An `L.P.G` cylinder contains `15kg` of butane gas at `27^(@)C` and 10 atm pressure It was leaking and its pressure fell down to 8 atm pressure after one day Calculate the amount of leaked gas .

Options

A. 1 kg
B. 2 kg
C. 3 kg
D. 4 kg

Answer

To solve the problem of how much butane gas leaked from the LPG cylinder, we can follow these steps:

### Step 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

Since the volume and temperature are constant, we can say that pressure is directly proportional to the number of moles of gas. Therefore, we can express this relationship as:
\[ P_1 \propto n_1 \]
\[ P_2 \propto n_2 \]
This leads to the relationship:
\[ \frac{P_1}{P_2} = \frac{n_1}{n_2} \]

### Step 2: Relate Moles to Mass
The number of moles \( n \) can be expressed in terms of mass \( W \) and molar mass \( M \):
\[ n = \frac{W}{M} \]
Since the molar mass of butane is constant, we can also say:
\[ \frac{n_1}{n_2} = \frac{W_1}{W_2} \]

### Step 3: Set Up the Initial and Final Conditions
From the problem:
- Initial pressure \( P_1 = 10 \) atm
- Final pressure \( P_2 = 8 \) atm
- Initial mass \( W_1 = 15 \) kg
- Final mass \( W_2 \) is unknown

### Step 4: Apply the Proportionality
Using the relationship derived:
\[ \frac{P_1}{P_2} = \frac{W_1}{W_2} \]
Substituting the known values:
\[ \frac{10}{8} = \frac{15}{W_2} \]

### Step 5: Solve for \( W_2 \)
Cross-multiplying gives:
\[ 10 \cdot W_2 = 15 \cdot 8 \]
\[ 10 W_2 = 120 \]
\[ W_2 = \frac{120}{10} = 12 \text{ kg} \]

### Step 6: Calculate the Amount of Leaked Gas
The amount of leaked gas is given by:
\[ \text{Leaked Gas} = W_1 - W_2 \]
Substituting the values:
\[ \text{Leaked Gas} = 15 \text{ kg} - 12 \text{ kg} = 3 \text{ kg} \]

### Conclusion
The amount of leaked gas is **3 kg**.