A vessel of volume 50 litre contains an ideal gas at `0^@ C`. A portion of the gas is allowed to leak out from it under isothermal conditions so that pressure inside falls by 0.8 atmosphere. The number of moles of gas leaked out is nearly:

Options

A. (a)` 1. 51" mole"`
B. (b)`1. 63 " mole"`
C. (c)`1.98 " mole"`
D. (d)`1.78 " mole"`

Explanation

To solve the problem step by step, we will use the ideal gas law and the information provided in the question.

### 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
- \( R \) = ideal gas constant
- \( T \) = temperature in Kelvin

### Step 2: Convert Temperature to Kelvin
The temperature given is \( 0^\circ C \). To convert this to Kelvin:
\[ T = 0 + 273 = 273 \, K \]

### Step 3: Calculate Initial Moles of Gas
We know the volume of the gas is \( 50 \, L \) and we need to find the initial number of moles. However, we need the initial pressure \( P \) to calculate this. Let's denote the initial pressure as \( P \).

Using the ideal gas law:
\[ n = \frac{PV}{RT} \]
Substituting the known values:
\[ n = \frac{P \times 50}{0.0821 \times 273} \]

### Step 4: Determine Final Pressure
The problem states that the pressure inside the vessel falls by \( 0.8 \, atm \). Therefore, the final pressure \( P' \) is:
\[ P' = P - 0.8 \]

### Step 5: Calculate Final Moles of Gas
Using the ideal gas law again for the final state:
\[ n' = \frac{P'V}{RT} \]
Substituting \( P' \):
\[ n' = \frac{(P - 0.8) \times 50}{0.0821 \times 273} \]

### Step 6: Calculate Moles of Gas Leaked
The number of moles of gas leaked out, \( n_{leaked} \), is given by:
\[ n_{leaked} = n - n' \]
Substituting the expressions for \( n \) and \( n' \):
\[ n_{leaked} = \frac{PV}{RT} - \frac{(P - 0.8)V}{RT} \]
Factoring out common terms:
\[ n_{leaked} = \frac{V}{RT} \left( P - (P - 0.8) \right) \]
\[ n_{leaked} = \frac{V}{RT} \times 0.8 \]

### Step 7: Substitute Values
Now substitute \( V = 50 \, L \), \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \), and \( T = 273 \, K \):
\[ n_{leaked} = \frac{50 \times 0.8}{0.0821 \times 273} \]

### Step 8: Calculate the Result
Calculating the denominator:
\[ 0.0821 \times 273 \approx 22.4143 \]
Now substituting back:
\[ n_{leaked} = \frac{40}{22.4143} \approx 1.78 \, moles \]

### Conclusion
The number of moles of gas leaked out is approximately \( 1.78 \, moles \).