1 gram `H_(2)` and 8g `O_(2)` were taken in a 10 liter vessel at 300 K. The partial pressure exerted by `O_(2)` will be

Options

A. 1.84 atm
B. 0.88 atm
C. 1.03 atm
D. 0.61 atm

Answer

To find the partial pressure exerted by \( O_2 \) in a 10-liter vessel at 300 K, we can follow these steps:

### Step 1: Calculate the number of moles of \( O_2 \)
The number of moles can be calculated using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
For \( O_2 \):
- Mass of \( O_2 = 8 \, \text{g} \)
- Molar mass of \( O_2 = 32 \, \text{g/mol} \) (since \( O = 16 \, \text{g/mol} \))

Calculating the moles:
\[ \text{Moles of } O_2 = \frac{8 \, \text{g}}{32 \, \text{g/mol}} = 0.25 \, \text{mol} \]

### Step 2: Calculate the number of moles of \( H_2 \)
Using the same formula for \( H_2 \):
- Mass of \( H_2 = 1 \, \text{g} \)
- Molar mass of \( H_2 = 2 \, \text{g/mol} \)

Calculating the moles:
\[ \text{Moles of } H_2 = \frac{1 \, \text{g}}{2 \, \text{g/mol}} = 0.5 \, \text{mol} \]

### Step 3: Calculate the total number of moles in the vessel
\[ \text{Total moles} = \text{Moles of } O_2 + \text{Moles of } H_2 = 0.25 \, \text{mol} + 0.5 \, \text{mol} = 0.75 \, \text{mol} \]

### Step 4: Calculate the mole fraction of \( O_2 \)
The mole fraction (\( \chi \)) is given by:
\[ \chi_{O_2} = \frac{\text{Moles of } O_2}{\text{Total moles}} = \frac{0.25 \, \text{mol}}{0.75 \, \text{mol}} = \frac{1}{3} \approx 0.33 \]

### Step 5: Calculate the total pressure using the ideal gas law
Using the ideal gas law:
\[ PV = nRT \implies P = \frac{nRT}{V} \]
Where:
- \( n = 0.75 \, \text{mol} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 300 \, \text{K} \)
- \( V = 10 \, \text{L} \)

Calculating total pressure:
\[ P = \frac{0.75 \, \text{mol} \times 0.0821 \, \text{L atm/(K mol)} \times 300 \, \text{K}}{10 \, \text{L}} = \frac{18.4725}{10} = 1.84725 \, \text{atm} \]

### Step 6: Calculate the partial pressure of \( O_2 \)
The partial pressure of \( O_2 \) is given by:
\[ P_{O_2} = \chi_{O_2} \times P \]
Substituting the values:
\[ P_{O_2} = 0.33 \times 1.84725 \approx 0.609 \, \text{atm} \]

### Final Result
The partial pressure exerted by \( O_2 \) is approximately \( 0.61 \, \text{atm} \).

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