A gas is enclosed in a vessel at a constant temperature at a pressure of 5 atmosphere and volume 4 litre. Due to a leakage in thhe vessel, after some time, the pressure is reduced to 4 atmosphere. As a result the

Options

A. volume of the gas decreased by `20%`
B. Average K.E. of gas molecule decreases by `20%`
C. `20%` of the gas escaped due to the leakage
D. `25%` of the gas escaped due to the leakage

Answer

To solve the problem, we will use the ideal gas law and the relationship between pressure, volume, and the number of moles of gas.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Initial pressure (P1) = 5 atm
- Initial volume (V) = 4 L
- Initial temperature (T) = constant (not given but we will assume it remains constant)

2. **Identify Final Conditions**:
- Final pressure (P2) = 4 atm
- Volume (V) remains constant at 4 L.

3. **Use the Ideal Gas Law**:
The ideal gas law is given by the equation:
\[ PV = nRT \]
Since the volume (V) and temperature (T) are constant, we can relate the initial and final states using the pressures and the number of moles (n):
\[ \frac{P_1}{P_2} = \frac{n_1}{n_2} \]

4. **Calculate the Ratio of Moles**:
- Substitute the known values into the equation:
\[ \frac{5}{4} = \frac{n_1}{n_2} \]
This implies:
\[ n_1 = \frac{5}{4} n_2 \]

5. **Calculate the Change in Number of Moles**:
- The decrease in the number of moles (Δn) can be expressed as:
\[ Δn = n_1 - n_2 = n_2 \left(\frac{5}{4} - 1\right) = n_2 \left(\frac{1}{4}\right) \]

6. **Calculate the Percentage Decrease**:
- The percentage decrease in the number of moles is given by:
\[ \text{Percentage decrease} = \left(\frac{Δn}{n_1}\right) \times 100 = \left(\frac{n_2 \left(\frac{1}{4}\right)}{n_1}\right) \times 100 \]
- Since \( n_1 = \frac{5}{4} n_2 \):
\[ \text{Percentage decrease} = \left(\frac{n_2 \left(\frac{1}{4}\right)}{\frac{5}{4} n_2}\right) \times 100 = \left(\frac{1}{5}\right) \times 100 = 20\% \]

### Conclusion:
The percentage of the number of moles that escaped due to the leakage is **20%**.