quantity of \( N_2 \) gas that leaked out from the flask, we will use the Ideal Gas Law, which states that:

\[ PV = nRT \]

Where:
- \( P \) = Pressure
- \( V \) = Volume
- \( n \) = Number of moles
- \( R \) = Universal gas constant
- \( T \) = Temperature in Kelvin

### Step 1: Convert Temperatures to Kelvin
First, we need to convert the temperatures from Celsius to Kelvin.

- Initial temperature \( T_1 = 57^\circ C = 57 + 273 = 330 \, K \)
- Final temperature \( T_2 = 27^\circ C = 27 + 273 = 300 \, K \)

### Step 2: Calculate Initial Number of Moles
Given the mass of \( N_2 \) gas is 28 g, we can calculate the initial number of moles \( n_1 \).

The molar mass of \( N_2 \) is approximately 28 g/mol.

\[ n_1 = \frac{\text{mass}}{\text{molar mass}} = \frac{28 \, g}{28 \, g/mol} = 1 \, mol \]

### Step 3: Set Up the Ideal Gas Equation for Initial and Final States
Using the Ideal Gas Law for the initial and final states, we have:

\[ \frac{P_1 V}{n_1 T_1} = \frac{P_2 V}{n_2 T_2} \]

Since the volume \( V \) remains constant, we can simplify the equation to:

\[ \frac{P_1}{n_1 T_1} = \frac{P_2}{n_2 T_2} \]

### Step 4: Substitute Known Values
We know:
- Initial pressure \( P_1 = 10 \, atm \)
- Final pressure \( P_2 = \frac{10}{2} = 5 \, atm \)
- \( n_1 = 1 \, mol \)
- \( T_1 = 330 \, K \)
- \( T_2 = 300 \, K \)

Substituting these values into the equation gives:

\[ \frac{10}{1 \times 330} = \frac{5}{n_2 \times 300} \]

### Step 5: Solve for \( n_2 \)
Cross-multiplying to solve for \( n_2 \):

\[ 10 \times 300 = 5 \times 330 \times n_2 \]

\[ 3000 = 1650 \, n_2 \]

\[ n_2 = \frac{3000}{1650} = \frac{60}{33} = \frac{20}{11} \, mol \]

### Step 6: Calculate the Quantity of Gas Leaked
The quantity of gas that leaked out is given by:

\[ \text{Leaked moles} = n_1 - n_2 = 1 - \frac{20}{11} = \frac{11}{11} - \frac{20}{11} = -\frac{9}{11} \, mol \]

### Step 7: Calculate Mass of Gas Leaked
Now, we convert the leaked moles back to mass:

\[ \text{Mass leaked} = \text{Leaked moles} \times \text{Molar mass} = \left(-\frac{9}{11}\right) \times 28 \, g \]

Calculating this gives:

\[ \text{Mass leaked} = -\frac{252}{11} \approx -22.91 \, g \]

Since we are looking for the absolute value of the leaked gas:

\[ \text{Mass leaked} \approx 22.91 \, g \]

### Final Answer
Thus, the quantity of \( N_2 \) gas that leaked out is approximately **22.91 grams**.

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