`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 step by step, we will use the Ideal Gas Law, which states that \( PV = nRT \), where \( P \) is the pressure, \( V \) is the volume, \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature in Kelvin.

### Step 1: Convert the initial temperature to Kelvin
The initial temperature is given as \( 57^\circ C \).
\[ T_1 = 57 + 273 = 330 \, K \]

### Step 2: Calculate the initial number of moles of \( N_2 \)
The molar mass of \( N_2 \) is \( 28 \, g/mol \). The mass of \( N_2 \) is given as \( 28 \, g \).
\[ n_1 = \frac{\text{mass}}{\text{molar mass}} = \frac{28 \, g}{28 \, g/mol} = 1 \, mol \]

### Step 3: Write down the initial conditions
- \( P_1 = 10 \, atm \)
- \( V = V \) (unknown volume)
- \( T_1 = 330 \, K \)
- \( n_1 = 1 \, mol \)

### Step 4: Use the Ideal Gas Law to find the volume
Using the Ideal Gas Law:
\[ P_1 V = n_1 R T_1 \]
Substituting the known values:
\[ 10 \, atm \cdot V = 1 \, mol \cdot R \cdot 330 \, K \]
Thus, we can express \( V \):
\[ V = \frac{R \cdot 330}{10} \]

### Step 5: Convert the final temperature to Kelvin
The final temperature is given as \( 27^\circ C \).
\[ T_2 = 27 + 273 = 300 \, K \]

### Step 6: Write down the final conditions
- \( P_2 = \frac{10}{2} = 5 \, atm \)
- \( T_2 = 300 \, K \)
- \( n_2 = ? \) (unknown number of moles after leakage)

### Step 7: Use the Ideal Gas Law for the final conditions
Using the Ideal Gas Law again:
\[ P_2 V = n_2 R T_2 \]
Substituting the known values:
\[ 5 \, atm \cdot V = n_2 \cdot R \cdot 300 \, K \]

### Step 8: Substitute the expression for \( V \)
From Step 4, we have \( V = \frac{R \cdot 330}{10} \). Substitute this into the equation:
\[ 5 \cdot \left(\frac{R \cdot 330}{10}\right) = n_2 \cdot R \cdot 300 \]
Cancel \( R \) from both sides:
\[ \frac{5 \cdot 330}{10} = n_2 \cdot 300 \]
Simplifying gives:
\[ \frac{1650}{10} = n_2 \cdot 300 \]
\[ 165 = n_2 \cdot 300 \]
\[ n_2 = \frac{165}{300} = \frac{11}{20} \, mol \]

### Step 9: Calculate the final mass of \( N_2 \)
Using the number of moles to find the final mass:
\[ \text{mass}_f = n_2 \cdot \text{molar mass} = \frac{11}{20} \cdot 28 = \frac{308}{20} = 15.4 \, g \]

### Step 10: Calculate the mass of \( N_2 \) that leaked out
The mass that leaked out is given by:
\[ \text{mass leaked} = \text{mass}_i - \text{mass}_f = 28 \, g - 15.4 \, g = 12.6 \, g \]

### Final Answer
The quantity of \( N_2 \) gas that leaked out is \( 12.6 \, g \).