Nitrogen gas `N_2` of mass 28 g is kept in a vessel at pressure of 10 atm and temperature `57^@ C`. Due to leakage of `N_2` gas its pressure falls to 5 atm and temperature to `26^@ C`. The amount of `N_2` gas leaked out is

Options

A. `(5)/(63) g`
B. `(63)/(5)g`
C. `(28)/(63) g`
D. `(63)/(28)g`

Answer

To solve the problem of how much nitrogen gas \( N_2 \) leaked out from the vessel, we can use the Ideal Gas Law and the relationship between pressure, volume, temperature, and the number of moles of gas.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Mass of \( N_2 \) (W1) = 28 g
- Initial Pressure (P1) = 10 atm
- Initial Temperature (T1) = 57°C = 57 + 273 = 330 K

2. **Identify Final Conditions:**
- Final Pressure (P2) = 5 atm
- Final Temperature (T2) = 26°C = 26 + 273 = 299 K

3. **Use the Ideal Gas Law:**
The Ideal Gas Law states that:
\[ PV = nRT \]
where \( n \) is the number of moles, \( R \) is the gas constant, and \( V \) is the volume (which remains constant in this case).

4. **Set Up the Equations:**
For the initial state:
\[ P_1 V = n_1 R T_1 \quad \text{(1)} \]
For the final state:
\[ P_2 V = n_2 R T_2 \quad \text{(2)} \]

5. **Divide Equation (1) by Equation (2):**
\[ \frac{P_1 V}{P_2 V} = \frac{n_1 R T_1}{n_2 R T_2} \]
Simplifying this gives:
\[ \frac{P_1}{P_2} = \frac{n_1 T_1}{n_2 T_2} \]

6. **Express the Number of Moles:**
The number of moles can be expressed as:
\[ n = \frac{W}{M} \]
where \( W \) is the mass and \( M \) is the molar mass of \( N_2 \) (approximately 28 g/mol).

7. **Substituting for Moles:**
Let \( W_2 \) be the final mass of \( N_2 \) after leakage:
\[ \frac{10}{5} = \frac{(W_1/M) \cdot T_1}{(W_2/M) \cdot T_2} \]
This simplifies to:
\[ 2 = \frac{W_1 \cdot T_1}{W_2 \cdot T_2} \]

8. **Plugging in Values:**
Substituting \( W_1 = 28 \) g, \( T_1 = 330 \) K, and \( T_2 = 299 \) K:
\[ 2 = \frac{28 \cdot 330}{W_2 \cdot 299} \]

9. **Solving for \( W_2 \):**
Rearranging gives:
\[ W_2 = \frac{28 \cdot 330}{2 \cdot 299} \]
Calculating this:
\[ W_2 = \frac{9240}{598} \approx 15.46 \text{ g} \]

10. **Calculate the Amount Leaked:**
The amount of \( N_2 \) leaked out is:
\[ \text{Leaked} = W_1 - W_2 = 28 - 15.46 = 12.54 \text{ g} \]

### Final Answer:
The amount of \( N_2 \) gas leaked out is approximately **12.54 g**.