Pressure of a mixture of 4 g of `O_(2)` and `2 g H_(2)` confined in a bulb of 1 litre at `0^(@)C` is

Options

A. 25.215 atm
B. 31.205 atm
C. 45.215 atm
D. 15.210 atm

Answer

To find the pressure of a mixture of 4 g of O₂ and 2 g of H₂ confined in a 1-liter bulb at 0°C, we can use the ideal gas law equation:

\[ PV = nRT \]

### Step-by-Step Solution:

**Step 1: Calculate the number of moles of O₂ and H₂.**

- For O₂:
- Given mass = 4 g
- Molar mass of O₂ = 32 g/mol
- Moles of O₂ = \(\frac{\text{mass}}{\text{molar mass}} = \frac{4 \, \text{g}}{32 \, \text{g/mol}} = \frac{1}{8} \, \text{mol}\)

- For H₂:
- Given mass = 2 g
- Molar mass of H₂ = 2 g/mol
- Moles of H₂ = \(\frac{\text{mass}}{\text{molar mass}} = \frac{2 \, \text{g}}{2 \, \text{g/mol}} = 1 \, \text{mol}\)

**Step 2: Calculate the total number of moles (n) in the mixture.**

- Total moles = Moles of O₂ + Moles of H₂
- Total moles = \(\frac{1}{8} + 1 = \frac{1}{8} + \frac{8}{8} = \frac{9}{8} \, \text{mol}\)

**Step 3: Use the ideal gas law to find the pressure (P).**

- Rearranging the ideal gas law gives us:
\[ P = \frac{nRT}{V} \]

- Where:
- \( n = \frac{9}{8} \, \text{mol} \)
- \( R = 0.0821 \, \text{L atm/(K mol)} \)
- \( T = 0°C = 273 \, \text{K} \)
- \( V = 1 \, \text{L} \)

- Substituting the values:
\[ P = \frac{\left(\frac{9}{8}\right) \times 0.0821 \times 273}{1} \]

**Step 4: Calculate the pressure.**

- First, calculate the product:
\[ P = \frac{9 \times 0.0821 \times 273}{8} \]

- Calculate \( 9 \times 0.0821 \times 273 \):
\[ 9 \times 0.0821 \approx 0.7389 \]
\[ 0.7389 \times 273 \approx 201.25 \]

- Now divide by 8:
\[ P \approx \frac{201.25}{8} \approx 25.15625 \, \text{atm} \]

### Final Answer:
The pressure of the mixture is approximately **25.22 atm**.

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