16 gm of `O_(2)` was filled in a container of capacity 8.21 lit. at 300 K.Calculae
(i) Pressure exerted by `O_(2)`
(ii) Partial pressure of `O_(2)` and `O_(3)` if 50% of oxygen is converted into ozone at same temperature.
(iii) Total pressure exertedby gases if 50% of oxygen is converted into ozone ( `O_(3)`) at temperature 50 K.

Answer

To solve the given problem, we will break it down into three parts as specified in the question.

### Given Data:
- Mass of \( O_2 \) = 16 g
- Volume of container = 8.21 L
- Temperature = 300 K
- Molar mass of \( O_2 \) = 32 g/mol (since \( O \) has a molar mass of 16 g/mol)

### (i) Calculate the Pressure exerted by \( O_2 \)

1. **Calculate the number of moles of \( O_2 \)**:
\[ n = \frac{\text{mass}}{\text{molar mass}} = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ moles} \]

2. **Use the Ideal Gas Law to find the pressure**:
The Ideal Gas Law is given by:
\[ PV = nRT \]
Rearranging for pressure \( P \):
\[ P = \frac{nRT}{V} \]
Where:
- \( R = 0.0821 \text{ L atm/(K mol)} \)
- \( T = 300 \text{ K} \)
- \( V = 8.21 \text{ L} \)

Substituting the values:
\[ P = \frac{0.5 \text{ moles} \times 0.0821 \text{ L atm/(K mol)} \times 300 \text{ K}}{8.21 \text{ L}} \]
\[ P = \frac{12.315}{8.21} \approx 1.5 \text{ atm} \]

### (ii) Calculate the Partial Pressure of \( O_2 \) and \( O_3 \) if 50% of \( O_2 \) is converted into \( O_3 \)

1. **Determine the moles of \( O_2 \) converted to \( O_3 \)**:
- 50% of \( O_2 \) means:
\[ \text{Moles of } O_2 \text{ converted} = 0.5 \times 0.5 = 0.25 \text{ moles} \]

2. **Use the reaction \( 3O_2 \rightarrow 2O_3 \)** to find moles of \( O_3 \):
- From the stoichiometry, for every 3 moles of \( O_2 \), 2 moles of \( O_3 \) are produced.
- Moles of \( O_3 \) produced from 0.25 moles of \( O_2 \):
\[ \text{Moles of } O_3 = \frac{2}{3} \times 0.25 = 0.167 \text{ moles} \]

3. **Calculate the remaining moles of \( O_2 \)**:
\[ \text{Remaining } O_2 = 0.5 - 0.25 = 0.25 \text{ moles} \]

4. **Calculate the partial pressure of \( O_2 \)**:
\[ P_{O_2} = \frac{n_{O_2}RT}{V} = \frac{0.25 \times 0.0821 \times 300}{8.21} \]
\[ P_{O_2} \approx \frac{6.1575}{8.21} \approx 0.75 \text{ atm} \]

5. **Calculate the partial pressure of \( O_3 \)**:
\[ P_{O_3} = \frac{n_{O_3}RT}{V} = \frac{0.167 \times 0.0821 \times 300}{8.21} \]
\[ P_{O_3} \approx \frac{4.0901}{8.21} \approx 0.50 \text{ atm} \]

### (iii) Total pressure exerted by gases if 50% of \( O_2 \) is converted into \( O_3 \) at temperature 50 K

1. **Calculate the total moles of gas**:
- Moles of \( O_2 \) = 0.25
- Moles of \( O_3 \) = 0.167
- Total moles = \( 0.25 + 0.167 = 0.417 \text{ moles} \)

2. **Use the Ideal Gas Law to find the total pressure at 50 K**:
\[ P = \frac{nRT}{V} \]
Where \( T = 50 \text{ K} \):
\[ P = \frac{0.417 \times 0.0821 \times 50}{8.21} \]
\[ P \approx \frac{1.707}{8.21} \approx 0.208 \text{ atm} \]

### Summary of Results:
- (i) Pressure exerted by \( O_2 \) = 1.5 atm
- (ii) Partial pressure of \( O_2 \) = 0.75 atm, Partial pressure of \( O_3 \) = 0.50 atm
- (iii) Total pressure at 50 K = 0.208 atm