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. ` 1. 51" mole"`
B. `1. 63 " mole"`
C. `1.98 " mole"`
D. `1.78 " mole"`

Explanation

To solve the problem, we will use the ideal gas law and the information provided in the question. Let's break it down step by step.

### Step 1: Understand the Initial Conditions
We have a vessel with a volume (V) of 50 liters containing an ideal gas at a temperature (T) of 0°C. We need to convert this temperature to Kelvin for calculations:
\[ T = 0°C + 273 = 273 \, K \]

### Step 2: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant = 0.0821 L·atm/(K·mol)
- \( T \) = temperature (in K)

### Step 3: Initial Pressure and Moles
Let’s denote the initial pressure as \( P \) and the initial number of moles as \( n \). The equation for the initial state is:
\[ P \cdot 50 = n \cdot 0.0821 \cdot 273 \]
This can be rearranged to find \( n \):
\[ n = \frac{P \cdot 50}{0.0821 \cdot 273} \]

### Step 4: Final Conditions After Leakage
After a portion of the gas leaks out, the pressure drops by 0.8 atm. Therefore, the new pressure \( P' \) is:
\[ P' = P - 0.8 \]

The equation for the final state (after leakage) becomes:
\[ P' \cdot 50 = n' \cdot 0.0821 \cdot 273 \]
Where \( n' \) is the number of moles remaining after leakage.

### Step 5: Calculate the Change in Moles
The change in the number of moles (the amount leaked) can be expressed as:
\[ n - n' = \text{leaked moles} \]

Substituting \( P' \):
\[ n - n' = \frac{P \cdot 50}{0.0821 \cdot 273} - \frac{(P - 0.8) \cdot 50}{0.0821 \cdot 273} \]

### Step 6: Simplify the Expression
Factoring out common terms:
\[ n - n' = \frac{50}{0.0821 \cdot 273} \left( P - (P - 0.8) \right) \]
\[ n - n' = \frac{50}{0.0821 \cdot 273} \cdot 0.8 \]

### Step 7: Calculate the Number of Moles Leaked
Now we can calculate the number of moles leaked:
1. Calculate the denominator:
\[ 0.0821 \cdot 273 \approx 22.4143 \]
2. Substitute into the equation:
\[ n - n' = \frac{50 \cdot 0.8}{22.4143} \]
\[ n - n' \approx \frac{40}{22.4143} \approx 1.78 \, \text{moles} \]

### Conclusion
The number of moles of gas leaked out is approximately **1.78 moles**.

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