12 gm He and 4 gm `H_(2)` is filled in a container of volume 20 litre maintained at temperature 300 K. the pressure of the mixture is nearly :

Options

A. 3 atm
B. 5 atm
C. 6.25 atm
D. 12.5 atm

Explanation

To solve the problem, we will use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = Pressure of the gas
- \( V \) = Volume of the gas
- \( n \) = Number of moles of the gas
- \( R \) = Universal gas constant (8.31 J/(mol·K))
- \( T \) = Temperature in Kelvin

### Step-by-Step Solution:

1. **Calculate the number of moles of Helium (He)**:
- The molar mass of Helium (He) is approximately 4 g/mol.
- Given mass of Helium = 12 g.
- Number of moles of He (\( n_{He} \)) can be calculated as:
\[ n_{He} = \frac{\text{mass}}{\text{molar mass}} = \frac{12 \, \text{g}}{4 \, \text{g/mol}} = 3 \, \text{mol} \]

2. **Calculate the number of moles of Hydrogen (H₂)**:
- The molar mass of Hydrogen (H₂) is approximately 2 g/mol.
- Given mass of Hydrogen = 4 g.
- Number of moles of H₂ (\( n_{H2} \)) can be calculated as:
\[ n_{H2} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \, \text{g}}{2 \, \text{g/mol}} = 2 \, \text{mol} \]

3. **Calculate the total number of moles (n)**:
- Total moles of the gas mixture is the sum of the moles of Helium and Hydrogen:
\[ n = n_{He} + n_{H2} = 3 \, \text{mol} + 2 \, \text{mol} = 5 \, \text{mol} \]

4. **Convert the volume from liters to cubic meters**:
- Given volume = 20 liters.
- To convert liters to cubic meters:
\[ V = 20 \, \text{liters} = 20 \times 10^{-3} \, \text{m}^3 = 0.020 \, \text{m}^3 \]

5. **Substitute values into the Ideal Gas Law to find pressure (P)**:
- Rearranging the Ideal Gas Law to solve for pressure:
\[ P = \frac{nRT}{V} \]
- Substituting the known values:
- \( n = 5 \, \text{mol} \)
- \( R = 8.31 \, \text{J/(mol·K)} \)
- \( T = 300 \, \text{K} \)
- \( V = 0.020 \, \text{m}^3 \)
\[ P = \frac{5 \times 8.31 \times 300}{0.020} \]

6. **Calculate the pressure**:
- Performing the calculation:
\[ P = \frac{12465}{0.020} = 623250 \, \text{Pa} = 6.2325 \times 10^5 \, \text{Pa} \]

7. **Convert pressure from Pascals to atmospheres**:
- Knowing that \( 1 \, \text{atm} = 1.013 \times 10^5 \, \text{Pa} \):
\[ P_{\text{atm}} = \frac{6.2325 \times 10^5 \, \text{Pa}}{1.013 \times 10^5 \, \text{Pa/atm}} \approx 6.15 \, \text{atm} \]

### Final Answer:
The pressure of the mixture is approximately **6.15 atm**.