A container of 1 L capacity contains a mixture of 4 g of `O_(2)` and 2 g `H_(2)` at `0 .^(@)C` . What will be the total pressure of the mixture ? (a) 50 . 42 atm (b) 25 . 21 atm (C) 15 . 2 atm (d) 12 . 5 atm

Options

A. 50 . 42 atm
B. 25 . 21 atm
C. 15 . 2 atm
D. 12 . 5 atm

Explanation

To find the total pressure of the mixture of gases in the container, we will use the Ideal Gas Law, which is given by the equation:

\[ PV = nRT \]

Where:
- \( P \) = pressure (in atm)
- \( V \) = volume (in liters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (0.0821 atm·L/(mol·K))
- \( T \) = temperature (in Kelvin)

### Step 1: Calculate the number of moles of \( O_2 \)

Given:
- Mass of \( O_2 = 4 \, g \)
- Molar mass of \( O_2 = 32 \, g/mol \)

Using the formula for moles:

\[ n_{O_2} = \frac{\text{mass}}{\text{molar mass}} = \frac{4 \, g}{32 \, g/mol} = 0.125 \, mol \]

### Step 2: Calculate the number of moles of \( H_2 \)

Given:
- Mass of \( H_2 = 2 \, g \)
- Molar mass of \( H_2 = 2 \, g/mol \)

Using the formula for moles:

\[ n_{H_2} = \frac{\text{mass}}{\text{molar mass}} = \frac{2 \, g}{2 \, g/mol} = 1 \, mol \]

### Step 3: Calculate the total number of moles of the gas mixture

\[ n_{total} = n_{O_2} + n_{H_2} = 0.125 \, mol + 1 \, mol = 1.125 \, mol \]

### Step 4: Convert the temperature to Kelvin

Given:
- Temperature = \( 0 \, ^\circ C \)

Convert to Kelvin:

\[ T = 0 + 273 = 273 \, K \]

### Step 5: Use the Ideal Gas Law to find the total pressure

Given:
- Volume \( V = 1 \, L \)
- \( R = 0.0821 \, atm \cdot L/(mol \cdot K) \)

Substituting the values into the Ideal Gas Law:

\[ P = \frac{nRT}{V} = \frac{(1.125 \, mol)(0.0821 \, atm \cdot L/(mol \cdot K))(273 \, K)}{1 \, L} \]

Calculating the pressure:

\[ P = \frac{(1.125)(0.0821)(273)}{1} \approx 25.21 \, atm \]

### Conclusion

The total pressure of the mixture is approximately \( 25.21 \, atm \).

### Answer

(b) 25.21 atm

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