A 2.24L cylinder of oxygen at 1 atm and 273 K is found to develop a leakage. When the leakage was plugged the pressure dropped to 570 mm of Hg. The number of moles of gas that escaped will be :

Options

A. 0.025
B. 0.05
C. 0.075
D. 0.09

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the initial number of moles of oxygen (N1) using the Ideal Gas Law.

The Ideal Gas Law is given by the equation:
\[ PV = nRT \]

Where:
- \( P \) = pressure in atm
- \( V \) = volume in liters
- \( n \) = number of moles
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

Given:
- Volume \( V = 2.24 \, L \)
- Initial Pressure \( P_1 = 1 \, atm \)
- Temperature \( T_1 = 273 \, K \)

Rearranging the Ideal Gas Law to find \( n \):
\[ n = \frac{PV}{RT} \]

Substituting the values:
\[ N_1 = \frac{(1 \, atm)(2.24 \, L)}{(0.0821 \, L·atm/(K·mol))(273 \, K)} \]

Calculating \( N_1 \):
\[ N_1 = \frac{2.24}{22.414} \approx 0.1 \, moles \]

### Step 2: Convert the final pressure (P2) from mmHg to atm.

Given:
- Final Pressure \( P_2 = 570 \, mmHg \)

To convert mmHg to atm:
\[ P_2 = \frac{570 \, mmHg}{760 \, mmHg/atm} \]

Calculating \( P_2 \):
\[ P_2 \approx 0.75 \, atm \]

### Step 3: Calculate the final number of moles of oxygen (N2) using the Ideal Gas Law again.

Using the same Ideal Gas Law:
\[ N_2 = \frac{P_2 V}{RT} \]

Substituting the values:
\[ N_2 = \frac{(0.75 \, atm)(2.24 \, L)}{(0.0821 \, L·atm/(K·mol))(273 \, K)} \]

Calculating \( N_2 \):
\[ N_2 = \frac{1.68}{22.414} \approx 0.075 \, moles \]

### Step 4: Calculate the number of moles that escaped.

The number of moles that escaped can be calculated by subtracting the final number of moles from the initial number of moles:
\[ \text{Moles escaped} = N_1 - N_2 \]

Substituting the values:
\[ \text{Moles escaped} = 0.1 \, moles - 0.075 \, moles = 0.025 \, moles \]

### Final Answer:
The number of moles of gas that escaped is **0.025 moles**.

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