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 the gaseous mixture of 1 mole of O₂ and 3 moles of H₂ 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, and we need to convert it to Kelvin using the formula:
\[ T(K) = T(°C) + 273 \]
\[ T = 125 + 273 = 398 \, K \]

### Step 2: Calculate the total number of moles
We have:
- Moles of O₂ = 1 mole
- Moles of H₂ = 3 moles

Total number of moles (n):
\[ n = n_{O_2} + n_{H_2} = 1 + 3 = 4 \, moles \]

### Step 3: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
We need to rearrange this to find pressure (P):
\[ P = \frac{nRT}{V} \]

### Step 4: Substitute the values into the equation
We will use:
- \( n = 4 \, moles \)
- \( R = 8.314 \times 10^{-2} \, \text{L bar K}^{-1} \text{mol}^{-1} \)
- \( T = 398 \, K \)
- \( V = 125.3 \, L \)

Now substituting these values into the equation:
\[ P = \frac{4 \times (8.314 \times 10^{-2}) \times 398}{125.3} \]

### Step 5: Perform the calculations
Calculating the numerator:
\[ 4 \times 8.314 \times 10^{-2} \times 398 = 1326.12 \, \text{L bar} \]

Now divide by the volume:
\[ P = \frac{1326.12}{125.3} \approx 10.59 \, bar \]

### Step 6: Final result
After calculating, we find:
\[ P \approx 1.056 \, bar \]

Thus, the total pressure of the gaseous mixture is approximately **1.056 bar**.

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