A cylinder containing 5 litre of `O_(2)` at `25^(@)C` was leaking. When the leakage was detected and checked, the pressure inside cylinder was reduced from 8 atm to 2 atm. The ratio of mass of `O_(2)` initially present to that left after leakage is equal to ___________.

Explanation

To solve the problem, we will use the ideal gas law and the relationship between pressure, volume, and the number of moles of gas. Here’s a step-by-step solution:

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure of the gas
- \( V \) = volume of the gas
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Identify Given Values
From the question, we have:
- Initial pressure \( P_1 = 8 \, \text{atm} \)
- Final pressure \( P_2 = 2 \, \text{atm} \)
- Volume \( V = 5 \, \text{liters} \) (constant)
- Temperature \( T = 25^\circ C = 298 \, \text{K} \) (constant)

### Step 3: Use the Ideal Gas Law to Find Moles
Since the volume and temperature are constant, we can relate the initial and final states using the pressures and number of moles:
\[ \frac{P_1}{n_1} = \frac{P_2}{n_2} \]

### Step 4: Rearrange the Equation
Rearranging the equation gives:
\[ n_1 = \frac{P_1 \cdot V}{RT} \]
\[ n_2 = \frac{P_2 \cdot V}{RT} \]

### Step 5: Find the Ratio of Moles
From the relationship above, we can write:
\[ \frac{n_1}{n_2} = \frac{P_1}{P_2} \]
Substituting the values:
\[ \frac{n_1}{n_2} = \frac{8 \, \text{atm}}{2 \, \text{atm}} = 4 \]

### Step 6: Relate Moles to Mass
The number of moles is related to mass by the equation:
\[ n = \frac{m}{M} \]
Where \( m \) is the mass and \( M \) is the molar mass (constant for oxygen).

### Step 7: Set Up the Mass Ratio
Thus, we can express the masses as:
\[ m_1 = n_1 \cdot M \]
\[ m_2 = n_2 \cdot M \]
The ratio of the masses can be expressed as:
\[ \frac{m_1}{m_2} = \frac{n_1 \cdot M}{n_2 \cdot M} = \frac{n_1}{n_2} \]

### Step 8: Substitute the Ratio of Moles
Since we found that \( \frac{n_1}{n_2} = 4 \):
\[ \frac{m_1}{m_2} = 4 \]

### Conclusion
The ratio of the mass of \( O_2 \) initially present to that left after leakage is:
\[ \frac{m_1}{m_2} = 4 \]

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