1 mole of sample of `O_2` and 3 mole sample of `H_2` are mixed isothermally in a 125.3 litre container at `125^@C` the total pressure of gaseous mixture will be

Options

A. 2.50 bar
B. 9.8 bar
C. 1.056 bar
D. 4.038 bar

Answer

To find the total pressure of a gaseous mixture of 1 mole of \( O_2 \) and 3 moles of \( H_2 \) in a 125.3-liter container at 125°C, we can follow these steps:

### Step 1: Convert the Temperature to Kelvin
The temperature is given in Celsius, so we need to convert it to Kelvin using the formula:
\[ T(K) = T(°C) + 273.15 \]
Substituting the given temperature:
\[ T = 125 + 273.15 = 398.15 \, K \]

### Step 2: Use the Ideal Gas Law
According to the Ideal Gas Law, the pressure \( P \) can be calculated using the formula:
\[ PV = nRT \]
Where:
- \( P \) = pressure in bar
- \( V \) = volume in liters
- \( n \) = number of moles
- \( R \) = ideal gas constant (8.314 × \( 10^{-2} \) bar·L/(mol·K))
- \( T \) = temperature in Kelvin

### Step 3: Calculate the Total Number of Moles
The total number of moles \( n \) in the mixture is:
\[ n = n_{O_2} + n_{H_2} = 1 + 3 = 4 \, moles \]

### Step 4: Substitute Values into the Ideal Gas Law
Rearranging the Ideal Gas Law to solve for pressure \( P \):
\[ P = \frac{nRT}{V} \]
Substituting the values:
- \( n = 4 \, moles \)
- \( R = 8.314 \times 10^{-2} \, \text{bar·L/(mol·K)} \)
- \( T = 398.15 \, K \)
- \( V = 125.3 \, L \)

\[ P = \frac{4 \times (8.314 \times 10^{-2}) \times 398.15}{125.3} \]

### Step 5: Calculate the Pressure
Calculating the numerator:
\[ 4 \times (8.314 \times 10^{-2}) \times 398.15 = 1326.57 \, \text{bar·L} \]
Now, divide by the volume:
\[ P = \frac{1326.57}{125.3} \approx 10.59 \, bar \]

### Step 6: Final Result
The total pressure of the gaseous mixture is approximately:
\[ P \approx 10.59 \, bar \]