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 of how many moles of oxygen gas escaped from the cylinder, we will follow these steps:

### Step 1: Understand the Initial Conditions
We have a cylinder containing oxygen gas with the following initial conditions:
- Volume (V) = 2.24 L
- Initial Pressure (P1) = 1 atm
- Temperature (T) = 273 K

### Step 2: Convert Initial Pressure to mmHg
Since the final pressure is given in mmHg, we need to convert the initial pressure from atm to mmHg:
- 1 atm = 760 mmHg
- Therefore, P1 = 1 atm = 760 mmHg

### Step 3: Determine the Final Pressure
The pressure after the leakage was plugged is given as:
- Final Pressure (P2) = 570 mmHg

### Step 4: Calculate the Change in Pressure
Now, we will calculate the change in pressure (ΔP):
- ΔP = P1 - P2
- ΔP = 760 mmHg - 570 mmHg = 190 mmHg

### Step 5: Calculate the Moles of Gas that Escaped
Using the ideal gas equation \( PV = nRT \), we can rearrange it to find the number of moles (n):
\[ n = \frac{PV}{RT} \]

We will use the change in pressure (ΔP) to find the moles of gas that escaped:
- P = ΔP = 190 mmHg
- V = 2.24 L
- R = 0.0821 L·atm/(K·mol) (we will convert pressure from mmHg to atm for this calculation)
- T = 273 K

First, convert ΔP from mmHg to atm:
\[ P = \frac{190 \text{ mmHg}}{760 \text{ mmHg/atm}} = 0.25 \text{ atm} \]

Now substitute the values into the equation:
\[ n = \frac{(0.25 \text{ atm}) \times (2.24 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)}) \times (273 \text{ K})} \]

### Step 6: Calculate n
Now we perform the calculation:
\[ n = \frac{0.25 \times 2.24}{0.0821 \times 273} \]
\[ n = \frac{0.56}{22.4133} \]
\[ n \approx 0.0249 \text{ moles} \]

### Step 7: Round the Result
We can round this to:
\[ n \approx 0.025 \text{ moles} \]

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

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