Pressure of a mixture of 4 g of `O_2` and 2 g of `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 bulb of 1 liter at 0°C, we can use the ideal gas equation:

### Step-by-Step Solution:

1. **Write the Ideal Gas Equation**:
The ideal gas equation is given by:
\[ PV = nRT \]
where:
- \( P \) = pressure of the gas
- \( V \) = volume of the gas
- \( n \) = number of moles of the gas
- \( R \) = universal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = absolute temperature in Kelvin

2. **Convert Temperature to Kelvin**:
The given temperature is 0°C. To convert this to Kelvin:
\[ T = 0 + 273 = 273 \, K \]

3. **Calculate Moles of O₂**:
The molar mass of O₂ is 32 g/mol. The number of moles of O₂ (\( n_1 \)) can be calculated as:
\[ n_1 = \frac{\text{mass of O₂}}{\text{molar mass of O₂}} = \frac{4 \, g}{32 \, g/mol} = 0.125 \, mol \]

4. **Calculate Moles of H₂**:
The molar mass of H₂ is 2 g/mol. The number of moles of H₂ (\( n_2 \)) can be calculated as:
\[ n_2 = \frac{\text{mass of H₂}}{\text{molar mass of H₂}} = \frac{2 \, g}{2 \, g/mol} = 1 \, mol \]

5. **Calculate Total Moles**:
The total number of moles (\( n \)) in the mixture is:
\[ n = n_1 + n_2 = 0.125 + 1 = 1.125 \, mol \]

6. **Substitute Values into the Ideal Gas Equation**:
We need to find the pressure \( P \). Rearranging the ideal gas equation gives:
\[ P = \frac{nRT}{V} \]
Substituting the values:
- \( n = 1.125 \, mol \)
- \( R = 0.0821 \, L·atm/(K·mol) \)
- \( T = 273 \, K \)
- \( V = 1 \, L \)

Thus,
\[ P = \frac{1.125 \times 0.0821 \times 273}{1} \]

7. **Calculate the Pressure**:
Now, performing the calculation:
\[ P = 1.125 \times 0.0821 \times 273 \approx 25.215 \, atm \]

8. **Final Answer**:
Therefore, the pressure of the mixture of 4 g of O₂ and 2 g of H₂ confined in a bulb of 1 liter at 0°C is approximately:
\[ P \approx 25.215 \, atm \]